Date: Mon, 08 Jan 90 16:55:06 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Re: Cubism for Fun Peter, I'm still interested in seeing the CFF table of contents, though I might be subscribing to it, because you write RUBIK'S CUBE IN 44 MOVES: HANS KLOOSTERMAN Does that article actually show how to solve the cube in 44 moves? Even if they count half-turns as single moves, it is significantly better than the 52-move Thistlethwaite solution in Singmaster. Also, Thistlethwaite was thinking of improving his method, and perhaps this is a report of it. Or maybe it's just more rumor and conjecture, but it's nice to hear after all this time. I was making a few patterns over the weekend for some kids, and thought of some stuff I was thinking of trying out. For instance, if you restrict a face to two colors, there are only about fifty different patterns, at least if you ignore handedness. I wonder how many of them can be put on every face of the cube. We know the ones with corners alternating colors are impossible. We have some experience with some of the patterns--the X's, Crosses, Spots, T's, and H's--but that still leaves a large number of possibilities. My Christmas present to myself this year was to order Rubik's Cubic Compendium. I hope to be able to report on that sometime soon. It's always possible we may have a Cubic renaissance, though I'm not holding my breath. Dan --- Newsgroups: comp.sys.sun From: hoey@aic.nrl.navy.mil (Dan Hoey) Date: 1 Feb 90 22:12:51 GMT Subject: .getwd and .getwdCNNNNN (summ I wrote: >X-Sun-Spots-Digest: Volume 8, Issue 228, message 10 of 18 >We at last have a way to prevent users from stepping on each other's >temporary files in 4.3BSD (and therefore SunOS 4.0.3). I do a ``chmod +t >/tmp'' so that /tmp/* can only be deleted by owner or superuser. >But we have been seeing problems with /tmp filling up with thousands of >files named .getwdCNNNNN, where C is a letter, either ``a'' or ``b'', and >NNNNN is a process ID. Several people explained that whenever /etc/mtab is newer than /tmp/.getwd, the getwd routine will create /tmp/.getwdCNNNNN and attempt to rename it to /tmp/.getwd. The rename fails when .getwd exists and is owned by another user. The bug is known (1011843) and is supposed to be fixed in 4.1. Until then, the general solution is to chmod -t /tmp. If you really have to have the t-bit set, a fairly ingenious solution came from Rob McMahon : > The cure is to put a `pwd' command in /etc/rc.local after > mounting all the NFS filesystems, to remember to use pwd as root after > mounting or unmounting any filesystems, and to put a `pwd' in root's > crontab for good measure. The rationale is that we force /tmp/.getwd to be newer than /etc/mtab. The reason for doing this as root is that root is always able to rename .getwdCNNNNN to .getwd, so once root calls getwd, .getwd will be owned by root and so root will be the only account able to change it. I should note that Rob's solution will work only marginally if you run the automounter, because the automounter will mount and unmount filesystems without giving you a chance to use pwd. If you have source, I guess you can get the automounter to call getwd. Of course, if root's crontab runs getwd often enough, only a few .getwdCNNNNN files can get created, and you can delete them with another crontab entry. Thanks to Guy Harris, Rob McMahon, Ken Smith, and Steve Rumsby for explaining the operation of .getwd and the status of the bug. Dan Hoey@AIC.NRL.NAVY.MIL --- Date: Thu, 22 Feb 90 19:16:07 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Language in Rubik's Cubic Compendium Well, my order of Rubik's Cubic Compendium came through. I ordered it through Reiter's (The DC technical bookstore) and paid $30 for it. It's basically six largely independent chapters translated from Hungarian, with foreword and afterword by Singmaster and a bibliography. There's definitely some neat stuff there. My favorite piece of the book is in Tamas Varga's ``The Art of Cubing'', which develops some interesting new additions to what we used to call ``Rubiksong'', the language we use to describe processes. He starts by renaming the Up face to be the Top, the advantage of which is to make all the face names consonants. He then uses vowels to indicate the direction of turn, "O" for 90 degrees fOrward (or clOckwise), "A" for 90 degrees bAckward (or Anticlockwise) and "I" for a 180 degree half-turn (twIce). This works out neatly to allow a process to be described with a syllable for each quarter- or half-turn. So Pons Asinorum can be done with FIBITIDIRILI and Laughter is 3 FOBOROLOs. But wait, there's more! Remember how Befuddler never was able to handle whole-cube moves neatly? In this notation, you append a "C" to a syllable to indicate that instead of turning the face, you turn the whole cube. So the way I usually do Laughter is really 6 ROLOTOCs. This notation is not as parsimonious, since FOC=BAC, TOC=DAC, and ROC=LAC, but it's better than having to stop in the middle and say ``then move the cube''. For instance, Jim Saxe's 28-qt Plummer's Cross can be done as "FOLIRIFO BOLIRIFO ROFIBIRO LOFIBIRO TIDI", but the way he originally described it (3 Dec 1980 00:50) was "FOLIRIFO BOLIRIFO TOC FOLIRIFO BOLIRIFO TIDI", but instead of TOC he had a couple of lines of text. He also has a way of talking about the slice moves, where you move the middle layer of the cube instead of the faces. For moving the middle, you append "M" to the syllable. So the way most people do a Spratt wrench is 4 TOROMs, and we can do the Plummer's Cross as FOLIMBO FOLIMFO TOC FOLIMBO FOLIMFO TIM. Of course, ROM=LAM, etc. (This could also work for Rubik's revenge, where ROM and LAM are different, being moves of the inner layers adjacent to the R and L faces). It's unfortunate that he doesn't extend the language past the point of appending "M" and "C". I would like to have a way of talking about slice moves where you move the faces rather than the middle. Of course, we could say ROLA, but I'd rather say something like ROS. This might interfere with the use of "s" for plurals (as they do in the book and I do above), but that could be fixed by pronouncing the pluralizing s as "z". Another idea is to append "N" to syllables for aNtislice moves. So Laughter would be six RONTOCs. I'm a little concerned though, that "M" and "N" might be difficult to distinguish. Another suggestion is to append "P" to allow "deeP" moves, where we do RO and ROM simultaneously by grabbing two layers of the cube and turning them while keeping the remaining face fixed. It might be nice to use "G" to denote the way we "Wring" the cube, as with ROPRO. So 6 ROGTOC's does an 8-Flip. To summarize, F,B,T,D,R,L -- faces Front, Back, Top, Down, Right, Left. O,A,I -- directions fOrward, bAckward, twIce. C,M,S,N,P,G -- extensions whole-Cube, Middle, Slice, aNtislice, deeP, wrinG All extensions but C are redundant, since ROM=ROCRALO ROS=ROLO RON=ROLA ROP=ROCLA ROG=ROCROLO I'm going over their list of pretty patterns, and hopefully I can find out which ones are improvements. I did notice they don't have Saxe's Plummer's cross process. Dan --- Newsgroups: comp.dcom.sys.cisco From: hoey@ai.etl.army.mil (Dan Hoey) Date: 28 Mar 90 21:27:44 GMT Subject: Re: Ideal router user interface s...