Newsgroups: rec.puzzles From: Dan Hoey Date: Tue, 05 Feb 2008 12:37:32 -0500 Subject: Re: Enigma 1470 - Dicey dice On Jan 21, 10:57am, johnjo wrote: > Chappy wrote: > > On Jan 16, 4:22 am, johnjo wrote: > > > Chappy wrote: > > > > Enigma 1470 - Dicey dice > > > > New Scientist magazine, 24 November 2007. > > > > by Peter Harrison. > > snip > > Correct. How did you work it out? snip > so a downward sequence of 1-5 is not possible. > The downward sequence for 2-6 is examined in the same way. there are > all three possibilities now since 3,6 does not occur. > But it falls out readily anyway. I don't know what you mean by "it falls out", but the dice orientation you reported is not the only orientation possible. We can have either your pattern of opposite pairs: 2:5 3:4 1:6 3:6 4:5 1:2 4:2 5:6 1:3 5:3 6:2 1:4 6:4 2:3 1:5 or the pattern with 2-6 in the opposite order: 6:3 5:4 1:2 5:2 4:3 1:6 4:6 3:2 1:5 3:5 2:6 1:4 2:4 6:5 1:3 But the puzzle asks only for the opposite pairs of the fourth die, which is the same in either case. Dan Hoey haoyuep@aol.com --- Newsgroups: talk.bizarre From: Dan Hoey Date: Tue, 12 Feb 2008 16:23:06 -0500 Subject: Re: attn scott dorsey Ace Lightning wrote: > nikolai kingsley wrote: >> http://venus.oracchi.com/Illustrator/mechanical/nagra.htm > that's impressive. well, the Nagra is impressive, and the drawing > indicates that someone has either severe OCD or way too much free > time. Or perhaps not quite enough free time, since the wireframe has so many artifacts that should be filled in by the rendering engine, like the diffuse reflections of the front knobs and the specular reflections off the tape. And what's with those screw heads that have their slots blanked out in the wireframe. Near the right side there's a reflection of a slotted screw on the tape, but the slot is missing on the screw itself. Imprécis. Dan --- Newsgroups: talk.bizarre, soc.retirement From: Dan Hoey Date: Tue, 12 Feb 2008 16:38:38 -0500 Subject: Re: Don't Treat The Old And Unhealthy, Say Progressive British Doctors (David P.) wrote: > Gilchrist wrote: [...] >> A nation should be judged on >> how it treats its poorest citizens. > A nation's poorest citizens are the > not-yet-born! They can't speak or vote! That's pretty pathetic, but I'd have to say that for poverty, the already-dead have them beat. The already-dead can't speak or vote, either, but they can't make a woman's innie into an outie, the way the not-yet-born can. Can you find any poorer citizens than the already-dead? Dan --- Newsgroups: talk.bizarre From: Dan Hoey Date: Thu, 21 Feb 2008 16:42:20 -0500 Subject: Re: attn scott dorsey Ace Lightning wrote: > Dan Hoey wrote: > >Ace Lightning wrote: > >> that's impressive. well, the Nagra is impressive, and the drawing > >> indicates that someone has either severe OCD or way too much free > >> time. > >Or perhaps not quite enough free time, since the wireframe has so > >many artifacts that should be filled in by the rendering engine, like > >the diffuse reflections of the front knobs and the specular > >reflections off the tape. And what's with those screw heads that > >have their slots blanked out in the wireframe. Near the right side > >there's a reflection of a slotted screw on the tape, but the slot is > >missing on the screw itself. > >Imprécis. > i suspect that the "wireframe" began with the photograph being > subjected to the "find edges" and/or "sharpen edges" filters in > Photoshop. My point exactly. The postprocessing, as I noticed was sloppy, in that screw heads were replaced by wireframes of cylinders without slots, and the reflections were not cleaned up. To get a good reverse- engineered wireframe requires someone with a bit more free time. Dan "Slowly I turned..." Hoey --- Newsgroups: rec.puzzles From: Dan Hoey Date: Fri, 22 Feb 2008 16:08:49 -0500 Subject: Re: Enigma 1474 - Nativity places I saw immediately that this was not possible as an ordinary Sudoku puzzle. Rather than put children atop each other, I tried modifying the letter forms by adding more ink. With the modifications I -> D I -> H I -> F -> E -> B C -> G I could not find a aolution. Perhaps she used H -> B or D -> B. Or even H -> A in block printing. Dan --- Newsgroups: rec.puzzles From: Dan Hoey Date: Fri, 22 Feb 2008 17:15:48 -0500 Subject: Re: 1 3760 (spoiler and worse) Kevin Stone wrote: > Don Del Grande wrote: >> Kevin Stone wrote: >>> 1 3760 5318008; 345 5335, 1 376616; 345 56075, 1 771! >> SPOILER SPACE [Spoiler space omitted] >> Slogs? >> -- Don > http://www.answers.com/slogs&r=67 > (v.tr #2) But the usage was not transitive, and anyway, what about that kind of 6075 51 771, 1 31738? The following is a possible expansion. Squick alert: Not for the sensitive of stomach! 1 335 3751 37738; 1 3760 618 5318008; 345 5335, 1 376616; 345 54615, "07734, 1 372215, 1 7735 '5137'!" 004008, 1 335 618 67108; 73507 53200 45188075 006; 1 38 05 771 1 3507 3718. 1 06 805, 553755178. I had to use the dictionary for 73507, but you presumably understand from context, unfortunately. I would have got it a lot faster if the modification suggested in other posts had been done, or my name isn't Dan Hoey haoyuep @ aol.com 334 334 --- Newsgroups: rec.puzzles From: Dan Hoey Date: Tue, 11 Mar 2008 13:30:40 -0400 Subject: Re: Bluffing Androids Evan wrote: > Yes, the queen of hearts was the intended answer, and darn, I think > I see your point. Oh well, I'll tweak it some more. Thanks for > playing. I think the problem is the addition of quantifiers extending beyond the present case. To be properly self-referential, Android 2 should say "Only two androids at this table are telling the truth" and Android 5 should say "Android 7 is not lying." You might as well have Android 7 say "The card is not a King, Queen, or Joker" so you don't need to mention it later. I think that makes the puzzle kosher. The problem of overquantification messes up the Liar Paradox, such as when Epimenides (a Cretan) said "All Cretans are liars." Not a paradox if true (since Epimenides might be a liar, telling the truth for a change) nor if false (since the non-lying Cretans might not include Epimenides. Dan Hoey haoyuep@aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Tue, 11 Mar 2008 13:35:08 -0400 Subject: Re: Poor men paradox Derek Holt wrote: > For n >= 1, with probability 1/2^n, I place 3^(n-1) and 3^n dollars in > the two envelopes, and you know that I have followed this rule. Now you > play the game as before. > You open one envelope and observe x = 3^m dollars. Of course, if x=1, then > you definitely swap. > Otherwise, you reason, that the n that I used must have been m-1 or m. > Since it is twice as likely to have been m-1 as m, the other envelope > contains x/3 with probability 2/3, and 3x with probability 1/3. > So your expected gain from swapping is -4x/9 + 2x/3 = 2x/9, and you should > always swap. A lot of paradoxical things happen when the money supply (or its expected value) is infinite. Dan --- Newsgroups: talk.bizarre From: Dan Hoey Date: Fri, 14 Mar 2008 16:42:45 -0400 Subject: Eliot Spitzer memorial scumfest Holier than thou Emperors take a pratfall. Reporters all come. It was interesting watching the local newscasters decide "high-class prostitute" wasn't politically correct, so they switched to "high-priced prostitute", then finally "high-level prostitute." Where's Gary Gygax when we need a saving throw against a ninth-level studded leather whore? Dan --- Newsgroups: rec.puzzles From: Dan Hoey Date: Fri, 14 Mar 2008 17:59:59 -0400 Subject: Re: Enigma 1475 - Lines of thought The puzzle is notable for how the matching constraints cut down the work. At first, it looks like we're searching through a solution space of 6^24 cases, but once we require that the red lines match up, we're down to 222 