Date: Tue, 04 Jan 94 19:05:25 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Combining conjugacy classes Last month Jerry Bryan posted a sequence of articles about counting the number of M-conjugacy classes of Rubik's cube positions. Having calculated the number of conjugacy classes of the corner and edge groups separately, his idea was to combine these to calculate the number of conjugacy classes of the entire group. Eventually, he withdrew the calculation, after realizing that he had not found enough information to determine the answer. This article is about how we might calculate the answer from separate searches of the edge and corner groups. My first idea is to formalize the concept of symmetry in the conjugacy classes that Jerry used in his searches. Recall that the conjugacy class of a position X is defined to be the set of all positions m'Xmc, where m is an element of M, the 48-element symmetry group of the cube, and c is an element of C, the 24-element subgroup of M consisting of rigid rotations of the cube in space. The reason some positions have more symmetry than others is that for some positions, there are nontrivial elements m and c such that m'Xmc=X. The way in which this arises can be formalized as a kind of symmetry group of X. For an edge-group position X, let CSymm(X) be the set of all f in M such that X'f'Xf is an element of C. First, I'll claim that CSymm(X) is a subgroup of M [see proof 1, below]. Second, I note that CSymm(X) is the set of all m in M such that there exists c in C with m'Xmc=X [proof: m'Xmc=X iff X'm'Xm=c']. Third, I'll claim that if m'Xmc=Y, then CSymm(X) and CSymm(Y) are conjugate subgroups of M [proof 2]. So when Jerry says that a position X has order-N symmetry, he is saying that CSymm(X) has 1152/N elements. But the identity of CSymm(X) has more information than just its size, and I believe this information is crucial if we are to combine symmetry groups. It looks to me as if it would be sufficient to record the conjugacy class of CSymm(X), and there are only 33 possibilities. Now the usual symmetry group of X, Symm(X), is defined to be the group consisting of all f in M such that X'f'Xf=I [or, equivalently, Xf=fX. Symm(X) is the largest group such that X is Symm(X)-symmetric, in the sense of the Symmetry and Local Maxima article]. The first step in combining the corner and edge sets is to calculate the symmetry groups of the rotations of a position X, AllSymms(X)={Symm(Xc) : c in C}. This corresponds to computing the symmetry groups of the edges-and- centers group from the symmetry groups of the edges group. I suspect there is a way of computing this from CSymm(X), but I do not know it. I am not even sure that AllSymms(X) is determined by CSymm(X). One useful experiment would be to calculate CSymm(X) and AllSymms(Xc) for all elements of the corner group and see what the correspondence is. Barring an ability to calculate AllSymms(X) from CSymm(X), we could calculate AllSymms(X) directly. This involves a great number of calculations, though: 24 symmetry group calculations for each element of the edge group. My first thought was to try to split the problem up further, to deal with the group of permutations separately from the group of orientations. But I abandoned this when I realized there is a problem that shows up when we try this with the corner group. The permutation of the corners that takes each cubie to its antipode is clearly M-symmetric, and no matter how we decide to measure orientation, there is a way to perform this permutation leaving the cubies in their `home' orientation. But there is no way to compose the two together in an M-symmetric way. I suspect the same problem arises in the edge group. But there may be some help from the edge search available in calcu- lating AllSymms(X). For take a position Y in the edges-and-centers group; Y is also a rotated position in the edges group, so Y=m'Xmc for some X in Jerry Bryan's list. So for f in Symm(Y), Y'f'Yf=I is an element of C, so f is in CSymm(Y). This says that Symm(Y) is a subgroup of CSymm(Y), which is a conjugate of CSymm(X). So if Symm(Y) is nontrivial, then CSymm(X) will also be nontrivial. So to find the symmetry groups of the edges-and-centers group we need only look at those positions that have nontrivial groups in Jerry's list (i.e. less than order-1152 symmetry), as all the others will have Symm(Y)=I. So, Jerry, do you have the data on how many positions of the edge group have less than order-1152 symmetry, and which positions those are? So, on to finding the symmetry groups of the Rubik's group positions. We need to calculate Symm(X) for every element X of the edges-and-cen- ters group and Symm(Y) for every element Y of the corners-and-centers group, while keeping track of the permutation parity of X and Y. (The permutation parity will be constant over each Symm(X), Symm(Y)). The symmetry groups in the Rubik's group will be the intersections of symmetry groups of edge and corner positions of the same parity. We need not keep track of the particular positions here, only the numbers for each parity and each (conjugacy class of) symmetry group. I have a program that could produce a table easily enough. Recently, I took a look in Paul B. Yale's _Geometry_and_Symmetry_ and it looks like this is the sort of problem we could use the Polya-Burnside theorem on. Unfortunately, I don't understand it yet, and it looks like the number of cases here might be too large to conveniently carry out by hand. So it would help to go after this problem computationally. The rest of this article has the proofs for the claims I mentioned in the second paragraph. ================================================================ Proof 1: Suppose f, g are elements of CSymm(X); it suffices to show that f'g is an element of CSymm(X). X'(f'g)'X(f'g)=X(g'f)X(f'g) =X'g'(Xgg'ff'X')fXf'g =(X'g'Xg) g'f (f'X'fX) f'g, =(X'g'Xg) (f'g)' (X'f'Xf)' (f'g). Since we assumed f, g in CSymm(X), X'g'Xg and X'f'Xf must be in C. (f'g)' and (f'g) are elements of M that are either both in C or neither. So the product is in C, so f'g is in CSymm(X). Therefore CSymm(X) is a group, QED. Proof 2: Suppose Y=m'Xmc. First let f be an element of CSymm(X), so that X'f'Xf is in C. I will first show that m'fm is an element of CSymm(Y). Y'(m'fm)'Y(m'fm)=(m'Xmc)'(m'fm)'(m'Xmc)(m'fm) =(c'm'X'm)(m'f'm)(m'Xmc)(m'fm) =c'm'(X'f'X)(mcm'fm) =c'm'(X'f'Xf)(f'mcm'fm) All of which are elements of M, with an even number in C. Therefore the expression is in C, so m'fm is in CSymm(Y). Now let g be an element of CSymm(Y), so that Y'g'Yg is in C. Let f=mgm', so f is an element of M such that m'fm is in CSymm(Y). I will show that f is an element of CSymm(X): X'f'Xf=(mc)(mc)'X'(mm')f'(mm')Xf(f'mcm'fm)(f'mcm'fm)' =(mc)(m'Xmc)'(m'fm)'(m'Xmc)(m'fm)(f'mcm'fm)' =(mc)Y'(m'fm)'Y(m'fm)(f'mcm'fm)' =(mc)Y'g'Yg(f'mcm'fm)', which is in C, so f is in CSymm(X). I've shown that for every element f of CSymm(X), m'fm is an element of CSymm(Y), and that every element of CSymm(Y) is m'fm for some f in CSymm(X). Therefore CSymm(Y)=m' CSymm(X) m, QED. ================================================================ Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Tue, 04 Jan 94 21:36:18 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Re: Which is the Real Start? Jerry Bryan has some more questions. Take a standard 3x3x3 Rubik's cube, and remove the corner and center labels to make an Edges Cube.... Scramble..., how will the cubemeister distinguish Start from Pons Asinorum? ... if you identify the identity with Start, you are in the disquieting situation of having a group with two distinct identities (grin!). The problem is that we would not be dealing with a _group_ then, but a collection of cosets of M. Just as in the edge `group', we deal with either 1) a less-symmetric group in which one of the edges never moves, or 2) a larger group in which we distinguish positions that differ by rigid motions of the cube, or 3) a non-group in which we consider cosets--equivalence classes of group #2, where group elements that differ by rigid motions are equivalent. You have got a lot of mileage out of working with group #2 to save duplication among symmetries, then reducing to non-group #3. But what you lose is the group structure of the object you are studying. Instead, you have to work in the large group and then deduce information about the cosets. All in all, though, I'm very glad of it, for the lost symmetries of group #1 were sorely missed. For most of the other questions, mouse@collatz.mcrcim.mcgill.edu provides satisfactory answers. However, strictly speaking we should not call an equivalence class to be a group element (unless it is a coset of a normal subgroup, and neither C nor M is normal in the large group). I'll admit I've also abused the term when considering distances in the ``edge group'', as if all 24 rotations of a position were the same element of some group. But when we start dealing with the distinction between fixed and movable cubes I think we need to start being more careful. [ mouse also mentions that quarter-turn ``usually doesn't include slice turns on the 3-Cube, but on the 4-Cube and higher, they must of necessity be included.'' I'll take that as an argument for eccentric slabism: a QT rotates any 1xNxN slab except a central slab of an odd-edged cube. As opposed to cutism, where a QT consists of a rotation of part of the cube with respect to the other. ] Other questions: > ...since Start and Pons Asinorum differ only by a simple reflection, > why had not my version of M-conjugation declared them to be equivalent? Your versino treats positions X,Y for which m'Xmc=Y (m in M, c in C) as equivalent. If you instead determine when m'Xmn=Y (m,n in M) you would find them equivalent. This is equivalent to changing the loop in your version of M-conjugacy. For j = 1 to 24 for k = 1 to 24 for m = 1 to 2 for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i))))) so that the two occurrences of Qm need not be the same. (I speak of "my version of M-conjugation", but the question is no different if you look at Dan Hoey's original M-conjugation). No, I didn't use M-conjugation except for a cube with a fixed orientation in space [or equivalently, with face centers]. So in the original concept of M-conjugation that Jim Saxe and I put together, Start and Pons Asinorum don't just differ by a reflection. I found Dan Hoey's postings about the four special states of the Edge Group to be delightful.... However, [without the results on distances] if we identified the group as being rectangular, would we be led to saying which of the four special states were diagonally opposed without the computer search? Without the search, I might be tempted to say that Start and Pons Asinorum were diagonally opposed. Well, really the `group' is in the shape of a sphenoid, a word I learned yesterday for a tetrahedron whose three pairs of opposite edges are equal. [Or equivalently, a tetrahedron whose edges are face diagonals of a rectangular prism.] But it might be more accurate to consider it as a large ball of string with a bunch of symmetries. Calling it a rectangle or sphenoid may lead us to ignore the structure that is not representable in Euclidean space. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Fri, 21 Jan 94 18:32:15 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Re: Some Proposed Terminology I welcome Jerry Bryan's efforts to improve the terminology of the groups associated with Rubik's cube. But there is some additional clarification I think is necessary. Let G be the standard cube group for the 3x3x3 cube.... Let GC be the corners with centers without edges group.... Let GE be the edges with centers without corners group.... That much will do, mod quibbles about what name is best. Let G\C be the corners with edges without centers group. I intend for the notation to indicate G reduced by C, where C is the rotation group for the cube.... Let GC\C be the corners without edges without centers group.... Let GE\C be the edges without corners without centers group.... First, these are not, strictly speaking, groups. Well, you can make them groups, by defining what the group operation is. But I don't know any way of doing that without losing the symmetrical nature of the problem. Second, I would suggest that G/C, GC/C, and GE/C are more standard names for these objects. The elements are nominally 24-element sets, each of which is an equivalence class when two positions are considered equivalent when they differ by their position with respect to the corners. The classes are called the cosets of C in G, GC, and GE, respectively. Let G\M be the set of M-conjugate classes for G..... Let GC\M be the set of M-conjugate classes for GC.... Let GE\M be the set of M-conjugate classes for GE.... The partition of a group into conjugacy classes is not at all the same as the partition into cosets. So I would prefer to use different symbology, like "\" for conjugacy and "/" for cosets, but.... Recall that B is the function which calculates the canonical form for a cube under the composed operations of M-conjugation plus rotation. My programs calculate equivalence classes under B. Let G\B be the set of B-classes for G [ and likewise for GE, GC ]. Well, if you are using "\" for a generic partition into equivalence classes, then we should really do something like G\Conj(M) for partitions into conjugacy classes. At least then you can say G/C=G\Cosets(C). Then, we have Gx\B=(Gx\C)\M=(Gx\M)\C. In English, we can decompose B into a multiplication by C and M (in either order). No, that's _multiplication_ by C and _conjugation_ by M. A good example of why it's important not to use confusing symbols. M and C are not at all treated the same, except inasmuch as they are used to induce partitions into equivalence classes. Say instead that Gx\B = (Gx/C)\Conj(M) = (Gx\Conj(M))/C. If I wanted to model GC\C, I would have had to either model only seven of the cubies, or else modeled all eight but moved only seven of them. Since what I really wanted was (GC\C)\M, and since what I had was GC, I had to invent this funny B thing, where GC\B=(GC\C)\M. If I had been clever enough to model GC\C in the first place, I never would have had to invent B. Similar comments apply to my model for the edges. Well, the part about moving only seven (corner) cubies is the approach that's been taken before on this list to deal with cubes that don't have face centers. It has the advantage that the object being treated is a group. But the problem is that the group is no longer cubically symmetrical (in some vague sense). This led me, at least, to lose track of the structure that would allow analysis of M-conjugacy. So I have to admire your tackling GE as a whole, instead of trying to stick to GE/C. At first blush, it looks like GE/C is 24 times smaller. But since GE/C\Conj(M) is almost 48 times smaller still, it's important to work in GE at least enough to be able to use the conjugation. Which is beside the point that I'm actually very interested in the structure of Gx/Conj(M) itself. And that is what I was really getting at in 1984 when I asked about how many positions there really are. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Mon, 24 Jan 94 19:15:15 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Concerning B, CSymm, and Symm In his message of Sat, 8 Jan 1994 08:46:20 EST, Jerry Bryan considers his use of the term "B" ``to indicate various aspects of the conjugacy class generated by m'Xmc.'' I don't think that's properly called a conjugacy class, but a different sort of equivalence class. A conjugacy class is a special kind of equivalence class (just as a coset is a special kind of equivalence class) but this B is a little bit of both, so I don't think it is correct to call it either. Let X be any cube. Then the set of B-conjugacy classes of X is the set of all m'Xmc for all m in M and all c in C. We denote this set as BClass(X). B is the function B(X)=min(BClass(X)). That's a little unfortunate--I'd prefer to use B(X) for the equivalence class, and min(B(X))--or repr(B(X))--for the canonical representative. That's because the representative is not the important thing here, it's just a convenient way to represent (!) the class in a computer. Note that we could have defined BClass(X) equivalently as the set of all mXm'c, or as the set of all cm'Xm, or as the set of all cmXm'.... This is the justification for the assertion in a previous note that Gx\B = (Gx\M)\C = (Gx\C)\M. Not quite. The justification for (Gx\Conj(M))/C = (Gx/C)\Conj(M) is that instead of m'Xmc we could choose m'Xcm, a possibility you didn't list. In his message of "Sat, 8 Jan 1994 10:52:22 EST", Jerry continues with discussion on combining conjugacy classes. We've exchanged some private email on the subject material, but in case anyone on the list is following this stuff.... There are only 10 distinct values for |BClass(X)| and for |BClass(Y)|, namely 24, 48, 72, 96, 144, 192, 288, 384, 576, and 1152. (By the way, I have never figured out why it is *exactly* the same set of values for both the corners and for the edges. It is easy to see why it is approximately the same set of values.... I'm not sure what kind of approximation you mean, but certainly those ten values are all that are possible: Proof: For if m1,m2 are in the same coset of M/CSymm(X), then (m1 m2') is in CSymm(X), so X' (m1 m2')' X (m1 m2') = c0 in C so m1' X m1 = m2' X c0 m2. It's then clear that { m1' X m1 c : c in C } = { m2' X m2 c : c in C } (*) are equal 24-element sets. The same manipulation in reverse shows that if (*) holds for some m1,m2 in M, then m1 and m2 are in the same coset of M/CSymm(X). So |BClass(X)|=24 |M/CSymm(X)|. |M/CSymm(X)| must be a divisor of |M|=48, QED. It wouldn't have been all that surprising to see one of the possible sizes of |CSymm(X)| fail to appear as a symmetry group of the corners or edges, but it's not surprising that they all do, either. [For the original approach] I needed to be able to prove that for a fixed m and n, that |(BClass(X)[m] * BClass(Y)[n]| had the same value for all X in GC[m]\B and all GE[n]\B. That is to say, that the sizes |CSymm(X)| and |CSymm(Y)| might determine {|Symm(X*Y)|} for X in GC\B, Y in GE\B, and so (X*Y) in G\Conj(M). It doesn't, but the situation is even worse. Jerry goes on to suppose that perhaps CSymm(X) and CSymm(Y) themselves might determine {|Symm(X*Y)|}, and even that isn't true. I've discovered this by a computer search of GC\B. (A search of GE\B is in progress, but for the current result we can take Y=I in GE\B). I have found that AllSymms(X) is not determined, even up to subgroup sizes, by CSymm(X). According to the search, the following are the only positions of GC\B for which |CSymm(X)|=16. X1 X2 X3 +---+ +---+ +---+ |F F| |B F| |F B| |B B| |F B| |F B| +---+---+---+ +---+---+---+ +---+---+---+ |R R|D D|L L| |L L|D T|R R| |L L|D T|R R| |R R|T T|L L| |L L|T D|R R| |L L|D T|R R| +---+---+---+ +---+---+---+ +---+---+---+ |B B| |F B| |B F| |F F| |B F| |B F| +---+ +---+ +---+ |T T| |T D| |T D| |D D| |D T| |T D| +---+ +---+ +---+ Coincidentally, I have been (privately) calling the CSymm(Xi) subgroups the "X subgroups" of M, an X subgroup being the subgroup that maps an orthogonal axis of the cube (in the above examples, the L-R axis) to itself. X1 is a notable position, in that each corner has been swapped with its opposite corner. Symm(X1) is an X subgroup as well, and there is another X subgroup in AllSymms(X1). There is, however, no 16-element subgroup in AllSymms(X2) or AllSymms(X3). (We have seen X2 before: it is the corners of the Laughter (or 4/) position). In fact, my program says that AllSymms(X1) contains two occurrences of 16-element X subgroups, two occurrences of the 8-element HX subgroup, two occurrences of 8-element R subgroups, two occurrences of 8-element S subgroups, eight occurrences of 4-element HS subgroups, and eight occurrences of the 2-element HV subgroup. AllSymms(X2) and AllSymms(X3) each contain two occurrences of 8-element CX subgroups, two occurrences of 8-element AX subgroups, two occurrences of 8-element P subgroups, two occurrences of 8-element Q subgroups, eight occurrences of 2-element ES subgroups, and eight occurrences of 2-element HW subgroups. The names of these groups are part of a taxonomy of the subgroups of M I've developed, which I won't go into just now. But the point that I find surprising here is that AllSymms(X1) and AllSymms(X2) are completely disjoint. While that can't happen all the time (smaller CSymm() groups have many occurrences of the one-element "I" subgroup) I think the tendency to disjointness is too pronounced to be simple anti-coincidence. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: news.admin.misc From: hoey@pooh.tec.army.mil (Dan Hoey) Date: 16 Feb 1994 15:08:11 -0500 Subject: Re: America Online isn't any better pmdatro...@aol.com (PMDAtropos) writes: >... As I have mentioned below, our database of mailing list >information is in the process of a complete overhaul. When it is >finished, it will bear only the most coincidental similarity to the >PAML, but will (if I am successful) provide more information on more >mailing lists and be in a format which is more comprehensible to the >"average" person. And will it make it easier or harder, or more or less likely, for (possibly well-meaning) nuisances like the "Computer user family" to send their factoids to every mailing list in creation, regardless of topic? For those who didn't see it, there's an article on ``Computers and Health'' that C...@AOL.COM sent to everyone from rec.woodworking to comp.protocols.appletalk, containing a combination of information and misinformation about ergonomics. As far as I can see, they are the lay equivalent of Clarence Thomas, except that AOL gives them an alias to hide behind. How long will it be possible to subscribe to special-interest mailing lists without turning your mailbox into multiple copies of everyone's favorite soapbox? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.lang.lisp From: hoey@pooh.tec.army.mil (Dan Hoey) Date: 25 Feb 1994 18:38:12 -0500 Subject: Re: WANTED: Card shuffling algorithm ka...@cs.uiuc.edu (Carl M Kadie) writes: >Here is the shuffle code I use.... >(defun SHUFFLE (list) ... > (let ((vector (apply #'vector list)) ... ^^^^^^^^^^^^^^^^^^^^^ That looks like a particularly bad way of making a vector out of a list, especially if CALL-ARGUMENTS-LIMIT is less than (length list). C-A-L can be as small as 50. Besides, that's exactly what (coerce list 'simple-vector) is for. > (let ((ith (svref vector i)) > (j (random (1+ i))) > ) > (setf (svref vector i) (svref vector j)) > (setf (svref vector j) ith) ... I think clarity is enhanced with (let ((j (random (1+ i))) (unless (= i j) (rotatef (svref vector i) (svref vector j)))) and quite possibly speed, as well. I can't tell from CLtL2 whether the test for (= i j) is necessary or not; it's certainly not desirable. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.folklore.urban From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 1 Mar 1994 23:14:36 GMT Subject: Re: Charles Drew story debunked on NPR da...@cleveland.Freenet.Edu (Richard N Kitchen) > The legend is that Dr. Drew, the developer of blood plasma, was > allowed to bleed to death in an all-white hospital in the South in > 1950 because he was black, and no one would treat him there. The NPR show also mentioned that Dr. Drew had earlier faced discrimination related to blood donation, in that African Americans were rejected as blood donors during World War II. The irony of this story led to its wide circulation before Dr. Drew's death. The historian on NPR conjectured that the spread of the death legend was enhanced by the earlier spread of the donation story. It should be noted that this legend is more demonstrably harmful than most. The African American community has a below-average level of blood donation, and this legend is often cited as the reason. I hope McGill's student newspaper at least printed a retraction of the recent story. An interesting side issue is that while ordinary blood and plasma transfusion is carried out without regard to race, bone marrow transplantation has much more stringent histocompatibility requirements: the donor and recipient are seldom compatible unless they have genetically similar ancestry. In particular, refusal by African Americans to register for bone marrow donation is directly responsible for reducing the availability of bone marrow transplants to other African Americans. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.lang.lisp, comp.lang.lisp.mcl, comp.lang.lisp.franz Followup-To: comp.lang.lisp.franz From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 2 Mar 1994 21:35:29 GMT Subject: Franz marketing sleaze In a message last week to comp.lang.lisp, I commented on some code to shuffle a list in Lisp. Today I got email from sarah@franz.com about my message ``regarding Apple's announcement to discontinue their support and development of MCL.'' That was odd, since I had not mentioned Apple or MCL. It assured me that my sentiment was ``shared by most all MCL users.'' I wonder which sentiment that was. Reading the message, it became clear that this was not even about that subject. It was an offer to send me sales information on Franz Inc.'s Lisp products for non-Macintosh platforms. I think it would be very nice to receive such an invitation if I had asked about how to get Lisp. But when it is sent to people who don't ask for software information, it qualifies as marketing sleaze. I don't want my mailbox full of junk mail every time I venture an opinion on a software topic. Sending marketing mail to people who don't ask for the information is a pretty slimy way of doing business, and I had thought Franz Inc. was a lot more reputable than that. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- From hoey Wed Mar 9 14:11:43 1994 To: Dave@Delta.NIS.Com,nis@netcom.com (National Information Systems) Subject: Re: Watch out for nis.com. Count me as one person who doesn't think posting on Usenet should result in junk mail from random hucksters. Sending information on something you can provide is useful if the poster asks for your product. But sending your marketing questions to people who don't ask is a reprehensible practice that damages the free exchange of information on the network. Stop it. Dan Heoy Hoey@AIC.NRL.Navy.Mil --- From hoey Wed Mar 9 16:30:01 1994 To: DAVE@nis.com Subject: Re: Watch out for nis.com. You write: > Thanks for your message. Unfortunately, you've been mislead > by the poster to news.admin.misc. Please see my response. According to the poster, you sent mail about your company's products to him after he posted somewhere on some other topic, not solicting such information. Your posting does not dispute that. Count me as one person who doesn't think posting on Usenet should result in junk email from random hucksters. Sending information on something you can provide is useful if the poster asks for your product. But sending your mailing list information to people who don't ask is a reprehensible practice that damages the free exchange of information on the network. Stop it. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Article 36577 of sci.math: Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Subject: Re: Irrationalism Sender: usenet@ra.nrl.navy.mil Organization: Naval Research Laboratory, Washington, DC Date: Thu, 10 Mar 1994 17:45:30 GMT mark@freenet.uwm.edu (Mark Hopkins) says: >Give me any indication or formula that will allow me to compute the >Nth decimal for any non-rational on any machine for any N at all >without exceeding the capacity of that machine... John Baez told you how this may not work if the number is rational. But as long as you can provide N--as a decimal number, say--it's easy to exhibit such an irrational. For instance, we may take the irrational number 0.1234567891111111111222222222233... whose Nth digit is the leading digit of N. You don't even need to look at all of N to figure out what the N'th digit is. I wouldn't be surprised if someone could tell me what the N'th digit is where N is floor(1.5^(10^10000)). But probably not where N is floor(1.5^(10^10000)) with its digits in reverse order. Unlike John, though, I probably won't respond to this guy's next post even if he includes the answer. Anyone who can't tell the difference between C!=- <=> U!=| and NP=Co-NP is just a little too irrational for me. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.lang.lisp.mcl, comp.lang.lisp.franz Followup-To: comp.lang.lisp.franz From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 10 Mar 1994 19:39:42 GMT Subject: Marketing sleaze, revisited In case my message last week got lost in the storm of duplicates to comp.lang.lisp.mcl, I'd like to alert MCLers to an unsavory network practice. I got email last week from sarah@franz.com that read in part: : I read your message on the newsgroup regarding Apple's announcement : to discontinue their support and development of MCL. As I'm sure : you realize, your sentiment is shared by most all MCL users. The rest of the email, in its entirety, was a form letter containing a marketing survey and an invitation to join a mailing list for Franz, Inc. products. I don't think posting an opinion about MCL or Apple should subject the poster to junk email from Franz and other software providers. Furthermore, I find the ingratiating platitudes quoted above to be an insincere ruse to make the form letter appear to be relevant to my Usenet post. (In my case that attempt failed hilariously, because my post didn't even mention MCL or Apple. That's the sort of clerical error that is easy to make when you send out form letters in bulk.) If this practice continues, any time you post to Usenet you may be subjected to mass mailings from any advertiser who thinks they might make money on you. It's as bad as postal junk mail. Worse, in fact: the lower cost to the huckster will lead to greater abuse. And in a medium where anyone can ask for the information they want, sending junk email to people who don't ask for information is gratuitously obnoxious. I call on Franz, Inc. and Sarah Molyneux to discontinue sending form letters to addresses they cull from Usenet posts, except when the poster asks for help finding their products. I think they should make their policy public so that their current and prospective customers can determine whether they are dealing with a reputable firm or a public nuisance. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 11 Mar 1994 22:45:06 GMT Subject: Re: More on Marilyn With respect to the so-often-repeated problem: >> Suppose you're on a game show, and you're given the choice of >> three doors. Behind one door is a car, the others, goats. You >> pick a door, say #1, and the host, who knows what's behind the >> doors, opens another door, say #3, which has a goat. He says to >> you, "Do you want to pick door #2?" Is it to your advantage to >> switch your choice of doors? f...@peregrine.Eng.Sun.COM (Ed Falk) writes: > ... The original problem states that the host KNOWS what's behind > which door, and that he DOES give the contestant a choice, not that > he sometimes give a choice. I assume you're saying "choice" for "chance to switch", and that the "original problem" you're talking about is the one above that you quoted in your post. You should realize that the problem statement doesn't include the word "always" any more than it includes the word "sometimes". It says the host gave you a chance to switch in the one observed case. It does not say he gives this opportunity as a rule. And since he knows what's behind the door you chose, he may have decided whether to give you that opportunity based on that knowledge. If you were on the show, and the host offered you a choice of doors, and you picked door #2 instead, and instead of offering you a chance to switch, he opened up door #2 and said "Ha, ha, loser, take your goat and go home, and here's a Spiegel catalog in case you want to buy a car with your own money," you wouldn't get anywhere claiming he'd broken the rules. If you have won money from hosts who did not realize they had this option, it doesn't mean they didn't have the option. And if you just happen to be dealing with a host who takes advantage of this option when he gets the chance, switching when you get the chance will always lose. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.usage.english From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Sat, 26 Mar 1994 05:58:54 GMT Subject: Re: The curse of the apostrophe da...@s0dl.exnet.com (Dave Lockwood) writes: > Normally, singular nouns prefix the possessive 's' with the > apostrophe, and plural nouns suffix it. The exception is a singular > noun already ending in 's' where suffixing is used.... And he...@ucthpx.uct.ac.za (Heidi de Wet) responds: > Wrong, wrong, wrong. Singular nouns take "'s", except in a few > cases where they end in "s" already ("James's book" is correct, but > "Jesus' words" is fairly standard.).... Wrong indeed, but if I were he I wouldn't take your word for it. Who says it's right? I won't take Dave Lockwood's word for it, either, especially if he's too careless to get "bizarre" right. In fact, both of your willingness to hand down your stylistic preference as authoritative without further justification is, to me, a reason to disregard your judgment on matters of style. But there are other ways of arriving at a decision. The Associated Press style book says to use "'s" for single-syllable words but a bare apostrophe for longer words--a bizarre rule. It made me wonder last summer when the New York Times reported that Andrew Wiles had proved Fermat's Last Theorem. They mentioned "Wiles'" proof on page 1 and "Wiles's" proof on page D22. I wondered which is the Times's style, or whether it might be a disagreement over the number of syllables in "Wiles". Or his age? Strunk and White make exceptions for "ancient proper names" without explanation. And is theirs a reasoned decision? Or just a cowardly unwillingness to take King James to task for the spelling of "Jesus'"? Fowler goes into more detail: It was formerly customary, when a word ended in "-s", to write its possessive with an apostrophe but no additional "s", e.g. Mars' hill, Venus' Bath, Achilles' thews. In verse and in poetic or reverential contexts, this custom is retained.... But elsewhere we now usually add the "s" and the syllable--always when the word is monosyllabic, and preferably when it is longer, Charles's Wain, St. James's Street, Jones's children, the Rev. Septimus's surplice, Pythagoras's doctrines. (So it's clear that the front page editor was simply being properly worshipful.) But Fowler still doesn't justify the syllabic criterion. Still, my curiosity is aroused as to where Dave Lockwood got his singularly bizarre idea. Is he one of those who follows Fowler's former custom? Is he immersed in a dialect that follows this custom? Or did he learn his style from the ragtag rabble of Usenet, who are quite willing to speak of "Oz' tin woodsman's ax' head" now that their schoolteacher isn't around to rap their knuckles for it? Dan Hoey - Hoey@AIC.NRL.Navy.Mil - Striking like a boat from the blue --- Article 830 of alt.humor.best-of-usenet: From: hoey@AIC.NRL.Navy.Mil Newsgroups: alt.humor.best-of-usenet Subject: [sci.math] Re: Stupid Question Date: 4 Apr 1994 02:47:00 GMT Approved: best@cc.ysu.edu Originator: doug@unix1.cc.ysu.edu Newsgroups: sci.math From: mmeyer@m2.rts.dseg.ti.com (Mark Meyer) Subject: Re: Stupid Question drobot@sjsumcs.sjsu.edu (Vladimir Drobot) said: vd>Try to test [0!=1] in your C program. It will tell you that it is true. On Tue, 22 Mar 94 20:24:29 GMT, jplee@galaxy.csc.calpoly.edu (Jason Lee) said: jl> What are you talking about? What C program? Could you be any jl> less specific? *sigh* THIS C program: #include main() { if (0!=1) printf("It's true!\n"); else printf("It's bogus!\n"); } Get it now? -- Postings to alt.humor.best-of-usenet reflect what the submittor considers to be the best in usenet humor, and the poster is responsible for the content. The moderator removes duplicates, copyrighted material, posts without headers, but does not drop articles based on content. See the group charter for more info. Sigs may be truncated. Moderator address: best@cc.ysu.edu --- Newsgroups: alt.fan.wodehouse From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 6 Apr 1994 22:42:13 GMT Subject: Brass rags and fish slices When the wedding is removed from the agenda, the formerly-affianced parties are said to have ``parted brass rags''. I'm smart enough to figure out why I can't find any bespoke tailors in this town, but I'm not smart enough to figure out where the brass rags come from. And is there any significance to the fact that after the reconciliation, Bertie always provides a fish slice? I mean, as opposed to any other habitual and relatively useless gift. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.fan.wodehouse From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 7 Apr 1994 22:44:09 GMT Subject: Fish slices, brass rags, sponge bags, and now silver rings Yesterday I posted to find out the significance of fish slices, but it turns out I should have just asked Anne Werkheiser. It seems that about the time Master Plum was in kneepants learning the ways of aunts, fish slices had been very recently invented. They were considered the height of elegance, and would make a perfect wedding gift. Of course, the constant repetition makes the point that to Bertie, one wedding is just like another. I don't recall a scene where Bertie says someone is getting married and asking Aunt Dahlia what should he do about it, and getting the answer ``buy them a fish slice'', but it's not hard to deduce it from the sequelae. [Inciden- tally, the tables later turned. By the 1920's, fish slices were the consigned to the depths of Non-U, along with abbreviations like 'phone and names like Desmond and Norman. But of course, Bertie and Jeeves remained in the previous century.] Roger.Coll...@Uk.Sun.COM notes that we have only heard of the intent to give a fish-slice, not its actual delivery. But have we ever heard of the non-delivery of the fish slice (in those cases that the wedding eventually occurs)? My dictionary tells me that a sponge bag is a waterproof case for carrying a bath sponge and toiletries, while the adjective ``sponge-bag'' means ``checked''. The source given for the latter term is the checked cloth of which sponge bags are often made. I still want to know why broken-up fiances are said to have ``parted brass rags''. I am able to tell whether my cat is in an adage without asking a veterinarian, but this brass rag thing is a complete mystery. Also, Anne and I want to know what a ``Silver Ring bookie'' is, or how the turf betting system works in general. The phrase is used so often, we're sure there's some amusing misconception involved. But we can't tell without knowing anything about the situation it's based on. Clues? Dan Hoey -- Hoey@AIC.NRL.Navy.Mil -- Like a boat from the blue --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 8 Apr 1994 00:07:43 GMT Subject: Re: SUMMARY: card shuffling algorithms vaugh...@cs.man.ac.uk (Vaughan Marks) offers the following card shuffling algorithm: > Card deck[52]; /* the deck of cards */ > void shuffle() > /* shuffle a deck of cards using random() */ > { >Card tmp, left=51, n; > while (left) { > n=random() % left; /* pick a rand card */ > tmp=deck[left]; > deck[left--]=deck[n]; /* move the chosen card */ > deck[n]=tmp; /* pick from those remaining */ > } > } m...@hubcap.clemson.edu (Matthew Saltzman) makes the small point that "n=random() % left" may be badly distributed if random() misbehaves. But there's a much more certain bug: the line in question should be "n=random() % (left+1)" or some other method of producing a random integer in {0,...,left}. As written, the program will not generate uniformly distributed shuffles. For instance, every permutation will be even, and the original last card will _never_ end up in last place. I'm amused by this bug, since I suspect it results from "repairing" a non-bug: the correct algorithm sometimes exchanges a card with itself. I wonder if there is any quantitative way of comparing the nonuniformity caused by this bug with the nonuniformity caused by the FCB* of using n=random()%52. jo...@rogue.Princeton.EDU (Joel Evan Rosenberg) notes that Vaughan is storing a bridge deal in 13 bytes, while the information should be compressible into 12 bytes. [ 2^96 > C(52,13).C(39,13).C(26,13), where C(n,k) denotes the binomial coefficient n!/(k!(n-k)!). ] But how, Joel asks, can this be done so that the cards in each hand can be determined quickly? If we can perform 96-bit arithmetic sufficiently quickly, this boils down to finding a bijective mapping from {0,..,C(n,k)} to the set of k-element subsets of {0,...,n-1}. This can be done as follows. Given as input integers 0<=k<=n and N in {0,..,C(n,k)}, 1. If k is zero, stop. 2. If N>=C(n-1,k), output n-1, decrease N by C(n-1,k), and decrement k. In this case, N will now be in {0,...,C(n-1,k-1)}. 3. Decrease n by 1 and go back to step 1. This will output the set in decreasing order. Now given an integer NN representing a bridge deal, first form a vector deck[52] containing the cards and set top=51. We will run the above algorithm once for each the first three players. When the above algorithm outputs a number t, we will deal deck[t] to the designated player and set deck[t]=deck[top--]. Then to deal the hand, 1. Run the above algorithm with N=NN%C(52,13), n=52, and k=13, dealing cards to the first player. 2. Set NN=Floor(NN/C(52,13)). 3. Run the algorithm with N=NN%C(39,13), n=39, and k=13, dealing cards to the second player. 4. Run the algorithm with N=Floor(NN/C(39,13)), n=26, and k=13, dealing cards to the third player. 5. Give all the remaining cards to the fourth player. The only drawback of this method is that the cards are not dealt in a useful order. It should be possible to deal them in order, at the expense of an increase in bookkeeping and a greater use of high-precision arithmetic. Dan Hoey -- Hoey@AIC.NRL.Navy.Mil -- * FCB = Frequently Committed Bug --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 8 Apr 1994 01:05:34 GMT Subject: Re: Not the famous game show problem, but similar. wcl...@remus.rutgers.edu (Bill Clark) writes: > "I have a sibling and one of us is a boy, but I don't remember which." > Then the probability *would* be 1/3 that both are boys. G'G, G'B, > B'G, and B'B are again the possible scenerios, all equally probable. > G'G is eliminated, and we are left with G'B (which was not possible > in the original case), B'G, and B'B, again all equally probable. > The probability that B'B (which is the *same* as BB') is the case > is exactly 1/3. Close, but not quite. The problem is, you are the same person volunteering this information as the one referred to. Now, if you and your sibling, after deciding on your posting, have submitted to randomized sex-changes until one of you is a boy, then I agree that you have selected your sex uniformly from {G'B, B'G, B'B}. I doubt you have done so. [And I do not accept "we did, and we were accepted at the outset." I insist you start with the operation, as proof that your intentions are as earnest as those girls with a sister who carry out this project.] If you are in fact talking about some other pair of siblings chosen by rejecting randomly sampled two-child families until you find one with a boy, or if you are talking about some hypothetical pair of siblings whose sex is generated in such a way, then that would be a uniform selection. But in that case you should have said so instead of using the first-person pronoun. But let us suppose you are talking about your real self and your real sibling, with your sex given in advance. I will consider factors such as that you post on this subject to sci.math, and that your name is Bill, and that the birth rate is not exactly 50-50 to be suspended from consideration here. I now ask, if you had been a girl with a sister and the same form of amnesia, what would you have posted? I propose that you would have made your point with the statement: "I have a sibling and one of us is a girl, but I don't remember which." ^^^^ Furthermore, I believe that if you have or had a different gender from your sibling, you would be as likely to make the statement about a girl as about a boy. Certainly I've seen this sort of problem posed for both sexes, and I don't see any reason why children of mixed families might prefer one statement to the other. So if we initially started out with 8n pairs of siblings, consisting of 2n each of the pairs {G'G, G'B, B'G, B'B}, all formulating their statement for sci.math about this topic, how many of them would volunteer that they come from a family with a boy? Certainly all 2n of the B'Bs would volunteer your statement, and I believe that n of the G'Bs and n of the B'Gs would do so as well. So there are 2n pairs of boys, and 2n mixed pairs: the probability of your sibling having the same gender as you is 1/2. I appreciate that you, and many of the other people posting otherwise are trying to make a valid statistical point about pairs of siblings chosen uniformly from two-child families with a boy. But please do not in that endeavor fail to accurately specify your sample space, and do not be too lazy to actually go and do the sampling. For you will likely end up proving the answer to the wrong problem, producing the wrong answer to the problem you posed. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Sat, 9 Apr 1994 19:56:20 GMT Subject: Re: Not the famous game show problem, but similar. Two days ago, I commented on the statement: > "I have a sibling and one of us is a boy, but I don't remember which." In reply, wcl...@romulus.rutgers.edu (Bill Clark) writes: > "X has a sibling and at least one of them is a boy, what is the > probability that both of them are boys?" which obscures an essential condition to my remarks, that the family was determined before he decided on what statement to make. > In order to use the correct sample space, you can only take one > representative from each pair of siblings. The B'B families would > *not* have twice as many representatives in the sample space.... > We are not supposed to double sample the B'B families, and so the > probability stays at 1/3. I've condensed Bill's post, since it all arises from him missing my point entirely. I was not double-sampling the B'B families, I was half-sampling the G'B and B'G families. If you decide what family the statement is about before you make up the statement, then for half of the mixed families you will end up making the analogous statement about there being a girl in the family rather than about there being a boy. In contrast, when you have a B'B family, you always make the statement about a boy, because the statement about a girl is false. This is what makes the probability 1/2. If you don't think that when referrinng to a mixed family, you will as often make the statement about a girl as about a boy, then consider this: when you _do_ make the statement about a girl, do you then consider the probability of there being two sisters _greater_ than 1/2, because for a mixed family you would more likely have mentioned a boy? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 19 Apr 1994 21:21:51 GMT Subject: Wu-Yi Hsiang's paper on Kepler's sphere-packing problem In 1990, Wu-Yi Hsiang announced a proof of Kepler's conjecture that the densest packing of spheres in three dimensions is achieved by the known hexagonal packings. In the Spring, 1994 issue of _Mathematical Intelligencer_, there is a letter from J.H. Conway, T.C. Hales, D.J. Muder, and N.J.A. Sloane (all of them sphere-packing researchers) stating their opinion that Hsiang's work does not constitute a proof of the conjecture. Hales will present a fuller exposition of the objections in the next issue of the _Intelligencer_. They note that Hsiang's paper will be published soon in _International Journal of Mathematics_. I hesitate to write of this in sci.math, since the local toxic waste will presumably respond to the mention of one of the mathematicians on his enemies list. I encourage all readers to ignore the expected Ludifoggery. Wrestling with a pig only widens its sty. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 21 Apr 1994 14:18:11 GMT Subject: Re: Weird little number puzzle d_no...@ondec.lonestar.org (David G. North, CCP) writes: > n-1 > i i P (x+i) > x 2 1 i=0 > S ... S S (1) = --------- > i =1 i =1 i =1 n! > n 1 0 > where "S" is summation and "P" is product. First off, your equation is wrong. For instance, when n=1,x=2, the left side has two summation signs and equals 3, while the right side equals 2. I expect you meant for there to be only n summation signs, for instance: n-1 i i P (x+i) x 3 2 i=0 S ... S S (1) = --------- i =1 i =1 i =1 n! n 2 1 Assuming this, on the left hand side, you are counting the number of nondecreasing sequences {i[1],i[2],...,i[n]} on {1,2,...,x} (I use [brackets] for subscripts). The way to count these is to use what I call the ``spreading-out trick'': We count the number of strictly increasing sequences {i[1],i[2]+1,i[3]+2,...,i[n]+n-1} on {1,2,...,x+n-1}. But the increasing sequences are in one-to-one correspondence with the subsets. So what we are counting is the number of n-element subsets of an (x+n-1)-element set. This is the binomial coefficient C(x+n-1,n) = (x+n-1)!/(n!(x-1)!), which is your right hand side. If you don't recognize the binomial coefficient (usually written as two numbers, one above another, in parentheses, and sometimes pronounced ``choose'') you can read about it in an introductory math book for high schoolers or possibly college freshmen (though I remember a mathematics graduate student who claimed never to have heard of them--it boggles the mind). The spreading-out trick is in the first chapter of combinatorics texts. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: news.admin.policy, news.admin.misc From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 21 Apr 1994 20:34:31 GMT Subject: Re: sngmedia@world.std.com spams comp.* kentb...@world.std.com (Kent Borg) writes: > Looks like the World is acting responsibily in turning off the > sngmedia account but I wish they would be more vocal about it.... So where are the cancels? According to Seth Breidbart, the sgnmedia people _want_ to cancel their message but don't know how. But the only cancel I've seen is from a moderator. Is there nobody at world.std.com to do this? I realize a lot of admins don't accept cancels, but for those who do--including those who check the batch and realize these are appropriate fodder--a little cancel script would get the ball rolling. And for a lot of low-traffic groups, an uncancelled spam can generate noise a week later. If we are not going to devote every usenet group to the promulgation of and debate over spam, news admins need to get their tools ready for generating the cancels. I would ask some other site to help out sgnmedia and cancel it for them, were I not afraid of triggering the keystone kancel phenomenon. Anyone who does take on the job had better at least start slow, and stop if they start seeing cancels from someone else coming their way. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- From: hoey@AIC.NRL.Navy.Mil Date: Wed, 27 Apr 94 15:44:33 EDT To: support@netcom.com Subject: Mail abuse by DAVE@DELTA.nis.com Dear Netcom, I'm sure you're aware of the mail solicitation being carried out by Dave@NIS.COM. When this was first reported to Usenet in March, I dropped him a line noting that I considered it abusive. On Monday, I received one of his solicitations. Please stop supporting that network nuisance with MX service. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles, sci.math Followup-To: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 3 May 1994 20:38:13 GMT Subject: Digital Invariants (was Factorions) kevin2...@delphi.com "Kevin S. Brown" writes that there are > ... only finitely many cases where an n-digit number is equal to the sum > of the nth powers of its digits. The known occurrences of this include > 153, 1634, 8208, 9474, .... up to 4679307774... but it isn't known if > 4679307774 is the largest ... According to David Wells's book _The Penguin Dictionary of Curious and Interesting Numbers_ these numbers are called digital invariants. They were mentioned on rec.puzzles in 1992, and Dik Winter provided a list from 1985 that I confirmed with a program I wrote. The following is a list of all 89 base-ten digital invariants. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, 54748, 92727, 93084, 548834, 1741725, 4210818, 9800817, 9926315, 24678050, 24678051, 88593477, 146511208, 472335975, 534494836, 912985153, 4679307774, 32164049650, 32164049651, 40028394225, 42678290603, 44708635679, 49388550606, 82693916578, 94204591914, 28116440335967, 4338281769391370, 4338281769391371, 21897142587612075, 35641594208964132, 35875699062250035, 1517841543307505039, 3289582984443187032, 4498128791164624869, 4929273885928088826, 63105425988599693916, 128468643043731391252, 449177399146038697307, 21887696841122916288858, 27879694893054074471405, 27907865009977052567814, 28361281321319229463398, 35452590104031691935943, 174088005938065293023722, 188451485447897896036875, 239313664430041569350093, 1550475334214501539088894, 1553242162893771850669378, 3706907995955475988644380, 3706907995955475988644381, 4422095118095899619457938, 121204998563613372405438066, 121270696006801314328439376, 128851796696487777842012787, 174650464499531377631639254, 177265453171792792366489765, 14607640612971980372614873089, 19008174136254279995012734740, 19008174136254279995012734741, 23866716435523975980390369295, 1145037275765491025924292050346, 1927890457142960697580636236639, 2309092682616190307509695338915, 17333509997782249308725103962772, 186709961001538790100634132976990, 186709961001538790100634132976991, 1122763285329372541592822900204593, 12639369517103790328947807201478392, 12679937780272278566303885594196922, 1219167219625434121569735803609966019, 12815792078366059955099770545296129367, 115132219018763992565095597973971522400, and 115132219018763992565095597973971522401. My program checked that there are no more up to 60 digits, and it is easy to prove that there can be none larger. As a devout zerophilist, I of course include zero on the list. While I believe that zero is, properly speaking, not a one-digit number but a zero-digit number (customarily written with a leading zero), it is a digital invariant in either case. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.folklore.urban, misc.legal Followup-To: misc.legal From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 4 May 1994 22:06:19 GMT Subject: Parking violations (was Re: Driving whilst barefoot) jkirk...@osiris.hmc.edu (Jose Kirkland) wrote of a guy who became drunk and loud in his ex-girlfriend's apartment building's parking lot. Police were called, noticed he was drunk, and told him that he was trespassing and must leave. According to the driver, they said ``that he must take the car he drove there with him or it would be impounded.'' When he drove out of the parking lot, the police arrested him for DUI. v140p...@ubvms.cc.buffalo.edu (Daniel B Case) quotes it and writes: >I heard a story once of someone who was pulled over on the side of >Pennsylvania avenue in Washington DC (very wide, for those of you >who haven't been there) with his engine running talking to a friend >on the sidewalk. Since it was a no-parking zone, some DC parking- >enforcement cop came up to write a ticket. The driver protested that >he wasn't parked as his engine was running. The cop asked to see >his registration. In order to do this, he had to open his glove >compartment, which was locked at the time. He took the key out of >the ignition and was ticketed for parking in a no-parking zone. I suspect this story has been worded to deceive--could it have come from a source with an axe to grind? I've heard [from a lady who was ticketed] that the criterion DC uses for whether a car is parked is how long the car has been stopped, not whether the engine is running or the key is in the ignition [OBUL: You can park anywhere as long as you leave the engine running. Take a spare key so you can lock the door]. It certainly has nothing to do with the width of the street. Dan Case signs as: >Dan "I believe entrapment would work as a defense in both cases" Case d...@gemini.gsfc.nasa.gov (Doug S. Caprette) replies: > Probably not in the second case. From what I read in the papers, DC > does not recognise defense for parking violations. Being cited is > taken as irrefutable proof of guilt. Either you are misrepresenting what you read, or you are reading papers that contain more blatant falsehoods than the ones I read. If you want to know how to contest a parking ticket, you can call 202-767-5000 and listen to DC's own voice mail instructions. > In the first case, I would guess that the cops were hoping the man > would articulate his reasons for refusing to drive. Then they could > point out that he admitted being drunk and run him in for public > intoxification. Since the arrest for DUI was plainly a result of > entrapment, I would guess that any subsequent evidence of > intoxication (e.g. blood test) would be inadmissible. Feh. I certainly don't believe it was entrapment. And I suspect that the police statement quoted above has gotten altered in the third-hand retelling so that it sounds more like an order to DUI than it really was. The police are not allowed to tell him to drive away, but they don't have to tell him not to. But after he does, it's their job to arrest him, and I'm sure it would stick. > Bet he spent the night in jail regardless (and didn't get into any > further trouble, I might add), which is clearly what the cops > wanted. Think again. The original story begins: > i met this fellow at about two in the morning at a bus stop, very > secure situation, you can be sure, and he was drunk, and told me a > story: apparently he had just broken up with a girlfriend.... Sounds like he was released (sans car) the same night. I expect he was required to attend his day in court, if only to plead guilty. That is, if any of the events of the story actually happened--it sure sounds like blarney to me. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: misc.legal From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Mon, 9 May 1994 23:04:35 GMT Subject: Re: Parking violations With regard to v140p...