@cisco.com (Greg Satz) writes: > Any comments, critiques or recommendations would be appreciated. I don't know if this is really an interface issue, but I noticed when setting up the access list for my LAN, that the access list mechanism tends to lead to unnecessarily many rules. Typically I'll prohibit one kind of packet for some hosts and another kind for other hosts. But both prohibitions end up in the same access list, and they both have to be checked for all hosts. When distinguishing between hosts takes N rules, and distinguishing between packets takes M rules, then we end up with about N*M rules that all the packets have to filter past. It can be worse: when distinguishing between N1 source hosts and N2 destination hosts, we end up with about N1*N2*M rules. I'm running every packet past a 23-rule extended access list, and I'm sure I'm paying for it in performance. Also, if things get more complicated, I have to add rules about five at a time. It would be nice if the action specified by the list, instead of being just "permit" or "deny", could be "go to access list N". Packets then go by N1 rules that determine which access list to use based on the source, then N2 rules to determine which access lists to use based on the destination, then M rules to determine a final permit/deny decision. So each packet needs to be tested against only N1+N2+M rules, a significant speedup. The storage for each rule would become slightly larger, and there would probably be a speed penalty for switching access lists, but I think it would pay off in significantly faster and simpler access lists. I don't know whether it's important, but it could help if the "go to" were more like a subroutine call, so that if you fall off the end of the called list, you return to the list it was called from. If you wanted to avoid the return, you could code in a wild-card permit/deny at the end of the rule. Another feature would be to allow a "return" action, which would allow you to return from the middle of an access list instead of falling off the end. These features give you a somewhat more powerful production system language at the expense of a more complicated processor. For these extensions, I'm not sure their use would pay off. Dan --- Newsgroups: comp.dcom.sys.cisco From: hoey@ai.etl.army.mil (Dan Hoey) Date: 28 Mar 90 22:27:33 GMT Subject: Re: Ideal router user interface The last time I tried to find the "ip broadcast-address" of an interface I ended up doing "write terminal". Show interfaces ought to provide this info. Maybe each configuration command ought to mention how to find out what the current setting is, and also how to set it to the default (I never could set the hostname to a default that would get omitted in a "write x" command. Also, "write erase" should ask for confirmation. I once typed "writ er" instead of "wri ter". Gack. Dan --- Newsgroups: sci.math From: hoey@ai.etl.army.mil (Dan Hoey) Date: 29 Mar 90 00:09:41 GMT Subject: Re: Determining whether n numbers are distinct gor...@cs.tamu.edu (Dan Gordon) writes: >>Given n numbers (real or integer), we want to determine whether they >>are distinct. Can this problem be solved in O(n) time? >This problem is known as the element uniqueness problem and it can't be done >in less than order nlogn time ... if you are only allowed to do comparisons. >If your numbers are integers in the range 1 to M, then you can solve the >problem in O(M) time. More generally, you can use hashing techniques too. In fact, on a RAM model of computation, where you can do indexing into arbitrarily large arrays in constant time, there is a fast algorithm for arbitrary numbers. It may be a real ram with floor, and you determine element uniqeness for real numbers. I read it in 1978 or 79, probably in JACM or CACM. Dan --- Newsgroups: rec.puzzles From: hoey@ai.etl.army.mil (Dan Hoey) Date: 3 Apr 90 20:03:09 GMT Subject: Re: Pythagorean Problem (SPOILER) amb...@acf5.UUCP (FJLevM{n[]Balamurali Ambati) writes: >Find all integer sided triangles whose area and perimeter are equal. Let the sides be a, b, c, and s=(a+b+c)/2 the semiperimeter. Note that by the triangle inequality, s > max(a,b,c). The area, by Hero's formula, is SQRT(s(s-a)(s-b)(s-c)). Therefore SQRT(s(s-a)(s-b)(s-c)) = 2s s(s-a)(s-b)(s-c) = 4 s^2 (s-a)(s-b)(s-c) = 4 s . The right hand side of the equation is an integer (even if s is not), so the left hand side must be an integer, so s must be an integer. Writing x=s-a, y=s-b, z=s-c, we have x+y+z = 3s - (a+b+c) = 3s - 2s = s. So our equation is now x y z = 4 (x+y+z) . Wlog let x <= y <= z . Considering z as a function of x and y, we write z(x,y) = 4 (x+y)/(xy-4) . We note first that xy > 4, which is to say that y >= max([5/x],x). Differentiating yields dz / dy = 4 ((xy-4) - x(x+y)) / (xy-4)^2 = -4 (x^2 + 4) / (xy-4)^2 which is negative when y>[5/x], so z is a strictly decreasing function of y and (by symmetry) of x. We therefore can find all solutions by checking all x from 1 increasing until x>z(x,x), for each such x checking all y from max([5/x],x) increasing until y>z(x,y), and reporting a triangle whenever z(x,y) is an integer. The triangles are then as follows: x y z(x,y) a=y+z b=x+z c=x+y A=P 1 5 24/1 29 25 6 60 1 6 28/2 20 15 7 42 1 7 32/3 not integer 1 8 36/4 17 10 9 36 1 9 40/5 < y 2 3 20/2 12 13 5 30 2 4 24/4 10 8 6 24 2 5 28/6 < y 3 3 24/5 not integer 3 4 28/8 < y 4 4 32/12 < x That's QED, folks. Dan --- Newsgroups: misc.misc From: hoey@ai.etl.army.mil (Dan Hoey) Date: 15 May 90 15:48:30 GMT Subject: Re: Roman Number Arcana In a March, 1990 issue of Cecil Adams's column, The Straight Dope, he fields the query: This is important! What are the Roman numerals for 1990? Possible solutions: 1) MXM, 2) MCMXC, or the cumbersome 3) MDCCCCLXXXX. Help! --Anonymous, Chicago, Ill. He missed his chance: The standard abbreviation for Illinois, IL, is a Roman numeral! Other Roman numeral states are ID and MD. (And is Nebraska NE or NB? The u*x quiz program thinks either is fine.) Other top-level domains include CL, MIL, and MX. (What are CL and MX? Nation codes? Weapons systems?) But NIC's list doesn't include ID, so there are probably other nation codes not listed as well. Cecil's response begins: By God, this *is* urgent. Even now sweaty movie moguls are undoubtedly wondering: What the hell are we going to do about the date at the end of the credits? Well much as I'd like to cash in selling Roman-numeral consulting services to Hollywood, this time you guys are on your own. There is not now nor has there ever been any universally accepted method of styling Roman numerals. For that matter, it's only been in the last few hundred years that there's been any general agreement on what symbols stand for which quantities. He continues, dropping little gems of information. To quote and paraphrase: Many authorities think it's only coincidence that the number M happened to look like the letter M (ditto for C). It's unlikely they stood for the Latin *mille* or *centum*. As often as not, the Romans indicated 1,000 not with M but with the lazy-8 infinity symbol or else something along the lines of (I)--that is, a vertical stroke framed by exaggerated parentheses. The so-called subtractive principle, i.e. that IV=5-1=4, was used only sporadically by the ancient Romans and their medieval successors and never in a systematic way. Old documents and inscriptions include LXL, 90; XXCIII, 83; LXXIIX, 78; and even IIIIX, 6. A popular German arithmetic textbook published in 1524 gives 99 as XCIX. He concludes: So where does this leave us? Well, if we are truly desperate for moral guidance, we may turn to the world of computers. Cecil happens to have a desktop publishing program known as Xerox Ventura Publisher, an amazing bit of software thought to have been used originally to torture heretics during the Inquisition. Among other things it will convert numbers up to 9,999 into Roman numerals for use as page numbers. Punching in 1990, we come up with MCMXC, an unsurprising and somehow comforting result. But if we then try 1999, we get MIM. Why MIM for 1999 and not MXM for 1990? Lord knows. Worse, if we enter 