cases, or 84 cases if we reduce by the symmetries of the rectangle. Requiring that the areas be simply-connected reduces this to 201 cases, or 74 using symmetry. The area constraint is satisfied exactly when there are three areas, yielding the following 21 cases (or 59 when symmetry is ignored). ..................................................................... : : : ______ : ___ ___ : : /\ /\ /\ : /\ /\ /\ : /\ / \ : / \ / \ : :| | | | | |: \/ | \/ |: \/ | ___/ : \___/ \___/ : :| | | | | |: | |: | / : : :| | | | | |: | |: ___/ | : ____________ : :| | | | | |: /\ | /\ |: / | /\ : / \ : : \/ \/ \/ : \/ \/ \/ : \______/ \/ : \____________/ : :................:................:................:................: : ______ : ______ : ___ ___ : : : /\ / \ : /\ / \ : / \ / \ : /\ /\ /\ : : \/ \ / : \/ | ___/ : \ | \___/ :| | \/ | |: : \ / : | / : \ | :| | | |: : / \ : | \___ : / | ___ :| \ / |: : /\ / \ : /\ | \ : / | / \ :| \ / |: : \/ \______/ : \/ \______/ : \___/ \___/ : \___/ \___/ : :................:................:................:................: : ______ : : ___ ___ : ______ : : /\ / \ : /\ /\ /\ : / \ / \ : /\ / \ : : \/ \___ |: \/ | | \/ : \___/ \ |:| | \______/ : : \ |: | | : \ |:| | : : ___/ |: ___/ \___ : ___ / |:| | ______ : : /\ / |: / \ : / \ / |:| | / \ : : \/ \______/ : \____________/ : \___/ \___/ : \/ \______/ : :................:................:................:................: : : : ______ : ______ : : /\ /\ /\ : /\ /\ /\ : /\ / \ : /\ / \ : : \/ | | | |: \/ \/ | |: \/ \______/ :| | \______/ : : | | | |: | |: :| | : : | \/ |: _________/ |: ____________ :| \ ___ : : /\ | |: / |: / \ :| \ / \ : : \/ \______/ : \____________/ : \____________/ : \___/ \___/ : :................:................:................:................: : ______ : ______ : : : : /\ / \ : /\ / \ : /\ /\ /\ : /\ /\ /\ : : \/ \ / : \/ \___ |: \/ | | | |:| | \/ | |: : \ / : \ |: | | | |:| | | |: : ___ | \ : ______ | |: ___/ | | |:| \___ | |: : / \ | \ : / \ | |: / | | |:| \ | |: : \___/ \___/ : \______/ \/ : \______/ \/ : \______/ \/ : :................:................:................:................: : ______ : : /\ / \ : : \/ \___ |: : \ |: : ___ / |: : / \ / |: : \___/ \___/ : :................: Dan Hoey haoyuep@aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Mon, 17 Mar 2008 10:26:53 -0400 Subject: Re: Enigma 1475 - Lines of thought Chappy wrote: > So what do you get as the answer to the question, "How many of the > combinations?" Same as everyone, 12+2v straight sections and 12-2v corners, where v is the number of inner corners, or "7s". By inspection, v is 0, 1, 2, 3, or 4. Dan --- Newsgroups: sci.math From: Dan Hoey Date: Tue, 18 Mar 2008 11:43:04 -0400 Subject: Re: How many trains it will meet? León-Sotelo wrote: > Trains leave from New York to Washington every hour on the > hour(1:00,2:00....).Trains leave from Washington to New York every > hour on the hour and half hour(1:00,1:30,2:0,2:30....). The real problem here is what to do with all the extra equipment in New York, and where to find cars in Washington. If the trains in New York go to Chicago, then (by a commodious vicus of recirculation) back to Washington you will still find that the direct trains from New York to Washington are twice as crowded as the ones from Washington to New York. That's why the railroads are in trouble. Dan --- Making Light: Phase one: collect underpants ::: March 19, 2008, 01:41 PM Did you know there's a lesbian bar in Washington, D.C. named Phase 1? I don't know about underpants, but I bet they have oats. --- Newsgroups: rec.puzzles From: Dan Hoey Date: Wed, 19 Mar 2008 16:36:28 -0400 Subject: Re: SuDoKu Mike Williams wrote: > Wasn't it nano bagonghi who wrote: >> fin wrote: >>> nano bagonghi kirjoitti 2008-03-19: [...] >>>> Felgenhauer & Jarvis >>>> determined that there are 6670903752021072936960 possible >>>> 9x9 Sudokus. >>> That is the number of solutions, not puzzles. It also doesn't take into account the fact that the puzzle is not essentially changed by uniform renumbering, by permuting the three triples of rows and columns, by permuting the rows and columns in each triple, or by rotating the puzzle. Ed Russell and Frazer Jarvis found that there are only 5,472,730,538 essentially different solution patterns when these symmetries are considered. [...] >> But, given a scheme with some filled cells, >> it is considered a puzzle [...] >> 3) when it can be completed to EXACTLY one solution. [...] > Even (3) isn't really sufficient to make it a Sudoku puzzle in the > normally accepted sense. For example, if I were to choose one of the > 6670903752021072936960 solution grids and remove one number then there'd > be exactly one solution, but it wouldn't be acceptable as a puzzle. There is another requirement that is usually imposed, which is that the solution can be found by the application of some set of accepted procedures. The list of accepted procedures may differ from site to site, but it generally does not include unbounded backtracking. (There is a one-level backtracking procedure called "penciling-in", but you're not supposed to need multiple pencils.) --Dan --- Newsgroups: sci.math, alt.math.recreational From: Dan Hoey Date: Mon, 24 Mar 2008 10:20:23 -0400 Subject: Re: The term "computer algebra system" Vladimir Bondarenko wrote: > The term "computer algebra system" is totally misleading. > What we have now is as much genuine computer algebra systems > as a glob of warm clay is Adam. > It requires to ensoul what we have with the breath of life. Putney says the Bohrmann-6 CAS is got to have soul. --- Newsgroups: sci.math, comp.graphics.algorithms From: Dan Hoey Date: Mon, 24 Mar 2008 11:54:57 -0400 Subject: Re: Covering rectangles in 2D I'm still a little mystified by the idea of a "staircase", since it looks to me like the hard cases happen when we have something more like a woven basket. Suppose the original rectangle is m x m, say m=O(n^(1-e)). We have O(n) 1xn rectangles, O(n^e) in each row; similarly O(n) nx1 rectangles, with O(n^e) in each column. I suspect you have to enter and leave O(n) of the rectangles about O(n^(1-2e)) times each, and I don't see how you avoid having to do O(n^(2-2e)) bookkeeping. Dan Hoey haoyuep at aol.com --- Newsgroups: sci.math, comp.graphics.algorithms From: Dan Hoey Date: Tue, 25 Mar 2008 16:26:44 -0400 Subject: Re: Covering rectangles in 2D James Waldby wrote: > On Mon, 24 Mar 2008 11:54:57 -0400, Dan Hoey wrote: >> I'm still a little mystified by the idea of a "staircase", since it >> looks to me like the hard cases happen when we have something more >> like a woven basket. Suppose the original rectangle is m x m, say >> m=O(n^(1-e)). We have O(n) 1xn rectangles, O(n^e) in each row; >> similarly O(n) nx1 rectangles, with O(n^e) in each column. I suspect >> you have to enter and leave O(n) of the rectangles about O(n^(1-2e)) >> times each, and I don't see how you avoid having to do O(n^(2-2e)) >> bookkeeping. > Would a concrete instance of your example be like the following? > Say e=.2, n=10. Let u = n^(1-e) ~ 6.3 and v = n^e ~ 1.6. Let m = > 12 ~ 2u and k = 3 ~ 2v. Then place 3 each 1x10 thin horizontal > rectangles in each of 12 rows, and likewise 3 each 10x1 thin vertical > rectangles in each of 12 columns. Ie, rectangle set R contains 30 > vertical and 30 horizontal rectangles. (Of course, by construction > all these numbers are O(n).) I was actually thinking that the horizontal and vertical strips would both be needed to cover the rectangle. A 15x15 rectangle with two strips in each line might look like this |---||||------| -||--+-+----||| -|||||-+-+--||| -|||||-+-+--||| -||||--+-+---|| |||-+--+++---|+ |||-||||++-+-|| ||--|||||||+-|| +|--+-+---||||| |---+-+---||+|| |---+-+---|||-| +||||-+-------| +||||-+-------- ||----||||----- |---------||-|| where