@ubvms.cc.buffalo.edu (Daniel B Case)'s story: >: >I heard a story once of someone who was pulled over on the side of >: >Pennsylvania avenue in Washington DC (very wide, for those of you who >: >haven't been there) with his engine running talking to a friend on >: >the sidewalk. Since it was a no-parking zone, some DC parking- >: >enforcement cop came up to write a ticket. The driver protested that >: >he wasn't parked as his engine was running. The cop asked to see his >: >registration. In order to do this, he had to open his glove >: >compartment, which was locked at the time. He took the key out of the >: >ignition and was ticketed for parking in a no-parking zone. prus...@netcom.com (Brian Fagan) writes: > ...the officer should have given a warning and sent him on his way. > Community policing is the buzzword nowadays. No. Pennsylvania Ave is not a community, it's a bunch of offices and monuments. And in D.C. we need the revenue. Even if it means constant bellyaching by people who think they should be allowed to park illegally for free. On the other hand, a_ru...@dsg4.dse.beckman.com (Arthur Rubin) writes: > Wrong. "Parking" means the engine is stopped. If it was marked "no > stopping", that would be another matter. (Like the "no stopping" > sign next to a maildrop.) Arthur would do better to provide information on a topic he knows something about. As I mentioned earlier in this thread, I believe that in this city, 'parking' refers to leaving your vehicle by the side of the road for more than some specified amount of time--I've heard it's 15 minutes. Whether the engine is running or the driver is present is irrelevant. This is a good thing--there are plenty of people who have noticed that gasoline is cheaper than parking lot fees. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 17 May 1994 16:02:17 GMT Subject: Re: Geometric Construction Puzzle mpra...@magnus.acs.ohio-state.edu (Manish S Prabhu) writes: > Given a straight line, a point and a circle (on the same side of > the line as the point), construct a circle that touches the line, > the circle and passes through the point. What makes this problem difficult is that circles are harder to work with than lines (they aren't as straight). But there is a powerful tool, called _inversion in a circle_, that allows us to turn circles into lines. Besides that, it's worth looking at inversion just because it's pretty. Let P+oo be the plane together with a point oo at infinity. Given any circle C in the plane with center c and radius R, the inversion in C is defined to be a mapping V from P+oo to itself as follows: V(c)=oo, V(oo)=c, and otherwise p and V(p) are points on the same ray from c for which d(c,p) d(c,V(p)) = R^2. It is immediate that V is a bijection from P+oo to itself and that V is its own inverse. What makes inversion useful is that every line or circle in P+oo is mapped to a line or circle in P+oo. (This can be verified easily by calculation; I don't know a geometric proof). The image of any line or circle containing the center c must be a line, since finite circles do not contain V(c)=oo. The mapping preserves tangency of lines and circles. Furthermore, it is easy to construct the image under V of a point, line, or circle. If a point p is inside the circle of inversion C, this may be done by raising the perpendicular from cp at p to intersect C at a point b, then raising the perpendicular to cb at b to intersect the ray cp at a point q; that V(p)=q can be seen from the similar right triangles cpb and cbq. If p is outside C, construct a circle of diameter cp to intersect C at a point b, then drop a perpendicular from b to cp (since cbp is a right angle, this is the previous case taken in reverse). Mapping of a circle or line can be constructed by taking a diameter (of the circle) or perpendicular (to the line) through c, mapping the endpoint(s), and constructing the line or circle on the resulting diameter or perpendicular. Note that the center of a circle is _not_ in general mapped to the center of the image of the circle. So to solve the given construction problem. I will describe the case of the given point and the constructed circle exterior to the given circle; similar methods work for other cases. We first construct a distance H, which is half the distance from the point to the circle. Construct a circle C1 about the given point with radius H; construct a circle C2 concentric with the given circle and with radius greater by H than that of the given circle; construct a line L3 parallel to the given line and H units away in the direction of the given circle and point. We will now construct a circle tangent to C1, C2, and L3. Note that C1 and C2, as constructed, are tangent. If L3 intersects the point of tangency, then consider the point of tangency to be a zero-radius circle tangent to the three of them. If not, take an arbitrary circle C0 about the point of tangency, and invert C1, C2, and L3 in C0. C1 and C2, since they are tangent at the center of C0, will be mapped to parallel lines L1', L2'; L3 will be mapped to a circle C3'. Let K be half the distance between L1' and L2'. Construct the circle concentric with C3' and with a radius greater by K; construct the the parallel line equidistant from L1' and L2'. The circles of radius K with centers at the intersection of the line and circle will be tangent to L1', L2', and C3'. Inverting these circles with respect to C0 yields circles tangent to C1, C2, and L3. We now have at least one circle tangent to C1, C2, and L3. Increasing the radius of such a circle by H will yield a circle, containing the given point and tangent to the given line and circle, Q.E.C. I hope there is a shorter construction; perhaps the inversion can be skipped entirely. But I've mostly bothered writing this so that more people can know about inversion in a circle, a useful and beautiful method. Interested readers might want to try proving that inversion preserves not only tangency, but perpendicularity of lines and circles. Also, consider finding those circles and lines that are invariant under inversion. And find a geometric interpretation for inversion, preferably one that illustrates the preservation of lines circles, and perpendicularity. am...@FreeNet.Carleton.CA (Brice Wightman) answered: > 1. Draw a line parallel to the given line (call this the "parallel") > 2. Drop the perpendicular from the given point to the given line. > 3. Draw a radius of the circle and produce outside of the circumference. > 4. With centre at the intersection of the circumference and the > radius and length equal to the perpendicular previously constructed > strike an arc on the radius produced. > 5. With centre at the centre of the given circle and radius equal to > the distance from the centre to the intersection constructed in (4) > draw a circle. > 6. With centre at the intersection of the circle constructed in (5) > and the "parallel" and radius equal to the length of the > perpendicular, draw a circle. This is the required circle. It is unfortunate for a construction to be given without any hint of why the author considers it a solution to the problem; it is most unfortunate when such a construction is incorrect, as Brice's is. (For instance, the parallel need not intersect the circle in step 6, and even if it does the final circle need not be tangent to, nor even intersect, the given line.) Due to the lack of explanation, I find it impossible to figure out whether the error is due to minor problems with a correct answer, or to a complete failure to understand the techniques and requirements of geometric construction. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.usage.english From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 17 May 1994 21:42:01 GMT Subject: I could care less (was Re: Grammar rules will live on) Quoting a Jack Smith column, sno...@netcom.com writes: > Pinker also defends "I could care less" when "I couldn't care > less" is clearly intended, arguing that the phrase is delivered in a > tone of sarcasm th makes its meaning obvious. I have read that ``I could care less'' arose earlier in this century as a contraction of the phrase, then in vogue, ``I could care less, but I'd have to try.'' I find the sentiment to be a sublime expression of apathy, and am distressed to see it it branded mistaken rather than ironic. Of course, most people who use the phrase today don't know these origins. But they don't know the origins of most words and phrases they use--does that mean their usage is incorrect? Unfortunately, I can't recall just where I found the account of the vogue phrase, or when it was popular. Does anyone have a source or a date? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 19 May 1994 15:48:52 GMT Subject: Re: Is a gross the largest square Fibonacci number? tycc...@banach.mit.edu (Timothy Y. Chow) quotes membrillo@vax.oxford.ac.uk (Fausto H. Membrillo): > The square Fibonacci numbers are {0,1,144}.... > Moreover Ray Steiner proved in 1967 that the cube Fibonacci numbers > are {0,1,8} When this was posted two years ago, I mentioned that we might want to consider the cubes F[-2]=-1 and F[-6]=-8 as well. Of course, F[-12]=-144 doesn't count because it isn't a square. We may wish to count F[-1]=F[1]=F[2]=1 three times. Or not. Is it known if there are other Fibonacci numbers the Nth power of an integer, N>1? I still haven't heard. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Article 6006 of rec.arts.sf.fandom: Newsgroups: rec.arts.sf.fandom From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Subject: Re: How to Run a Con Suite? Date: Tue, 24 May 1994 15:35:03 GMT dwbutler@mtu.edu (Daniel W. Butler-Ehle) writes: > Locally, a 4.5 gallon (17L) cannister of premix soda is $12 > [stuff deleted] > But with a little price shopping, I can get brand X in 2-liter > bottles for about half that. Two-liters don't take an hour to set > up and two hours to clean. Fountain just doesn't make sense. Great logic up until the last sentence (if your fen like Brand X). But the right conclusion is that pre-mix just doesn't make sense. A CO2+syrup system saves a lot of money: the estimate is that Disclave's system paid for itself in four years. And you get to serve the real Brand C stuff (I've heard it lacks both coca and cola these days; perhaps the C stands for caramel, caffeine, and calories). And I, at least, prefer the taste of carbonation to that of phosphation. The downside (other than finding the labor for the care and cleaning) is that you need a source of running water. This year, it's looking like we may not to be able to swing it. Our contingency plan is for bottled soda. In case anyone doesn't know, Disclave is this Friday to Monday at the Sheraton Premier at Tysons Corner, Virginia. It's the big crystal thing on Route 7 at the toll road. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.games.abstract From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Mon, 20 Jun 1994 21:18:49 GMT Subject: Re: The Game of Y Last month there were several messages here about the game "Y", which was reviewed in the June _Games_ magazine. I missed that issue on the newsstands, but I've managed to put together a little information on the game. The game is played on a roughly triangular board that looks something like the following: ____ / \____ / __/ \__ \__/ \__ \____ / \ \____/ \__ / / __/ \__ \____ \ \__/ \__ \____/ \__ / / \ \____/ \__ \__ \__/ / __/ \__ \____/ \__ / \ \__/ \__ \____/ \ \__ / / / \ \____/ \ \____/ \__ \ \__/ / __/ \ \____/ \ \ / / \ \__/ \__ \____/ \ \__ \__ \__/ / / \ \__/ \ \____/ \__/ \__ / \ \__/ / / \ \____/ \ \ \ / / / \ \____/ \____/ \ \__ \__ \__ \ \__/ \__/ \____/ \ \____/ \__/ \__/ \__ / / \ / \ / \ \____/ \ \ \ \ \__/ \__/ \__/ \____/ \__ \__ \__ \__ \__ / \ / \ / \ / \ / \__/ \__/ \__/ \__/ \ \ \__/ \__/ \____/ \__/ \ \ \ \ \ / / \ / \ / \ / \ __/ __/ __/ __/ / \__/ \__/ \__/ \____/ \__/ \__/ \__/ \__/ \__/ / \ / \ / \ / \____/ / / / / \ \__/ \__/ \____/ / \____/ __/ __/ __/ / / \ / \____/ \____/ / \__/ \__/ \__/ \ \ \__/ / \ / \____/ / / / \__/ / \ \ \__/ / \____/ __/ __/ / \ \ \__/ __/ \____/ / \__/ \__/ \ \__/ / \__/ / \____/ / / / / \ \ \____/ / \____/ __/ \ \ \__/ __/ \____/ / \__/ \__/ / \__/ __/ \____/ __/ / \ \ \____/ __/ \__/ \ \__/ __/ \____/ __/ / / \__/ __/ \____/ \ \ \____/ __/ \__/ __/ \____/ / \__/ __/ \ \____/ \____/ The spaces approximate ninety hexagons and three pentagons. Players alternate claiming spaces, and the player wins who claims a connected area touching all three sides of the triangle. Each corner is considered to touch both incident sides. k...@panix.com (Kevin Maroney) suggests that the game "Y" is the same as "Poly-Y", and in a sense it's a special case. Poly-Y is played on a field with a boundary consisting of an odd number of sides. For each pair of adjacent sides, the winner of the pair is the player who forms a connected area touching those two sides and an arbitrary third side. The winner of the game is the player who wins a majority of the pairs. I think the generalization from Y to Poly-Y is somewhat artificial, so I prefer to investigate the former. I'm told that Y and Poly-Y were described in the book _Mudcrack Y and Poly-Y_, by Craige Schensted and Charles Titus, which is now out of print. I understand the section on Y is being revised and will be sent to purchasers of the commercial Y game. I've never played the game, but it has several features that are theoretically satisfying. First, like Hex, the game is drawless in a strong sense. Even if play doesn't always alternate, and even if play continues after a win, when the board is filled there will be one and only one winner. A second feature that makes this nicer than Hex is that the winning configurations are common to both players. A third nice feature is that the board can be considered to be half of an icosahedron, with the sides divided into triangles. The visible vertices of the icosahedron are the three pentagons, the three corners, and the midpoints of the three sides. Note that the board has only six symmetries, though, as opposed to the 120 symmetries of the icosahedron. What would be _really_ nice is a drawless game with common goals that has more symmetry. My first idea was to extend the Y board to the far side of the icosahedron, and look for a family of winning configurations. Unfortunately if the winning positions have the symmetry of the icosahedron, this can't work: The second player could always play opposite the first player, covering the board with congruent patches. This would end up with either zero or two winners, violating drawlessness. I thought of working on some partition based on a tetrahedron, in which there is no inversion mapping to opposite points. But that won't work, either: The rotated inversion mapping yields a draw. Right now, I'm considering whether there might be a way to create a family of winning configurations for the projective icosahedron, in which opposite spaces are identified with each other. It's like playing Y, only every time you play on the boundary, you get the opposite point of the boundary for free. But the family of winning positions of Y is not symmetric enough: There should be sixty symmetries. The idea is that the boundary of the Y board should not have a special role. Another possibility is the chiral icosahedron, where the mirror image of a winning configuration may not be a winning configuration. In either case, I haven't been able to construct a drawless family of configurations, but I haven't proven it's impossible, either. #include /*/ /*/ /*/ /*/ /*/ /*/ /*/ /*/ /*/ /*/ /*/ /*/ /*/ In case you're /*/ /*/ interested in /*/ main /*/ different sizes /*/ (n,v)char** /*/ of Y boards, I've /*/ v;{int m,r,s,t,x /*/ written a small /*/ ,y;if(n!=2||(n=atoi /*/ obfuscoid to /*/ (v[1]))<1)exit(fprintf /*/ print them. I'm /*/ (stderr," Usage: n\n",*v /*/ sorry it's not /*/ ));for(y=5*n;y>=-5*n;--y){ /*/ quite right for /*/ for(x=y-1-6*n;abs(7*y+2)<30* /*/ n=1. /*/ n+10-3*x+3*y;++x){r=x>>1;s=y;m= /*/ /*/ 0;do{for(t=0;t<6;t++&1?r+=s=-s:( /*/ As always, I /*/ s+=r=-r))r>(r+((r==2*n-2&&s<2)^s)+4*n)%6+( /*/ substantive /*/ r&1)*6:3*r+2*s>15*n+5?0:3*r-s>6*n /*/ followups, /*/ -2?409>>(s-(r>s)+6*(r-(r-s<5*n)+ /*/ especially /*/ n))%9:228>>(s+(r-s<=2*n-3)+5* /*/ answers to the /*/ r+6*n)%8)&1);s=r-s+y+y-(x>>1 /*/ open questions. /*/ );r=1+s-r+(x&-2)-y;m*=2;} /*/ Requests for /*/ while(r>x>>1);putchar( /*/ clarification and /*/ " \\ / __??"[ /*/ expressions of /*/ m|x&1]);}putchar /*/ admiration or /*/ ('\n');}exit /*/ other emotions /*/ (0);} /*/ are better sent /*/ /*/ by email. /*/ /*/ Dan Hoey /*/ /*/ Hoey@AIC.NRL.Navy.Mil /*/ --- Article 18222 of rec.puzzles: Newsgroups: sci.math,rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Subject: Kaprekar's process (was Re: Help! Pattern Driving Me Nuts!) Followup-To: rec.puzzles Sender: usenet@ra.nrl.navy.mil Organization: Naval Research Laboratory, Washington, DC Date: Wed, 22 Jun 1994 22:16:51 GMT lhom@OCF.Berkeley.EDU (Louis Hom) writes: > I saw this little activity in a _pre-algrebra_ book.... > If you start with any 4 digit number (with at least one digit > different from the others...) say 3486, then subtract the smallest > number you can make with those digits (i.e., 3468) from the largest > number you can make (i.e., 8643), and then repeat the process with > the resulting difference, and so on down the line, then you > eventually get to 6174, which will continue to yield 6174:.... According to a couple of curious and interesting books by David Wells, the process is called Kaprekar's process, after D. R. Kaprekar; the fixed point is called Kaprekar's constant. > Why does this work and why does it converge on 6174?!? It works because we have five fingers on each hand. That is to say, the only fixed points of the process are in number bases divisible by five (except for the three exceptional cases 0111[2], 1001[2], and 3021[4]); the fixed point has digits 3R/5,R/5-1,4R/5-1,2R/5 in base R. I conjecture that the reason it _always_ works is that the number of our hands is half a power of four, but I don't know why that might hold. For pre-algebra students, this is probably intended as an incentive to practice subtraction (or to purchase a calculator). The way to prove it for all 4-digit numbers is by exhaustive testing, but trying 10,000 numbers would be too exhausting. You really only need to work it for 30 numbers, using tricks that might even be accessible to the bright pre-algebra student with a strong grasp of arithmetic. The first thing to notice is that we don't have to work out every four-digit number, as long as we work out every number that can result from one step of the process. Now if ABCD is a four-digit number, ABCD - DCBA = (A00D-D00A) + (BC0-CB0) = (A-D)*999 + (B-C)*90. Multiples of 999 and 90 are of the form X99Y and XY0, where X and Y are digits summing to 9. If the multiple of 90 is nonzero, the sum is a number of the form STUV with S+V=10 and T+U=8. Note that the next step of the process does not depend on the order of (S,V) or of (T,U). So immediately we have reduced ourselves to considering five numbers of the form X99Y (modulo the order of (X,Y)) and 25 numbers of the form STUV. It's a short exercise to find that they all lead to 6174. I think all the above could be done with arithmetical reasoning on the level of cryptarithms. Now if we want a fixed point in an arbitrary base R, it will be a number of the form X99Y or STUV (here "9" refers to the digit R-1) with X+Y+1 = S+V = T+U+2 = R. If a>=b>=c>=d are the digits in descending order, we may solve one of the systems of equations {b=c,a-d+Y=R} or {a-d=S, b-c=T-1} together with one of the 24 assignments of {a,b,c,d} to {X,R-1,R-1,Y} or {S,T,U,V}, respectively. We usually find that the solutions do not lie among {0,...,R-1} or violate a>=b>=c>=d, but there are the cases a=Y=R/2, b=R-1, c=R-1, d=X=0 for R=2 (0111[2]), a=S=3, b=U=R-2, c=V=R-3, d=T=0 for R in {4,5} (3021[4], 3032[5]), a=S=3-R, b=V=2r-3, c=T=R-2, d=U=0 for R=2 (1001[2]), a=S=1, b=V=R-1, c=U=R-2, d=T=0 for R=2 (1001[2]), a=V=1, b=S=R-1, c=U=2R-4, d=T=2-R for R=2 (1001[2]). a=V=R-1, b=S=1, c=T=0, d=U=R-2 for R=2 (1001[2]). a=U=4R/5-1, b=S=3R/5, c=V=2R/5, d=T=R/5-1 for R divisible by 5. So the only fixed points are 0111[2], 1001[2], 3021[4] and (3R/5),(R/5-1),(4R/5-1),(2R/5) in base R. The existence of a Kaprekar's constant for a given base does not, however, imply that all four-digit numbers eventually reach that constant. For instance, in base 4, some of the numbers reach 3021[4], while others end up cycling between 1332[4] and 2022[4]. And in base 15, there is the fixed point 92B6[15] and the cycles (D672[15], B0D4[15]), (50DA[15], D492[15], B494[15], 7498[15]), (B854[15], 72B8[15], 90D6[15], D2B2[15]), (9496[15], 52BA[15]), (D852[15], B2B4[15], 9676[15], 30DC[15]), and (D0D2[15], DA32[15], B674[15], 70D8[15]). I wrote a program that tells me the only unavoidable Kaprekar constants for bases less than 100 occur with bases 5, 10, and 40. I have since checked that the Kaprekar constants for bases 160 and 640 are unavoidable, while base 320 has cycles. So I will conjecture that bases of the form 10*4^n may have unavoidable Kaprekar constants (with base 5 being an exception). But I have no idea of how one might prove it. Followups are set to rec.puzzles, where such number puzzles are customary. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.games.abstract From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 23 Jun 1994 23:32:18 GMT Subject: Re: The Game of Y In reply to my consideration of > ... whether there might be a way to create a > family of winning configurations for the projective icosahedron, w...@math.canterbury.ac.nz (Bill Taylor) writes: > However, it seems to me there's an obvious candidate rule sticking > out like a sore thumb. Namely, a winning position is one which > includes a closed loop which is non-contractable-to-a-point > (topologically). Yes! It is the obvious choice, but it takes someone like Bill, who knows what to look for in projective planes, to see it. For people who aren't comfortable with contractability, recall that we are playing on a Y board, with the proviso that any move on the boundary will claim two opposite spaces, halfway around the boundary from each other. The winner is the player who can form a closed loop of spaces in which 1. each pair of consecutive spaces in the loop is either a pair of adjacent spaces or a pair of opposite boundary spaces, and 2. there are an odd number of the latter pairs in the loop. In the following diagram, for instance, the loop 1,2,...,14 is a win (three opposite pairs 3-4, 6-7, 11-12) but neither A,B,C,D (two opposite pairs A-B, C-D) nor Q,R,S,T,U,V (none) is. ____ / \____ / 12__/ \__ \__/ \__ 3 \__ / \ \____/ \__ / / __/ 2 \ 7 \__ \ 13\__/ \__ \____/ \__ / / \ \__/ \ B \ \__/ 14/ 1 / \__ 8\__ \__ / \ \____/ R / \__/ \__/ \ \ \__/ Q \__/ \ 9 \ C \ / D / \____/ \ S __/ __/ / \__/ / \ \__/ \__/ \__/ / \ \ V \__/ T / / \ \__/ __/ \____/ 10__/ / A / \__/ U / \__/ \ \ 5 \____/ 11__/ \__/ __/ \__/ / \__/ 4 __/ \ 6 \____/ \____/ I think this might make a really good game. > >Another possibility is the chiral icosahedron, > OK, you've got me at last! What on earth is this? ``Chiral'' is Lord Kelvin's word for ``having handedness'' or ``being distinguishable from its reflection''. You can use ``enantiomorphic'' if you prefer. What I meant is that you play on the full icosahedron (two Y boards joined at the boundary), but the mirror image of a winning position is not always a winning position--the symmetry group is the group of _proper_ rotations of the icosahedron, not involving mirror-reflection. This would prevent the copycat strategy by having the copycat end up with a mirror image of the desired position. I don't know of a satisfying set of winning positions, though--it's just a possibility. I'm so happy with your solution for the projective icosahedron, though, that I'm not particularly concerned about it. By the way, I've updated my obfuscoid to deal with n=1, and fixed its error message. Enjoy. #include /****** Dan */ int m,r,s,t,x,y;main /* ____ ***/ (n,v)char**v;{if(n==2&&(n= atoi(v[1]))>0)for(y /* / \__ ***/ =5*n;y>=-5*n;--y){for(x =y-1-6*n;abs(7*y+2 /* / __/ \__ ***/ )<30*n+10-3*x+3*y;++ x){r=x>>1;s=y;m=0; /* \__/ \ \__ */ do{for(t=0;t<6;t++ &1?r+=s=-s:(s+=r=- /* / \ \____/ \ */ r))r>(r+(( /* \ \__/ \ \ */ r==2*n-2&&s<2&&n> 1)^s)+4*n)%6+(r&1) /* / / \____/ / */ *6:3*r+2*s>15*n+5 ?0:3*r-s>6*n-2?409 /* \__/ / \__/ */ >>(s-(r>s)+6*(r-( r-s<5*n)+n))%9:228 /* / \__/ __/ */ >>(s+(r-s<2*n-2||n <2)+5*r+6*n)%8)&1) /* \ \__/ ***/ ;s=r-s+y+y-(x>>1);r= 1+s-r+(x&-2)-y;m*=2 /* \____/ ***/ ;}while(r>x>>1);putchar (" \\ / __??" /* ***/ [m|x&1]);} putchar('\n'); }else fprintf(stderr, /******/ "Usage: %s n\n",*v);exit(0);} Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 30 Jun 1994 22:05:47 GMT Subject: Multiplicative Persistence (was: Solve this funny problem...) fra...@davina.inria.fr (Franck ) writes: > Take one number, 47 for example, and multiply its digits [together], > until you have only one digit : > 4*7 = 28 > 2*8 = 16 > 1*6 = 6 > This number is said to have a peridicity of 3, because we have > processed 3 multiplications before having one digit. .. > A researcher tells me that i can't find a number with periodicity > 12, 13 .... As p...@eurocontrol.fr (Philip Gibbs) noticed, Franck used both `peridicity' and `periodicity' in his message. I have no idea how the first might have been derived, and the second is a rank misnomer. Fortunately, the term `multiplicative persistence' is already established in the literature for this concept. David Wells's _The Penguin Dictionary of Curious and Interesting Numbers_ cites N.J.A. Sloane, `Multiplicative Persistence', _Journal of Recreational Mathematics_, vol. 6. Wells [presumably summarizing Sloane] states, ``No number less than 10^50 has a higher multiplicative persistence [than 11] and it is conjectured that there is an upper limit to the multiplicative persistence of any number.'' Philip noted that the only nonzero numbers that can arise from multiplying the digits of a number have all their prime factors less than ten. I will call such numbers `low-pf numbers.' I also wrote a program to find the low-pf numbers of high persistence. Up to 10^100, the only ones with persistence greater than 6 are the following: MP(338688) = MP(2^8 3^3 7^2) = 7 MP(826686) = MP(2 3^10 7) = 7 MP(2239488) = MP(2^10 3^7) = 7 MP(3188646) = MP(2 3^13) = 7 MP(6613488) = MP(2^4 3^10 7) = 7 MP(14224896) = MP(2^9 3^4 7^3) = 7 MP(3416267673274176) = MP(2^6 3^27 7) = 7 MP(6499837226778624) = MP(2^24 3^18) = 7 MP(4478976) = MP(2^11 3^7) = 8 MP(784147392) = MP(2^6 3^6 7^5) = 8 MP(19421724672) = MP(2^21 3^3 7^3) = 8 MP(249143169618) = MP(2 3^2 7^12) = 8 MP(717233481216) = MP(2^9 3^5 7^8) = 8 MP(438939648) = MP(2^12 3^7 7^2) = 9 MP(231928233984) = MP(2^33 3^3) = 9 MP(4996238671872) = MP(2^19 3^4 7^6) = 10 MP(937638166841712) = MP(2^4 3^20 7^5) = 10 These include several numbers missing from Philip Gibbs's table, and omit those numbers in his table that contained the digit zero. [I don't know why Philip included the latter. He said something about allowing the digit zero, but if the digit zero is ignored, then the conjecture is false: for instance MP0(2^270)=12, MP0(2^872)=13.] The smallest numbers of large persistence are then MP(2677889)=8, MP(26888999)=9, MP(3778888999)=10, and MP(277777788888899)=11. The first two improve Philip's results--I suspect he didn't consider using the digit 6. I have a stronger conjecture than the one Wells provided, though I suspect Sloane had it. Note that the numbers I listed are all much less than 10^100. I noticed that the low-pf numbers of persistence 3, 4, 5, and 6 also seem to dwindle well before 10^100. If no more appear, then the following are the maximum low-pf numbers for their persistence. MP(1438916737499136) = MP(2^24 3^6 7^6) = 6 MP(36381499733311488) = MP(2^35 3^2 7^6) = 5 MP(6863918177676863471616) = MP(2^59 3^5 7^2) = 4 MP(268476843674491723183742239912919243314992) = MP(2^4 3^17 7^38) = 3 The low-pf numbers of persistence 2 continue past 10^100, but appear to be dwindling as well. This is not surprising: The number of N-digit low-pf numbers increases quadratically with N, but (assuming the digit "5" appears with positive density) the probability that the product of the digits is nonzero decreases exponentially. Thus the expected number of low-pf numbers with persistence greater than one converges. So I conjecture: Of the positive integers whose prime factors are less than ten, only finitely many have multiplicative persistence greater than one. or equivalently Of the positive integers whose decimal representation lacks the digit "1", only finitely many have multiplicative persistence greater than two. If this conjecture is true, then it is may be one of those that is theoretically, but not practically, provable. wieck...@deca.cs.umn.edu (Zbigniew Wieckowski) said that Banach spaces are somehow related here. I would be interested to hear an explanation of the connection, unless it is as superficial as his remarks about Kaprekar's process. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 1 Jul 1994 21:31:08 GMT Subject: Re: Rubik Nightmare (a.k.a. Rubik's Revenge) Ben.Fow...