9,999 we get what appears to be IZ. I have scoured my reference books in vain for any indication that Z was ever used for 10,000, which moves me to write the whole thing off as the product of malicious computer geekery, an impression that actually trying to *use* Ventura will certainly confirm. No doubt all this numerological uncertainty is distressing. But look on the bright side: It also gives us a strange and terrible freedom. You can use any damn notation for 1990 you want to, and no one will be able to say you're wrong. It may not give you the same rush as dancing on the Berlin Wall, but in post-Reagan America you make do with what you get. In response to an earlier version of this message, Alan Bawden reminded me that Common Lisp specifies a Roman numeral printer option. In the ones I know of, each digit is represented by the Roman numeral for its value (i.e., MCMXC for 1990). Now that we know there are no standards, it is clearly beneficial to use MXM and MIM, since they are shorter and snazzier than the conservatively crippled MCMXC and MCMXCIX, and there is no ambiguity. But more shortening is possible if we risk ambiguity with more general subtraction. My question is whether 1989 should have been represented as MIXM or MXIM. I at first thought that XIM should be preferred over IXM for 989, since the former cannot be parenthesized wrong. However, this does not extend to MXIM vs MIXM, so I have come to discard misparenthesizability as a criterion. If the parsing algorithm is ``process numerals from right to left, subtracting if the numeral is less than the currently accumulated value, otherwise adding'', either MXIM or MIXM will work. If instead the test is ``... subtracting if the numeral is less than the previous numeral'', then MIXM works and MIXM doesn't. The friendliness principle therefore dictates MIXM and the first algorithm. Dan --- Newsgroups: rec.puzzles From: hoey@ai.etl.army.mil (Dan Hoey) Date: 18 May 90 16:42:43 GMT Subject: Re: Palindromes (was Re: No rhyme or reason - word games ??!!) Reingold-Nicho...@cs.yale.edu (Nicholas Reingold) writes: > A man, a plan, a cat, a canal; Panama? That's by Jim Saxe. It was in his CMU plan file on 9 October 1983. >I recall a much longer version constructed by some people at CMU (I >think). Does anyone remember this? A man, a plan, a cat, a ham, a yak, a yam, a hat, a canal--Panama! (Guy Jacobson) This was in Guy Jacobson's CMU plan file, probably also in 1983. A man, a plan, a caret, a ban, a myriad, a sum, a lac, a liar, a hoop, a pint, a catalpa, a gas, an oil, a bird, a yell, a vat, a caw, a pax, a wag, a tax, a nay, a ram, a cap, a yam, a gay, a tsar, a wall, a car, a luger, a ward, a bin, a woman, a vassal, a wolf, a tuna, a nit, a pall, a fret, a watt, a bay, a daub, a tan, a cab, a datum, a gall, a hat, a fag, a zap, a say, a jaw, a lay, a wet, a gallop, a tug, a trot, a trap, a tram, a torr, a caper, a top, a tonk, a toll, a ball, a fair, a sax, a minim, a tenor, a bass, a passer, a capital, a rut, an amen, a ted, a cabal, a tang, a sun, an ass, a maw, a sag, a jam, a dam, a sub, a salt, an axon, a sail, an ad, a wadi, a radian, a room, a rood, a rip, a tad, a pariah, a revel, a reel, a reed, a pool, a plug, a pin, a peek, a parabola, a dog, a pat, a cud, a nu, a fan, a pal, a rum, a nod, an eta, a lag, an eel, a batik, a mug, a mot, a nap, a maxim, a mood, a leek, a grub, a gob, a gel, a drab, a citadel, a total, a cedar, a tap, a gag, a rat, a manor, a bar, a gal, a cola, a pap, a yaw, a tab, a raj, a gab, a nag, a pagan, a bag, a jar, a bat, a way, a papa, a local, a gar, a baron, a mat, a rag, a gap, a tar, a decal, a tot, a led, a tic, a bard, a leg, a bog, a burg, a keel, a doom, a mix, a map, an atom, a gum, a kit, a baleen, a gala, a ten, a don, a mural, a pan, a faun, a ducat, a pagoda, a lob, a rap, a keep, a nip, a gulp, a loop, a deer, a leer, a lever, a hair, a pad, a tapir, a door, a moor, an aid, a raid, a wad, an alias, an ox, an atlas, a bus, a madam, a jag, a saw, a mass, an anus, a gnat, a lab, a cadet, an em, a natural, a tip, a caress, a pass, a baronet, a minimax, a sari, a fall, a ballot, a knot, a pot, a rep, a carrot, a mart, a part, a tort, a gut, a poll, a gateway, a law, a jay, a sap, a zag, a fat, a hall, a gamut, a dab, a can, a tabu, a day, a batt, a waterfall, a patina, a nut, a flow, a lass, a van, a mow, a nib, a draw, a regular, a call, a war, a stay, a gam, a yap, a cam, a ray, an ax, a tag, a wax, a paw, a cat, a valley, a drib, a lion, a saga, a plat, a catnip, a pooh, a rail, a calamus, a dairyman, a bater, a canal--Panama. (Dan Hoey) I found this in 1984. I believe a little work on a search algorithm could make it several times as long. I was at CMU once, but not when I constructed this. Dan --- Newsgroups: comp.misc From: hoey@ai.etl.army.mil (Dan Hoey) Date: 30 May 90 20:13:22 GMT Subject: Re: C obfuscator st...@taumet.com (Stephen Clamage) writes: >... For you youngsters out there .... I punched my first deck of cards in 1968. I was a high school junior in a summer intern program at Goddard Space Flight Center. I had just read this book about the SDS 930 assembler language and I was excited about trying to write a real program. After I wrote it down on a coding form* I went from the building my office was in to a separate building where they had a noisy, hot, stuffy room with this card punching machine that had a keyboard that I could use my recently-learned touch typing on as long as I was typing alphabetic letters, but for numbers and punctuation it used this sort of shift key and the numbers and punctuation were all over the letters and it was hunt-and-peck. The most tortuous part was that I had to look away from the keyboard for the touch typing to work, and back to the keyboard to do hunting and pecking, and back to the coding form to see what I was supposed to be typing, and if I thought I might have made a mistake but wasn't sure, I couldn't see the character I had just punched, it was hidden under the puncher. I could type space, and look at it, and backspace, but the backspace wasn't a key, it was a big button under the punch station and I had to take my hand off the keyboard and put it back and find the home row again afterwards. I sweated over that thing for two hours. A lot of rejected cards went into the pile, and I needed to look them over, but I wanted to get out of the torture chamber and back to the peaceful office. You know the Beatles song, ``When the rain comes, they run and hide their heads. They might as well be dead.'' In the lobby a crowd was waiting for a summer shower to finish. Laughing at these poor worker ants I ran across the lawn to the building with my office. I got in the door and shook the water off my head and tried mopping my glasses dry with my shirt, humming the song, and my office mate came by and said, ``You know, those cards won't go through the reader.'' The deck was soggy and warped and pretty much useless. Eventually I got them dried out and managed to use the card punch to duplicate some of them onto good flat cards. Eventually I learned to keypunch without undue stress. Eventually I even got to write a program that worked. So I guess that moment of looking at those precious, ruined cards was the low point of my career. Dan *For you youngsters, a coding form is sheet of paper with eighty spaces marked on each line, so you can write down which card column each character is going to go into. It used to matter. --- Newsgroups: sci.math From: hoey@ai.etl.army.mil (Dan Hoey) Date: 30 May 90 21:02:05 GMT Subject: Re: exponential diophantine eqn stei...