the "-" denotes part of a horizontal strip, "|" denotes part of a vertical strip, and "+" denotes overlapping parts of strips. > Recall that the method copies R into priority queue X, ordered first > by the x_l coordinate of each rectangle, then by the y_l of each. So after processing half the rectangles, you have covered an area like | || | || | || | || | | || | | | | || | | | || || | | | ||||| |---+-+---||+|| |---+-+---|||-| +||||-+-------| +||||-+-------- ||----||||----- |---------||-|| (or maybe I've swapped x and y). So does D contain one entry for each exposed rectangle, or one entry for each segment of a exposed rectangle's side? Dan --- Newsgroups: rec.puzzles From: Dan Hoey Date: Wed, 26 Mar 2008 16:44:17 -0400 Subject: Re: Dear Friends, do you know this puzzle, his history and solution? Alan Morgan wrote: > ... I was assuming that the laptop figured into the puzzle somehow > (actually, I didn't assume there was a puzzle at all at first). I'm > wondering if this is a puzzle where you have to balance the 6 objects > on the head of the 7th (I've seen something like this with nails > before, where one is hammered into a board and you have to balance > the rest of them on top of the first). Neat. It looks like what you get when you Google "balance six nails". Spoiler warning: this page shows a solution. http://www.fcmuk.org/freebies/nails/nails.htm Dan --- Wikipedia Talk:Pseudoforest Minors I'm less familiar with the literature, but I'm used to running into the term "graph" meaning "simple graph". I've added a footnote to the definition of graph to the effect that this sort of graph is a generalization, often called a multigraph or pseudograph. I hope this doesn't bog things down, or give a mistaken impression of what to expect from current literature, so I welcome correction from people who read more. -Dan Hoey 12:41, 31 March 2008 (UTC) --- Newsgroups: sci.math, comp.graphics.algorithms From: Dan Hoey Date: Mon, 31 Mar 2008 16:19:25 -0400 Subject: Re: Covering rectangles in 2D James Waldby wrote: > quasi wrote: >> So does that mean that you are, for now, withdrawing your previous >> claim that the problem has an O(n^(3/2)) (or better) resolution? > No; as I mentioned in a post on 22 Mar, line sweep methods can be > done in O(n log n) time using a maintained interval binary search > tree. The post of 22 Mar uses terminology such as "the newly introduced corner" which seems to refer to the algorithm you later agreed was faulty. Furthermore, the 22 Mar article claims to offer only an incomplete description and cites lack of time for your unwillingness to make it complete. Under such circumstances, I must doubt any claims of the algorithm's correctness and performance. Please spare us the graphs of your program's behavior, which suggest only that you don't know what kind of an answer would be convincing. A full description of your proposed algorithm, together with at least a sketch of the proof of its performance and correctness, are what might repair this sorry state. Dan --- Making Light: Thoroughly spoiled Little Brother ::: May 07, 2008, 05:16 PM I knew I had to read this when Patrick announced the review copies, and I'm glad Cory put up CC versions so I don't have to wait for the bookstore. I thought the book was very good, but maybe not as much of a paradigm shift as Down and Out in the Magic Kingdom. I guess you that's the price of present-day relevance. I can't say DHS got off too hard. They had the politico and haircut woman who were pretty flat evil, but they also had Masha, who at least couldn't rationalize it forever. And you have to expect there were a bunch more who tried to come in from the cold and got caught. I'm more concerned that DHS was shown as too incompetent--I don't think a real-life Marcus could stay free. Realism would come out more like 1984, only with less wiggle room. The only real groaner typo I saw was "proscribed" for "prescribed", though I might quibble about some of the hyphenated terms. --- Making Light: Thoroughly spoiled Little Brother ::: May 09, 2008, 12:22 AM Count me as one who doesn't think the Iraq mission was a punishment in any sense. I read it as relocation of a valuable tool to a position where it couldn't be compromised by subpeona or interrogation or coercion. I'm concerned that my previous comments were too negative. I personally found too much of my goshwow diluted with stuff I've already done goswow with--public key crypto, man in the middle, RFIDs, hardware tampering, keystroke monitoring, flash crowds, key signing parties, cold boots... I won't say I didn't learn some interesting things. I really wonder how well the doot-doot bugfinder works, and I'm intrigued in imagining the mechanism of the microwave grape plasma (having cracked Swanwick's luminous pickle, I believe). But I hope and suspect this is a concentrated bit of goshwow to people who haven't had quite the exposure. So this may be a very, very important eye-opener to the proto-geeks. --- Making Light: Thoroughly spoiled Little Brother ::: May 09, 2008, 12:34 AM PJ@91--There was the (description of?) video of the DHS people sitting around dissing the [Bay] area. I recall they weren't so much sincerely dissing as discussing how to use the culture division to drive a wedge between anti-DHS factions and the mainstream. --- Making Light: Little Brother on the New York Times bestseller list ::: May 15, 2008, 01:43 PM Neat! Now I just have to get to the damn bookstore and buy my copy. First YA I've seen with that word on the cover. Dan --- Newsgroups: sci.math From: Dan Hoey Date: Mon, 19 May 2008 15:57:33 -0400 Subject: Re: Very hard combinatorics-problem. lundslaktare@yahoo.com would have written > This problem have not been solved! if he traded in that ASR33. However, some finite values are known. See http://www.research.att.com/~njas/sequences/A000988 . Dan --- From Hoey@aic.nrl.navy.mil Wed May 21 13:43:31 2008 Date: Wed, 21 May 2008 13:43:14 -0400 (EDT) From: Dan Hoey To: "Keith F. Lynch" Subject: Re: Paul Parsons Hello, Keith, Thanks for reporting Paul's death to rasff. I had considered posting it myself, but I had less information (not being on the PRSFS list.) In case you haven't noticed, there's a news article at www.eveningsun.com/ci_9309775 that seems to be about Paul's accident. It's a little confusing, since Carlisle Pike seems to go through Huntington Township, not Tyrone, according to the maps I've found on line. Dan --- From Hoey@aic.nrl.navy.mil Thu May 22 11:56:13 2008 Date: Thu, 22 May 2008 11:56:08 -0400 (EDT) From: Dan Hoey To: "Keith F. Lynch" Subject: Re: Paul Parsons > What newsfeed did you see my rasff posting on? It never appeared on > Google Groups, and it got no replies, so I wonder if it ever left Panix. It got to NRL, through bloom-beacon.mit from panix. I also found a Usenet feed on the web with your article, but I don't see it there now. > ... the Washington Post (though I haven't checked the actual > hardcopy newspaper, just the online version). I got the hardcopies Tuesday and Wednesday; I'll pick up another today. But I think the death notice was probably delayed by the autopsy, since the cause of death is so ambiguous. > Monica says that Kyle says that Martin Morse Wooster will be doing > something at Balticon. Will you be there? You don't seem to be > pre-registered. I'm not, but I'm planning to attend, mainly for this reason. I'm glad Martin is arranging things; I was thinking of dropping a line to Balticon's program people about getting a room for it. > I didn't know you read rasff. It's been mostly politics, as usual, > but there has been some recent discussion of encryption that you > might have something to contribute to. I mostly don't read it; I was mostly