@launchpad.unc.edu (benjamin peter fowler) writes: >On seeing it again, I decided to solve it, and did so, save for the >fact that one of the (edge) red faces was left on the orange side, and >one of the orange faces (naturally) on the red side. There was no >(obvious) way to interchange them. a...@jupiter.cs.swin.oz.au (Alan Christiansen) replies: > This description sounds like you are saying that one physical object > (an edge) is flipped. The edge is one of the two edges between the > red and the orange face. This is indeed impossible even with disassembly. No, that's not at all what he said. He said that two faces have have one unmatched edge colortab each. Without considering the what is mechanically possible with the cube, there could be two explanations of the observation. One explanation is that an edge piece common to the two faces is flipped. Another explanation is that there is a third face adjacent to the first two, and that an edge piece common to the first and third is exchanged with an edge piece commmon to the second and third, with the colortabs from the third face remaining on the third face. The third face then appears solved, since it is a solid color. Since the former situation is mechanically impossible on the 4^3, we must be dealing with the latter. There are two possible situations even then, since the first and second faces may be either opposite or adjacent. So the situation on the third face will be one of the following (or the mirror image), with the swapped edges marked (*): c e e c c e * c e f f e e f f e e f f * e f f e c * e c c * e c adjacent opposite There are three other pairs of edges (up to symmetry) but swapping them would put the wrong colortabs on the third face. A conceptually easy way of solving either of these positions is: 1. Put both edge pieces on the same inner slab of the cube. Remember what turns you use. 2. Turn the slab 90 degrees. 3. Use commutators to undo the 90-degree turn of the face centers of the slab. 4. Use commutators to perform the permutation of the edge pieces of the slab to make step 5 do the right thing. 5. Do the inverse of step 1. It's probably a ten- or twenty-turn process, but I don't have the time to figure it out right now. Also, I think there is a shortcut for step 3 that I forget. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 1 Jul 1994 21:49:06 GMT Subject: Re: Multiplicative Persistence (was: Solve this funny problem...) Yesterday I wrote: > The number of N-digit low-pf numbers increases quadratically with > N, but (assuming the digit "5" appears with positive density) the > probability that the product of the digits is nonzero decreases > exponentially. I should of course have spoken of the digit "0". Also, I have extended the search to 10^200, and it appears that the largest low-pf number of multiplicative persistence greater than one is the 140-digit number 2^25 3^227 7^28. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Mon, 4 Jul 1994 01:15:29 GMT Subject: Re: Rubik Nightmare (a.k.a. Rubik's Revenge) Two days ago I wrote of the problem of exchanging the following pairs of edge pieces on one face of a 4^3 Rubiks cube. > c e e c c e * c > e f f e e f f e > e f f * e f f e > c * e c c * e c > adjacent opposite I've put together processes for these now. For people unfamiliar with my notation: 0. Moves consist of turning a 1x4x4 slab of the cube by ninety degrees with respect to the rest of the cube. 1. Clockwise turns of the outer slabs are labeled by the first letters of the words Back, Front, Top, Down, Left, and Right. 2. Clockwise turns of the adjacent inner slabs are suffixed "i". 3. Anticlockwise turns (and inverses generally) are suffixed "'". 4. Half turns are suffixed "^2", repeated operations generally are suffixed "^n". Let us suppose the pictures refer to the F face in an upright position. The processes can be done as follows: > 1. Put both edge pieces on the same inner slab of the cube. For the adjacent case, RT; for the opposite case FRF'T. This puts both in the Li slab, at the DF and TF positions. > 2. Turn the slab 90 degrees. This is Li. > 3. Use commutators to undo the 90-degree turn of the face centers > of the slab. As I mentioned, there is a shortcut for this: (Li' T^2)^6. Actually, the shortcut is a five-cycle of pairs of face centers, but it doesn't matter unless like-colored centers are distinguished. > 4. Use commutators to perform the permutation of the edge pieces of > the slab to make step 5 do the right thing. I use T'R'T Li T'RT Li T'R'T Li^2 T'RT. > 5. Do the inverse of step 1. That's T'R' or T'FR'F'. I've chosen conjugates of some of these processes so there is cancellation to form the 29-qt process RT' (Li' T^2)^4 Li' TR'T Li T'RT Li T'R'T Li^2 T' for the adjacent case. The 35-qt process FRF'T' (Li' T^2)^4 Li' TR'T Li T'RT Li T'R'T Li^2 T' RFR'F' for the opposite case is not so felicitous; I'm sure there's a better way for this (call it an ObPuzzle). If I hadn't been willing to allow permuting like-colored face centers in step 3, a cyclic permutation of the adjacent case would work. What goes around, comes around. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 12 Jul 1994 21:42:38 GMT Subject: Re: COMMON DIVISORS Barry Paul (barry.p...@hofbbs.com) writes: > Find a sequence of N consecutive positive integers such that each > member of the sequence shares a common divisor with at least one > other member of the sequence.... gd...@cl.cam.ac.uk (Gareth Rees) writes: > Spoilers > Choose N = 17 and let the first number of the sequence be 1284. Did you transpose two digits of your answer just to provoke interest? No, that would be too coy--but it worked! > I have a rather shaky argument that suggests that unless there's a > solution for N < 5 (which there isn't by examination) then we want to > pick N such that > N N N N N N N N > - + - + - - - + - - -- + -- + ... +/- --- < N > 2 3 5 6 7 10 11 N-1 > where the division is integer, the denominators are those numbers > greater than or equal to 2 which have no repeated prime factors, and the > sign of each term is positive if the denominator is prime, negative > otherwise. Well, you're right about it being shaky. For instance, the formula holds for N=7 and other numbers less than 17. Also, since it seems the left hand side is counting the number of sequence elements that share a factor with a different element, the sense of the equality should be reversed--but then it holds for N=6. Unless we want strict inequality, in which case it doesn't hold for any number less than 1000. Unless "integer division" means with rounding, in which case it holds for N=8. But I think the whole idea is wrong--I think you need a long case analysis of the possible residues of the sequence modulo primes less than N to prove it one way or another. I've written a program that does such a case analysis, and indeed the answer is N=17. Modulo 2*3*5*7*11*13, the sequence is unique up to negation. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 15 Jul 1994 20:48:46 GMT Subject: A turn on the carouselsius (Re: Move one digit ...) cl...@cnj.digex.net (Chris Long) wrote: > When Waldo recently did a conversion of a positive integral Celsius > temperature c=275 to its Fahrenheit equivalent f (which turned out > to be 527), he noticed to his amazement that he could have simply > moved the last digit of c to the front to obtain f. Doing some > intense calculations he failed to discover the next largest such > example. Does one exist, and if so, what is it? This has been answered, but no one gave the (easy) proof that all such temperatures have been described. First, notice that for the Fahrenheit temperature to be an integer, the Celsius temperature must be divisible by 5, so the units digit must be 5. Suppose we are dealing with N+1-digit numbers. Then there is an N-digit number A such that the Celsius temperature is 10A+5 and the Fahrenheit temperature is 5x10^N + A. So the conversion equation 9C+160=5F gives us 17A + 41 = 5x10^N. Modulo 17, this requires 5x10^N==41==7; multiplying both sides by 7 gives 10^N==35x10^N==49==-2. Clearly this holds for N=2; Fermat's little theorem shows that this holds for N=16K+2, and since it doesn't hold for N=8+2 that is all the solutions there can be. So the Celsius temperature is 5+10x(5x10^(16K+2)-41)/17 = (5x10^(16K+3)-325)/17 and the Fahrenheit temperature is 5x10^(16K+2) + (5x10^(16K+2)-41)/17 = (9x10^(16K+3)-41)/17. ................ It's no coincidence that the solutions [5]294117647058823527[5] ................ are so reminiscent of 5/17 = 0.2941176470588235 and ................ 9/17 = 0.5294117647058823. If Waldo were to convert Celsius to Fahrenheit by moving the _first_ digit to the _end_, the situation is a little trickier, since he could not immediately deduce the identity of the digit. If the Celsius temerature is Dx10^N + A and the Fahrenheit is 10A+D, then 9(Dx10^N + A)+160 = 5(10A+D), so (9x10^N - 5)D=41A-160, so we need to find solutions of (9x10^N - 5)D==-160 (mod 41). Since 10^5==1(mod 41), we need only consider N (mod 5). We want (9x10^N - 5)D==-160==4; multiplying both sides by -10 gives -10(9x10^N - 5)D==-40==1. The following table shows the work (mod 41): N(mod 5) 0 1 2 3 4 10^N 1 10 18 16 -4 9x10^N 9 8 -2 21 5 9x10^N-5 4 3 -7 16 0 -10(9x10^N-5) 1 11 -12 4 0 D 1 15 17 -10 NaR (Not a Residue) The last line, taking a reciprocal mod 41, can be done by trial and error or by Euclid's algorithm. Noticing that only one value of D is a digit, the result is that N=5K, the Celsius temperature is (5x10^(5K+1)+155)/41 and the Fahrenheit ..... temperature is (9x10^(5K+1)+1591)/41; or [1]2195121955[1], where ..... ..... 5/41=0.12195 and 9/41=0.21951. For an easy ObPuzzle, try this with negative temperatures, where the digits cycle but the minus sign stays in place. For a harder ObPuzzle, make the minus sign cycle with the digits. About a year ago, rec.puzzles searched for cases of a Fahrendrome, whose Celsius equivalent has the same digits in reverse. I believe that for the case of positive integers, it was shown there are none. Suppose we allow decimal fractions, and require that the digits from the first nonzero digit through the last nonzero digit be reversed? I didn't find any of them, either, but I'm not sure I looked carefully enough. What if we allow negatives? Certainly -40F=-40C, but are there others? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 20 Jul 1994 15:44:56 GMT Subject: Re: Game analysis sought ber...@fantasyfarm.com (Bernie Cosell) writes about the solitaire game played on graph paper: > You start with the following layout: > * * * > * * > * * * * * * > * y * > * * * * * * x > * * > * * * [ I've added the letters x and y ] > The rule is simple: you are to make as many lines in any of the > eight orthogonal/diagonal directions that are exactly > five-dots-long by adding a single dot to the grid making a run of > four into a run of five. And so you could take: * * X * * and > put your mark as 'X' and draw the line-of-5. In the opening grid, > the only legal move is to extend a diagonal from the middle of one > arm to the middle of an adjacent arm. I think you're talking about the move to that adds "x". It's also possible to add "y" to make a vertical line. > You cannot use a particular > pair of adjacent points in more than one line [and so * * * * > can only be extended to the left _or_ to the right, but once you do > one of those you can't do the other. Note that you can use a *dot* > in more than one line [even two in the same direction] --- it is > the line segment connecting two points than cannot be done twice. What I know about this game comes from a short paper by Wlodzimierz Holsztynski that he posted to sci.math in 1990. ``Students at the Poznan Technology Institute did not waste their time during lectures. Instead they played Poznan Solitaire.'' The game he describes differs from yours in two ways. First, in addition to being able to add one dot to make a line of five, he allows non-incrementing moves--drawing a line through five existing dots, adding none--as long as no segment is reused. Second, his classical starting position is just 50% larger than yours. : : : : : : : : : : ..:..:..:..o..o..o..o..:..:..:.. : : : : : : : : : : ..:..:..:..o..:..:..o..:..:..:.. : : : : : : : : : : ..:..:..:..o..:..:..o..:..:..:.. : : : : : : : : : : ..o..o..o..o..:..:..o..o..o..o.. : : : : : : : : : : ..o..:..:..:..:..:..:..:..:..o.. : : : : : : : : : : ..o..:..:..:..:..:..:..:..:..o.. : : : : : : : : : : ..o..o..o..o..:..:..o..o..o..o.. : : : : : : : : : : ..:..:..:..o..:..:..o..:..:..:.. : : : : : : : : : : ..:..:..:..o..:..:..o..:..:..:.. : : : : : : : : : : ..:..:..:..o..o..o..o..:..:..:.. : : : : : : : : : : > The question is: is the game bounded or not? The paper proves it is. A starting position of N dots can yield a population of at most A = (8 N^2 + 7)/14 dots, and can last at most A^2-Sqrt(A) moves. This produces a bound of 310 moves for your 24-point starting position, and 713 for his 36-point one. Of course, disallowing the non-incrementing moves can't make the game last any longer. He says the best scores (number of moves) of the Poznan masters were under 200. I've got a copy of the paper, partially edited (the original used a lot of messy DOS characters that got munged) but I doubt most rec.puzzles readers would appreciate 8 pages of dense math. So anyone who wants a copy should drop me a line by email. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: news.admin.policy, news.admin.misc, alt.current-events.net-abuse Followup-To: alt.current-events.net-abuse From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 16 Aug 94 16:48:56 GMT Subject: Infoseek.com net abuse (was Anyone else recieve this?) In alt.current-events.net-abuse, fiel...@rintintin.Colorado.EDU (Jeanette A Fielden) wrote: > I had the following waiting in my mailbox this morning. I > have no idea where they got my name. > ************* > Date: Wed, 10 Aug 1994 15:22:06 -0700 > From: Steve Kirsch > To: nob...@infoseek.com > Subject: 140 Computer Publications searchable on the Internet FREE > Reply-To: i...@infoseek.com > Hello. Your name was given to us as being a computer user who might be > interested in a ***FREE*** trial of a new service that our company, > InfoSeek, is offering to Internet users. [ Ad for ``***FREE***'' offer omitted. ] I got one of those too, but it was longer. Addressed to "friends-of-infoseek-us...@infoseek.com", my version said "One of our current users gave us your e-mail address as someone who might be interested....", and "As a courtesy to our current users, we've sent mailings to their friends." Apparently they have noticed their offer is not universally appreciated, because they have a section on ``If you object to receiving "unsolicited" e-mail''. Cute. It even shows their commitment to privacy: > Unfortunately, we did not keep records of which of our users gave us > your name. It was one (or more) of our current users, but we do not > give out email lists of users in order to protect their privacy as > I'm sure you can appreciate if you are reading this. Yeah, sure. So they are harassing me on behalf of some user of their infomercial services, and shielding that user so I can't send complaints. Or are they just lying about digging my name off some database or previous posting? Well, maybe they will explain. By the way, is Dave Wagner of NIS.COM still sending e-junk to everyone who posts to Usenet? The second time I got one, I complained to his DNS server, PSI.COM, but of course got no indication that they were going to stop. There's a junk-mail organization that has set up a receipt-of-solicitation-for-pay service. They inform junk mailers that they will receive their solicitations for $50 per letter. Then they do business with someone who sells addresses, and and send bills and bring claims for whatever solicitations show up. I've heard it is at least seriously annoying to the paper-spam business. Has anyone tried something similar for e-junk? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Thu, 25 Aug 94 14:43:13 -0400 (EDT) From: Dan Hoey To: Cube Lovers Subject: Re: Updated Upper Limits, Q-turns Jerry Bryan was looking at some formulas I had in the archives: Dan's recursion formula is: > P[n] <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4] ... I just rechecked Dan's original note of 9 January 1981. He specifically says the formula is good for n > 4. Mea culpa. However, I still do not fully understand *why* this should be the case. I thought I put the bound in.... I've been meaning to look that up and explain it, but time's been short. The analysis is done by breaking up the minimal processes into "syllables", where a syllable is a maximal sequence of commuting turns. So for each pair (x,y) in {(F,B),(T,D),(L,R)} there are four syllables of length 1: x, x', y, and y'; six syllables of length 2: xx, xy, xy', x'y, x'y', and yy; four syllables of length 3: xxy, xxy', xyy, and x'yy; one syllable of length 4: xxyy. (It's not really a coincidence that this is most of the fifth line of Pascal's triangle.) Now for the first syllable, we can pick any of the three pairs for (x,y). But for succeeding syllables, we must pick a pair that is not equal to the preceding pair. So each term in the recurrence refers to the length of the last syllable: Length of last syllable=1 2 3 4 n=1 P[n] = 4*3 P[n-1]; n=2 P[n] = 4*2 P[n-1] + 6*3 P[n-2] n=3 P[n] = 4*2 P[n-1] + 6*2 P[n-2] + 4*3 P[n-3] n=4 P[n] = 4*2 P[n-1] + 6*2 P[n-2] + 4*2 P[n-3] + 1*3 P[n-4] n>4 P[n] = 4*2 P[n-1] + 6*2 P[n-2] + 4*2 P[n-3] + 1*2 P[n-4] The second part of each coefficient is 2, except that when the length of the last syllable is equal to n (so that we are counting the first syllable), the second part of the coefficient is 3. In response to my description: > >The recurrence on which this bound relies is due to the > >relations F^4 = FBF'B' = I (and their M-conjugates.) It may be > >possible to improve the recurrence by recognizing other short > >relations. Exhaustive search has shown that there are none of > >length less than 10. Jerry continues: > ... there must be cases where X1 X2 ... Xn = Y1 Y2 ... Ym, where > the sum of the length of the sequences is less than 10, and where > the equality is not explained by the relations F^4 = FBF'B' = I. > Otherwise, Dan's calculations would yield exact values rather than > upper limits close to Start, and the "new calculations" for upper > limits would equal the "old calculations". No. The bounds fail to be exact when we have a relation r=s with |r|=|s|=n. This corresponds to a relation r s'=I of length 2n. The shortest relations of length >4 are the ones of length 12 (as I reported on 22 March 1981) so my bounds become inexact at length 6. Chris Worrell listed the length-12 relations on 08/02/81, and I reported that his list was complete on 14 August 1981 0111-EDT. The 12-qtw identities (up to M-conjugacy) are: I12-1 FR' F'R UF' U'F RU' R'U I12-2 FR' F'R UF' F'L FL' U'F I12-3 FR' F'R UF' UL' U'L FU' As Allan C. Wechsler noted on 17 August 1981, any two of them can be combined to form the third. Consider relations such as LRLRLRLR = I or RR'RR'RR'RR' = I. The first is a consequence of the relations L^4=R^4=LRL'R'=I. The second is a consequence of group theory; no relations are needed. The recurrence deals with these: it models the freeest group specified by the given relations. I have tried unsuccessfully to create a recurrence that will deal with the 12-qtw identities, but it's complicated. For instance, repeatedly putting I12-1 in the center of another I12-1 yields identities of the form: F (R'F' RU)^n F'U' (FR U'R')^n U There are a bunch of other cases, too. Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.arts.sf.fandom Followup-To: alt.folklore.computers From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 25 Aug 94 22:06:25 GMT Subject: Re: Neofan's Glossary/FAQ ahr...@linnea-grind.stacken.kth.se (Ahrvid Engholm) wrote: > FOO > Another fannish Ghod, from the 50's (?). Inspired by a comics > character. (Connection to the "Foobar" of hackers?) crack...@io.com (Casey Hamilton and E. A. (Ed) Graham, Jr.) comments: > Actually, "foobar" is a corruption of the military slang "fubar" which > is an acronym for (euphemism time) "fouled up beyond all recognition." Actually, that's just a popular bit of misinformation. While BAR was probably chosen because of FOO and FUBAR, the FOO predates it and comes from the ``Smokey Stover'' and ``They'll do it every time'' comic strips. Is that what ``Inspired by a comics character'' means? Smokey Stover had no character named Foo, it was just used as an interjection, perhaps akin to Yiddish ``feh''. See the The New Hacker's Dictionary for details. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.comics.alternative, rec.arts.comics.misc From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 26 Aug 94 15:50:53 GMT Subject: Re: Ellison book? Again, la...@mack.rt66.com (Lazlo Nibble) claims that Christopher Priest's pamphlet is about Harlan Ellison's laziness. But again, Lazlo's arguments fail to connect the concept of laziness with anything that appears in the pamphlet. For instance, he says > Priest makes it clear that the only thing keeping the book from > being released is the fact that Ellison hasn't bothered to sit down > in front of his typewriter and write the damned introductions.... That is plain wrong. First, Priest writes at length of other impediments to publishing the book. Second, the difference between ``hasn't'' and ``hasn't bothered to'' is the difference between what appears in the pamphlet and what appears in Lazlo's imagination. Third, Lazlo goes on to quote Ellison at length, but there's nothing about laziness in the quotes--not that Ellison's admission to laziness would have constituted such a charge by Priest. I am driven to the conclusion that Lazlo is unable to distinguish what is in the pamphlet from his own beliefs about Ellison. Lazlo has already said that he enjoyed the pamphlet because he likes "seeing Ellison get kicked in the balls every now and then." It seems he got so caught up in his fantasies about Ellison's balls that he failed to notice what the words in the pamphlet were saying. Whatever floats his boat, but I think the readers of these groups should realize they are reading reviews of Lazlo's brain, not of The Book on the Edge of Forever. So if you think it's not worth printing a book for name-calling or vicarious testicular assault, be assured that's not what has been done. That being the only relevance this discussion has to these groups, I don't propose to debate it any futher. Anyone who feels I have misrepresented Lazlo's comments should e-mail me for a copy of the messages in this thread. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 26 Aug 94 16:17:18 GMT Subject: Re: Tuition FEES David C. Pheanis wrote: >If we use F, E, and S to represent three different digits, we can >write the tuition fee for a certain university as FEES in base ten. >Remarkably, we can use the same digits in other bases to arrive at the >following equation: FEES(10) = FEES(9) + FEES(6) where (j) indicates that >the value is in base j. What is FEES in base ten? Cute puzzle. Chris Franz (fr...@enuxsa.eas.asu.edu) analyzed it : (10^3)F + (10^2)E + (10)E + (1)S = (9^3)F + (9^2)E + (9)E + (1)S + : (6^3)F + (6^2)E + (6)E + (1)S : => : 1000F + 100E + 10E + S = 729F + 81E + 9E + S + : 216F + 36E + 6E + S : => : Since the only digit that can be added to itself and yield itself is zero, : assume S = 0 and eliminate it from the equation. So, but that's not valid reasoning--it looks like he expects to be able to perform multiple-base addition on a digit-by-digit basis. r...@pge.com (Richard Irving) noticed the reasoning was wrong, but did not mention counterexamples. The equation FEES(3)+FEES(6)=FEES(7) has a solution in which S is not zero, and that FEES(5)+FEES(7)=FEES(8) has a solution in which none of the letters is zero. Take them as ObPuzzles if you like. For a harder ObPuzzle, determine the rest of the sequence 10,11,14,16,19,... of bases Z which admit a puzzle FEES(X)+FEES(Y)=FEES(Z) with a solution such that S=0. Is Z=6 the largest base that admits no soluble puzzle? Richard Irving, like Jonathan Haas a while earlier, also noticed that FEES(6)+FEES(9)=FEES(10) amounts to solving 0 = -55 F + 22 E + S for 0 <= F,E,S < 6, but they go into unnecessary case analysis. The easy way to solve this is to notice that S must be divisible by 11, and so S=0. This leaves us with F/E = 2/5, so F=2, E=5. It's harder to solve when none of the digits is zero. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 26 Aug 94 16:42:40 GMT Subject: Re: Numbers of subsets of N closed under addition. g...@arp.anu.edu.au (Greg Restall) asks: > How many subsets of N are closed under addition? > Of course there are at least countably many. Each {n,n+1,n+2,...}, > is a candidate. But are there uncountably many?... No. For let G be the GCD of such a set; G is realized as the GCD of a finite subset, and the fundamental theorem of arithmetic shows that the closure of that finite subset under addition includes all sufficiently large multiples of G. So every closed subset consists of a finite set together with all the larger multiples of its GCD. There are only countably many finite subsets of N, so only countably many closed sets, QED. [Also e-mailed] Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 29 Aug 94 22:20:45 GMT Subject: Re: filling the surface with 3,4,6-gons (spoiler and new puzzle) ho...@sci.kun.nl (Mark van Hoeij) writes: > One can fill the surface R^2 with triangles, squares and hexagons as > shown in the postscript file included below. Or, in case not everyone can hack postscript, it's not too hard to do it in ASCII. ____ ____ ____ ____ ,-'| |`-. ,-'| |`-. ,-'| |`-. ,-'| |`-. /`-.|__|,-'\ /`-.|__|,-'\ /`-.|__|,-'\ /`-.|__|,-'\ / / \ \____/ / \ \____/ / \ \____/ / \ \ |`-./ \,-'| |`-./ \,-'| |`-./ \,-'| |`-./ \,-'| |,-'\ /`-.|__|,-'\ /`-.|__|,-'\ /`-.|__|,-'\ /`-.| \ \____/ / \ \____/ / \ \____/ / \ \____/ / \,-'| |`-./ \,-'| |`-./ \,-'| |`-./ \,-'| |`-./ /`-.|__|,-'\ /`-.|__|,-'\ /`-.|__|,-'\ /`-.|__|,-'\ / / \ \____/ / \ \____/ / \ \____/ / \ \ |`-./ \,-'| |`-./ \,-'| |`-./ \,-'| |`-./ \,-'| |,-'\ /`-.|__|,-'\ /`-.|__|,-'\ /`-.|__|,-'\ /`-.| \ \____/ / \ \____/ / \ \____/ / \ \____/ / \,-'| |`-./ \,-'| |`-./ \,-'| |`-./ \,-'| |`-./ /`-.|__|,-'\ /`-.|__|,-'\ /`-.|__|,-'\ /`-.|__|,-'\ / / \ \____/ / \ \____/ / \ \____/ / \ \ |`-./ \,-'| |`-./ \,-'| |`-./ \,-'| |`-./ \,-'| |,-'\ /`-.|__|,-'\ /`-.|__|,-'\ /`-.|__|,-'\ /`-.| \ \____/ / \ \____/ / \ \____/ / \ \____/ / \,-'| |`-./ \,-'| |`-./ \,-'| |`-./ \,-'| |`-./ `-.|__|,-'\ /`-.|__|,-'\ /`-.|__|,-'\ /`-.|__|,-' \ \____/ / \ \____/ / \ \____/ / \,-'| |`-./ \,-'| |`-./ \,-'| |`-./ `-.|__|,-' `-.|__|,-' `-.|__|,-' Too bad ASCII doesn't have a backward comma. > What is the ratio between the 3, 4, and 6-gons if the surface is > filled with this pattern? The easy way is to erase some of the lines, to form a monohedral tiling by seashells. ____ ____ ____ ____ ,-' `-. ,-' `-. ,-' `-. ,-' `-. / \ / \ / \ / \ / \____/ \____/ \____/ \ `-. ,-' `-. ,-' `-. ,-' `-. ,-' \ / \ / \ / \ / \____/ \____/ \____/ \____/ ,-' `-. ,-' `-. ,-' `-. ,-' `-. / \ / \ / \ / \ / \____/ \____/ \____/ \ `-. ,-' `-. ,-'| |`-. ,-' `-. ,-' \ / \ /`-.|__|,-'\ / \ / \____/ \____/ / \ \____/ \____/ ,-' `-. ,-' `-./ \,-' `-. ,-' `-. / \ / \ / \ / \ / \____/ \____/ \____/ \ `-. ,-' `-. ,-' `-. ,-' `-. ,-' \ / \ / \ / \ / \____/ \____/ \____/ \____/ ,-' `-. ,-' `-. ,-' `-. ,-' `-. / \ / \ / \ / \ / \____/ \____/ \____/ \ `-. ,-' `-. ,-' `-. ,-' `-. ,-' \ / \ / \ / \ / \____/ \____/ \____/ \____/ As one in the the center shows, each shell is composed of one hexagon, two triangles, and three squares. ObPuzzle: Is there a tiling in which every two congruent tiles are equivalent, and which cannot be converted into a monohedral tiling. By ``equivalent'', I mean that there is a symmetry of the tiling taking one tile to the to the other. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 29 Aug 94 22:40:00 GMT Subject: Re: Q: What is FreeCell? Thanks to bl...@gpx01.d39.lilly.com for posting the rules to FreeCell. There seems to be a typo, though: > ====From the help screens==== > 3. You can move a card from a free cell or from the bottom of a > column to the bottom of another column as long as the rank of the > card is less than the rank of the card you will place it on, and the ^^ ^^^^ > colors of the cards are different. For example, you can move a > black three onto a red four. Any card can be moved to an empty > column. I'm sure ``is less'' should be ``is one less''. That is, you can't move a black three onto a red six. tur...@cs.utexas.edu (Russell Turpin) said as much when he introduced the problem last month: > ...the usual usual solitaire rule holds: the moved element must be > one less and a different color than the card onto which it is moved. Otherwise the rules are pretty much like I expected from the discussion so far. That said, what is the deal 11982 that seems to be baffling the experts? Has it been _proved_ unsolvable yet? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.arts.movies From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 31 Aug 94 21:49:53 GMT Subject: Eat Drink Man Woman Question [No spoiler in original article] I saw Eat Drink Man Woman over the weekend. A beautiful and entertaining movie with a wonderful sense of humor and a very attractive view of life. Full of delicious surprises, so be careful what you read about it. And pay attention, because there are a lot of riddles that are answered when the surprises come out. It's also full of delicious food. I've seen at least one reviewer suggest you avoid seeing it on an empty stomach for that reason. But I understand the director advises just the reverse, on the theory that hunger makes the best spice. For me, I saw it without breakfast, and it was piquant but not unpleasant. I have a question that doesn't spoil any surprises, but I don't know if it can be _answered_ without spoiling surprises. So for followups, please change the subject to read "[spoilers]" if your message spoils a surprise. People who haven't seen the movie should beware of any followup that says "[No spoiler in original article]" because it is from someone unable or unwilling to change the subject line. Please change the subject line to "[no spoilers]" if your reply really has no spoilers (and if you put spoilers in a followup to one of _those_ without changing the subject line, Lord have mercy on your soul). That said, my question: there's a scene where the cook stops himself just before he kills a fish (just as he previously killed a fish before cleaning and cooking it). Why doesn't he kill the second fish? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 2 Sep 94 20:46:59 GMT Subject: Re: physics/milk.and.coffee dwr2...@tam2000.tamu.edu (Dave Ring) writes: > Assume Newton's law of cooling (the heat exchange rate is proportional > to the temperature difference). Assume the milk temp stays constant until > added. Let room temp be 0, initial coffee temp be 1 and initial milk > temp be Tm. Assume the milk volume is small. I'm going to do this with two parameters. Let V be the ratio of the milk+coffee volume to the coffee volume, and let H be the amount of heat in the milk (relative to the room temperature). So Tm=H/(V-1). This means that adding milk to coffee at temperature T will change the temperature to (T+H)/V. Note that the temperature of the milk is not dependent on when it's added--this is why it the problem is not symmetrical between coffee and milk. > Assume same density, heat capacity, thermal conductivity, etc... > Assume temp always uniform throughout the liquid. Assume the cup is > an open-ended insulating cylinder (so no change in surface area). I think we need to assume that the air above the coffee is continuously agitated, so that a warm layer doesn't form. I suspect that the presence of such a layer, and its disturbance when the cream is added, are the most serious departures of this model from reality. > Assume the time period is such that without the milk, the coffee would > cool to T=1/e . I'm going to call that temperature L, so I can solve for other final temperatures. > Puzzle: Find the range of Tm at which you should add the coffee > (before/during/after) to get the warmest coffee. Note that the original puzzle did not provide "during" as an option. In fact, I don't know how to solve that problem, so I'm simply going to compare before and after. Solution follows: Without milk, the coffee at time t will have temperature T(t)=exp(t log L)=L^t. At time 1 it will reach temperature L, and if the cream is then added then the temperature will be (L+H)/V. Adding milk first, the temperature will start at (1+H)/V and (due to the increased volume) will drop at a rate 1/V as fast for any given temperature. So the temperature at time t will be ((1+H)/V) exp((t log L)/V). So at time 1 the temperature will be ((1+H)/V) L^(1/V). Let R=L^(1/V), so this temperature is (1+H)R/V. The milk should be added first when (1+H)R/V > (L+H)/V , or H < (R-R^V)/(1-R). This is positive, so the FAQ is correct that chilled milk should also be added first. But room temperature milk should also be added first. The critical milk temperature above which the milk should be added last is Tm = (R-R^V)/((V-1)(1-R)), or (L^(1/V)-L) / ( (1-L^(1/V)) (V-1) ). As V approaches 1, the critical temperature approaches L log L/(L-1). In the case of L = 1/e ~ 0.368, the answer is that you add the milk last when Tm > 1/(e-1) ~ 0.582. That is pretty warm milk. In fact, the limiting critical temperature is always warmer than the temperature of the coffee that has been cooled without milk, and I think that's the case even when V>1. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 3 Sep 94 02:23:23 GMT Subject: Re: Vos Savant:0 Math Profs.:1 Gary Peterson (f...@cc.gatech.edu) writes: : She in no way "triumphed" over the Math Profs. : Anybody can get (most) any mathematician to answer a problem wrong : if they are vague and ambiguous stating the problem. dcunn...@netcom.com (Dave Cunningham) writes: > Indeed, the problem was unclearly stated; however, the responses > from the math profs indicate they interpretted the problem as > intended. Well, I will interpret your indication of the problem "as intended" to mean the variant in which Monty always offers the contestant a chance to switch doors. But there is no reason to believe that that was the variant intended by the person who posed the problem to vos Savant. It certainly was not specified in the problem as she printed it. I don't believe that leaving unspecified conditions unspecified constitutes an unclear statement of a problem. Certainly if I posed it, I would pose the variant in which Monty's method of deciding whether to offer a swap is clearly unspecified, by virtue of not being mentioned. And if you then solved the problem as if Monty always offered a second chance, you would get the wrong answer. If instead you mean that she did not specify that Monty is trustworthy--that any contestant who picks a door and refuses any offer of switching may take home what's behind that door--then I agree. That was indeed never stated, only implied. But no one seems to be arguing that. > It is obvious from reading their arguments that their flawed > analysis was due to their own faulty reasoning, not the wording of > the problem. Their own arguments were logically inconsistent. I have just reread the responses she published in her 17 Feb 1991 _Parade_ article. Your statement is not an accurate summary of them. Seven of the ten excerpts say she's wrong; one calls her a goat, one makes an arguably sexist comment, and one says she's right. But none of the seven excerpts includes one word about the reasoning in the letters, or hint at what the argument might be. It is impossible to tell whether the criticisms advanced in those and other letters were accurate or not, nor whether her excerpts were representative of the best or the worst of the correspondence. But given her blatant lack of integrity in dealing with Andrew Wiles's proof, I trust the parts of the letters I have never seen more than I trust her synopsis of them. And given _your_ wild inaccuracy in writing about the excerpts, I wonder whether you have even read them. > That they were wrong due to the wording of the problem is > revisionism. They were wrong because they were stupid. That they > now attempt to reframe the discussion when their sophmoric, > illogical rantings are so well documented for all to laugh at is > further evidence of their stupidity. It is certainly not revisionism to respond to the problem as it was posed in the column you are writing to. For all I know, some of them may have mentioned that the problem has other variants than the one she posed. (Does anyone have a copy of one of those long-ago posts that are cited to establish which variant arose first? I've heard they came out in the early 80s, but I'll take anything before 1988). Not knowing the arguments of the people whose letters were excerpted, so I can't tell what they intended. But my criticism of the wording of the problem arose in rec.puzzles prior to the vos Savant publication (in response to the "unspecified" variant as posed there). So who's engaging in revisionism? Dan Hoey La sophomorique est comme la sophomorique Hoey@AIC.NRL.Navy.Mil rante. -- Fermat Gump mere --- Newsgroups: alt.folklore.urban From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 8 Sep 94 00:14:30 GMT Subject: Re: The Referee's Death jf...@acpub.duke.edu (Joel K. Furr) writes: > ... I was reminded of an urban legend I heard often when I was in > elementary school in the early-to-mid 1970's. [ Soccer spectators incinerate referee with metallized programs. ] > Obviously a UL, but one which we students at Harding Avenue > Elementary often heard and retold. It reminds me of the urban legend of the baseball team that sold their souls to the devil to win the World Series. That is to say, it's an old piece of fiction. Odd to see it get turned into a legend. Those of us who once read Arthur C. Clarke's ``A Slight Case of Sunstroke'' got a big deja vu hit when the Colombian soccer player got shot for losing in the World Cup, so of course it was on rec.arts.sf.written. cr...@superior.carleton.ca (Tom Hamill) mentioned: > I found it in a collection of Clarke's stories called 'Tales of Ten > Worlds', published in paperback by Dell in 1964, although the > copyright of the original story is shown as 1958. There was an > interesting acknowledgement on the copyright page. Clarke credits > Dr. John Pierce of Bell Labs for the idea, which he advanced in > Chapter 10 of "Electrons, Waves, and Messages". Several others mentioned it being in his _Tales from the White Hart_ collection, too. I can't recall where I saw it. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 8 Sep 94 15:47:07 GMT Subject: Re: The Monty Hall problem Daniel Cohen writes: >When stated in this way, we don't know whether Monty has opened a >door at random which just happens to reveal nothing of consequence, >or whether he has deliberately chosen to open a door which concealed >nothing of consequence. ris...@xerver.icl.fi (Risto Lankinen) analyzes: > The fact that we don't know it calls for a mixed strategy. Assume > chance '1-p' that Monty made a random choice and 'p' that he knew > which door to open. In the first case switching the doors makes > no difference to the expected payoff, but in the second case the > expected payoff is increased by switching the doors. Here's the > payoff matrix: > swap doors don't swap > Monty knows 2/3 1/3 > Monty doesn't know 1/2 1/2 > ---------------------- > Monty knows with > probability 'p' 1/2+p/6 >= 1/2-p/6 So close.... But you missed the point that if Monty knows where the prize is (and acts on it) then presumably he also knows whether you picked the prize in the first place, and can base his decision on that. So the payoff matrix is: swap doors don't swap Monty doesn't know, opens unchosen door at random 1/2 1/2 Monty knows, always opens an unchosen nonprize door 2/3 1/3 Monty knows, opens unchosen nonprize door only if you picked a nonprize 1 0 Monty knows, opens unchosen nonprize door only if you picked a prize 0 1 The saddle-point strategy is for you to never swap, because that is the only way to ensure you don't give away your prize in strategy 4. > People who say that "the question was stated improperly" always > seem to forget this. In other words, they haven't really solved > the problem, at least in the form they themselves claim it was > originally stated. I don't know what people you are talking about. There are two common variants of this problem, and neither of them is "improperly" stated, unless you mean to state the other one. In one variant, it is given that Monty always opens an unchosen nonprize and offers a chance to swap. In the other variant, there is no statement of the circumstances under which Monty will offer to swap, save that those circumstances occur in the given trial. The mistake people keep making (on Usenet and in the newspaper) is that they state the latter and solve the former. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 8 Sep 94 21:01:27 GMT Subject: Re: Tower of Hanoi Formula micha...@ix.netcom.com (Michael O'Connor) asks: > I know the algorithm for determining what disc in the Tower of > Hanoi puzzle to move on a certain turn, but is there a way to find > an algebraic formula, given the number of the turn, to find the > number of the disc (where the lightest one is 1, second lightest is > 2 so forth) for an infinite Tower of Hanoi puzzle, and if so, what > is it? I don't know if you consider this algebraic, but on turn K you move disc D(K) = Log[2] ( 1 + ((K-1) XOr K )) where Log[2] is the logarithm to the base 2, and XOr is the bitwise exclusive Or of the two numbers written in binary. This is just another way of writing "one plus the number of trailing zeroes in the binary representation of K", as simen.ga...@math.uio.no (Simen Gaure) noted. As pet...@cco.caltech.edu (Peter T. Wang) pointed out, each odd-numbered turn moves disc 1 clockwise. That is, you move it from peg (K-1)/2 to peg (K+1)/2, modulo 3. In general, odd-numbered discs will be moved clockwise and even-numbered discs will be moved anticlockwise, so on move K you move disc D(K) from peg (K (-2)^(1-D(K)) + (-1)^D(K))/2 to peg (K (-2)^(1-D(K)) - (-1)^D(K))/2, modulo 3. Does anyone have a proof that if you put disc I on disc J, then J-I is odd? It's funny, I watched a demonstration in which the even and odd discs were colored differently, and colors alternated in each pile throughout the procedure. It took me a long time to just to notice the phenomenon, but it's quite surprising when you think about it. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 9 Sep 1994 15:40:39 GMT Subject: Re: Tower of Hanoi Formula micha...@ix.netcom.com (Michael O'Connor) asked for a formula for which disc to move on a given turn K in the Towers of Hanoi. Yesterday I gave it as: D(K) = Log[2] ( 1 + ((K-1) XOr K )) and I gave expressions (mod 3) for the discs to move: From peg (K (-2)^(1-D(K)) + (-1)^D(K))/2 To peg (K (-2)^(1-D(K)) - (-1)^D(K))/2. Thinking about it, I was able to see that the last is much too complicated. Using the fact that 2==-1 (mod 3), and swapping pegs 1 and 2, the rule is: From peg K + (-1)^D(K) To peg K - (-1)^D(K). This allows us to express the Hanoi algorithm extremely simply: Move a disc between pegs K+1 and K+2 (mod 3) in the unique legal direction. Also, this can be used to prove the alternating-color property, by noticing that after move K the top disc of peg k+2 (mod 3) will be even, and the others will be odd (counting the peg number if there is no disc on it). Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 9 Sep 1994 21:07:47 GMT Subject: Unfolding polyhedra and polytopes (was Re: How many hexominoes...) ad...@super.org (Jeffrey P. Adams) showed 11 ways of unfolding the surface of a cube into hexominoes (20 if you count reflections of the chiral ones). > Now, I'm thinking about octohedra, dodecahedra, and icosahedra. I > wonder, is there an easier way to go about this? Does anyone know > whether this has already been done? I did it for the octahedron a while ago. There is a nice correspondence between the unfoldings of the cube and the octahedron. Note that an unfolding of any polyhedron is completely determined by the set of edges that are cut to leave a simply-connected set of faces. If two polyhedra are dual, then the set of edges cut for an unfolding of the first is dual to the set of edges _not_ cut for an unfolding of the other. This forms a chirality-preserving bijection between the sets of unfoldings. So the duality between cube and octahedron induces a bijection between their unfoldings. Some bijection will occur between the dodecahedron and icosahedron. The tetrahedron is self-dual and duality induces the identity bijection between its two distinct unfoldings (three if you count reflections). Notice that when you unfold a non-convex polyhedron and flatten it out, the surface may overlap itself. For instance, the following depicts one form of the unfolded surface of the L-tricube: +--+-----+-----+ | | | | +--+ +--+--+-----+-----+ | | |XX| | | +--+--+--+ | | | | | +-----+--+-----+ The XX square is actually two overlapping squares, one joined at the bottom and the other joined at the right. Can an overlapping unfolding be found for _every_ nonconvex polyedron? Can one ever occur with a convex polyhedron? If so, can it occur on the dodecahedron or icosahedron? I do know it doesn't happen on the tesseract, and I think it doesn't happen on a hypercube of any dimension, but I haven't got a proof. d...@mirage.esl.com (Dana Albrecht) asked: > Related question: If one "unfolded" n-dimensional cubes, how many > "different" unfoldings are there? ... I wrote on this subject in 1992. There are 261 octocubes you can get from unfolding the surface of a tesseract, or 355 if you count reflections of the chiral ones. Then I found a paper by Peter Turney in J Rec Math V17(1), 1984, in which he shows the problem to be equivalent to counting the number of 8 vertex graphs that consist of a tree together with a differently-colored pairing of the 8 vertices, disjoint from the tree (the tree represents face adjacency in the unfolding, and the pairing represents opposite pairs of faces in of the solid. This would plainly extend to the problem of unfolding the surface of a 5-dimensional hypercube into a dekatesseract, but I don't know if it's been done. Also, I don't know how to figure out which ones are chiral. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 14 Sep 94 20:42:20 GMT Subject: Re: Unfolding polyhedra and polytopes (was Re: How many hexominoes...) Is there an overlapping unfolding for a convex polyhedron? c...@cus.cam.ac.uk (Chris Thompson) writes: > Croft/Falconer/Guy, UPiG, B21 say re [2] that the snub dodecahedron > has an overlapping net "if we cut perversely". This seems right (the > thing damn near overlaps even without perversity) but I don't think > I am up to an ASCII drawing of it at the moment! At first I thought you meant the truncated dodecahedron (12 pentagons, 20 triangles) and I wrote a program that found there were no overlapping unfoldings. But I've learned the snub dodecahedron is actually a 92-gon (12 pentagons, 80 triangles) which is beyond my program at present, and I haven't found it doodling (perhaps I'm just not perverse enough). Can you write down directions? Necessary and sufficient is a path from one face to another that overlaps it. Numbering the exits from a each face clockwise from its entrance, can you say what turns are taken? fuk...@gssm.otsuka.tsukuba.ac.jp (Komei Fukuda) writes: > Another example by Namiki is the simplest possible. It is a long > tetrahedron. If anyone is interested in this example, I will upload > the Mathematica notebook.... Not necessary: Knowing it's possible makes it easy to find. Here's an example (not to scale). Identify the edges and vertices marked a,B,c,D,e to fold the tetrahedron. (33,56) B / / // // D (18,35) // \. / / \\. / / \\. / / \ \./ / \./\./ /\ /\. / \/ \. / /\. \. a / / \ \. / / \ \. c / / c \. \. / / e \ \. / / \ \. / / e \.(45,9) \. /(8,3)/ _____------*.. \. / ,D----- ,...'''' ``.. \. / ,.' ,...'''' ``.. \. /,.:.'''' ``..\. *------------------------------------------B (0,0) a (65,0) I wonder what the smallest example with integer coordinates is. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 16 Sep 94 23:54:16 GMT Subject: Re: Instant Insanity angil...@bach.seattleu.edu (Angi Long) writes: > I thought that there was no way the right number of colors could > appear on the outside of the stack and not be solvable. But it's > been brought to my attention that this is indeed possible, so if > you have cubes that differ from mine, you might have to go back > to step #1 and choose another set of faces for the axis. Yes, you might. Might you also have to go back to step 2 if step 3 or 4 fails? (The answer is no, but why?) > This is one advantage of my technique over the graphic one somebody > else described -- this one will definitely give *all* solutions. Well, if you are going to go for _all_ solutions, then you had better be explicit about backtracking over different choices in steps 2, 3, and 4 as well as different choices for step 1. Also, if you think that O'Beirne's method, as described by k...@cs.cornell.edu (David A. Karr), does not give all solutions, then you haven't solved the ObPuzzle he posed: Generalize step (5) so it works with any set of cubes. The solution to that is to allow, in place of a Hamiltonian cycle, any system of disjoint cycles that visit all four colors. As with the Hamiltonian cycles, we will require that arcs from all four cubes be used. If you do it that way, you get definitely get all solutions. Using this method, I found all the solutions for your cubes: Sol I Sol II Sol III Sol IV Sol V cube 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 top R B G W B R W G B R W G B G R W R G B W down B W G R R G B W R G B W R W B G B W R G front G R W B W G R B W G B R R W G B R W G B back W G B R G W B R G W R B W B G R W B G R Of course, all of these can be modified trivially by permuting the sides {top,down,front,back} in any way that preserves opposites sides. Also, they can be modified by exchanging the roles of cubes 1 and 4, which are colored identically. Under this system, the solutions found by eua...@eua.ericsson.se (Goran Wicklund) are Sols I' and V, while the ones you found are Sols IV', I, II, and III', where the prime marks those in which cubes 1 and 4 have been swapped. You also said: > This algorithm will always work, no matter how the colors appear on > your cubes (as long as there are at least four faces of each color). You shouldn't make such claims when you are just guessing. In this case, you are guessing wrong. The presence of at least four faces of each color, while necessary, is not even close to sufficient for there to be an Instant Insanity solution. For instance, in each of the following sets of cubes, you are allowed to mark the Green, Blue and White faces any way you like. Each set is insoluble, in spite of having at least four red faces, and plenty of space for the others. The reasons for insolubility are pretty obvious, but rot13'd in case you want to consider them an ObPuzzle. cube 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 top R R x x R R x x R R R R R x R x down R R x x R R x x x R R R x x x x front R R x x R R x x R x x R x x R x back R x x x R x x x x x x R x x x x left R x x x R R x x x x x x x x R x right R x x x x x x x x x x x x x x x Why: Gbb znal Ef Gbb znal Ef Bqq ahzore bs Ef Gbb srj Ef If we generalize the problem to an arbitrary number of cubes, then determining whether there is a solution is NP-complete [see Garey and Johnson]. This does not prove it's all that hard for four cubes, but it suggests there will not be a trivial criterion. In fact, I'll pose the ObPuzzles: 1. How many possible sets of Instant Insanity cubes are solvable? What percentage of all possible sets? (It is necessary to carefully define what you consider to be "different" sets of cubes. For instance, if one cube is reflected in a mirror, does that make it, and its set, "different"? Are you counting the order of the cubes with each set? ) 2. What is the average number of solutions for those sets that are solvable? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 19 Sep 94 16:26:52 GMT Subject: Re: POINT PUZZLE geko...@sgige70.sdrc (Vish Kohli) paraphrased the old chestnut: ==> geometry/construction/5.lines.with.4.points.p <== Arrange 10 points so that they form 5 rows of 4 each. As above, it's in the archive, with the barely adequate solution: ==> geometry/construction/5.lines.with.4.points.s <== Draw a 5 pointed star, put a point where any two lines meet. The problem is first that it's not the only solution, and second that it does not demonstrate optimality. On demonstrating optimality, b...@ss.titech.ac.jp (Sermsak Uatrongjit) starts a fair job of a proof: Numbering the (first five) lines L0, L1, L2, L3, and L4, then line Li can have at most i points in common with previous lines, and so introduces at least 4-i new points. This is an upper bound, too, since that number will include all ten points. Unfortunately, he finishes abruptly with ``we cannot construct any line furthur since it will cause overlap.'' He offers no justification for the last statement, and I see no simple way of proving it. My method of proving optimality is first to consider the case that three of the lines meet at a point. Those three lines contain three new points each, so all ten points occur on those three lines. Any line of four in the arrangement must then include two points from one of the three lines, and so must coincide with that line. So at most three lines can be formed in this way. In the other case, if at most two lines meet at each point, then each point will be in line with at most six other points. This adds up to at most 60 ordered pairs of distinct points in the lines; using them up at twelve pairs per line means we can only have five lines. Now for finding a solution: Finding solutions is in a sense easier than finding non-solutions. Pick five lines at random. Unless you hit the probability-zero jackpot, each pair of lines will intersect in a different point, making ten points, QEC. But a star need not occur, as in the following two configurations: * / \ / \ *.------*---*-------.* / \ `. \ / ,' *. ,* `. * ,' / `. ,' \ `. / \ ,' / ,*. \ *. ,*' *---,*---*.---* / ,*. \ / ,' `. \ /,' `.\ / ,' `. \ *' `* /,' `.\ *' `* For an ObPuzzle, characterize the shapes of possible solutions, with proof. The FAQ and the previous respondent are not alone in offering barely adequate solutions, though. _Lewis Carroll's Games and Puzzles_ from Dover offers a case where Alice has ten tiny cakes in a two-by-five rectangular array. By moving four of them she is to form five rows of four. Carroll mentions, ``There is more than one solution to this problem. See if you can find other solutions.'' Only one solution is listed, and it is mentioned that ``other solutions use the possibility of stacking cakes on top of one another.'' I see no way of making such an attack work, and I don't think it is at all what Carroll had in mind. An Obpuzzle is to (1) characterize, and (2) count, the ways of moving four of the cakes to other locations in the plane so as to make five rows of four. Treat the cakes as points, and no making rows of more than four. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 19 Sep 94 21:15:28 GMT Subject: Painted cubes I don't see this in the archive, though I'm pretty sure I've seen it here before. Looking through _Lewis Carroll's Games and Puzzles_, I saw it again: Given six pots of paint and a supply of wooden cubes, paint each side of a cube a different color. How many ways can this be done, up to rotation? The problem also asks that we develop a method of coding and recording the different cubes. The counting puzzle is very easy, but I won't spoil it, except to say that it ends up as a number that is immediately recognizable as the number of a certain combinatorial object constructed from the six colors. The interesting question, for which I have not found an answer, is to provide a natural (or at least simply described) one-to-one correspondence between the combinatorial objects and the colorings. Most satisfying would be a rule for placing the cubes so we could in some sense see the combinatoric. As another puzzle, we can ask into what shape of box we can pack our set of colored cubes with like colors abutting. What if we require the colors on the top face of the box to match the bottom? What if we require all the opposite pairs of faces to match? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 19 Sep 94 21:36:40 GMT Subject: Re: 4-D puzzles bco...@acsu.buffalo.edu (Bram Cohen) writes: > Here are two puzzles which were in an old Martin Gardner column: The first, at least, was reprinted in Gardner's _Mathematical Circus_, and is touched on in Rudy Rucker's _The Fourth Dimension_. > (1) The surface of a cube can be 'unfolded' in the following way: > _ > _|_|_ > |_|_|_| > |_| > |_| > There are other ways as well. How many ways can a 4-D cube be > 'unfolded' as a bunch of connected cubes in 3-space? I wrote about this on sci.math in 1992 and again earlier this month. There are 261 octocubes you can get from unfolding the surface of a tesseract, or 355 if you count reflections of the chiral ones. I am quite certain of this, having it both in manual analysis and from a program to generate them. Then I found a paper by Peter Turney in J Rec Math V17(1), 1984, in which he shows the problem to be equivalent to counting the number of 8-vertex graphs that consist of a tree together with a differently- colored pairing of the 8 vertices, disjoint from the tree (the tree represents face adjacency in the unfolding, and the pairing represents opposite pairs of faces in of the solid.) This would plainly extend to the problem of unfolding the surface of a 5-dimensional hypercube into a dekatesseract, but I don't know if it's been done. Also, I don't know how to figure out which ones are chiral. > Martin Gardner commented that he received about 10 responses to > each of these. Unfortunately, he got as many different answers as > responses! For that reason, please ... specify what the unfolded > patterns are for (1). I could give you twenty pages of diagrams, but how would that help? If you can't solve the problem, how could you verify exhaustiveness and nonduplication in a list? I suggest you see Turney's proof, instead. There's an interesting ObPuzzle here. Notice that when you unfold a non-convex polyhedron and flatten it out, the surface may overlap itself. For instance, the following depicts one form of the unfolded surface of the L-tricube: +--+-----+-----+ | | | | +--+ +--+--+-----+-----+ | | |XX| | | +--+--+--+ | | | | | +-----+--+-----+ We know that doesn't happen for the tetrahedron, cube, or octahedron. Prove it doesn't happen for the dodecahedron or icosahedron. It doesn't happen for the tesseract, either: Can you prove that unfolding the N-dimensional cube never overlaps? What about the other regular N-dimensional polytopes? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Article 20761 of rec.puzzles: From: hoey@sun13.aic.nrl.navy.mil (Dan Hoey) Newsgroups: rec.puzzles Subject: Re: C puzzle Date: 23 Sep 1994 17:48:37 GMT Organization: Navy Center for Artificial Intelligence In-reply-to: sethb@panix.com's message of 22 Sep 1994 18:00:17 -0400 sethb@panix.com (Seth Breidbart) writes: > It's quite simple in APL. > '''[1 1,26RI13]'[1 1,26RI13] Easier, in fact: 1U23R11R'''1U23R11R''' or 1|^HO22R11R'''1|^HO22R11R''' Where R is rho, U is drop, and |^HO is rotate. Also, don't forget that Lisp has the Lisplet, the Lambdamatic, and the Formatic. (let ((let '`(let ((let ',let)) ,let))) `(let ((let ',let)) ,let)) ((lambda (lambda) `(,lambda ',lambda)) '(lambda (lambda) `(,lambda ',lambda))) (format t "~a~s ~2:*~s ~s)" "(format t " "~a~s ~2:*~s ~s)") I think Mike McMahon is the author of the Lisplet. Traditionalists hold with the variableless Fortran IV version. PRINT 1 1 FORMAT(6X,7HPRINT 1/ 2(36H 1 FORMAT(6X,7HPRINT 1/ 2(36H .), 2(/5X,1H.,66H), 2(/5X,1H.,66H .), 2(/5X,1H.,66H), 2(/5X,1H.,66H .) /6X,4HSTOP/6X,3HEND) .) /6X,4HSTOP/6X,3HEND) STOP END If you want a C version that prints with the trailing newline, assuming '\n'==10 is bletcherous. Instead, use: main(){char*a="main(){char*a=%c%s%c,q='%c',b='%c%c'; printf(a,q,a,q,q,b,b,b,'%cn');}%c",q='"',b='\\'; printf(a,q,a,q,q,b,b,b,'\n');} (ignoring the internal line breaks). Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 30 Sep 1994 15:10:24 GMT Subject: Re: An interesting riddle posit...@Starbase.NeoSoft.COM (Jonathan Haas) writes: > You have twenty-nine thousand, five hundred twenty-three coins. > One is counterfeit and is either lighter or heavier than the others. > Using only *ten* weighings on a balance scale, determine which > is counterfeit, and whether it is lighter or heavier. I don't know why he didn't bother to mention this is a FAQ problem. The archive solution, while elegant, isn't too explicit about the actual working of the process, though. Also, the archive answer ends up with a lot of cases of putting unnecessary coins on the scales. eua...@eua.ericsson.se (Goran Wicklund) tried to answer the problem explicitly, but he got it a little wrong in the details, so that his method will sometimes require eleven weighings instead of ten. In particular, if the first eight weighings test equal, he will eliminate (9837+9837)+6561+2187+729+243+81+27+9 = 29511 coins, leaving 12. Then as he says ``three weighings left = the original problem!'' But after 8 weighings he doesn't have three weighings left, only two. The following is an explicit explanation with all the gory details, derived using the archive solution, of just how the coins may be weighed. You may prefer to skip to the ObPuzzle at the end. The first step, given W weighings to measure the odd coin among (3^W-3)/2 unknown coins, is first to weigh (3^W-3)/6 coins against (3^W-3)/6 others. Here we will have 9841 coins on each pan. If they test equal, we will have 9841 suspect coins left and continue; if not we will skip to the If the first weighing tests equal, we repeatedly test 3^(W-1) of the suspect coins against the same number of known good coins, as long as they test equal. Table 1: Continue as long as balance tests equal. SETUP : TEST : RESULT Weighing Number of: Number of :Number of Number of number suspect : suspect coins :suspect coins suspect coins coins : to weigh :left if equal if unequal : : 2 9841 : 6561 : 3280 6561 3 3280 : 2187 : 1093 2187 4 1093 : 729 : 364 729 5 364 : 243 : 121 243 6 121 : 81 : 40 81 7 40 : 27 : 13 27 8 13 : 9 : 4 9 9 4 : 3 : 1 3 10 1 : 1 :won't happen 1 The reason we can measure 13 coins in the last three weighings is that we have a supply of good coins to weigh them against, so when inequality occurs we know whether the bad coin is heavy or light. Goran outlined how to identify the bad coin out of 3^(W-1), using W-1 weighings. If the first weighing tests unequal, then we have 9841 coins in the light set L and 9841 coins in the heavy set H, together with a set of known good coins. In general, we have (3^W-1)/2 in each set to solve in W weighings. Our procedure is on each weighing to put (3^(W-1)-1)/2 coins from H and 3^(W-1) coins from L in the left pan, and (3^(W-1)-1)/2 coins from L and 3^(W-1) good coins in the right pan. If they test equal, we have completely eliminated L, and have only the other 3^(W-1) other coins from H to test, as above. If the left pan tests light, then we have the 3^(W-1) right-pan coins L in the left pan to test, as above. If the left pan tests heavy, then we have the (3^(W-1)-1)/2 left coins from H and the (3^(W-1)-1)/2 right coins from L to test, using this procedure. Table 2: Continue as long as left pan tests heavy. SETUP : TEST : RESULT WeighingSize:Left pan Right pan:Equal: Left Left number of : H L L good: Size Light: Heavy: Size L = H:coins coins coins coins: of H Size of L of L, H : : 2 9841: 3280 6561 3280 6561 : 6561 6561 3280 3 3280: 1093 2187 1093 2187 : 2187 2187 1093 4 1093: 364 729 364 729 : 729 729 364 5 364: 121 243 121 243 : 243 243 121 6 121: 40 81 40 81 : 81 81 40 7 40: 13 27 13 27 : 27 27 13 8 13: 4 9 4 9 : 9 9 4 9 4: 1 3 1 3 : 3 3 1 10 1: 0 1 0 1 : 1 1 won't happen > The different outcomes of 10 weighings are 3^10=59049. This can > point out 59049/2 = 29524.5 i.e. 29524 considering that the false > coin can be lighter or heavier. The problem contains 29523 coins, > i.e. one less than the theoretical maximum. Or put it in other > words, the solution will contain three redundant states (outcome > of weighings). These occur in weighing 10, were, for some of the > possibilities, the result "in balance" can not occur. The theoretical maximum is actually easily seen to be 29524, because any scheme that admits the possibility of always testing equal will, in that case, not determine whether the bad coin is heavy or light. I don't know an obvious way of showing that the odd coin can't be measured from (3^W-1)/2 coins in W weighings. In fact, we know it can be found from Table 1, provided that we have at least 3^(W-1) known good coins. But is that the minimum number of good coins needed? ObPuzzle: What is the minimum number of known good coins needed to find the one odd coin from (3^W-1)/2 suspect coins, and to find whether it is heavy or light, in W weighings on a balance scale? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math, sci.physics From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 30 Sep 1994 22:18:47 GMT Subject: Re: irrestible forces (was: MARILYN VOS SAVANT ...) In response to Bo.Par...@msfc.nasa.gov (Bo Parker)'s comment on Marilyn vos Savant: > >| And then, of course, her writings on Wiles' (currently > >| incomplete) proof of FLT are so far off the wall and reveal such > >| a misunderstanding of the nature of mathematical reasoning that > >| they are just laughably wrong. The best example IMO of this > >| misunderstanding is her apparent belief, revealed in her > >| writings, that Euclidean and Noneuclidean geometries are somehow > >| "in competition" for the status of "the _real_ geometry," rather > >| than just having been derived from different sets of axioms. kda...@oak.math.ucla.edu (Kevin Davey) writes: > Oh come on, surely you have to read her with a little liberty. What > she OBVIOUSLY means is that there is some history to the question of > which geometry is most appropriate for description of the real world > - and this point is completely true, there is a long history to this > debate both in physics and philosophy. I'm sure she doesn't think > that raw axiomatic systems viewed without any real world semantics > attached to them have any sorts of truth values. I find the > behaviour of people who twist her meanings to attribute such > perversities to her especially anal. Well, I'm sure she can't tell the difference between a raw axiomatic system, real world semantics, and her elbow. In case you missed it, she was writing about Fermat's last theorem, not the shape of the real universe. You and I know that when a result of geometry is used to prove a result of number theory, it is done by a transformation that depends on the axiomatic foundation of the geometry involved. That is why Bolyai's "squaring of a circle" in hyperbolic geometry and the impossibility of squaring the circle in Euclidean geometry can be transformed into true statements about the integers without rendering number theory inconsistent. The different axiomatic bases of Euclidean and hyperbolic geometry mean that the "same problem" in different geometries is rendered as different statements about numbers: One set of equations has a solution, a different one doesn't, and there is no contradiction. If the non-Euclidean transformation from non-Euclidean geometry proves the theorem about numbers, the proof is valid, even though the Euclidean transformation from Euclidean geometry may disprove a different theorem. [ The interpretation that she suspects that a result from non-Euclidean geometry was mistakenly transformed using Euclidean axioms is not supported by her writing, especially in that she voices not a suspicion but a rejection of the proof. That would be a most scurrilous way of dealing with a work that she not only didn't read, but couldn't understand if it were placed before her. ] If she'd asked enough questions of a mathematician, she would have had axiomatic sytems explained to her. If she understood axiomatic systems, she would know how different geometries relate to numbers. But she doesn't understand and she doesn't ask questions, she just publishes her mis-explanations as fact. This isn't philosophical musing, it's uninformed nonsense. And if you think I'm twisting her meanings, here's a quote from her Parade article: Bearing all this in mind, what would we think if it were discovered that Janos Bolyai, one of the three founders of hyperbolic geometry, managed to "square the circle"--but only by using his own hyperbolic geometry? Well, that's exactly what happened. And Bolyai himself said that his hyperbolic proof would not work in Euclidean geometry. So one of the founders of hyperbolic geometry (the geometry used in the current proof of Fermat's last theorem) managed to square the circle?! Then why is it known as such a famous impossibility? Has the circle been squared, or has it not? Has Fermat's last theorem been proved, or has it not? I would say it has not; if we reject a hyperbolic method of squaring the circle, we should also reject a hyperbolic proof of Fermat's last theorem. In a generous interpretation, she doesn't know what she's talking about. I suspect, though, that she now realizes this is nonsense, and just isn't motivated to admit it. Nor, I suppose, is she motivated to admit that the use of hyperbolic geometry in Wiles's proof is an unfounded rumor. Now _that's_ perversity. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 30 Sep 1994 23:03:14 GMT Subject: Re: An interesting riddle hoey@aic.nrl.navy.mil (I) wrote: > ObPuzzle: What is the minimum number of known good coins needed > to find the one odd coin from (3^W-1)/2 suspect coins, and to find > whether it is heavy or light, in W weighings on a balance scale? Jim Saxe was good enough to point out to me that this is in fact covered in the FAQ. The answer is one. I suppose I need to trot out a more challenging ObPuzzle. Suppose we are instead charged 1 unit for the each coin we put on the balance, each time we weigh it. What is a minimum-cost solution for weighing the bad coin then? What if are charged 1 unit per coin, plus K units for each weighing? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 05 Oct 1994 15:57:08 GMT Subject: Re: The Money Pump Paradox ch...@questrel.com (Chris Cole) writes: > This question is in the rec.puzzles archive: ... > ==> decision/envelope.p <== > Someone has prepared two envelopes containing money. One contains twice as > much money as the other. [ well-known paradox ] > In a variant of this problem, you are allowed to peek into the envelope > you chose before finally settling on it. Suppose that when you peek you > see $100. Should you switch now? > ==> decision/envelope.s <== [ analysis of the variant. ] > OK, so always switching is not the optimal switching strategy. Surely > there must be some strategy that takes advantage of the fact that we > looked into the envelope and we know something we did not know before > we looked..... "Surely"? I don't think this is an obvious result. Surely you don't think that information about any problem will always result in an improvement in the strategy for dealing with it. For instance, suppose we were told the color of the herring! > Well, if we know the maximum value $M that can be in the smaller envelope, ... > What if we do not know the maximum value of the pdf? You can exploit > the "test value" technique to improve your chances. The trick here is > to pick a test value T. If the amount in the envelope is less than the > test value, switch; if it is more, do not. This works in that if T happens > to be in the range [M,2M] you will make the correct decision. Therefore, > assuming the unknown pdf is uniform on [0,M], you are slightly better off > with this technique. > Of course, the pdf may not even be uniform, so the "test value" > technique may not offer much of an advantage. If you are allowed to > play the game repeatedly, you can estimate the pdf, but that is > another story... Sadly, this discussion obscures the extremely counterintuitive result that makes this problem interesting. The question of whether the pdf is uniform is irrelevant. The "test value" technique works for _any_ pdf, and even when the ratio between the higher amount H and the lower amount L is not restricted to 2. In general (assuming you pick the initial envelope at random) the technique wins when T is in (L,H] and otherwise breaks even. The payoff matrix is: : Initial envelope : Probability of : Payoff Choice of T : L H : correct decision : ---------------+-------------------+------------------+---------- T <= L < H : L H : 1/2 : (L+H)/2 L < T <= H : H H : 1 : H L < H < T : H L : 1/2 : (L+H)/2 All that is needed is a positive a priori probability that the second case occurs, i.e. that T is in (L,H]. For this to work in the case that the pdf is a single accumulation point, all possible intervals (L,H] must be covered; in particular, T must be picked from an open-ended distribution. One way is to pick T=N with probability 6/(Pi^2 N^2), for N=1,2,3,.... Using this technique, you are guaranteed to make the correct decision more than half the time (and to win more than half the money) whenever there is a positive probability that the envelopes contain different amounts. It is indeed the case that the improvement may not be very much (whether or not the pdf is uniform). What is amazing is that an improvement over even odds can be realized at all. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 05 Oct 1994 16:13:48 GMT Subject: Re: 8 bishops puzzle from "The 7th Guest" mum...@Aztec.co.ZA (Michael Rolfe) posted the puzzle: four black bishops at one end of a 4 x 5 chessboard must exchange places with four white bishops at the other end, with neither side exposed to capture. eua...@eua.ericsson.se (Goran Wicklund) writes: > You do not say if the moves have to be white-black-white etc. I have > assumed not! That assumption is unnecessary. Take any non-alternating solution for the bishops on white squares. There is an isomorphic solution for the bishops on the black squares with the colors of the bishops reversed. In each two moves, take one move of the white-square solution and one move of the black-square solution. Because the colors of the bishops involved in those two moves are different, alternating order can be chosen. Michael Rolfe mentioned that 18 moves on each square-color is necessary and sufficient; my program agrees. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 06 Oct 1994 15:37:13 GMT Subject: Re: Pudgiest Mobius band asi...@wk223.nas.nasa.gov (Daniel A. Asimov) writes: >>Consider a rectangular strip of paper of width = 1 and as long as you wish. >>If it's long enough, then it can be twisted in 3-space to form a Mobius >>band by giving the rectangle a half-twist and then matching opposite edges >>of width 1 with each other. ... >>Note: We are assuming that the paper has no thickness and can be bent, >>but it cannot be stretched at all. >>(An old Martin Gardner column offered a solution based on pleating >>the paper. But this will not work under these assumptions.) k...@cs.cornell.edu (David Karr) claims to have done this with a square of side 1. I'd like to know how, unless it violates the assumptions like: jim...@ATHENA.MIT.EDU (Jim Hsu) writes: : If you fold the paper accordian-like into a 1 by 1/5 strip, it can be : easily twisted into the moebius position and taped. One edge looks : like: (when viewed end on) 2 3 : 1\ /\ /\ : \ / \ / \ : \/ \/ \4 : The other like: : /\ /\ /8 : / \ / \ / : 5/ \/ \/ 6 7 I imagine that's the old Martin Gardner solution. The problem is, as Dan Asimov said when he posed the problem, that pleating doesn't work under these assumptions. If you leave the pleats open as you've drawn them, the paper can't be twisted without stretching it. If you fold them absolutely flat, then the two sides of a fold occupy exactly the same location. If we are going to allow such a self-intersecting piece of paper, I think we must at least view it as a 1 by 1/5 rectangle, not a 1 by 1 rectangle. The closer you get to an absolutely flat fold, the less the paper would have to stretch to be twisted. But if we are going to prohibit stretching it _at all_, then the fold can't be open _at all_. This is why a 1 by Sqrt(3) piece of paper is not the smallest Moebius-foldable rectangle, but the largest non-Moebius-foldable rectangle. If you fold it exactly into a triangle, it is a triangle, not a rectangle; if you fold it slightly less than exactly it doesn't quite reach. (At least, I don't think so.... I haven't got a proof). The other Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 11 Oct 1994 21:02:20 GMT Subject: Re: Pudgiest Mobius band When Daniel A. Asimov wrote: > >I don't have a proof of this, but I doubt that any solution based > >on folding an almost-square on its two almost diagonals, then > >joining the two edges of length = 1, can possibly work. isaac...@OCF.Berkeley.EDU (Isaac Ji Kuo) writes: > You are wrong. You obviously deleted my solution without carefully > reading it and considering the solution.... Perhaps he did. But perhaps he did look at your "solution"--as I did--where it uses "a small amount e" and "an even smaller amount a< Rather than just state your belief, look at the solution presented > and try to find any stretching. There is none. Rather than just state that we are not looking at your solution, why don't you try to prove it correct. I am not going to prove that there are _no_ values of a, e, the "_very_ slight shift" and the dihedral angles involved such that a non-overlapping Moebius band is formed. But then, I'm not claiming I have a solution. You're the one who's claiming you have a solution. Put up or shut up. > Look, I went through the trouble to figure out exactly how to fold the > paper so that no overlapping was needed, so you could at least have > the decency to pick apart the presented solution if you're going to > say you don't "believe" it will work. If you calculate the exact three-space coordinates of the seven vertices of the Moebius surface you claim to have constructed, I'll endeavor to figure out whether it works. Until then the problem is open. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 20 Oct 1994 16:55:07 GMT Subject: Re: The sum of the first n squares can not be a perfect square. r...@emu.pmms.cam.ac.uk (Richard Pinch) writes: > MAUC...@CSPVX1.CSP.IT (Mauceri) writes: > >"The sum of first n squares can not be an perfect square, > >except that for n=24." > Richard Guy, "Unsolved problems in number theory", Springer, 1981, > refers to this problem in section D3.... Guy says that Mordell > asked whether there was an elementary solution. _Unsolved Problems in Number Theory, Second Edition_ has just hit the shelves. ``Mordell asked if there was an elementary proof, and affirmative answers have been given by Ma, by Xu & Cao, by Anglin, and by Pinter.'' Ma De-Gang, An elementary proof of the solution to the Diophantine equation 6y^2=x(x+1)(2x+1), _Sichuan Daxue Xuebao_, 4(1985) 107-116; MR 87e:11039. Z. Y. Xu & Cau Zhen-Fu, On a problem of Mordell, _Kexue Tongbao_, 30(1985) 558-559. W. S. Anglin, The square pyramid puzzle, _Amer. Math. Monthly_, 97(1990) 120-124; MR 91e:11026. Akos Pinter, A new solution of two old diophantine equations, Technical Report No. 92(1993), Department of Mathematics, Kossuth University, Debrecen, Hungary. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: news.admin.policy, alt.current-events.net-abuse Followup-To: alt.current-events.net-abuse From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 21 Oct 1994 17:17:12 GMT Subject: Spam: Net Games - New Book from Creators of Net Guide "Michael Wolff" and his company have spammed the fifty-odd rec.games.* groups, rec.puzzles, rec.puzzles.crosswords, and perhaps more, with an ad for his book. It seems to be relevant to two or three of them, but even then in pretty poor taste for all but the marketplace groups. No crossposting was used. This sort of low-level spam should get cancelled too, though it's probably going to get to be annoyingly frequent. Dan Hoey Hoey@AIC.NRL.Navy.Mil ================================================================ ] From: edit...@ypn.com (Michael Wolff) ] Newsgroups: rec.games.abstract ] Date: 20 Oct 1994 18:43:57 -0400 ] Organization: Michael Wolff & Company, Inc. ] Sender: n...@ypn.com ] Nntp-Posting-Host: bill.ypn.com ] ] We've just finished our new book Net Games--a guide to the world of online ] games. It's stuffed with addresses and hints for playing games over the ] Net from chess to Doom to FurryMUCK. We've put up a sampling of pages from ] the book as GIF's that can be downloaded from ftp://www.ypn.com/pub/pages. ] Feel free to upload these pages anywhere on the Net. For more information ] about Net Games, our first book Net Guide, or our upcoming books Net Chat, ] Net Money, Net Trek, Net Sports, and Net Tech check out our Web site at ] http://www.ypn.com/. 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(There are more than 400 virtual world games which, at any hour ] of the day, have as many as 200 players!); ] ] - Play-by-mail strategy classics like rotisserie baseball and Henry ] Kissinger's favorite game, Diplomacy, that travel over email lines. ] ] By some estimates more than half of online time is spent playing ] interactive games. Indeed, games are proliferating around the Net at ] such an astonishing rate and are gaining so much popularity that some ] universities have begun to control the bandwidth games take up on their ] mainframes. On commercial services, games are one of the fastest growing ] offerings. GEnie players of Air Warrior, a flight simulation game, have ] spent up to a $1,000 a month. It is no coincidence that shortly after ] Rupert Murdoch bought Delphi, he also bought Kesmai, one of the largest ] producers of online games. Besides the games people play online, Net Games ] also covers the thousands of free computer games and demos that people ] download from online archives and the modern-day mutual aid societies that ] share cheats, hints, and walk-throughs for mastering home and arcade hits ] ranging from Mortal Kombat and NBA Jam to Sim City 2000 and Doom (and even ] the daily crossword puzzle). ] ] Net Games, the second book in the Net Books Series, is by the creators of ] Net Guide, the best-selling guide to Cyberspace that Wired editor Louis ] Rossetto called "the TV Guide to Cyberspace," and USA Today called "the ] liveliest most readable online guide yet!" In addition to Net Guide and ] Net Games, the Net Books Series includes: ] ] - Net Chat, a map of salons and meeting places in Cyberspace, to be ] published in November; ] ] - Net Money, a handbook for using online personal finance resources, to ] appear in time for tax season in January; ] ] - Net Trek, a directory of the online Trekkie universe, to arrive in ] bookstores in March. ] ] -- ] Michael Wolff & Company, Inc., 1633 Broadway, 27th floor, New York, NY 10019 ] Vox: 212-841-1572 Fax: 212-841-1539 Email: edit...@ypn.com ] To order books or find out more about our new Internet service, YPN-- ] Your Personal Network, call toll-free 1-800-NET-1133 ================================================================ --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 24 Oct 1994 15:59:45 GMT Subject: Re: Rubik's Cube impossible states chris.pel...@octel.com writes of taking a Rubik's Cube apart: > ... There are twelve different ways to put it back together > (referred to as the "12 universes"). Only one will lead back > to START. The impossible modes are: > 1 corner twisted > 1 edge flipped > 2 corners swapped > 2 edges swapped > and 7 other combinations of the above 4 states Well, you seem to have a vaguely correct idea of this, but the details of your list are somewhat confused. First they are called "orbits", not "universes". Second, the orbit with two corners swapped is the same as the orbit with two edges swapped; it is more properly called the odd permutation orbit. Third, there are two different orbits with "one corner twisted", depending on which way you twist it. Of course, speaking of only one corner or edge being twisted or flipped is misleading--they are the orbits of nonzero *total* twist or flip (modulo 3 or 2). So the twelve orbits are formed by orthogonal combinations of the three twist orbits, the two flip orbits, and the two permutation parity orbits. The ones you list are the ones in which all but one component is the home orbit. > Of course the other way is if somebody removes the stickers > and places them on incorrectly (i.e. making two opposite colors > adjacent on certain pieces!) You mean "e.g.", not "i.e."--there are a lot of ways to place the stickers incorrectly without having opposite colors on the same piece. Like having two of the *same* color on a piece. Or having duplicate pieces. Or having duplicate face centers. Or having the wrong order of colors on a corner piece. I think that's all the ways. ObPuzzles: How many ways can you place the stickers so that the cube is not solvable? How many of them have a piece with a pair of opposite colors adjacent? How many have duplicate pieces? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 25 Oct 1994 23:06:07 GMT Subject: Re: Rubik's Cube impossible states chris.pel...@octel.com writes (in entirety): > Dan Hoey writes: > >Well, you seem to have a vaguely correct idea of this, but the details > >of your list are somewhat confused. First they are called "orbits", ... > >You mean "e.g.", not "i.e."--there are a lot of ways to place the > I stand corrected! Sheesh... I must protest the dishonesty of your editing my reply to suggest that I discussed only these trivial errors in your description of the impossible states of Rubik's Cube. Your rudeness in following up with an article that has nothing to say about puzzles speaks for itself. As for the ObPuzzles I posed: How many ways can you place the stickers so that the cube is not solvable? How many of them have a piece with a pair of opposite colors adjacent? How many have duplicate pieces? I think these may be quite challenging questions. The first is easily seen to be 54!/(9!)^6 - 6! 8! 12! 2^10 3^7, if we ignore the position of the cube in space. But figuring out how many ways there are if we ignore whole-cube moves is a neat problem. The others are quite difficult even ignoring symmetry. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Fri, 28 Oct 94 11:38:15 -0400 (EDT) From: Dan Hoey To: Cube Lovers Subject: Cube colors and face names Singmaster designates the faces as Up, Down, Right, Left, Front, and Back. The names are chosen so that no two of the faces start with the same letter. There have been some latter day efforts to rename Up as Top so that all the faces have names beginning with consonants. Yes, this is the main reason for using Top, because of the Rubiksong introduced by Varga that I described (unfortunately with many typos) on 22 Feb 90 ( is a URL that I hope works--anyone who is actually able to point and click on this, please let me know). But there's another reason. Remember the annoying feature that the color assignments to faces were never standardized? The first cube I bought had red opposite yellow, blue opposite white, and orange opposite green (I think). Even though in later days most cubes are manufactured with opposite faces ``differing by yellow''--red opposite orange, blue opposite green, and yellow opposite white--there does not seem to be a standard for the handedness of the coloring. This has long been a problem on cube-lovers, where everyone starts out asking ``I've got my cube solved except a blue sticker on the white face, a white sticker on the green face, and a green sticker on the blue face,'' and the puzzle becomes trying to figure out where those faces are. (This was fixed in Square 1, where they printed a full-color instruction book coordinated with the puzzle). My modest proposal is to define the Standard Earth-Tone Cube, which has the faces in standard and easily remembered places. The colors are taupe, dun, fawn, beige, loam, and roan. This supports the use of Top over Up, because ``taupe'' is so much more evocative than ``umber''. ``Dun'' is also a major win, and I wish I had better names for the other faces. I have yet not tried painting such a cube, because I can't figure out which color is which. Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 29 Oct 1994 01:47:29 GMT Subject: Re: Rubik's Cube impossible states d...@cwi.nl (Dik T. Winter) wrote that the number of ways of arranging the stickers on a Rubik's cube with different colors on the face centers is 2080262517294685958156352617272800000. > It is 6! . 48!/(8!)^6 (i.e. number of ways to give the centers > different colors multiplied by the number of ways to color the > remainder). Yes, I figured that out eventually. It's interesting that the ratio of the total number of ways of stickering the cube to the number of ways of stickering without repeating face centers is 9^6 / C(54,6). There ought to be an intuitive way of seeing this. > And indeed, to get different cubes up-to whole-cube moves this must > be divided by 6!. No, you divide it by 24 (the number of proper isometries of the cube). You would divide it by 6! if you wanted the number of cubes up to permutations of the colors--but not exactly by 6!, as there are some recoloring symmetries. At any rate, I've since counted the ways of arranging the stickers so that no edge or corner piece has two colors the same, calling a violation of this criterion a ``collision''. This is a lot harder. A way of doing this is with generating functions. The edge generating function: E(x1,x2,...,x6) = Sum 2 xi xj 0 To: Cube Lovers Subject: The real size of cube space In January of this year, Jerry Bryan and I wrote of counting the number of M-conjugacy classes of Rubik's cube. In the sense that (for instance) there is really only one position 1 QT from start, even though that QT may be applied in twelve different ways, this task amounts to counting the true number of positions of the cube. The earlier discussion centered on calculations involving computer analysis of large numbers of positions. However, a look in Paul B. Yale's book _Geometry and Symmetry_ gave me a clue: the Polya-Burnside theorem is a tool that allows us to perform this calculation by hand. The Polya-Burnside theorem describes a relation between a finite group J and a _representation_ of the group. For our purposes, a represen- tation is a homomorphism of J into a permutation group, R: J -> S[X]. Here, S[X] refers to the group of all permutations of a set X; the image of J, called R(J), need not be the whole of S[X], but R(J) will be a subgroup of S[X]. The _orbits_ of R(J) are the equivalence classes of X under the relation x~y, said to be true if there is some permutation p in R(J) for which p(x)=y. The _fixed points_ of a permutation p in R(J) are the elements x of X for which p(x)=x. The Polya-Burnside theorem states that the average number of fixed points of permutations in R(J) is equal to the number of orbits of R(J). That is, |R(J)| |Orbits(R(J))| = Sum[p in R(J)] |FixedPoints(p)|. The average may also be taken over J: |J| |Orbits(R(J))| = Sum[j in J] |FixedPoints(R(j))|, a nontrivial distinction, since R may not be one-to-one (though it is for our application). The Polya-Burnside theorem is not very inaccessible nor hard to prove, but I will not prove it here. For our purpose, we take the group J to be M, the 48-element group of symmetries of the cube. X will be the set of all cube positions, which we usually call Gx (for GE, GC, or G, depending on whether we consider edges, corners, or both; we are considering the positions relative to fixed face centers in all three cases). And the repre- sentation R is the operation of M-conjugation: (R(m))(g) = m' g m. Verifying that R is a homomorphism is an exercise in associativity that Jim Saxe and I carried out in the Symmetry and Local Maxima paper, in the archives [cube-mail-1, 14 December 1980]. R has been so chosen because we wish to calculate the number of M-conjugacy classes of Gx, |Gx\Conj(M)|, which is be the number of orbits of R(M). To apply the Polya-Burnside theorem for this, we need to calculate, for each element of m of M, the number of fixed points of R(m). That is the number of elements g of Gx for which m' g m = g. Multiplying by m, this becomes g m = m g: the fixed points we wish to count are just those elements g of Gx that commute with m. There are several tools to make the counting easier. First, I'll note that just as there are M-conjugacy classes of Gx, there are M-conjugacy classes of M itself. The number of fixed points of R(m) is the same for any m in a given conjugacy class. So to calculate the total number of fixed points over R(M), we need only calculate the number of g in Gx commuting with each of the ten classes of cube symmetry and multiply by the size of the class. The fundamental principle we use in finding whether g commutes with m can be found by examining the cycles of m. Suppose m permutes a cycle (c1,c2,...,ck), so that c2=m(c1), c3=m(c2),...,ck=m(c[k-1]),c1=m(ck). For g to commute with m, we have g(c2)=m(g(c1)), g(c3)=m(g(c2)), ..., g(ck)=m(g(c[k-1]), and g(c1)=m(g(ck)). So (g(c1),g(c2),...,g(ck)) is also a cycle of m. Thus g must map each k-cycle of m to another k-cycle of m, and in the same order. Conversely, if g acts thus on cycles, then g will commute with m, and so g is a fixed point of R(m). Suppose that m has j different k-cycles of cubies. There are j! k^j possibilities for g's action on the cubies in those k-cycles: j! permutations of cycles, and for each g:(c1,c2,...,ck)->(d1,d2,...,dk), k choices for g(c1) among {d1,...,dk}. It turns out to be a fairly easy exercise to show that half of those possibilities are even permutations and half odd, though the partition by parity is surprisingly different depending on whether k is even or odd. This will allow us to combine the results for GE and GC simply by multiplying together and dividing by two. Now consider orientation of cubies. This is similar to the case of permutation, in that the orientation that g imposes on a cubie is a constant for all cubies in a cycle. I will first discuss the edge orientation, which is fairly straightforward, and continue to corner orientation, which has some surprising features. For edge orientation, if all the cycles have even length, then g's orientation parity is zero over each cycle, and so zero over the entire cube. So we can choose the orientation of imposed by c1->g(c1) for each cycle (c1,...,ck) in 2^j ways. If there are odd-length cycles, then half of the orientations will have nonzero orientation parity, and only 2^(j-1) possible orientations can be achieved. For corners, we might expect there to be 3^(j-1) orientations, except 3^j for cycles of length a multiple of three, and this is often so. But there are two important exceptions. First, if m is a reflection (i.e., not a proper rotation in C) then alternate cubies in each cycle must be given the opposite orientation by g. If the cycle has even length, this conserves orientation, so there will be 3^j possibili- ties. If the cycle has odd length, this implies that the orientation of each cubie must be its own opposite (i.e., zero twist). Thus, there there is only one possible orientation of the 1-cycles in the diagonal reflections. The second exception, an even bigger surprise, occurs when m is either the 120-degree rotation or the 60-degree in- verted rotation. It turns out that the orientation constraint forbids any permutation that exchanges the two 1-cycles in these positions. (This constraint on permutations would throw off the equality between even and odd permutations, except that these classes of m have other corner cycles that restore the balance.) The impossibility of m commuting with an exchange of the two corners can be verified by examining the possible orientations, but I haven't got any good way of characterizing when it would be be a problem in general. In fact, I did not notice it until I investigated discrepancies with the exhaustive computer analysis. Using the above analysis, we may carry out the calculation as in the three tables below. The first two tables count the number of fixed points of R(m) for an element m of each class, multiply by the class size, and divide by |J|=48 to get the number of orbits as in the Polya-Burnside theorem. The third table calculates the number of fixed points by combining the results of the first two tables, divided by the class size (which was multiplied in both for edges and for corners), and divided by 2 (because only half the combined positions have matching permutation parity). Counting M-conjugacy classes of the edges of Rubik's cube. M class Cycles Total F.P. Numeric (class size) of m Perms Oris in class Total/48 ============== =========== ====== ====== ========== =========== Identity (1) 12 1-cycles 12! 2^12/2 12! 2^11 20437401600 Axis Rot/2 (3) 6 2-cycles 6! 2^6 2^6 6! 3 2^12 184320 Rot/3 (8) 4 3-cycles 4! 3^4 2^4/2 4! 3^4 2^6 2592 Diag Rot/2 (6) 5 2-cycles 5! 2^5 2^5 2 1-cycles 2 2^2/2 5! 3 2^13 61440 Rot/4 (6) 3 4-cycles 3! 4^3 2^3 3! 3 2^10 384 Inv Rot/4 (6) 3 4-cycles 3! 4^3 2^3 3! 3 2^10 384 Diag Ref (6) 5 2-cycles 5! 2^5 2^5 2 1-cycles 2 2^2/2 5! 3 2^13 61440 Inv Rot/6 (8) 2 6-cycles 2! 6^2 2^2 2! 3^2 2^7 48 Axis Ref (3) 4 2-cycles 4! 2^4 2^4 4 1-cycles 4! 2^4/2 4! 3^2 2^14 73728 Inversion (1) 6 2-cycles 6! 2^6 2^6 6! 2^12 61440 ----------- | GE\Conj(M) | = 20437847376 Counting M-conjugacy classes of the corners of Rubik's cube. M class Cycles Total F.P. Numeric (class size) of m Perms Oris in class Total/48 =============== ========== ====== ===== =========== ======= Identity (1) 8 1-cycles 8! 3^8/3 8! 3^7 1837080 Axis Rot/2 (3) 4 2-cycles 4! 2^4 3^4/3 4! 3^4 2^4 648 Rot/3 (8) 2 3-cycles 2! 3^2 3^2 2 1-cycles 1 3^2/3 3^5 2^4 81 Diag Rot/2 (6) 4 2-cycles 4! 2^4 3^4/3 4! 3^4 2^5 1296 Rot/4 (6) 2 4-cycles 2! 4^2 3^2/3 3^2 2^6 12 Inv Rot/4 (6) 2 4-cycles 2! 4^2 3^2 3^3 2^6 36 Diag Ref (6) 2 2-cycles 2! 2^2 3^2 4 1-cycles 4! 1 4! 3^3 2^4 216 Inv Rot/6 (8) 1 6-cycle 6 3 1 2-cycle 1 3 3^3 2^4 9 Axis Ref (3) 4 2-cycles 4! 2^4 3^4 4! 3^5 2^4 1944 Inversion (1) 4 2-cycles 4! 2^4 3^4 4! 3^4 2^4 648 ------- | GC\Conj(M) | = 1841970 Counting M-conjugacy classes of the entire Rubik's cube M class Edge Corner Corner times edge (class size) F.P. F.P. / (96*class size) =============== ========== ========= ======================= Identity (1) 12! 2^11 8! 3^7 901,083,401,551,872,000 Axis Rot/2 (3) 6! 3 2^12 4! 3^4 2^4 955,514,880 Rot/3 (8) 4! 3^4 2^6 3^5 2^4 629,856 Diag Rot/2 (6) 5! 3 2^13 4! 3^4 2^5 318,504,960 Rot/4 (6) 3! 3 2^10 3^2 2^6 18,432 Inv Rot/4 (6) 3! 3 2^10 3^3 2^6 55,296 Diag Ref (6) 5! 3 2^13 4! 3^3 2^4 53,084,160 Inv Rot/6 (8) 2! 3^2 2^7 3^3 2^4 1,296 Axis Ref (3) 4! 3^2 2^14 4! 3^5 2^4 1,146,617,856 Inversion (1) 6! 2^12 4! 3^4 2^4 955,514,880 ----------------------- | G\Conj(M) | = 901,083,404,981,813,616 These results have been corroborated and expanded by use of combinatorial computer programs, to be described in a later message. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.fan.cecil-adams From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 06 Nov 1994 22:01:28 GMT Subject: Re: Natural Law Party? aat...@aol.com (Aatrix) writes of :...Doug Henning (of magic fame) spouting the Natural Law Party line, :about how they would get a 100 gurus levitating over the capital in :Ottawa, which would solve all the country's problems. sj...@aol.com (SJF 35) writes: > Am I missing something here?? They have a magician, that is, a > PROFESSIONAL ILLUSIONIST, trying persuade people that levitation...? But that may not be quite as silly as the Gugu Mahahaha Jijiji's traditional approach to levitation, which involves: 1. Sitting around cross-legged. 2. Hopping a couple of inches from that position. 3. Calling that ``levitating''. 4. Using meditation to increase their ``hang time.'' Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 07 Nov 1994 17:39:41 GMT Subject: Re: Conway tiling mc...@alcor.concordia.ca (John McKay) writes: > There may be an interest in a new tiling based on a right > triangle with sides 1:2:sqrt(5) which is non-periodic. > It bears comparison with the Penrose tiling but is generated > by a single tile. I'm afraid that statement is somewhat misleading. According to John Conway, the new tiling uses a very large number -- say 10^100 -- of tiles derived from the right triangle by "kinking the edges" in a way found by Radin. The kinks enforce the stricture that any tiling can be "deflated" by combining sets of five smaller triangles into a larger one as follows: . |\ | \ | \ | \ |.-' \ |\-. \ | \ `-.\ | \ `\ | \.-' \ |_-'______\ An aperiodic tiling by a set of tiles that does not admit a periodic tiling is not new. The early ones found by Wang were based on squares with slightly modified edges, and the Penrose tiles are a two-tile set with this property. What is important about the new tiling is that the orientations of the tiles are evenly distributed: The number of tiles with orientation-parameter between A and B approaches B-A in the limit. The most important open question in aperiodic tilings is whether there exists a _single_ tile that tiles aperiodically, but not periodically. Conway believes an aperiodic monotile exists, and has some ideas on how to find it, but this is not that discovery. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.fan.cecil-adams From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 07 Nov 1994 22:28:41 GMT Subject: Re: Natural Law Party? Yesterday hoey@aic.nrl.navy.mil (I) wrote: > But that may not be quite as silly as the Gugu Mahahaha Jijiji's > traditional approach to levitation, which involves: > 1. Sitting around cross-legged. > 2. Hopping a couple of inches from that position. > 3. Calling that ``levitating''. > 4. Using meditation to increase their ``hang time.'' Thinking it over, I think I have mixed up two different gurus. The Transcendental Meditation/Natural Law/levitation/square root of one per cent cultists are led by the Maharishi Mahesh Yogi. The Guru Maharaj Ji was the "thirteen-year-old perfect master" who had a different cult in the 1970s. I don't know what happened to him afterwards--maybe he grew up. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.arts.disney From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 07 Nov 1994 22:29:33 GMT Subject: Swahili names in the Lion King? I have heard two reports of a letter from Pat Oliver in the british newspaper _The Guardian_. Dave Budd quotes it as follows: I took my children to see Walt Disney's cartoon The Lion King and was showing off my slight knowledge of Swahili by translating the names of the characters - "Simba" meaning lion, and "Shenzi" meaning barbarian. My son asked me what "Pumba" (the name of the warthog) meant, and I had to look that one up in the dictionary. It said "excretion from under the foreskin". Clearly someone is being cute, but is it Pat Oliver pulling the newspaper's leg, or some nameless Disneydroid trying to one-up the Little Mermaid? Does someone have an authoritative translation of Pumbaa? For that matter, how about Nala, Musafa, Sarabi, Zazu, Rafiki, and Timon? I assume Scar, Banzai, and Ed are not Swahili. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Tue, 08 Nov 94 17:59:31 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Re: The real size of cube space Wow, I didn't realize this sort of calculation had been automated. Martin.Schoenert@math.rwth-aachen.de writes: The way I view this is as follows. The entire cube group C is a permutation group group on 6*9 points, generated by the six face turns U, D, L, R, F, B; the three middle slice turns M_U, M_L, M_F; and the reflection S. This group has a subgroup M of symmetries of the cube (of order 48), generated by U M_U D', L M_L R', F M_F B', and S. Another subgroup is G, generated by the six face turns, which has index 48 in G. G is a normal ^ divisor of C, G is the semidirect product of M and G. The same is ^ true for GE and GC. I think two of those G's are supposed to be C's, right? What is the difference between a direct product and a semidirect product? ... [conjugation by] M fixes the set of the generators of G and their inverses. M is fact the largest subgroup of the outer autmorphism group with this property, which makes it rather important. In a 1983 Cubic Circular article (of which I know only Stan Isaacs's summary) David Singmaster observed that the group is larger for larger cubes, provided we work what I call the ``theoretical invisible group''. That is, we solve not only the surface of the cube, but the hypothetical interior (n-2)^3 cube, and all the smaller (n-2k)^3 cubes as well. I blithered at length about this in my article of 1 June 1983 archived (I think I've got it right this time) at . The idea is that a mapping called evisceration allows us to permute the layers of the cube. On the 4x4x4 cube, this for instance allows us to exchange each inner slab with its adjacent outer slab. It also allows us to conjugate each inner slab move by central inversion, while leaving the outer slab moves alone. In general, evisceration of a d-dimensional cube by f maps each feature (cubie, colortab, or face-center arrow) at coordinates (x[1],x[2],...,x[d]) to (f(x[1]),f(x[2]),...,f(x[d])), where f is a permutation of the intervals between the cleavage coordinates of the cube. I believe that if f commutes with the central inversion, then conjugation by evisceration is an outer automorphism of the Rubik's cube group. (I think I have proved this for d=3, and I think the proof in higher dimensions should not be difficult given the right notation.) The group of all eviscerations includes the central inversion; we can of course augment it by the rotation group in d-space. Is this the maximum outer automorphism group that respects generators of the Rubik's cube? For this we take the generators to be turns of slabs between adjacent cleavage planes. (Turns are direct d-1-dimensional isometries.) I was already familiar with this augmented symmetry group because it also induces automorphisms on d-dimensional tic-tac-toe. (In fact, it may be the maximal automorphism group on all tic-tac-toe boards of side greater than two. I know it's been proven for 4^3, but I don't know of any larger results). Do you know anything more about this group, like whether it has been named or studied? Since G's structure is very similar to a symmetric group (or more accurately the direct product of two symmetric groups), it allows to describe the centralizer of an element in G. The more a group differs from a symmetric group the less this analysis helps (for those that know what I'm talking about: the more a group differs from the symmetric group, the worse a backtrack computation using cycle structure analysis is). But no, G's structure is actually similar to the direct product of two _wreathed_ symmetric groups. Does this interfere with the backtracking as much as it interferes with my manual analysis? Do you know of any good treatments of finding centralizers of outer automorphisms of wreath products? In particular, I would very much like to know under what conditions the centralizer of the wreath product fails to cover the centralizer of the permutation factor, as we saw with the corners. As for when I wrote M class Edge Corner Corner times edge (class size) F.P. F.P. / (96*class size) ^^^^^^^^^^^^^^^^^^^^^^ That's not a typo. I was just saying that column 4 is equal to column 2 times column 3, divided by column 1, divided by 96. Perhaps I should have factored column 1 out of columns 2 and 3 first to avoid this confusion. gap-3.4 -b -g 4m gap> Sum( ConjugacyClasses( M ), > c -> Size( Centralizer(G,Representative(c)) ) / 48 * Size(c) ); 901083404981813616 Well, call me John Henry. Say, do you have gap libraries for other magic polyhedra? For higher-dimensional magic? Dan Hoey Hoey@AIC.NRL.Navy.MIl --- Date: Thu, 10 Nov 94 08:43:35 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Re: Symbolic cube program Since I received many requests for it, here it is. It is a uuencoded, compressed, tar file. The cube lovers archives contain over two megabytes of readable discussions on the cube. In the interests of keeping them readable, I ask that binary files and programs be distributed using some other channel: private email, ftp, web, or whatever. Please let Cube-Lovers remain a _discussion_ list for cube topics. I am aware that alternative distribution methods are not supplied by Alan Bawden in his maintenance of the Cube-Lovers list and archives. Please consider that encouragement to provide such methods, rather than an excuse to use this list for the purpose. I am also aware that the recent program was not much larger than some of the discussion articles on the list. Nonetheless, it does not contribute to the discussion, and imposes a burden on those of us who subscribe for the discussion. The revised versions and larger programs to follow would impose a growing burden on the list. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@pooh.tec.army.mil (Dan Hoey) Date: 11 Nov 1994 18:44:13 GMT Subject: Re: A Combinatorics Problem. Kanad Chakraborty (ka...@colorado.eecs.umich.edu) wrote: > Prove that the maximum number of infinitely long cylinders (in > R^3, i.e. the 3D real space) of the same radius such that each > cylinder touches each of the other cylinders, is 6. Where I come from, we call that a geometry problem. And oddly enough, I saw it in Croft and Guy's _Unsolved problems in Geometry_ earlier today. I'm pretty sure he said it's one of the unsolved ones. Seven can be done if the cylinders are finite, though. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 16 Nov 1994 23:07:02 GMT Subject: Re: 0 ^ 0 = ? ed...@math.ohio-state.edu (Gerald Edgar) writes: > 0^0 = 1, as has been noted. But 0.0^0.0 = ??? > Most of the arguments for 0^0=1 [presented here, in the FAQ, > and in Knuth's paper] apply to the case when the exponent > is the integer zero. They do not apply when the exponent > is floating-point zero. Well, sci.math and its FAQ are about the zero that occurs in the integers, a subset of the real numbers. As far as most mathematicians are concerned, the real number 0 is the same 0 as the integer 0. We don't usually deal with approximations, such as floating point, that are mostly of use to computer implementations and programming languages. That's what comp.arch.arithmetic is for. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 17 Nov 1994 16:32:57 GMT Subject: Re: 0 ^ 0 = ? When I wrote: > >As far as most mathematicians are concerned, the real number 0 is > >the same 0 as the integer 0. mckni...@ix.netcom.com (Lawrence McKnight) replied: > Actaully, you seem to be a bit confused (at best). Strictly > speaking, the real number 0 and the integer 0 ae different objects. > The real number 0 is either (take your choice) a Dedekind cut, or an > equivalence class of Cauchy sequences of rationals. Now, there is > an isomorphism of the integers into the reals, and when no confusion > can arise, it is common to treat the integers and their isomorphic > images as 'identical'. Well, you seem to be a little confused about what I was talking about. I was talking about most mathematicians, not about students doing foundational exercises. For most mathematicians, the ideal generated by the real unit is the actual integers, not an isomorphic copy. "Take those points in R^n with integer coordinates" means exactly what it says. You seem to have forgotten the last part of the construction of real numbers, where we say that the set R* of Dedekind cuts (or equivalence classes of sequences, or whatever) is only isomorphic to the real numbers. We get the real real numbers by finding the subset Q* of R* isomorphic to the rationals Q, and let R=Q union (R* minus Q*) (just as we did when we constructed Q from Z, and just as we will do when we construct C from R). This would be a silly, worthless exercise, except that it forestalls silly, worthless quibbles about whether some real numbers are rational. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: misc.education.science, alt.amateur-comp, sci.math Followup-To: sci.math From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 29 Nov 1994 17:50:30 GMT Subject: Re: Scientific American and Packing Spheres apsm...@u.washington.edu (Arthur Smith) writes: > > I was very surprised to see assertions that this particular problem > > (the optimal 3-dimensional packing of spheres) had not yet been solved. t...@maths.qmw.ac.uk (Thomas D Bending) writes: > I thought this was proved by a Japanese mathematician quite recently - > in the last couple of years. Unfortunately I can't remember his name > and don't have a reference - anyone else? In 1990, Wu-Yi Hsiang (at Berkeley) announced a proof of Kepler's conjecture that the densest packing of spheres in three dimensions is achieved by the known hexagonal packings. In the Spring, 1994 issue of _Mathematical Intelligencer_, there is a letter from J.H. Conway, T.C. Hales, D.J. Muder, and N.J.A. Sloane (all of them sphere-packing researchers) stating their opinion that Hsiang's work does not constitute a proof of the conjecture. They noted that Hsiang's paper will be published soon in _International Journal of Mathematics_. I do not know if it has come out yet. The Summer, 1994 issue of _Mathematical Intelligencer_ contained a more detailed paper by Hales describing a number of differences between what Hsiang claims to prove and what he actually proved. The summary is that Hsiang has significantly advanced the program for proving the Kepler conjecture, but it is difficult to tell how far. His work is marred by pervasive failures to distinguish between what he has proved and what he considers obvious but has not proved. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 05 Dec 1994 16:46:10 GMT Subject: Re: number theory stuff rstim...@silver.ucs.indiana.edu (robert and stimets) wrote: > >>can someone post describe all of the > >>numbers n for which n! is evenly divisible by 4? > >>Those that are 6,7,10,11,12,13,18,19,20,21,.... > >>Those that aren't 1,2,3,4,5,8,9,14,15,16,17,..... and later clarified that he meant: > ... that n! be divisible by an even number of factors of two, i.e. > n! = (4^m) * k for some odd k. In that case, you have misclassified 1, and we may as well throw in zero: certainly 0!=1!=(4^0)*1 is of the requested form. These are the numbers whose binary representation contains an even number of one bits, not counting the units place. An easy induction shows that the two characterizations are equivalent. This is an important set: it is the "sparse space" of rare G-values for Grundy's game (In Grundy's game players alternately split a heap of stones into two unequal piles; the player to move loses when all heaps have 1 or 2 stones.) The G-values have been calculated up to G(10^9), and only 1273 in that range are rare, of which G(82860)=108 is the last. If, as seems likely, there are only finitely many rare G-values, then the sequence G(n) will be periodic for sufficiently large (but perhaps very, very large) n. To read about this, see _Winning Ways_ by Berlekamp, Conway, and Guy. There is also information about the problem in an AMS monograph collection edited by Guy--the title is something like _Combinatorial Games_. ObPuzzle: Describe the numbers whose factorials are (27^m)*k for k not divisible by 3. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 07 Dec 1994 17:28:56 GMT Subject: Re: Dogs Mead In response to the "Dogs Mead" puzzle br...@Cadence.COM (Bruce Chiriatti) posted, whu...