@bgsuvax.UUCP (Ray Steiner) writes: >In an old paper of Pillai one finds the following statement: >If (a,b)=d and d>1 then it is obvious that the diophantine >eqn a^x-b^y=c has at most one solution for x and y (here a,b,c >are given integers). The obvious answer is that the lesser of x and y must be the largest k for which d^k|c . Closer examination shows that it is obvious that there are at most three solutions: k=xval...@b.gp.cs.cmu.edu (Raul Valdes-Perez) writes: >>What is the stoichiometry of the schematic mechanism >>below, where the starting materials are A,B,C and the >>target product is T? > 1 ( A + B => Y + N ) > + 2 ( B + C => M ) > + 2 ( M + N => Z + Y ) > + 1 ( B + Z => N + T ) > ------------------------- > A + 4B + 2C => T + 3Y + Z Why not 0 ( A + B => Y + N ) + 1 ( B + C => M ) + 1 ( M + N => Z + Y ) + 1 ( B + Z => N + T ) ------------------------- 2B + C => T + Y ? It produces just as much T as the other, using strictly less raw material. The reaction is catalyzed by N. Dan --- Newsgroups: rec.arts.sf-lovers From: hoey@ai.etl.army.mil (Dan Hoey) Date: 13 Jun 90 12:19:54 GMT Subject: Will the real Mike Resnick please... The day after reading of the uniqueness of Mike Resnick, regardless of disparate writing styles, I was calmly watching late night TV when Doctor Michael Resnick appeared! To let me know about a totally new way to raise my libido! After engaging in aerobic exercise, thirty percent of women experience increased sex drive! Over one quarter report an improvement in their ability to climax! Operators are standing by have your credit card handy void where prohibited! Is this the degenerate huckster side of the artistic prostitution side of the potboiler side of ... ? of .... ? Say it ain't so! Fortunately, *I* was at Disclave 1990, the Washington Area's Premier Science Fiction Convention, with Writer Guest of Honor Mike Resnick, Artist Guest of Honor Dawn Wilson, Movie Mogul Guest of Honor Somtow Sucharitkul, featuring the Discave Con Suite Fantastique, I could go on and on, but the point is I was at Disclave so I *know* Mike Resnick and that flack on the telly was definitely *not* *him*. Mike Resnick may have written some potboilers in his time, I say some of his works may have been inspired more by his mortgage than his muse, but he is not reduced to flogging jigglecize in the early morning hours. Ghuo Ghratia! Speaking of commercial announcements, though, you might be interested in a signed and numbered limited edition of *Through Darkest RESNICK With Gun & Camera* by Mike Resnick, published by WSFA Press, 1990, 200pp hc, dj by Todd C. Hamilton! You missed the convention now get the book! So new it's not on JWenn's list! Featuring the Hugo[SM]-winning story Kirinyaga! And numerous other short works of fact and fiction set in mysterious Africa (where numerous = 12)! Operators ought to be standing by! But I don't have an order form handy! So s^H/u^H/e^H/ m^H/e^H/ if you really want this baby you should find a specialty bookstore or a convention or I guess if you pester me I'll go find out how the mail order works and let you know. Oh, and the dust jacket says it'll set you back thirty-five simoleons, so if you're not a high-rolling fan with an appreciation for fine editions with a price tag to match, you may rather just sulk enviously and hope it comes out in paperback. Dan ---------------- [SM] means Hugo is a Service Mark (I think that's what it is) of the World Science Fiction Society (not affiliated with Krafft-Ebing). --- Newsgroups: sci.med, sci.electronics From: hoey@ai.etl.army.mil (Dan Hoey) Date: 13 Jun 90 14:06:44 GMT Subject: Omphaloskepsis (was Re: make money, help mankind) p...@pepsi.amd.com (Phil Ngai) writes: > ... My wife thinks it would be fun to be able to look inside whenever she >wants.... my device would use a TV as an output device and therefore you could >tape it on your VCR. This isn't intended to be a medicine quality diagnostic >device, just something for curious people. You don't even have to be pregnant for this. Look in your guts. Look in your brain. Look in your shoes. Look in your cat. Look in your turkey. Look in your Halloween candy or fruit. Oh, sure, babies are among the neatest surprise packages going, but they aren't the only ones. Dan --- Newsgroups: rec.humor.d From: hoey@ai.etl.army.mil (Dan Hoey) Date: 14 Jun 90 23:20:37 GMT Subject: Re: Re: Subject line ID's (was Re: Dan KoGai and his .sig) j...@hp-ptp.HP.COM (John_Fereira) writes: >geo...@purplehaze.Central.Sun.COM (Geoff Miller) writes: >>Who the fuck cares how long anyone's .sig file is? >1. People reading the group with a terminal connected to a 1200 baud >(or 300) modem. That's what kill files are for. >2. Systems administrators with a shortage of disc space. They should just rig it to throw away anything he sends. It's not like he posts anything worth while. (Or does he? All I've seen has been most vile.) Dan --- Newsgroups: rec.arts.sf-lovers From: hoey@ai.etl.army.mil (Dan Hoey) Date: 11 Jul 90 18:41:38 GMT Subject: Re: Orson Scott Card on "Hypocrites of Homosexuality" In the *Sunstone* article (as quoted by Tom Maddox), Card wrote >This applies also to the polity, the community of citizens at large. >Laws against homosexual behavior should remain on the books, not to be >indiscriminately enforced against anyone who happens to be caught violating >them, but to be used when necessary to send a clear message that those who >flagrantly violate society's regulation of sexual behavior cannot be permitted >to remain as acceptable, equal citizens within that society. ma...@dgsi.UUCP (Mark Bernstein) writes: >... The laws Card referred to are LDS laws, and the people he refers to as >hypocrites are those who claim to be both actively homosexual and believing >Mormons. I don't think he said anything about the world in general outside >his church. t...@hoptoad.UUCP (Tim Maroney) writes: >Perhaps the problem is that you don't know the word "polity"? Perhaps the problem is that Tim doesn't know the word ``polity''. I thought I did, but just to make sure, I looked it up. The dictionaries I checked make it clear that ``polity'' may refer either to civil government or to ecclesiastical government, the latter being the laws of a church. So as far as the word ``polity'' goes, the passage in question is ambiguous with respect to Mark's position. The appositive phrase ``the community of citizens at large'' may be seen as clarifying the matter. Still, I suppose that a sufficiently weasely mind could interpret this to refer to a community of lay LDS members, and that only church laws are being proposed. I think that interpretation is very strained, but Card can be very subtle when he wants to be. Writing on theology to a group that is markedly more conservative than he could tempt him to use language that leads the reader to overestimate the conservatism of his argument. There are four other uses of the word ``polity'' in the essay, and all seem to refer to civil government. I would still feel a lot better about disparaging Card's views if he had used a term like ``civil'' or ``secular'' that would more unambiguously extend his argument beyond the LDS polity. I am also very uncomfortable with Card's lack of specificity on the enforcement of such laws. He asks that they be enforced only to deter ``those who flagrant violate society's regulation of sexual behavior.'' I would like to know just how flagrant he's talking about. After all, there are laws about public decency that regulate both heterosexual and homosexual acts if they are sufficiently flagrant. The question is what sanctions he proposes against homosexual behavior that would not extend to heterosexual behavior. Make no mistake: My best reading of the essay is that Card calls for the state to impose sanctions against some forms of homosexual behavior. However, I also believe that he is doing his best to avoid saying what behavior and what sanctions. I think he is trying to appear to support one position to the Mormons, while reserving some deniability for the rest of us. Dan --- Newsgroups: rec.arts.anime From: hoey@ai.etl.army.mil (Dan Hoey) Date: 13 Jul 90 18:55:23 GMT Subject: Re: Baycon (was Re: Cal-Animage) ew...