searching for anything about Paul. I also avoid entering in rasff discussions not specific to fandom; many rasff people think there is nothing off-topic, but I disagree. Dan --- From Hoey@aic.nrl.navy.mil Thu May 22 12:04:49 2008 Date: Thu, 22 May 2008 12:04:48 -0400 (EDT) From: Dan Hoey To: "Keith F. Lynch" Subject: Re: Fwd: [PRSFS] Additional news about Paul Parsons Thanks for the info, Keith. I also notice that your first notice has finally made it into Google Groups. Possibly that's why they don't show the web copies any more. Dan --- Making Light: Open thread 108 ::: May 23, 2008, 11:49 AM For those who hadn't heard... D.C. area fan Paul Parsons died of a heart attack while driving in Pennsylvania on Sunday May 18. Paul was a lead organizer of Unicon, a Maryland convention held in the 1970s and 80s, managed programming at the 2003 World Fantasy Con, and was central in the Potomac River Science Fiction Society, which has been meeting for over thirty years. He is survived by his wife, Aly Parsons, also an active fan. Paul struck me as one of the wisest fans I know, and he read and retained more written SF than anyone. There will be a memorial gathering at Balticon at 10AM Saturday (tomorrow) in the Garden Room. --- Article 970173 of rec.arts.sf.fandom: From: Dan Hoey Newsgroups: rec.arts.sf.fandom Subject: Re: Paul Parsons Date: Fri, 23 May 2008 13:26:36 -0400 Organization: Naval Research Laboratory, Washington, DC More recent news: Paul's death was determined to be due to a heart attack. The coroner also found that he had recently experienced at least one other heart attack. There will be a memorial for Paul at Balticon (www.balticon.org) at 10AM Saturday in the Garden Room. His family also plans to hold a Celebration of Life at a later date. Dan Hoey haoyuep@aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Fri, 06 Jun 2008 17:32:43 -0400 Subject: Re: Partition a square into rectangles Risto Lankinen wrote: > no.glamour@gmail.com wrote: >> Hello, >> Prove/disprove that a 3000X3000 square can be partitioned into 5X9 >> rectangles (the rectangles can be mixed, either 5X9 or 7X9). and Risto disproves it. There are over 30 proofs of the fact that if we tile a rectangle with rectangular tiles, where each tile has at least one integer side, then the rectangle will have at least one integer side. Thus it is impossible to tile a (1000/3)x(1000/3) square with (5/9)x1 and (7/9)x1 rectangles. One proof is to color the rectangle in a (1/2)x(1/2) checkerboard pattern with one corner of the rectangle on the checkerboard grid. Each tile will cover the same area of each color, while the rectangle will not have the same area of each color unless it has an integer side. Dan Hoey haoyuep a t aol.com --- Making Light: Open thread 110 ::: June 10, 2008, 07:10 PM Nancy @97: That's "ternary", not "trinary". I might even know why if I knew Latin, but I don't. Dan --- From Haoyuep@aol.com Sun Jun 22 11:20:29 2008 Date: Sun, 22 Jun 2008 11:20:16 -0400 From: Dan Hoey To: "Keith F. Lynch" Subject: Re: Followup Thanks a lot for the interesting conversation on the ride, and for the followup info. I've been thinking a bit about your circular string permutation problem. > There's no limit to the size of the alphabet, but as you can see, > I've never needed more than four letters so far. Curiously, 9 takes > 4 letters to maximize the period, but 10 only takes 3. How do you determine the maximum number of needed letters? If you try all length-10 strings with four letters and see no increase over three letters, do you try with five, six, seven, and eight letters? > I've proven that the period cannot exceed 2N unless at least one > letter appears at least three times. We discussed this. As you explained, a letter appearing once or twice imposes a reversal composed with a (possibly trivial) shift, so you can never get any permutation outside the dihedral group of order 2N. In fact, with an even number of distinct letters, you end up in the cyclic group of order N. > Speaking of periods, the last N decimal digits of successive powers of > 2 repeat endlessly every 4, 20, 100, 500, 2500, etc. This is Sloane's > A005054, in which it's mentioned that this is also the same thing for > successive Fermat numbers. Thanks for checking. I had misremembered this. It might be interesting to consider the sequence a(N), where every sequence of N digits appears in the *lowest* a(N) digits of a power of two. > I look foward to seeing you at the next PRSFS meeting. I hope I can make it. Dan --- From Haoyuep@aol.com Tue Jun 24 20:15:47 2008 Date: Tue, 24 Jun 2008 20:15:15 -0400 From: Dan Hoey To: "Keith F. Lynch" Subject: Re: Followup Keith F. Lynch wrote: >> How do you determine the maximum number of needed letters? If you >> try all length-10 strings with four letters and see no increase over >> three letters, do you try with five, six, seven, and eight letters? > No, I simply tried all eight-letter combinations. (Eight since I > knew there had to be at least three of at least one letter unless the > maximum period was 20 or less, which it wasn't.) But you do try five, six, and seven, right? > Since it doesn't > matter to the repetition rate where the string begins, I said it might > as well begin with "A." That's the only speedup. Given that my > program has been grinding away on 11 for 13 days now, perhaps I should > have considered more speedups. For instance the last letter of a > string to be tested should never be A, since then a rotation of the > string would be lexically previous to what I'm about to test. More > importantly, instead of just checking to see if I've got the original > string back, I should also check to see if I have a rotation of it. > And if the ordering of letters doesn't make any difference (which I > haven't yet proven, but haven't yet tried really hard to prove) then > the new letters of a string to be tested might as well always be > introduced in order. > Can you think of any other speedups? Not yet, but I'll consider it. In case you didn't know, checking for a rotation of the original takes linear time, using an algorithm like Knuth-Morris-Pratt. > I can't see any point in allowing more and more distinct letters until > I no longer see an increase in period, since it's not obvious to me > that there couldn't be an increase in period with an Xth letter when > there hadn't been one with an X-1th letter. Yes, it's not obvious to me either, but I wonder if there's a counterexample. >> We discussed this. As you explained, a letter appearing once or >> twice imposes a reversal composed with a (possibly trivial) shift, >> so you can never get any permutation outside the dihedral group of >> order 2N. > Right. >> In fact, with an even number of distinct letters, you end up in the >> cyclic group of order N. > You mean with no duplicate letters? Because my nine-letter string, > aabababcd, contains an even number of distinct letters, but repeats > with a period of 27, not 9. I meant, if no letter appears more than twice, and there are an even number of distinct letters, then the period is at most the length of the string. >> It might be interesting to consider the sequence a(N), where every >> sequence of N digits appears in the *lowest* a(N) digits of a power >> of two. > I'm not sure what you mean. > There is no number such that the last N decimal digits of its > successive powers take all possible values.... That just shows that a(N)>N. Consider that every digit appears in the lowest two digits of some power of two, so a(1)=2. Every two-digit string appears in the lowest three digits of some power of two, so a(2)=3. Similarly a(3)=5. I'm curious what the growth rate of a(N) is--probably linear. Dan --- From Haoyuep@aol.com Sat Jun 28 16:33:24 2008 Date: Sat, 28 Jun 2008 