@cco.caltech.edu (Wei-Hwa Huang) writes: > Fortunately, this puzzle is reprinted in James Fixx's "Games for the > Superintelligent." > And the reason poor Bruce couldn't solve it is because of the > many transcription errors.... > I realize that it may be possible that there are two very similar > puzzles... but as stated, I cannot get 15 across times 9 down to be > 16 across...the digits just don't work out. If James Fixx published the puzzle you describe, then there must indeed be two very similar puzzles. The puzzle Bruce posted is number 448 in David Wells's book _The Penguin Book of Curious and Interesting Puzzles_. Wells attributes it to _The Strand Problems Book_ by W. T. Williams and G. H. Savage. Does James Fixx say where he got his puzzle? As you noticed, it is impossible for two two-digit numbers to have a two-digit product; Bruce's version has "times" for "minus" in the clue for 16 Across. A more surprising observation is that that was Bruce's _only_ significant error. Bruce's version changes the original spelling of "Little Pigley", and omits to mention that it is the name of the farm of which Dog's Mead is a field. His version also gives Ted's name as Edward, and mentions Mary's relative youth in 11 Across instead of the book's 3 Down. The hints were also reworded, accidentally failing to mention that there is only one pair of equal answers. But the given relations between the numbers were unchanged. As for Bruce's original plea: > I posted a request for the answer to this puzzle (which I once > used to own). And I want it again!!! Anyone? Wells gives a solution (with misprint that will mislead only the least wary). ISBN 0-14-014875-2, Penguin Books, 1992. Go get it. Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 08 Dec 1994 17:28:07 GMT Subject: Re: Dogs Mead Ken_Duisenb...@qmgate.arc.nasa.gov (Ken Duisenberg) writes: > [Wei-Hwa made many changes to the following, basically creating a > different puzzle.] Wei-Hwa Huang attributed his version to James Fixx's _Games for the Superintelligent_, so I would assume the changes are Fixx's. Or perhaps the changes were David Wells's (Where did you get the puzzle?). Or perhaps the _Strand_ updated and reissued the puzzle four years after the original. > [The year is 1935, see title. This could perhaps be determined, but > I didn't work it that way.] It can be determined if you assume that the current year is the sum of a person's birth-year and age. That's dubious, since in reality you may get the previous year (unless this was a New-Year's Eve puzzle. I'm not sure whether they intend it that way. On the one hand, determining Mrs. Grooby's age doesn't require the assumption; on the other hand, you can't complete the puzzle diagram without it. Dan Hoey@AIC.NRL.Navy.MIl --- Newsgroups: alt.fan.cecil-adams, sci.med Followup-To: alt.fan.cecil-adams From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 12 Dec 1994 16:27:22 GMT Subject: Re: Inhaling helium is safe? d...@CS.Arizona.EDU (Dave Schaumann) writes: > All in all, it's hard to see how breathing helium is much worse for you > than holding your breath for the equivalent length of time. I've read in several places that it can be much worse for you, because it fools the breathing response, which is triggered by the presence of carbon dioxide, not the absence of oxygen. If you hold your breath, carbon dioxide builds up in the blood, which triggers an involuntary breathing response. If you are able to hold your breath long enough to pass out, you will start breathing immediately thereafter. But when you breathe an inert gas, the carbon dioxide is ventilated from your blood, and the breathing response is not triggered. You feel comparatively fine--you don't have the wracking spasms as your body tries to suck in air. As ahi...@nmt.edu (Aaron Hicks) writes: > Ya inhale it, you pass out, and the balloon goes flying across the room. As he neglected to mention: then you lie there unconscious and die of respiratory arrest, because your body doesn't start breathing automatically, and you aren't conscious to breathe voluntarily. I don't have a source for this (other than vague second-hand support of a memo from ``The Inert Gas Society of America (or something like that)''); I would appreciate something more authoritative. I've also read that there is also a low-oxygen response that can trigger breathing, but that for some reason it's unwise to rely on it alone. Anyone who wants to offer a report based on reading something from a safety study or a biology textbook would be helpful. Aimless theorizing based on subjective conceptions of how breathing works is likely to be ``harmful or fatal if swallowed.'' Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.misc From: hoey@pooh.tec.army.mil (Dan Hoey) Date: 12 Dec 1994 18:24:28 GMT Subject: Re: Conversion from inches to cm Hovig Heghinian (ho...@tubman.cs.uiuc.edu) wrote: : Isn't a meter defined as one part in 3e8 of the distance light travels : in a second? I understand that it used to be that the speed of light : was approx 3e8 m/s, but that the meter was recently redefined so that : the speed of light would be precisely 3e8. No, that would have changed too many gauges. They changed it so the meter is 1/299,792,458 of a light-second. I kid you not. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 12 Dec 1994 22:14:23 GMT Subject: Re: Those darn dice problems barry.p...@hofbbs.com (Barry Paul) writes: > Christopher Arthur asks for help with a homework problem: > > A die is rolled repeatedly until the sum of the numbers is 200. > > What is the probability that the die is rolled 66 times or less? > If there is a trick to work this out in under 3 months I don't know > what it is. So, out of curiousity, I rolled a die 66 million times..... > For each 66 rolls, I noted whether the sum of the rolls exceeded 200. > (This is exactly the same as Christopher's problem.) The number of > successful trials was 986,039 out of a million or 98.60% One way of getting it done faster is to let a computer expand the generating polynomial. For instance, in Mathematica: In[89]:= (Expand[PolynomialQuotient[x^7-x,x-1,x]^66]/6^66/.x^(_?(#>200&))->1)/. x->0 124928439612307108937585492254970113927880921238995 Out[89]= --------------------------------------------------- 126680573325946555412324573893623773219792923656192 In[90]:= N[%] Out[90]= 0.986169 That agrees pretty well with your number. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math, rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 20 Dec 1994 16:11:25 GMT Subject: Re: Cube inside other cube d...@sjfc.edu (Dan Cass) writes (excerpts in reverse order): > what does "ObPuzzle" mean... Obligatory puzzle. Some of us consider it de rigueur. On newsgroup X, you will often see ObX as a sop to being on-topic; on rec.puzzles, we seem to be pretty well confined to on-topic posts and spam. > [Q: what is the general solution of 1/a + 1/b + 1/c = 0 in integers?] > Is that related to Egyptian fractions, or a simple case of it? As p...@eurocontrol.fr (Philip Gibbs) pointed out, the general solution is a = tu(u+v), b = tv(u+v), c = -tuv It's easy to prove that this is the general form, and that any solution with c < 0 < b <= a has a unique such parameterization with (t,u,v)=1, 0 < v <= u, 0 < t. "Egyptian fractions" are just reciprocals of integers (possibly together with 2/3), which these certainly are. > I lost the post,... davi...@spcvxb.spc.edu (Dave Davis) asked: > One square nests neatly in another, one vertex of the inner on each > side of the outer. But with cubes it's not so neat.... find the > maximum number of points on the surface of a cube, at most one per > surface, such the vertices which complete it to a cube lie entirely > within. Your idea, to find > three vectors (a,b,c), (b,c,a), (c,a,b) which were mutually > orthogonal, and then use these three, along with (0,0,0) and the > four sums of 2 or 3 of them, as the vertices. is very nice. It amounts to taking a copy of the original cube, rotating it about the x=y=z axis, then shrinking it to fit inside the original. With the above parameterization, the ratio of the edge of the internal cube to that of the external cube is sqrt(a^2+b^2+c^2)/(a-c) = (u^2 + u v + v^2)/(u^2 + 2 u v) This approaches 1 as v/u approaches 0 or 1. Its infimum of sqrt(3/4) is approached as v/u approaches sqrt(3/4)-1/2. ObPuzzle]: Find the smallest integer cube-in-an-integer cube touching six sides and less than 2/3 of the volume of the larger cube. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.infosystems.www.providers From: hoey@aic.nrl.navy.mil (Dan Hoey) Subject: HTTPD 1.3 patch for automount symlinks Date: 21 Dec 1994 00:27:51 GMT Organization: Navy Center for Artificial Intelligence I sent this to the httpd developers and got the automated "we're too busy to read your message" message. I'd appreciate any comments. I had a problem installing NCSA/1.3 httpd for Unix. It led me to patch my copy of the source code. I don't believe I am alone, or even rare, in having this problem, so you may want to install something similar in later versions. Users on this system have directories on automounted partitions, and to get to an automounted partition you have to follow a symlink. I was not willing to enable FollowSymLinks, because users will put links to places without thinking of where they are going. SymLinksIfOwnerMatch had the problems that 1. I am concerned that some setuid programs might not mind creating symlinks owned by a privileged user. The ownership of a symlink is not designed to have any effect on the accessibility of the thing linked to. 2. It did not solve the problem, because some of the automounted partitions might not be owned by the owner of the link. I considered making httpd not check the ownership of links traversed to get to users' home directories, but that seemed to require more modification to the link-checking code than I was willing to perform. I decided to solve the problem by adding the "AutomountPoint" directive to httpd. AutomountPoint specifies a directory that is maintained as a logical mirror of the system name space. Specifying AutomountPoint /tmp_mnt enables httpd to follow links of the form /p1/p2/.../pn -> /tmp_mnt/p1/p2/.../pn even if following symbolic links is usually disabled. I consider this relatively secure, since the ownership of /tmp_mnt prevents adding any file /tmp_mnt/p1/p2/.../pn that shouldn't normally accessible as /p1/p2/.../pn. I am somewhat concerned that some circumvention of this may be possible by the use of race conditions, but I think using such a race condition would require the intentional collusion of a local user, which I am not so concerned about--that user would be able to simply copy materials into the public_html directory, anyway. Here are context diffs of the files I changed. I hope you find this useful. Dan Hoey Hoey@AIC.NRL.Navy.Mil WWW page =================================================================== RCS file: RCS/httpd.h,v retrieving revision 1.1 diff -c5 -r1.1 httpd.h *** /tmp/RCSAa10124 Tue Dec 20 17:41:23 1994 --- httpd.h Mon Dec 12 17:39:37 1994 *************** *** 234,243 **** --- 234,246 ---- #define RESOURCE_CONFIG_FILE "conf/srm.conf" /* The name of the MIME types file */ #define TYPES_CONFIG_FILE "conf/mime.types" + /* The name of the NFS automount point */ + #define AUTOMOUNT_POINT "" + /* The name of the access file */ #define ACCESS_CONFIG_FILE "conf/access.conf" /* Whether we should enable rfc931 identity checking */ #define DEFAULT_RFC931 0 *************** *** 401,410 **** --- 404,414 ---- extern char *server_hostname; extern char server_confname[MAX_STRING_LEN]; extern char srm_confname[MAX_STRING_LEN]; extern char access_confname[MAX_STRING_LEN]; extern char types_confname[MAX_STRING_LEN]; + extern char automount_point[MAX_STRING_LEN]; extern int timeout; extern int do_rfc931; #ifdef PEM_AUTH extern char auth_pem_decrypt[MAX_STRING_LEN]; extern char auth_pem_encrypt[MAX_STRING_LEN]; =================================================================== RCS file: RCS/http_config.c,v retrieving revision 1.1 diff -c5 -r1.1 http_config.c *** /tmp/RCSAa10127 Tue Dec 20 17:41:56 1994 --- http_config.c Mon Dec 12 17:39:08 1994 *************** *** 22,31 **** --- 22,32 ---- char *server_hostname; char srm_confname[MAX_STRING_LEN]; char server_confname[MAX_STRING_LEN]; char access_confname[MAX_STRING_LEN]; char types_confname[MAX_STRING_LEN]; + char automount_point[MAX_STRING_LEN]; int timeout; int do_rfc931; #ifdef PEM_AUTH char auth_pem_decrypt[MAX_STRING_LEN]; char auth_pem_encrypt[MAX_STRING_LEN]; *************** *** 52,61 **** --- 53,63 ---- server_hostname = NULL; make_full_path(server_root,RESOURCE_CONFIG_FILE,srm_confname); /* server_confname set in httpd.c */ make_full_path(server_root,ACCESS_CONFIG_FILE,access_confname); make_full_path(server_root,TYPES_CONFIG_FILE,types_confname); + strcpy(automount_point,AUTOMOUNT_POINT); timeout = DEFAULT_TIMEOUT; do_rfc931 = DEFAULT_RFC931; #ifdef PEM_AUTH auth_pem_encrypt[0] = '\0'; *************** *** 159,168 **** --- 161,174 ---- cfg_getword(w,l); if(w[0] != '/') make_full_path(server_root,w,types_confname); else strcpy(types_confname,w); } + else if(!strcasecmp(w,"AutomountPoint")) { + cfg_getword(w,l); + strcpy(automount_point,w); + } else if(!strcasecmp(w,"Timeout")) timeout = atoi(l); else if(!strcasecmp(w,"IdentityCheck")) { cfg_getword(w,l); if(!strcasecmp(w,"on")) =================================================================== RCS file: RCS/http_access.c,v retrieving revision 1.1 diff -c5 -r1.1 http_access.c *** /tmp/RCSAa10121 Tue Dec 20 17:40:50 1994 --- http_access.c Mon Dec 12 17:39:21 1994 *************** *** 139,149 **** if((!(opts[x] & OPT_SYM_LINKS)) || (opts[x] & OPT_SYM_OWNER)) { struct stat lfi,fi; lstat(d,&lfi); if(!(S_ISDIR(lfi.st_mode))) { ! if(opts[x] & OPT_SYM_OWNER) { char realpath[512]; int bsz; if((bsz = readlink(d,realpath,256)) == -1) goto bong; --- 139,149 ---- if((!(opts[x] & OPT_SYM_LINKS)) || (opts[x] & OPT_SYM_OWNER)) { struct stat lfi,fi; lstat(d,&lfi); if(!(S_ISDIR(lfi.st_mode))) { ! if(opts[x] & OPT_SYM_OWNER || *automount_point) { char realpath[512]; int bsz; if((bsz = readlink(d,realpath,256)) == -1) goto bong; *************** *** 153,165 **** strcpy(t,"../"); strcpy(&t[3],realpath); make_full_path(d,t,realpath); getparents(realpath); } ! lstat(realpath,&fi); ! if(fi.st_uid != lfi.st_uid) ! goto bong; } else { bong: sprintf(errstr,"httpd: will not follow link %s",d); log_error(errstr); --- 153,170 ---- strcpy(t,"../"); strcpy(&t[3],realpath); make_full_path(d,t,realpath); getparents(realpath); } ! if(strncmp(realpath,automount_point,strlen(automount_point)) ! || strcmp(realpath+strlen(automount_point),d)) { ! if(!(opts[x] & OPT_SYM_OWNER)) ! goto bong; ! lstat(realpath,&fi); ! if(fi.st_uid != lfi.st_uid) ! goto bong; ! } } else { bong: sprintf(errstr,"httpd: will not follow link %s",d); log_error(errstr); ================================================================ --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 23 Dec 1994 17:02:47 GMT Subject: Re: Cutting up "nice" 3D objects Naeem Sheikh wrote: >> We all know that a hollow cube can be cut up along its edges in >> a manner which allows the [connected] covering to be spread out in >> a plane. Obpuzzle: did you know that it _can't_ be so cut in a way that the covering overlaps when spread out? Same for the octahedron and the dodecahedron. Can you prove it for the icosahedron? >> Is it possible to cut up every hollow "nice" 3D object such that >> the skin can be spread out in one plane.... >> By "nice", I mean that all the faces of the object are parallel >> to one of the 3 axis-planes (namely z=0, x=0, and y=0 planes). se...@panix.com (Seth Breidbart) answers: > If you don't allow overlaps when it's laid down, no. At least, not > with your definition of "nice". > Hint: genus That's a somewhat misleading hint. Sure, the object you're hinting at is not unfoldable, but we can add a wart to form a nicely unfoldable object with the same hinted quality. The warty torus. +--+--+--+--+ | | | | | +--+ +--------+--+--+ +--+ +--+--+--+ | +| |`+--+ | | | | | | | | |+--+ | |`+ | +--+ | | | | | +--+--+--+ | `+ `+ | | | | | | | | | | | | | | | |`+| | | | +--+ | | +--+ | +--+--+--+ | +--+ | | | | | | | | | | | +--------+--+--------+--------+--+ +--+ +--------+.| | | | | `+--------+ +--------+--------+--------+ So the problem with the torus is not so much its genus, but the inability to cut up its faces. On the other hand, I haven't found a ``nice'' genus-zero object that is unfoldable. Obpuzzlers? Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.usage.english From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 23 Dec 1994 23:00:29 GMT Subject: Re: Definition Please: "Housing Charette" diam...@govonca.gov.on.ca (Gerald Diamond) writes: > >Does anyone know what this - "housing charette" - means? A charette (or charrette) the final project of architecture students, traditionally a large design performed under intense time pressure. Charette is French for a cart, which was used to transport the drawings. I believe a "housing charette" refers to an intensive design effort by professional architects to supply housing on a _pro bono_ or emergency basis. Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 23 Dec 1994 23:24:40 GMT Subject: Re: Cutting up "nice" 3D objects (oops) > Naeem Sheikh wrote: > >> By "nice", I mean that all the faces of the object are parallel > >> to one of the 3 axis-planes (namely z=0, x=0, and y=0 planes). hoey@aic.nrl.navy.mil (Dan Hoey) wrote: > I haven't found a ``nice'' genus-zero object that is unfoldable. > Obpuzzlers? I of course meant that I haven't found a ``nice'' genus-zero object that is not unfoldable. Sorry, I didn't mean to misle anyone. I still haven't found one, or figured out that they can't exist. I'd like to know if anyone has information on the status of that problem. Is the problem even solved for non-``nice'' polyhedra? Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 27 Dec 1994 16:36:55 GMT Subject: Re: Cutting up "nice" 3D objects (oops) dwr2...@tam2000.tamu.edu (Dave Ring) writes: > I keep getting confused with objects made of cubies which can be cut > along the cubie edges. I guess that can be an Obpuzzle. Is there an > un-unfoldable one? I considered Obposing that one, too. I don't have any examples that are not unfoldable, nor any idea of how to prove there are none. As to > > a ``nice'' genus-zero object that is not unfoldable. > What's wrong with the warty 3cube? A 3x3x3 with a wart in the center > of one face. Am I still confused? No, that will do fine. I thought of another over the weekend: The 3-cube with a wart _removed_ from the center of a face. But of course, both objects have the same skin. Those examples suffer, though, from having non-genus-zero faces, a condition I hadn't thought of last week. Here's one whose faces are all genus zero, but which cannot be unfolded into fewer than three pieces. It is a 4x4x4 cube from which a 14-cubie keyhole has been excavated. +----f---+--e--+-d+ | | | | | | | i a | | | | | | +--j--+ +-----------+-----+ +--+-----------+ | | | c | | | | | | +--------+-----+--------+ | | g | | | | | | | | f | | b | | | +-h+ | | | | | | | d | | | +--------+ | | +--e--+ | | | | | | +-----------+-----------+-----------+-----------+-----------+ | | | | | +-c+ | | | +--l--+-k+ | k | | | | | | | +--l--+ b | | | g | j | | | | | | | +----a---+ | | | +-----+--+ | | | | h | +-----------+-----------+-----------+ +--i--+ Dan Hoey@AIC.NRL.Navy.MIl --- Newsgroups: alt.usage.english, alt.flame Followup-To: alt.flame From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 28 Dec 1994 17:09:48 GMT Subject: cUNNINGHAM pUBLICLY sELF-aBUSED Young Bobby Cunningham awoke one morning to discover that he had been transformed into a wailing infant with a bad case of diaper rash and an urgent need to vent. Having been taken to task for unfairly criticising the note: "everyday (adj.) One word." he spews foam and froth alleging "failure to try to understand" his attempted communication, "careless reading", and "generally inappropriate" remarks. Adjectives such as "rude" and "insolent" appear in his missive, but are curiously used to describe motes, when beams abound. It is true that Fowler does not point out all of the non-idiomatic non-adjectival uses that may lead the word "every" to be followed by the word "day", but it is certainly the case that usage of "every day" as an adjective is incorrect. That is all the book says, and all it need say, on the subject. To the extent that this terseness has truly confused anyone, it might even be unfortunate, but I suspect that further explanation would add bulk to the book but not enlightenment to any readers. To the extent that tortured misinterpretation of the note has given the Cunningham an excuse to complain, it is one small light in a galaxy of Cunningham-galls. As for: > I would appreciate it if you would refrain from commenting on > any future posting of mine unless you can see fit to be more > courteous about it. You have no right to upset people with your > insufferable insolence. I would appreciate it if you would Learn to type your subject lines in a tasteful mix of capital and lower-case letters, perhaps even using the subject line of the thread you are responding to, so that people can tell what you're raving about, Use a newsreader that can include References: lines so people can find the posts you are whining about (if only to revel in their comparative coherence and civility), Mature to the point where you can respond to the differing opinions of other writers without indulging in abusive slurs, ill-mannered ascription of motives, and insufferable insolence (indeed!), and Refrain from commenting on any future posting, present posting, past posting, or any other subject, until you have accomplished the above. Remember, they also serve who only crawl back beneath their damp rock. Do. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.infosystems.www.providers From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 29 Dec 1994 21:49:41 GMT Subject: Re: Writing a 10 line httpd server in C or Perl d...@polarmet1.mps.ohio-state.edu (Doug Stevenson) writes: > ... the URL is what the user > (browser) enters on the very first line of the request! Like this: > GET /some/URL/here HTTP/1.0 Well, I tried % echo 'GET /home/services_docs/html-extensions.html HTTP/1.0' \ | mconnect -p 80 mcom.com and got connecting to host mcom.com (198.93.93.10), port 80 connection open HTTP/1.0 400 Bad Request Server: Netsite-Communications/1.0 Your browser sent a non-HTTP compliant message. It worked when I left off the 'HTTP/1.0'. Pity they couldn't say anything about what their complia\\aint is. Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.usage.english From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 30 Dec 1994 16:20:13 GMT Subject: Re: The English Language With All Forms of "To Be" Omitted With respect to Korzybski's "false-to-fact", dta...@pitt.edu (David M. Tate) writes: > The denotation is simply "false"; the connotation is "...by virtue > of being nonsensical or logically fallacious". Had Chomsky predated > Korzybski, they probably would have used a phrase like "syntactically > ill-formed". That is odd. I would expect that he intended the denotation to include what you call the connotation (false by virtue of internal inconsistency) else why redundify? Does he have a term for the other kind of falseness--false by virtue of simple failure to conform to reality (in spite of self-consistency)? I imagine that the later borrowers have debased the usage to the point where it has come to mean simply "false", with the vestigial connotation of philosophical pretentiousness. In fact, when I encountered it in my reading, I thought it meant the _second_ kind of falseness--false "to the facts", rather than false to itself. I seem to recall finding such usages very confusing, probably in part because of this false[2] interpretation. > [Sorry this reply took so long; I wanted to double-check my sources.] Thanks, it's good of you to do so. Can you suggest a source for me to become well-informed, too? Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.usage.english From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 30 Dec 1994 16:55:08 GMT Subject: Re: Adoption of English as the official language of U.S. ... According to St...@chubsoft.demon.co.uk (Steve Anderson), according to kpvhe...@nescio.cs.ruu.nl (Kilian Hekhuis), (was that in a.u.e?) arama...@mailhost.ecn.uoknor.edu (Arun Ramasamy) writes: > > |> I'd appreciate if anyone can give references to published work > > |> authenticating the following fact: > > |> English was chosen as the official language of United > > |> States of America over German by a single vote. From the alt.folklore.urban faq, available by FTP: * F.*German was once within one or two votes of becoming the official US language The F means it's false. The * means there's more information in the cathouse.org archives: * For people who can't get it, in file "german.us.official.lang", ci...@lise.unit.no (Cindy Kandolf) cites _Cambridge Encyclopedia of Language_, David Crystal, Cambridge University Press, (c)1987, p. 365, ISBN 0 521 26438 3. ] _A planning myth_ ] Probably the best-known myth in the history of language planning is ] the story that German nearly became the national language of the ] US in the 18th century, losing to English by only one vote in the ] legislature (the "Muhlenberg" legend). In fact, all that was ] involved was a request, made by a group of Virginia Germans, to have ] certain laws issued in German _as well as_ in Englih. The proposal ] was rejected by one vote, apparently cast by a German-speaking ] Lutheran clergyman, Frederick Muhlenberg (1750-1801). But the ] general status of English as the majority language was never in ] doubt. (After S.B. Heath and F. Mandabach, 1983.) Referring to ] Heath, S.B., and Mandabach, F. 1983. Language status decisions and ] the law in the United States. In J. Cobarrubias and J.A. Fishman ] (eds.), _Progress in language planning: international perspectives_ ] (Berlin: Mouton), 87-105. In another file, "the.story.of.english", Cindy also supports the interesting T.*Hebrew was considered as a candidate for official language of the US. [Guess what happened?] Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.lang.teco From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 30 Dec 1994 17:36:02 GMT Subject: Re: teco m...@marvin.df.lth.se (Magnus Olsson) writes: > Warren F. Seltzer wrote: > >I used the TECO editor nearly 25 years ago, and I'm wondering if > >that PDP-10 system, running on teletypes, is the same as the topic > >of this newsgroup... > Well, yes, except that you needn't run it on a PDP-10 using teletypes - > there are TECO's for Unix and MS-DOS as well. I saw a Unix TECO on the unix-sources (or could it be info-vax) mailing list sometime in the 80s--do you know where it is now? My apologies if "where do you get sources for TECO" is a FAQ, I can't seem to find the TECO FAQ list on RTFM. Dan --- Newsgroups: alt.usage.english From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 30 Dec 1994 20:46:31 GMT Subject: Re: Spend a penny? In regard to the euphemism "to spend a penny", k...@cus.cam.ac.uk (K. Edgcombe) writes: > >You had to pay only to have a dump, mind you; urination was free..." > It has been pointed out that this applied only to men. That demystifies the situation. I had heard the euphemism in connection with a Victorian male heterosexual urination fetishist (I kid you not), who sought an encounter in which his paramour would "spend a penny". The other case I recall is in the Cockney alphabet: A for horses, B for mutton, C for the Highlanders, D for dumb, E for brick, F for pheasant, G for police, H for it, I for Novello, J for orange, K for rancis, L for leather, M for sis, N for eggs, O for the rainbow, P for a penny, Q for the pictures, R for mo, S for you, T for two, U for mism, V for La France, W for a bob, X for breakfast, Y for girlfriend, Z for breezes. Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.folklore.urban From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 30 Dec 1994 21:21:12 GMT Subject: Re: Six Floors -- Fatal Drop? Eharvey <74354.3...@CompuServe.COM> writes: > cows might be different than a 70 kg male... Quite a bit so. A man thrown off a tall building will break. A cow thrown off a tall building will _splash_. Or so I'm told by a friend; I'm not sure who told him. Dan "all over the highway in Mystic, Connecticut" Hoey@AIC.NRL.Navy.Mil ObUL: The children's rhyme about "Hey Diddle Diddle" originated in the early days of the space program, in which a cow actually was sent into space. --- Newsgroups: sci.math From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 30 Dec 1994 22:25:51 GMT Subject: Re: 1995 a very interesting number Reiner.Eu...@psychol.uni-giessen.de (Reiner Euler) writes: > Which 4-digit year "abcd" solves the conditions > ab|cd|abcd (condition A) > abc|bcd (condition B) ? > ... [except for a=b=c=d], I found only four numbers containing both > conditions: 1248, 1664, 1995, 4998. Reiner notes that all the numbers are very close to 10000/n, for n=8,6,5,2. This is mostly because of condition B. If k(abc) = (bcd), then k(abcd) = 10(bcd) + k d then (10-k)(abcd) = 10000a + k d then (abcd) = 10000/((10-k)/a) + (k d)/(10-k) Since the first term is large, the question is why should (10-k)/a be an integer? In most cases a=1 which suffices; certainly a <= 4 . There are a few numbers (excluding k=0) satisfying condition B that do not satisfy condition A: 1250, 1426, 2498, 2500, 2855, 3748, and 3750. Only the last three have (10-k)/a not an integer. Is there a radix in a number satisfying both conditions has (r-k)/a not an integer? I've only tested up to base 12, unsuccessfully. Happy New Year. Dan Hoey@AIC.NRL.Navy.Mil ---