@lightning.Berkeley.EDU (Edward Wang) writes: >Another thing that bugged me, was that they were closely watching >the badges at the entrances to anime rooms. That was one of the biggest >gripes I had last year. They were even more restrictive this year. Good for them! I sure hope they made the parasites feel unwelcome. Dan --- Newsgroups: sci.math.num-analysis From: hoey@ai.etl.army.mil (Dan Hoey) Date: 19 Jul 90 16:43:49 GMT Subject: Re: probability problem r...@edat.UUCP (Brian Douglass) writes: >The first problem is straight forward, I need the answer to: >((2^64-1)/2^64)^96252316123628899849 As Richard Pavelle notes, it is easy to find the answer to this question if you apply some very elementary mathematics. But Brian continues >... trying to run this calculation on bc or as a C program causes >my Compaq to choke (with no FPU mind you). No rounding is allowed, >I must have all significant digits. There are over 6x10^21 significant digits. I don't believe there has been enough media manufactured yet to store the answer. >The second problem deals with the probability of an event repeating >given a specific sample size out of a population size of 2^64. It sounds like you want to generalize the above result. The approach is the same. >I have this problem currently written in C with the guts being only >about 3 lines long. However, after have it run for 2 weeks on >another Compaq.... It sounds like another case of excessive computer power being expended on a misguided approach to the problem. >I tried calling Cray, who wouldn't answer their phones the first two >times I called, but then haven't gotten back to me since. Perhaps they are not convinced you know what you are doing. Getting asked for Cray time to solve a pocket calculator problem can give people that impression. Dan --- Newsgroups: sci.math From: hoey@ai.etl.army.mil (Dan Hoey) Date: 14 Aug 90 21:33:25 GMT Subject: Re: Projection question. na...@stat.fsu.edu (B. Narasimhan) writes: > Consider two line segments in R3 ... project these line > segments on the XY plane ... computing the intersection point of the > projected segments (if there is one) .... > Now suppose I have a sequence of points (x(1),y(1),z(1))....(x(n),y(n),z(n)) > such that a line segment connects (x(i),y(i),z(i)) and > (x(i+1),y(i+1),z(i+1)). The last point is connected to (x(1),y(1),z(1)). > If I use the above approach, I would have an O(n^2) algorithm. merr...@bucasb.bu.edu (John Merrill) writes: > The poster asks: is there a more efficient algorithm? The answer is > no. It suffices to observe that there exists a configuration of > endpoints such that the total number of occlusions is O(n^2).... Well, there's no way out of an Omega(n^2) worst case algorithm, when you have Omega(n^2) intersections. But you can do better when there are fewer. There is an O((n+k)log n) algorithm for finding the k intersection points of n line segments. You will find this result in Preparata and Shamos's *Computational Gometry* on page 277. By time-sharing you can make the algorithm run in O(MIN((n+k)log n,n^2) time. Preparata and Shamos also mention later improvements by Chazelle and Mairson and Stolfi. Dan --- Newsgroups: sci.math From: hoey@ai.etl.army.mil (Dan Hoey) Date: 14 Aug 90 22:06:23 GMT Subject: Re: Euclid's proof tor...@sics.se (Torkel Franzen) quotes from Euclid: > Let A, B, C be the assigned prime numbers; I say that there are more > prime numbers than A, B, C. > For let the least numbered measured by A, B, C be taken, and let it be > DE; let the unit DF be added to DE. > Then EF is either prime or not. > First, let it be prime; then the prime numbers A,B,C,EF have been found > which are more than A,B,C. > Next, let EF not be prime; therefore it ... [has a prime factor] G. > I say that G is not the same with any of the numbers A,B,C.... Torkel continues: >The formulation seems to be nearer the constructive one, i.e. for any >finite set of primes a larger one can be found.... which misstates the result. Euclid has shown for any finite set of primes a *different* one can be found. He does not (erroneously) prove that ABC+1 is prime. But when ABC+1 is composite, he shows that any prime factor of ABC+1 must be different from A, B, and C. cos...@bbn.com (Bernie Cosell) writes: >you can *not* use the same logic to say anything about >the subsequent product A*B*C*(ABC + 1) + 1. Wrong. If ABC+1 is prime, then A*B*C*(ABC + 1) + 1 is either prime or has a prime factor different from A,B,C,(ABC+1). If not, then ABC+1 has a prime factor G, and ABCG+1 is either prime or has a prime factor different from A,B,C,G. But, originally, i...@Gang-of-Four.Stanford.EDU (Ilan Vardi) asked: >It seems interesting whether Euclid considered his proof of an >infinite number of primes as a proof by contradiction only, or >whether he actually thought about it as a constructive proof. Lest you infer that the proof is constructive, it should be noted that the rest of Euclid's proof--that G (the prime factor of ABC+1) is not among the previously found primes A,B,C--is in fact a proof by contradiction. I don't know if that proof could be turned into a constructive one. Dan --- Newsgroups: comp.lang.misc From: hoey@ai.etl.army.mil (Dan Hoey) Date: 17 Aug 90 20:37:30 GMT Subject: Re: The Forbidden (previously misspelled ``Forbiden'') gude...@cs.arizona.edu (David Gudeman) writes: >... It's really quite trivial to prove that recursion is >impossible in most non-recursive programs. The only programs where >non-recursion cannot be proven automatically are those which have a >recursive call in a section of code that will never be executed, but >where the compiler cannot prove that the code will never be executed. >Since such programs are almost guaranteed to be not what the >programmer intended, I feel confident in saying that non-recursion can >be automatically proven in correct non-recursive programs. That's true of simple recursion, but mutually recursive program units (functions or subroutines) are more problematical. It is permissible in (nonrecursive) Fortran to have unit A contain a call to unit B and vice versa as long as A never calls B when A was called by B (and v.v.). Such ``lexically recursive'' units are useful, and their failure to perform the forbidden dynamically recursive calls cannot be verified before run time. Dan --- Newsgroups: rec.puzzles From: hoey@ai.etl.army.mil (Dan Hoey) Date: 30 Aug 90 20:30:18 GMT Subject: Re: Rubik Cube Notation for > 3x3x3 bru...