16:33:15 -0400 From: Dan Hoey To: "Keith F. Lynch" Subject: Re: Followup Keith, you wrote: > Since it doesn't > matter to the repetition rate where the string begins, I said it might > as well begin with "A." That's the only speedup. I must have been low on oil not to have noticed that this speedup can be extended to a factor of almost 2N. You just have to make sure your string is the canonical representative of the set of its rotations and reversals. Starting with A is part of the work of choosing the lexicographically least as the canonical representative, but you waste at least half your tests, and probably more like 80-90%. You can use a "loser table" like in Knuth-Morris-Pratt so that you only generate strings that are lexicographically least among their sets of rotations, and then check in linear time that there is no rotation of their reversal that is lexicographically less. I don't know if you've ever used this data structure before, but I can go into it more if the material on the web isn't clear. There's a page somewhere out there on string searching algorithms that describes what's going on, but let me know if you want help. Dan --- From Hoey@aic.nrl.navy.mil Tue Jul 1 15:24:55 2008 Date: Tue, 1 Jul 2008 15:24:44 -0400 (EDT) From: Dan Hoey To: "Keith F. Lynch" Subject: Speeding up the mixing period algorithm Hi, Keith, I've thought some more about the problem of computing the maximum period of your mixing algorithm. Your initial observation that the period is independent of rotations and reflections has led to consideration of "bracelets", which are equivalence classes of strings under rotations and reflections. More concretely, a bracelet may be defined as a string which is the lexicographically least of such an equivalence class. The literature tends to consider labelded k-ary bracelets (bracelets over k-element alphabet) and "unlabeled k-ary bracelets" that are lexicographically least over permutations of the alphabet. Neither is quite what is needed here, since we would like to consider labeled bracelets that use every letter of the alphabet under consideration, or equivalently, an initial subset of the alphabet on n letters. This might be done by starting with unlabeled bracelets and assigning a permutation of letters, but a lot of permutations lead to the same bracelet. Still, a bracelet-based approach might speed things up a lot. The idea is to get some structure, probably a hash table, that can hold all the bracelets we are concerned with. It might even be a separate table for each alphabet size, if space is a problem. Then we need efficient algorithms to (1) generate all bracelets under consideration, and (2) determine the bracelet that a given string belongs to. The idea is to calculate the period of a given bracelet as follows: apply the mixing algorithm, and then find the bracelet of the resulting string. If we have gotten back to the original bracelet, then the period is at most 2N times the number of steps, and we can probably calculate the actual period quickly by collapsing the previous steps into a single dihedral operation. Then to find the maximum period, we step through all the bracelets, calculating periods of bracelets we haven't seen, and marking all the bracelets that occur in the period calculation as having been seen. Dan --- Newsgroups: rec.puzzles From: Dan Hoey Date: Thu, 31 Jul 2008 13:49:32 -0400 Subject: Re: Hound Catches Rabbit Peter Heichelheim wrote: >>> Dan Hoey wrote: >>>> Peter Heichelheim wrote: >>>>> Here is a problem by J.A.H. Hunter. >>>>> A hound sees a rabbit 6 yards away. The rabbit, which is >>>>> directly between the hound and its burrow, and 40 yards from >>>>> that burrow, sees the hound at the same moment. The rabbit makes >>>>> for its burrow with leaps of 5 feet; at the same instant the >>>>> hound gives chase with leaps of 9 feet. The rabbit makes three >>>>> leaps to every two leaps of the hound. Assuming that each animal >>>>> spends 5 times as long, pausing on the ground between leaps, as >>>>> the time it is in the air making each leap, how far from its >>>>> burrow is the rabbit when it is caught? Please give the answer >>>>> and the method to get the answer. . . . > All J.A.H. Hunter gives is the problem one day and the answer the > next day. Also 50 feet is not obviously wrong. Hoey > has given a method to get this answer. There has been a debate among > the posters if the hound can shorten his last jump if it means > catching the hare. Apparently J.A.H. Hunter thought that the hound > could. In my message five days ago, I also showed how the hound can catch the rabbit at 50 feet without shortening any jumps. On its penultimate jump the hound can leap nine feet to a point sqrt(3381) feet from the rabbit's burrow. As long as the rabbit continues on a straight line toward the burrow, the hound can then leap nine feet to make the capture. I read the phrase "makes for its burrow" as requiring the rabbit to run in a straight line, while to me the phrase "gives chase" connotes allowing the hound some leeway. This is a solution that I am pleased to label with the usually repugnant term "lateral thinking". Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Thu, 31 Jul 2008 14:03:04 -0400 Subject: Re: Chess - What's the most non-pawn pieces...? dgates wrote: >>>> dgates said: >>>>> Chess question/puzzle - What's the most non-pawn pieces you can >>>>> have on the board at the same time? > Now... having gone from 8 to 12 pawns being able to queen, I wonder if > there isn't some way to go higher. My intuition tells me there isn't, > but that's the same intuition that previously told me you could only > get 8 queens out of the pawns without giving up an equal number of > other pieces to get more. In each file, the only way to clear the pawn opposition is for a pawn to capture out of the file, or for a pawn in the file to be captured. Each capture can do at most one of each. Therefore clearing the eight files requires four captures, leading to at most 28 pieces. For completeness, note that failing to clearing a file is worse, costing us two unpromoted pawns. Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Thu, 07 Aug 2008 16:33:29 -0400 Subject: Re: Enigma 1499 - Not quite right Chappy wrote: > Enigma 1499 - Not quite right > New Scientist magazine, 21 June 2008. > By Susan Denham. > I took an addition sum and replaced digits > consistently by letters, with different > letters used for different digits. I gave > the result to the printer who produced: > N O T + > Q U I T E > --------- > R I G H T > Unfortunately, the printer has got it not > quite right as he has made one misprint. > I was also going to tell you that TEN is > bigger than ONE, and that NINE is divisible > by 9 but NOT is not. That would have > enabled you to work out what each letter > represented. Please send in QUITE soon. s p o i l e r s p a c e s p o i l e r s p a c e s p o i l e r s p a c e s p o i l e r s p a c e s p o i l e r s p a c e s p o i l e r s p a c e s p o i l e r s p a c e I guess the printer left a letter out before "NOT". It would have to be one of the ten letters already used, and the only letters that yield unique solutions (with the given extra clues) are "I" and "O". But if the letter were "O", then the extra clues about "NOT" and "TEN" would have been unnecessary. With the extra clues, the unique solution to I N O T + Q U I T E = R I G H T is 2 8 6 7 + 4 9 2 7 0 = 5 2 1 3 7, so QUITE is 49270. Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Sun, 10 Aug 2008 21:39:05 -0400 Subject: Re: Planarity Steven Jones wrote: > The multi-agent software Netlogo now includes a puzzle game > called Planarity as one of its example models. The puzzle > involves a graph with seemingly randomly connected nodes. The > object is to move the nodes so that none of the links overlap. > Is there a general solution for this game? I've made it > to level 6. Fáry's theorem ( http://en.wikipedia.org/wiki/F%C3%A1ry%27s_theorem ) states that any planar graph has a straight-line embedding. There is software available for constructing such an embedding; google for "planar graph" "straight line embedding" . Presumably Netlogo generates a planar graph and randomizes the placement its vertices, or even searches heuristically for a vertex placement maximizing the number of crossings. Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Sun, 10 Aug 2008 22:04:51 -0400 Subject: Re: Enigma 1499 - Not quite right Chappy wrote: > Dan Hoey wrote: >> I guess the printer left a letter out before "NOT".... > There is also a solution that replaces one letter with > another letter (no omitted or added letters), although > it produces the same value for QUITE that you got. My exhaustive search for such solutions agrees with Ilan Mayer that only one such replacement has a unique solution (with the extra clues). I also verified that all the extra clues are necessary to produce uniqueness. I don't have any basis to prefer either "INOT" or "REGHT" over the other as a pseudo-word. While there is an Inot River in Romania, most Google hits seem to arise from typos of "INTO". "REGHT" yields a surprising number of hits as a typo in the song title "Relight my fire". It is also a surname, and appears in the OED as an obsolete form of the word "RIGHT". Neither seems to be a contemporary non-capitalized word. Perhaps Susan Denham intended this ambiguity, since she asks only for the value of QUITE. Dan Hoey haoyuep at aol.com --- Article 978897 of rec.arts.sf.fandom: From: Dan Hoey Newsgroups: rec.arts.sf.fandom Subject: Re: This newsgroup was touted in a Denvention newsletter Date: Wed, 13 Aug 2008 20:48:14 -0400 Organization: Naval Research Laboratory, Washington, DC To: "Keith F. Lynch" Keith F. Lynch wrote: [...] > Is there a newer and better version of pdftotext? If so, I'll nag > Panix to install it. Keith, if your pdftotext understands the "-layout" switch, it is possible to find where the columns are. Unfortunately, Denvention's PDFs use Unicode hyphens, quotes, and spaces, so you also need some Unicode-aware tools if you want to see the punctuation right; in particular, your pdftotext needs to understand "-enc UTF-8". You also need to deal with the resulting 145-character lines (which most text-only systems find a little awkward) in order to determine which rectangles of text contain your article. It seems that "cut -c" operates differently depending on mysterious factors (at least, mysterious to me), but the shell command ( pdftotext -enc UTF-8 -layout newsletter9.pdf - \ | cut -c 54- | head -3 ;\ pdftotext -enc UTF-8 -layout newsletter9.pdf - \ | cut -c 54-96 | tail +4 | head -52 ;\ pdftotext -enc UTF-8 -layout newsletter9.pdf - \ | cut -c 102- | tail +4 | head -21 \ ) | cat -v \ | sed -e 's=M- *= =g' \ -e 's=M-\^@M-\^[]\]="=g' \ -e 's=M-\^@M-\^P=-=g' \ -e 's=M-\^@M-\^Y='\'=g \ -e 's=M-[bB]\([- "'\'']\)=\1=g' \ -e '/^$/d' with some minor variations can produce output not entirely unlike > Morning Edition > Make the Wonder of Worldcon Last the Whole Year > Denvention will soon be over. But > the fun does not have to end. There > are many ways to enjoy SF fandom > year-round. > Most regions of the U.S. (and > other countries too) have their own > regional conventions. Leaflets ad- > vertising many of these conventions > can be found in the Korbel lounge > near the newsletter rack. Conven- > tions are also listed in the back of > Asimov's and Analog magazines. > Many of these conventions are held > by local fan groups that also have > regular meetings. Contacting the > convention will usually put you in > touch with their spon- > sors or you can google > "science fiction" club > and your state or area. > Another way to stay > connected to fandom > is through fanzines. > These personal maga- > zines are available for > a nominal cost or > sometimes just a self- > addressed stamped > envelope. They wel- > come letters of com- > ment for true two-way > communications. A > list of this year's Hugo > nominees includes > several fanzines for > you to try, or visit the > fanzine lounge. > Of course the Inter- > net offers tons of op- > tions. The Usenet bul- > letin board service > offers rec.arts.sf.writ- > ten as well as rec.arts. > sf.fandom. Blogs and > Live Journals are also > a modern way to keep > in touch. Many SF > writers, editors, and artists have their > own blogs and/or websites available > for a visit year-round. Major publish- > ers like Baen and Tor offer extensive > online communities. > If no club exists near you, consider > starting one. Many bookstores pro- > vide space for book groups. Local > libraries and colleges also may make > space available. You can contact ex- > isting clubs for help getting started. > Finally, the week of Worldcon > takes considerable preparation. > Please consider becoming involved. > Contact Anticipation at their table or > online. (Copyright 2008 Denvention 3) which seems pretty readable (and well-written) to me. Your mileage will almost certainly vary; mine sure does. Good luck staying in the text lane, Keith. Dan Hoey haoyuep@aol.com --- Article 978912 of rec.arts.sf.fandom: From: Dan Hoey Newsgroups: rec.arts.sf.fandom Subject: Re: This newsgroup was touted in a Denvention newsletter Date: Wed, 13 Aug 2008 23:59:18 -0400 Keith F. Lynch wrote: > Thanks, but I find it much easier to just read the surreal-looking > paragraphs and get the gist of what's being said. How could I have > gotten the correct column numbers, anyway? Repeated trial and error? > I'm not particular troubled by long lines per se, as I can easily > rewrap them in Emacs. [...] In Emacs, the easiest thing to do is turn off line wrapping, and navigate your viewport around the output of "pdftotext -layout" to find what you want. It's also a good way to find the column numbers, but you can more easily use Emacs's rectangle commands to extract what you want to quote. Emacs can also handle the replacement of UTF-8 characters, but AKICIF knows why you can't search and destroy the "\324"s. You can, however, replace "\324-" with "-", etc. Dan Hoey haoyuep@aol.com --- Article 978973 of rec.arts.sf.fandom: From: Dan Hoey Newsgroups: rec.arts.sf.fandom Subject: Re: This newsgroup was touted in a Denvention newsletter Date: Fri, 15 Aug 2008 12:09:22 -0400 Organization: Naval Research Laboratory, Washington, DC Keith F. Lynch wrote: > Dan Hoey wrote: >> Emacs can also handle the replacement of UTF-8 characters, but AKICIF >> knows why you can't search and destroy the "\324"s. You can, however, >> replace "\324-" with "-", etc. > You certainly can search and destroy a \324 character (a lowercase a > with a hat over it) or anything else in Emacs. What makes you think > otherwise? Like I said, your mileage will almost certainly vary. In Emacs running in a shell window in a MacOSX Terminal.app session, the '\324's are different from the 'a circumflex'es, and you can't query-replace the former. Like I said, AKICIF knows why, but I haven't heard back. This is probably the penalty for gafiation. It would probably be better to use "mimencode -u" rather than "cat -v" anyway. I didn't, because MacOSX doesn't provide the latter (just as it doesn't provide pdftotext), but I may be driven to get it. This is probably a good opportunity for me to correct a statement I made upthread, that it was "unfortunate" that Denvention's PDFs use Unicode for their spaces, hyphens, and quotes. On further thought, I guess it's just unfortunate that pdftotext deletes them instead of turning them into ASCII equivalents. That is a large part of what makes the output so surreal when you don't use the "-layout" and "-enc UTF-8" switches, with all the solecisms like "ad vertising" and "twoway" and "this years Hugo" that make it look careless and unedited. If you really want to read PDFs, you might try "pdftotext -layout" on File 770 or the bus schedules and see if the columns come out right, so you can read them in Emacs. If you'd rather just admire the surreality and kvetch about the difficulty of PDFs, you'll have to count me out. I'm working on the antithetical problem of dealing with them more clearly and easily in a text-only environment. Not that I confine myself to a text-only environment, I just find it more comfortable for textual work. Dan --- Newsgroups: rec.puzzles From: Dan Hoey Date: Fri, 15 Aug 2008 17:43:43 -0400 Subject: Re: A puzzling bedside story hagen wrote: > Once upon a time there was this curious duck... He ducked under the lake? Dan Hoey haoyuep at aol.com --- From Hoey@AIC.NRL.Navy.Mil Fri Aug 15 21:28:02 2008 Date: Fri, 15 Aug 2008 21:27:55 -0400 From: Dan Hoey To: "Keith F. Lynch" Subject: Z2 bus schedule Keith, I've put together a pdftotext translation table that can handle the Denvention newsletters, File770, and at least the S1-2-4 and Z2-6 bus schedules. I include the last below, in case you find it worth looking at. Unfortunately, WMATA uses a lot of dingbats in their schedules, so every time I find a new one I have to add it to the table. Let me know if you'd like the configuration files I used to get this out of the PDF schedule. Dan How to use this timetable English-Espa~ol => Use the map to find the stops closest to where you will get on and off the bus. Z2 => Select the schedule (Weekday, Colesville-Ashton Line Saturday, Sunday) for when you will travel. Along the top of the schedule, Z6 find the stop at or nearest the point Calverton-Westfarm Line where you will get on the bus. Fol- low that column down to the time you want to leave. => Use the same method to find the times the bus is scheduled to arrive at the stop where you will get off the bus. => If the bus stop is not listed, use the time shown for the bus stop before it as the time to wait at the stop. => The end-of-the-line or last stop is listed in ALL CAPS on the schedule. Serves these locations- Brinda servicio a estas ubicaciones Como Usar este Horario Olney (Z2) => Use este mapa para localizar las Montgomery General Hospital (Z2) paradas mas cercanas a donde se Ashton (Z2) subira y bajara del autobus. Cloverly (Z2) Blake High School (Z2) => Seleccione el horario (Entre semana, Colesville (Z2) sabado, domingo) de cuando viajara. Burtonsville Crossing Park & Ride Lot (Z6) A lo largo de la parte superior del Old Columbia Pike (Z6) horario, localice la parada o el punto Castle Blvd. (Z6) mas cercano a la parada en la que se Briggs Chaney Park & Ride Lot (Z6) subira al autobus. Siga esa columna Briggs Chaney Woods (Z6) hacia abajo hasta la hora en la que Calverton (Z6) desee salir. Westfarm (Z6) => Utilice el mismo metodo para locali- White Oak zar las horas en que el autobus esta Four Corners programado para llegar a la parada Silver Spring station en donde desea bajarse del autobus. => Si la parada del autobus no esta listada use la hora que se muestra en la parada anterior como la hora de espera en la parada. => El final de la ruta o la ultima parada del autobus aparece en letras MAYUSCULAS en el horario. Schedule 12-30-07 Reprinted 3-18-08 Washington Metropolitan Area Transit Authority A District of Columbia, Maryland and Virginia Transit Partnership Page 1 of 7 ^LPage 2 of 7 ^LZ2 Colesville-Ashton Line Z6 Calverton-Westfarm Line Colesville-Ashton Line Weekday Southbound -- Entre semana con direccion al sur Lock- New New New wood Hamp- Hamp- Hamp- Dr. & Colesville shire New shire Ave. shire Ave. New Rd. & Spartan Norwood Ave. & Hamp- & & Ran- Hamp- Univer- Colesville & Mont- Rd. & Olney- shire Bonifant dolph shire sity Rd. SILVER Buehler gomery Doctor Sandy Ave. & Blake Rd. Rd. Ave. Blvd. & SPRING Route Rds. General Bird Spring Rd. Cloverly High (Coles- (Coles- (White (Four Spring Number (Olney) Hospital Rd. (Ashton) St. School ville) ville) Oak) Corners) St. AM Service -- Servicio matutino Z2 - - - - - - - 5:32 5:37 5:44 5:49 5:52 Z2 - - - - - - - 6:02 6:07 6:14 6:19 6:22 Z2 5:57 6:02 6:09 6:14 6:22 - - 6:29 6:35 6:42 6:47 6:52 Z2 6:30 6:35 6:42 6:47 6:55 - - 7:02 7:08 7:15 7:20 7:25 Z2 6:57 7:02 7:08 7:13 7:21 - - 7:28 7:37 7:44 7:50 7:55 Z2 7:23 7:28 7:34 7:39 7:47 - - 7:54 8:03 8:10 8:16 8:21 Z2 7:46 7:51 7:58 8:02 8:10 - - 8:17 8:25 8:31 8:37 8:42 Z2 8:08 8:13 8:20 8:24 8:32 - - 8:39 8:47 8:53 8:59 9:04 Z2 8:36 8:41 8:48 8:52 9:00 - - 9:07 9:15 9:21 9:27 9:32 Z2 - - - - - - 9:38 9:42 9:48 9:53 9:58 10:02 Z2 - - - - - - 10:08 10:12 10:18 10:23 10:28 10:32 Z2 - - - - - - 10:38 10:42 10:48 10:53 10:58 11:02 Z2 - - - - - - 11:08 11:12 11:18 11:23 11:28 11:32 Z2 - - - - - - 11:38 11:42 11:48 11:53 11:58 12:02 PM Service -- Servicio vespertino Z2 - - - - - - 12:08 12:12 12:18 12:23 12:28 12:32 Z2 - - - - - - 12:38 12:42 12:48 12:53 12:58 1:02 Z2 - - - - - - 1:08 1:12 1:18 1:23 1:28 1:32 Z2 - - - - - - 1:38 1:42 1:48 1:53 1:58 2:02 Z2 - - - - - - 2:08 2:12 2:18 2:23 2:28 2:32 Z2 2:07 2:12 2:18 2:23 2:32 - - 2:37 2:44 2:50 2:56 3:02 Z2 2:37 2:42 2:48 2:53 3:02 - - 3:07 3:14 3:20 3:26 3:32 Z2 3:07 3:12 3:18 3:23 3:32 - - 3:37 3:44 3:50 3:56 4:02 Z2 3:29 3:34 3:40 3:45 3:54 * 4:00 - 4:07 4:14 4:20 4:26 4:32 Z2 3:59 4:04 4:10 4:15 4:24 * 4:30 - 4:37 4:44 4:50 4:56 5:02 Z2 4:26 4:31 4:37 4:42 4:51 * 4:57 - 5:04 5:11 5:17 5:23 5:29 Z2 5:07 5:12 5:18 5:23 5:32 - - 5:37 5:44 5:50 5:56 6:02 Z2 5:43 5:48 5:54 5:59 6:08 - - 6:13 6:19 6:25 6:32 6:37 Z2 6:08 6:13 6:19 6:24 6:33 - - 6:38 6:44 6:50 6:57 7:02 * -- Trip diverts to serve Blake High School on days when school is open. When school is closed, trip leaves Olney and serves stops north of New Hampshire Avenue & Briggs Chaney Road 6 minutes later than times shown. No change to times at stops between New Hampshire & Briggs Chaney and Silver Spring station. Page 3 of 7 ^LZ2 Colesville-Ashton Line Z6 Calverton-Westfarm Line Colesville-Ashton Line Weekday Northbound -- Entre semana con direccion al norte Lock- wood New New Olney- Colesville Dr. & Hamp- Hamp- Sandy Rd. & New shire Ave. shire Ave. New Spring Rd. Norwood Colesville Univer- Hamp- & & Hamp- & New Rd. Spartan Silver Rd. sity shire E.Randolph Bonifant shire Hamp- & Mont- & Spring & Blvd. Ave. Rd. Rd. Blake Ave. & shire Doctor gomery Buehler Route Spring (Four (White (Coles- (COLES- High Cloverly Ave. Bird General Rds. Number St. Corners) Oak) ville) VILLE) School St. (Ashton) Rd. Hospital (OLNEY) AM Service -- Servicio matutino Z2 5:59 6:01 6:07 6:11 6:19 - - 6:25 6:32 6:35 6:40 6:44 Z2 6:18 6:20 6:26 6:30 6:38 - - 6:44 6:51 6:54 6:59 7:03 Z2 6:43 6:45 6:52 6:57 7:06 - - 7:13 7:21 7:26 7:31 7:35 Z2 7:01 7:03 7:10 7:15 7:24 - - 7:31 7:39 7:44 7:49 7:53 Z2 7:31 7:34 7:41 7:47 7:55 - - 8:02 8:10 8:14 8:20 8:24 Z2 [[]] 8:01 8:04 8:11 8:17 8:25 - - - - - - - Z2 [[]] 8:31 8:34 8:41 8:47 8:55 - - - - - - - Z2 9:01 9:04 9:11 9:17 9:25 9:29 - - - - - - Z2 9:31 9:35 9:40 9:45 9:53 9:57 - - - - - - Z2 10:01 10:05 10:10 10:15 10:23 10:27 - - - - - - Z2 10:31 10:35 10:40 10:45 10:53 10:57 - - - - - - Z2 11:01 11:05 11:10 11:15 11:23 11:27 - - - - - - Z2 11:31 11:35 11:40 11:45 11:53 11:57 - - - - - - PM Service -- Servicio vespertino Z2 12:01 12:05 12:10 12:15 12:23 12:27 - - - - - - Z2 12:31 12:35 12:40 12:45 12:53 12:57 - - - - - - Z2 1:01 1:05 1:10 1:15 1:23 1:27 - - - - - - Z2 1:31 1:35 1:40 1:45 1:53 1:57 - - - - - - Z2 2:01 2:05 2:10 2:15 2:23 - - 2:29 2:36 2:40 2:46 2:50 Z2 2:31 2:35 2:40 2:45 2:53 - - 2:59 3:06 3:10 3:16 3:20 Z2 3:01 3:04 3:10 3:16 3:24 - - 3:31 3:38 3:42 3:47 3:50 Z2 3:31 