@tekgen.BV.TEK.COM (Bruce Cheney) writes: >Is there a standard notation for Rubik cube moves on cubes that are larger >than 3x3x3 ? David Singmaster's notation (in Notes on Rubik's Magic Cube) >seems to be the standard for the 3x3x3, but there are several possible >extensions to larger dimensions. In *Rubik's Cubic Compendium* there is an article by Tamas Varga with an interesting modification to the language for the 3x3x3. First, he changes the "U" (for Up) to a "T" (for Top) so that every face is named by a consonant. Then he assigns vowels to the possible motions of a face: "O" for 90 degrees fOrward (or clOckwise), "A" for 90 degrees bAckward (or Anticlockwise), and "I" for a 180 degree half-turn (twIce). Then he makes syllables out of a consonant and a vowel, and words out of the syllables. So Pons Asinorum (the simple 6X) can be done with FIBITIDIRILI and Laughter (\/\/) is 3 FOBOROLOs. But wait, there's more! He also allows you to append a "C" to the syllable to indicate that instead of turning the face, you move the whole cube. the notation is not as parsimonious, since FOC=BAC, TOC=DAC, and ROC=LAC, but it's better than having to stop in the middle and say ``then move the cube''. For instance, Jim Saxe's 28-quarter-turn Plummer's Cross (6+) can be done as "FOLIRIFO BOLIRIFO ROFIBIRO LOFIBIRO TIDI" but it's easier to remember it as FOLIRIFO BOLIRIFO TOC FOLIRIFO BOLIRIFO TIDI". Varga also has a way of talking about the slice moves, where you move the middle layer of the cube instead of the faces. For moving the middle, you append "M" to the syllable for a parallel face. So the Spratt wrench (4-flip) is 4 TOROMs, and we can do the Plummer's Cross as FOLIMBO FOLIMFO TOC FOLIMBO FOLIMFO TIM. Of course ROM=LAM=ROCRALO. Unfortunately, Varga stopped there. I decided to continue. For instance, it can be useful to have a way of talking about slice moves where you move the faces rather than the middle. Of course, we could say RAMROC, but I'd rather say something like ROS. Another idea is to append "N" to syllables for aNtislice moves. So Laughter would be six RONTOCs. We could also append "P" to allow "deeP" moves, where we do RO and ROM simultaneously by grabbing two layers of the cube and turning them while keeping the remaining face fixed. Finally, I use "G" to denote the way we "wrinG" the cube, as with ROPRO. So six ROGTOCs does an 8-flip. To summarize, F,B,T,D,R,L -- faces: Front, Back, Top, Down, Right, Left. O,A,I -- directions: fOrward, bAckward, twIce. C,M,S,N,P,G -- extensions: whole-Cube, Middle, Slice, aNtislice, deeP, wrinG. All extensions but C are redundant, since ROM=ROCRALO ROS=ROLA RON=ROLO ROP=ROCLO ROG=ROCROLO (I sent most of the above to the Cube-Lovers mailing list in February. This corrects most of the mistakes, I hope.) m...@oddjob.uchicago.edu (Matt Crawford) writes: >I never saw one [extension to the larger cubes] except the one I made up when >I first got a 4x4x4 cube. In mine, the letters u d r l f b represent >clockwise rotation of an outer 4x4x1 slab and the letters U D R L F B >represent clockwise rotation of half the cube.... >For an NxNxN cube, I would suggest elementary moves u1, u2, ... to >represent rotation of the uppermost 1, 2, ... slabs together. This is the way I did it at first, choosing a cutting plane and rotating the part of the cube on one side of the plane wrt the rest of the cube. I call this philosophy Cutism, because you cut the cube. However, I later decided to espouse Slabism, where a primitive move turns an NxNx1 slab wrt the rest of the cube. In fact, I go for Eccentric Slabism, where turning the central slab of a cube of odd edge is not considered primitive. The fundamental credo behind Eccentric Slabism is that if you consider the NxNxN cube to contain an (N-2)x(N-2)x(N-2) cube inside it, then any two quarter-turns are conjugate. That is, there is an isomorphism of the cube that takes one quarter-turn to the other. (I discussed Eccentric Slabism and its relation to Qubic in a Cube-Lovers message back in June, 1983). We might as well label the slabs with the nearest face. I think the outer slabs might best be treated as in the 3x3x3. Inner slabs can be done with a diphthong or something. Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.physics From: hoey@ai.etl.army.mil (Dan Hoey) Date: 5 Sep 90 18:44:33 GMT Subject: Re: Beer Can Physics >/ hplred:sci.physics / m...@cup.portal.com (Mark Robert Thorson) / >3:16 pm Sep 3, 1990 / I've noticed that when I open a can of beer >within about 4 to 5 feet of my TV, I get a glitch on the picture. >This isn't totally repeatable, it only works about 25% of the time. >Clapping my hands or stamping my foot doesn't do it, so I don't >believe the effect is acoustic.... Old remote controls used sound pitched higher than human hearing, and some old TVs out there still respond to noises at just the right frequency (even if the remote control has been lost). Probably the beer is pitched just right, while your hands and feet aren't. Cecil Adams recently ran a column about a TV that would change channels when you play with a Slinky (metal coil toy) nearby. Dan --- Newsgroups: rec.puzzles From: hoey@ai.etl.army.mil (Dan Hoey) Date: 25 Sep 90 18:07:22 GMT Subject: Re: Finite Automata puzzle dschm...@athena.mit.edu (Dan Schmidt) writes: >The alphabet for this finite automaton consists of pairs of binary >digits, e.g. 0/0 or 0/1 or 1/0 or 1/1. The bit before the slash is from a binary number x and the bit after the slash is from a binary number y. >... The digits are read with the least significant bit first. Thus if x were >6 [110 in binary] and y were 25 [11001 in binary], the input sequence would be >0/1 1/0 1/0 0/1 0/1 >The problem: construct a finite automaton with the given alphabet >which accepts an input sequence only if y = 5x. Solution follows: Let the states be 0, 1, 2, 3, 4, and F. State 0 is the initial and the only accepting state. The transition from state S on input xi/yi is to state (S + 5 xi - yi)/2 provided the latter is an integer. All other transitions, including those from state F, are to state F. For the numerically impaired, the transition table is Input: 0/0 0/1 1/0 1/1 State 0 0 F F 2 1 F 0 3 F 2 1 F F 3 3 F 1 4 F 4 2 F F 4 F F F F F The inductive invariant needed to prove this recognizes the proper language is that initial strings x0 and y0 of length n leave the machine in state (5 x0 - y0)/2^n if the latter is an integer, otherwise state F. Dan --- Newsgroups: comp.lang.lisp From: hoey@ai.etl.army.mil (Dan Hoey) Date: 8 Nov 90 17:49:58 GMT Subject: Reference to THE TEMPEST in CLtL2 In the index of CLtL 2nd Edition there are entries for Shakespeare and THE TEMPEST referring to page 429, but I don't recognize anything on page 429 relevant to either subject. If anyone can explain the situation, please do so by mail. I'll summarize if there is interest. Thanks. Dan Hoey Navy Center for Applied Research in Artificial Intelligence Hoey@AIC.NRL.Navy.Mil --- Date: Fri, 09 Nov 90 15:02:48 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Rubik's Cube reassembly problem and solution In rec.puzzles, greg@math.berkeley.edu (Greg Kuperberg) writes: Consider a standard Rubik's cube. Disassemble it and put it back together at random. Find, with proof, the probability that it can be solved. It depends on how you take it apart. If you just pull out the corner and edge pieces then put them back in without respect to color, the probability is one in 12 that you will put it back into the right orbit. I won't bore you with yet another proof of this; if you spent the last decade in a box see the archives, Singmaster's NOTES ON RUBIK'S MAGIC CUBE, J. A. Eidswick's article in the March 1986 Math Monthly, or even Hofstadter's METAMAGICAL THEMAS. Now if you take the face centers off and scramble them, then there is only one chance in 60 of getting it right. Of the 720 permutations of the six face centers, only 24 can be generated by rigid motions of the cube. But half of these 24 permutations are odd, and leaving the cube in an unsolvable orbit. If you put the face centers on in the ``standard'' configuration with opposite faces ``differing by yellow'' (i.e., white opposite yellow, red opposite orange, and blue opposite green), your chances go up to one in four--half the time you will get an odd permutation, and half the time you will get a mirror-reversed configuration. But wait, if you took the face centers off you probably noticed that the corners and edges don't stay on very well. So, say you scrambled all three kinds of pieces. You will be able to solve the resulting cube if you could solve the corner/edge permutation and the face- center permutation. But if the only thing keeping you from solving the corner/edge permutation and the face-center permutation is that both permutation parities were odd, then you will be able to solve the two of them together. Therefore your chances of success are one in 360 (= (1/12)*(1/60)*2), or one in 24 if you preserved opposite pairs of face centers. Now suppose you peeled off the 54 colored stickers and stuck them back on at random (carefully keeping them out of the reach of children, as there are rumors the paint contains lead, especially on the cheap Taiwanese knockoffs), what is the probability of getting a solvable cube? This question was posed years ago (in Singmaster?) but I believe it is still open. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Mon, 12 Nov 90 18:38:49 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Re: Rubik's Cube reassembly problem and solution This problem of counting the number of solvable restickerings seems to be a lot easier than I had thought, but a lot trickier than you might think. Haym Hirsh sent in a buggy analysis, then corrected himself, but not quite enough. The fix was to account for cases where the stickers corresponded to a cube recoloring, but he just multiplied by 30 (cube recolorings up to rotational symmetry) rather than by 720 (total cube recolorings). We are dividing by 54!, which includes positions differing only by a rotation, so when figuring how many are solvable you have to count such positions also. Another way of figuring this is 6! ways of coloring the face centers, then (8! 3^8 12! 2^12)/12 ways of coloring the rest of the cube, then 9!^6 ways of arranging stickers among identically-colored faces, out of 54! ways of arranging stickers randomly. So the probability that a random restickering will be solvable is 71107747385251871439716429038520049557443706880000000000 ------------------------------------------------------------------------ 230843697339241380472092742683027581083278564571807941132288000000000000 40122452017152 = ------------------------------ ~ 3.0803 X 10^-16. 130253249618151492335575683325 It seems odd to me that this is not the reciprocal of an integer, but I guess that's because we are dealing with color cosets rather than some cube group. Haym Hirsch also asked me how to figure out the minimum number of stickers to fix to make an unsolvable stickering solvable. Sounds hard to me. His question arises in the same way that I recall the original problem arising: trying to clean up after someone who tried to solve the cube by restickering. Since the adhesive isn't designed for moving the stickers around, this leads rapidly to Dik Winter's problem: dealing cubes that have lost some of their stickers. Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.chem From: hoey@ai.etl.army.mil (Dan Hoey) Date: 23 Nov 90 20:29:03 GMT Subject: Re: liter vs. dm**3 (was Re: ln Keq and mathematical niceties) bran...@wa4mei.UUCP (Brandon Rhodes) writes: >If I remember correctly, a meter is the distance traveled by a light >wave in .000000003335640952 seconds as measured by a celsium clock. That's only an approximation. A meter is defined to be 1/299792458 of the distance traveled by EM in one second. It could be given as a repeating decimal, of course, but the repeating part of it is 3,520,056 digits long. Dan --- Newsgroups: rec.arts.sf-lovers From: hoey@ai.etl.army.mil (Dan Hoey) Date: 28 Nov 90 22:35:13 GMT Subject: SF Clubs and Organizations (was SF Club/Orginization) rgomb...@MCS.KENT.EDU writes: >Could anyone out there who knows of the following SF clubs eitherpost >me an e-mail address or a regular mail address so I could get in >contact with them... > World SF Society (USA) That's international, not USA. And I'm not sure it has an address, or even an existence, for 51 weeks of the year (though it does have a perennial committee). Perhaps you want to get in touch with Chicon, Magicon, ConFrancisco, or the Louisville or Winnepeg bids. Or the vestigial organizations of ConFiction or Noreascon (did NolaCon have any organization? Don't answer that.) Others you left out: Washington SF Assn, Philadelphia SF Soc, Hampton Roads SF Assn, Baltimore/Washington Worldcon Assn for Fandom (or something like that), and Washington Area Convention Org. All of these are East Coast USA fanorgs. Also, there's a Polish club called something like Gdansk Club Fantastikyi. I think the Fan Association of Central Texas is still out there too. The Discon III Worldcon bid assembled a list of clubs, which I've been meaning to send to the net for some time. Kevin R. Grazier (graz...@physics.purdue.edu or graz...@gphys1.geo.purdue.edu) compiled a list of college-based clubs a couple of months ago. Mike Sandler (sand...@vax1.acs.udel.edu) did the same thing last year. Richard Dansky (rdan...@eagle.wesleyan.edu) may have compiled something; at least he can tell you about Wesleyan Science Fiction, of which he was president about a year ago. There's a book called the Fandom Directory that's put out yearly or so, though it seems more geared to media fandom. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.crypt From: hoey@ai.etl.army.mil (Dan Hoey) Date: 30 Nov 90 20:01:20 GMT Subject: Re: Password Probabilities pal...@tallis.enet.dec.com (Colonel Mode) writes: >To derive all possible n bit binary values from an ascii password string, XOR >each byte of the string into the least significant byte of the n bit field and >rotate the field 7 bits to the left before XORing the next byte in. Actually, you should rotate the field by an amount that is at least 7 (so that all extremely short passwords are distinct) and relatively prime to n (so that the characters are shifted to every possible position within the field). Since n=56, the smallest such number is 9. But the really good way of mixing characters is is to notice that the Unix password scheme is losing out by always using 0 as the plaintext to (iterated) DES. What you ought to do is to use each character as a key for encoding *the result from the previous character*. Or you can use substrings of eight characters as the key, if you like, but the important thing is to chain the results together so you don't have the pathetic 56-bit restriction now endemic to Unix. Notice that the XOR method can never take you past 56 bits of key, it just makes some of the keys easier to type. Even better is to let the number of DES iterations increase at each stage, so you can even protect against increases in processor speed to some extent. I got the basic form of this idea off the net a couple of years ago but I forget who came up with it. I just wish someone would start using it, rather than just ignoring all characters past the eighth. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math, rec.puzzles Followup-To: poster From: hoey@ai.etl.army.mil (Dan Hoey) Date: 11 Dec 90 19:27:46 GMT Subject: Re: Monty Hall problem: a precise statement bh...@nswc-wo.navy.mil (Brian R. Hunt) writes: >Let me offer an attempt at a reasonably precise statement of a Monty >Hall scenario in which the "switching wins 2/3 of the time" answer is >correct. >"You will play a game with a fellow named Monty Hall, and he has to >play by the rules I'm about to describe.... and goes on to describe such a scenario. The problem is that in the real world we are never given a set of rules that Monty must follow. Well, actually he says there's a prize behind one of those doors, so I guess that's a rule. And he says he'll give it to us if we pick it. But this bit about being given a second choice has never been promised. I suppose a contestant who has just been selected to play the game might ask whether Monty is going to open an empty, unchosen door and offer a chance to switch. Monty might just say, ``you're not the contestant we're looking for'' and take the next contestant. The audience is full of people who would like a chance to win a prize. I've heard that Monty has an evil twin who sometimes stands in on the show. The evil twin offers you a choice of three doors, behind one of which is a prize. If you pick an empty door, he opens up the door with the prize (or the door you chose, or both) and says, ``Hard cheese, you lose.'' If you pick the door with the prize, he opens one of the empty doors and offers you a chance to switch. This is a scenario in which switching never wins, and not switching wins one third of the time. Then there's Saint Monty, who offers you a choice of doors and only offers you a chance to switch when you didn't pick the prize. When Saint Monty runs the show, switching wins every time, and not switching wins one third of the time. The problem is that you can't tell which Monty you've got. Has anyone got firm statistics on how often the contestants on Let's Make a Deal got a second chance, correlated to the accuracy of their first guess? Was there any correlation with the value of the prize? Did Monty's behavior change when the contestant had been through a few rounds, perhaps in response to the contestant's strategy? I solicit reliable answers to these questions via email. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@ai.etl.army.mil (Dan Hoey) Date: 14 Dec 90 18:38:28 GMT Subject: Re: Swimming pool riddle (SPOILER) yo...@cbnews.att.com (Yossi A. Nygate) writes: >X wanted to build a rectangle swimming pool with flooring of white >and blue tiles. The floor would have a white rectangle in the middle >surrounded by one band of blue tiles. After the tiles arrived X >decided to change the shape and have the rectangle blue and the >surrounding stripe white. The pool was then a different shape and no >tiles were left over. >Question. What is the smallest pair of pools that this could occur. Let there be B blue and W white tiles, and assume B<=W wlog. Since they form rectangular annuli, B and W must both be even. Now let the final blue rectangle be IxJ, with I<=J. There will be W=2I+2J+4 white tiles around the rectangle of B=IJ blue tiles. Now if I>=2, B <= W IJ <= 2I+2J+4 (I-2)(J-2) <= 8 (I-2)^2 <= 8 since 2<=I<=J I-2 <= 2 I <= 4 so I is among {1,2,3,4}. Now let the original white rectangle be KxL with K even. Then letting K=2n we combine our formulas for W to yield 2I+2J+4 = W = 2nL. Combining formulas for B yields IJ = B = 4n + 2L + 4 IJn = 4n^2 + 2nL + 4n = 4n^2 + 2I + 2J + 4 + 4n J(In-2) = 4n^2 + 4n + 2I + 4 I^2 J = (4I^2n^2 + 4I^2 n + 2I^3 + 4I^2)/(In-2) = 4In + 4I+8 + (2I^3 + 4I^2 + 8I + 16)/(In-2) I^2 J - 4In - 4I - 8 = (2I^3 + 4I^2 + 8I + 16)/(In-2). Let d=2 if I is odd, and d=4 if I is even. Now we noted that B=IJ is even, so I^2J is divisible by d. Thus the entire left hand side of the above equation is an integer divisible by d, so the right side must also be, so In-2 must divide (2I^3+4I^2+8I+16)/d. Consider this condition for the four possible values of I. I=1: n-2 must divide 15, so n-2 must be among {1,3,5,15}. I=2: 2n-2 must divide 16, so n-1 must be among {1,2,4,8}. I=3: 3n-2 must divide 65, so 3n-2 must be among {1,5,13,65}. I=4: 4n-2 must divide 60, so 2n-1 must be among {1,2,3,5,6,10,15,30}. Note that (in the I=3 case) 3n-2 is congruent to 1 mod 3, and (in the I=4 case) 2n-1 is odd, removing half the possibilities from those two cases. The rest of the solutions are below. The smallest is the 64-square pool. I,n IxJ = B W = KxL B+W 1,3 1x54 = 54 114 = 6x9 168 1,5 1x42 = 42 90 = 10x9 132 1,7 1x46 = 46 98 = 14x7 144 1,17 1x82 = 82 170 = 34x5 252 2,2 2x16 = 32 40 = 4x10 72 2,3 2x14 = 28 36 = 6x6 64 2,5 2x16 = 32 40 = 10x4 # 2,9 2x23 = 46 54 = 18x3 100 3,1 3x18 = 54 46 = 2x23 * 3,5 3x10 = 30 30 = 10x3 + 4,1 4x10 = 40 32 = 2x16 * 4,2 4x6 = 24 24 = 4x6 + 4,3 4x6 = 24 24 = 6x4 +# 4,8 4x10 = 40 32 = 16x2 *# * Removed, because B<=W is violated. + Removed, because the two pools have the same shape. # Removed, because it duplicates a previous answer (our assumption that K is even sometimes doesn't distinguish K from L). Dan Hoey --- Newsgroups: rec.puzzles From: hoey@ai.etl.army.mil (Dan Hoey) Date: 14 Dec 90 18:54:19 GMT Subject: Re: Swimming pool riddle (SPOILER) (patch) Hoey@AIC.NRL.Navy.Mil (I) wrote: > 1,3 1x54 = 54 114 = 6x9 168 The 9 should have been a 19, but I think the rest of the article is all right. Dan --- Newsgroups: rec.puzzles From: hoey@ai.etl.army.mil (Dan Hoey) Date: 19 Dec 90 14:00:20 GMT Subject: A new and better digit puzzle How many 9-digit primes have 9 different digits? For each digit d, how many of the primes contain all digits but d? Can you explain the distribution, at least approximately? What happens if we extend the problem to include 8-digit primes with a leading zero (which then must not occur in tne other 8 digits)? What about bases 3 through 16, say, or more? In each case we ask for a base B number with B-1 distinct digits. Fans of extremely short limericks may be interested in considering the case for B<3, but I don't want to hear the answers. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.skeptic, sci.math Followup-To: sci.skeptic From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 20 Dec 90 17:42:27 GMT Subject: Re: Paradox (?) bora...@ogicse.ogi.edu (M. Edward Borasky) writes: >>a...@gtx.UUCP (Alan Filipski) writes: >>>The antinomy that really bothers me is the one about "The least >>>positive integer that cannot be specified in less than 50 English >>>syllables." >Are you sure this is stated correctly? My impression is that ALL >positive integers CAN be specified in less than 50 English syllables. Certainly, if you include specifications like ``the number of seconds it takes to say `Aaa...aah,' '' though most of them are aborted by the human speakers when they run out of breath. Better, perhaps is ``the number of seconds between now and [pause] now,'' though again most cannot be said in a lifetime, or before the Sun goes out, etc. Dan --- Newsgroups: rec.puzzles From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 20 Dec 90 21:01:32 GMT Subject: Re: Another 10-digit puzzle (no spoilers) pa...@cbnewsd.att.com (paul.h.nelson) writes: >Find the corresponding answer, for base 16, i.e., find a 16-digit number: That is trivial if you know the ten-digit one, which has been posted many times in the last week (though without proof of uniqueness). How about: Describe all sequences (d[0], d[1], ..., d[n]) such that d[i] is the number of i's in the sequence. There is an infinite set containing one such sequence for every n>3, and in particular including the ten-digit solution I forbear to post. Don't bother responding unless you either (1) know one not in that infinite set or (2) can prove that's all there are, at least for some values of n. Dan --- Newsgroups: sci.math From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 20 Dec 90 21:19:37 GMT Subject: Boneheadism continued boo...@network.ucsd.edu (Booker bense) writes: >How about 8-colors? Are there any proofs for general graphs >that 8 colors is enough.... Of course, the complete graph on N vertices takes N colors, and in fact N colors is sufficient for any N-vertex graph. Now, there is a graph with E edges that takes about floor(sqrt(2E)) colors, namely the complete graph on floor(sqrt(2E)) vertices, with an extra vertex and some edges added to bring the number of edges up to E. Can every graph with E edges be colored with O(sqrt(E)) colors? Dan ---