3:34 3:40 3:46 3:54 - - 4:01 4:08 4:12 4:17 4:20 Z2 4:01 4:04 4:10 4:16 4:24 - [V] 4:32 4:40 4:48 4:52 4:57 5:00 Z2 4:31 4:34 4:40 4:46 4:54 - [V] 5:02 5:10 5:18 5:22 5:27 5:30 Z2 4:57 5:00 5:06 5:11 5:22 - - 5:29 5:37 5:41 5:46 5:49 Z2 5:25 5:28 5:34 5:39 5:50 - - 5:57 6:05 6:09 6:14 6:17 Z2 5:52 5:56 6:02 6:08 6:17 - - 6:24 6:32 6:36 6:40 6:44 Z2 6:19 6:23 6:29 6:35 6:44 - - 6:51 6:59 7:03 7:07 7:11 Z2 6:47 6:51 6:57 7:03 7:12 - - 7:19 7:27 7:31 7:35 7:39 Z2 7:14 7:18 7:24 7:30 7:39 - - 7:46 7:54 7:58 8:02 8:06 [V] -- Trip diverts to serve Blake High School on days when school is open. When school is closed, trip serves stops between New Hampshire Avenue & Briggs Chaney Road and Olney 6 minutes earlier than times shown. No change to times at stops between Silver Spring station and New Hampshire & Briggs Chaney. [[]] -- Passengers may remain on bus to New Hampshire Ave. & Bonifant Rd. Page 4 of 7 ^LZ2 Colesville-Ashton Line Z6 Calverton-Westfarm Line Colesville-Ashton Line Saturday Southbound -- Saturday Northbound -- En sabados con direccion al sur En sabados con direccion al norte Lock- Lock- New wood wood Hamp- New New Dr. & Colesville Colesville Dr. & New shire Hamp- Hamp- New Rd. & Rd. & New Hamp- Ave. shire shire Ave. Hamp- Univer- Colesville Colesville Univer- Hamp- shire Ave. & Ave. & & shire sity Rd. SILVER Silver Rd. sity shire & Bonifant Bonifant Randolph Ave. Blvd. & SPRING Spring & Blvd. Ave. E.Randolph Rd. Route Rd. Rd. (White (Four Spring Route Spring (Four (White Rd. (COLES- Number (Colesville)(Colesville) Oak) Corners) St. Number St. Corners) Oak) (Colesville) VILLE) AM Service -- Servicio matutino AM Service -- Servicio matutino Z2 8:58 9:01 9:07 9:12 9:17 9:22 Z2 8:31 8:33 8:38 8:42 8:50 8:55 Z2 9:28 9:31 9:37 9:42 9:47 9:52 Z2 9:01 9:03 9:08 9:12 9:20 9:25 Z2 9:58 10:01 10:07 10:12 10:17 10:22 Z2 9:31 9:33 9:38 9:42 9:50 9:55 Z2 10:28 10:31 10:37 10:42 10:47 10:52 Z2 10:01 10:03 10:08 10:12 10:20 10:25 Z2 10:58 11:01 11:07 11:12 11:17 11:22 Z2 10:31 10:33 10:38 10:42 10:50 10:55 Z2 11:28 11:31 11:37 11:42 11:47 11:52 Z2 11:01 11:03 11:08 11:12 11:20 11:25 Z2 11:58 12:01 12:07 12:12 12:17 12:22 Z2 11:31 11:33 11:38 11:42 11:50 11:55 PM Service -- Servicio vespertino PM Service -- Servicio vespertino Z2 12:28 12:31 12:37 12:42 12:47 12:52 Z2 12:01 12:03 12:08 12:12 12:20 12:25 Z2 12:58 1:01 1:07 1:12 1:17 1:22 Z2 12:31 12:33 12:38 12:42 12:50 12:55 Z2 1:28 1:31 1:37 1:42 1:47 1:52 Z2 1:01 1:03 1:08 1:12 1:20 1:25 Z2 1:58 2:01 2:07 2:12 2:17 2:22 Z2 1:31 1:33 1:38 1:42 1:50 1:55 Z2 2:28 2:31 2:37 2:42 2:47 2:52 Z2 2:01 2:03 2:08 2:12 2:20 2:25 Z2 2:58 3:01 3:07 3:12 3:17 3:22 Z2 2:31 2:33 2:38 2:42 2:50 2:55 Z2 3:28 3:31 3:37 3:42 3:47 3:52 Z2 3:01 3:03 3:08 3:12 3:20 3:25 Z2 3:58 4:01 4:07 4:12 4:17 4:22 Z2 3:31 3:33 3:38 3:42 3:50 3:55 Z2 4:28 4:31 4:37 4:42 4:47 4:52 Z2 4:01 4:03 4:08 4:12 4:20 4:25 Z2 4:58 5:01 5:07 5:12 5:17 5:22 Z2 4:31 4:33 4:38 4:42 4:50 4:55 Z2 5:28 5:31 5:37 5:42 5:47 5:52 Z2 5:01 5:03 5:08 5:12 5:20 5:25 Z2 5:58 6:01 6:07 6:12 6:17 6:22 Z2 5:31 5:33 5:38 5:42 5:50 5:55 Z2 6:28 6:31 6:37 6:42 6:47 6:52 Z2 6:01 6:03 6:08 6:12 6:20 6:25 Z2 6:31 6:33 6:38 6:42 6:50 6:55 Saturday service is operated by RIDE-ON. Information 240/777-7433 (Touchtone), 240/777-5869 (TDD) 240/777-5871 (Rotary), 1/800/732-3327 (http://www.montgomerycountymd.gov) Website No Sunday Service- No hay servicio los domingos Page 5 of 7 ^LZ2 Colesville-Ashton Line Z6 Calverton-Westfarm Line Calverton-Westfarm Line Weekday Southbound -- Entre semana con direccion al sur Stewart Dixon & Burtonsville La. Lockwood Colesville Wayne Crossing Old Colum- & Dr. & Rd. & Colesville Aves Park bia Pike & Briggs Broadbirch Old New University Rd. (SILVER & Green- #14000 Chaney Dr. & Plum Columbia Hampshire Blvd. & SPRING) Route Ride castle Castle Park & Ride Orchard Dr. Pike Ave. (Four Spring Number Lot Rd. Blvd. Lot (Westfarm) (White Oak) (White Oak) Corners) St. AM Service -- Servicio matutino Z6 - - 5:03 5:07 5:14 5:21 5:25 5:32 5:37 5:42 Z6 - - 5:33 5:37 5:44 5:51 5:55 6:02 6:07 6:12 Z6 - - 5:55 5:59 6:06 6:16 6:21 6:28 6:33 6:38 Z6 - - 6:23 6:27 6:34 6:44 6:49 6:56 7:01 7:06 Z6 - - 6:44 6:49 6:56 7:06 7:11 7:20 7:26 7:31 Z6 - - 7:02 7:07 7:14 7:24 7:29 7:38 7:44 7:49 Z6 - - 7:20 7:25 7:32 7:42 7:47 7:56 8:02 8:07 Z6 - - 7:42 7:47 7:54 8:04 8:09 8:18 8:24 8:29 Z6 - - 8:13 8:18 8:25 8:33 8:39 8:45 8:51 8:56 Z6 - - 8:39 8:44 8:51 8:59 9:05 9:11 9:17 9:22 Z6 8:56 9:02 9:10 9:15 9:24 9:32 9:37 9:42 9:47 9:52 Z6 9:26 9:32 9:40 9:45 9:54 10:02 10:07 10:12 10:17 10:22 Z6 9:56 10:02 10:10 10:15 10:24 10:32 10:37 10:42 10:47 10:52 Z6 10:26 10:32 10:40 10:45 10:54 11:02 11:07 11:12 11:17 11:22 Z6 10:56 11:02 11:10 11:15 11:24 11:32 11:37 11:42 11:47 11:52 Z6 11:26 11:32 11:40 11:45 11:54 12:02 12:07 12:12 12:17 12:22 Z6 11:56 12:02 12:10 12:15 12:24 12:32 12:37 12:42 12:47 12:52 PM Service -- Servicio vespertino Z6 12:26 12:32 12:40 12:45 12:54 1:02 1:07 1:12 1:17 1:22 Z6 12:56 1:02 1:10 1:15 1:24 1:32 1:37 1:42 1:47 1:52 Z6 1:26 1:32 1:40 1:45 1:54 2:02 2:07 2:12 2:17 2:22 Z6 1:56 2:02 2:10 2:15 2:24 2:32 2:37 2:42 2:47 2:52 Z6 2:28 2:33 2:41 2:46 2:54 3:01 3:05 3:11 3:17 3:22 Z6 2:58 3:03 3:11 3:16 3:24 3:31 3:35 3:41 3:47 3:52 Z6 3:28 3:33 3:41 3:46 3:54 4:01 4:05 4:11 4:17 4:22 Z6 3:58 4:03 4:11 4:16 4:24 4:31 4:35 4:41 4:47 4:52 Z6 - - 4:41 4:46 4:54 5:01 5:05 5:11 5:17 5:22 Z6 - - 5:11 5:16 5:24 5:31 5:35 5:41 5:47 5:52 Z6 - - 5:41 5:46 5:54 6:01 6:05 6:11 6:17 6:22 Z6 - - 6:15 6:20 6:26 6:33 6:37 6:43 6:50 6:55 Z6 - - 6:47 6:52 6:58 7:05 7:09 7:15 7:22 7:27 Z6 - - 7:22 7:26 7:32 7:38 7:42 7:47 7:52 7:57 Z6 - - 7:52 7:56 8:02 8:08 8:12 8:17 8:22 8:27 Z6 8:11 8:15 8:22 8:26 8:32 8:38 8:42 8:47 8:52 8:57 Z6 8:41 8:45 8:52 8:56 9:02 9:08 9:12 9:17 9:22 9:27 Z6 9:08 9:12 9:18 9:22 9:28 9:34 9:38 9:43 9:47 9:52 Z6 9:38 9:42 9:48 9:52 9:58 10:04 10:08 10:13 10:17 10:22 Page 6 of 7 ^LZ2 Colesville-Ashton Line Z6 Calverton-Westfarm Line Calverton-Westfarm Line Weekday Northbound -- Entre semana con direccion al norte Lockwood Stewart BURTONS- Dixon & Colesville Dr. La. VILLE Wayne Aves. Colesville Rd. & & & Old Colum- Crossing (Silver Rd. University New Old Broadbirch Briggs bia Pike & Park Spring) & Blvd. Hampshire Columbia Dr. & Plum Chaney #14000 Green- & Route Spring (Four Ave. Pike Orchard Dr. Park & CASTLE castle Ride Number St. Corners) (White Oak) (White Oak) (Westfarm) Ride Lot BLVD. Rd. Lot AM Service -- Servicio matutino Z6/ 5:33 5:35 5:41 5:45 5:50 5:54 6:01 6:08 - - Z6/ 6:08 6:10 6:16 6:20 6:25 6:29 6:36 6:43 - - Z6/ 6:36 6:38 6:45 6:50 6:55 7:00 7:08 7:12 - - Z6/ 7:06 7:08 7:15 7:20 7:25 7:30 7:38 7:42 - - Z6 7:36 7:39 7:46 7:52 7:58 8:03 8:11 8:20 8:24 8:33 Z6 8:11 8:14 8:21 8:27 8:33 8:38 8:46 8:55 8:59 9:08 Z6 8:41 8:44 8:51 8:57 9:03 9:08 9:16 9:25 9:29 9:38 Z6 9:11 9:14 9:21 9:27 9:33 9:38 9:46 9:55 9:59 10:08 Z6 9:41 9:45 9:50 9:55 10:02 10:07 10:15 10:22 10:25 10:33 Z6 10:11 10:15 10:20 10:25 10:32 10:37 10:45 10:52 10:55 11:03 Z6 10:41 10:45 10:50 10:55 11:02 11:07 11:15 11:22 11:25 11:33 Z6 11:11 11:15 11:20 11:25 11:32 11:37 11:45 11:52 11:55 12:03 Z6 11:41 11:45 11:50 11:55 12:02 12:07 12:15 12:22 12:25 12:33 PM Service -- Servicio vespertino Z6 12:11 12:15 12:20 12:25 12:32 12:37 12:45 12:52 12:55 1:03 Z6 12:41 12:45 12:50 12:55 1:02 1:07 1:15 1:22 1:25 1:33 Z6 1:11 1:15 1:20 1:25 1:32 1:37 1:45 1:52 1:55 2:03 Z6 1:41 1:45 1:50 1:55 2:02 2:07 2:15 2:22 2:25 2:33 Z6 2:11 2:15 2:20 2:25 2:32 2:37 2:45 2:52 2:55 3:03 Z6 2:41 2:45 2:50 2:55 3:02 3:07 3:15 3:22 3:25 3:33 Z6/ 3:11 3:14 3:20 3:26 3:33 3:38 3:46 3:49 - - Z6/ 3:41 3:44 3:50 3:56 4:03 4:08 4:16 4:19 - - Z6/ 4:11 4:14 4:20 4:26 4:33 4:38 4:46 4:49 - - Z6/ 4:40 4:43 4:49 4:54 5:01 5:07 5:16 5:19 - - Z6/ 5:11 5:14 5:20 5:25 5:32 5:38 5:47 5:50 - - Z6/ 5:39 5:42 5:48 5:53 6:00 6:06 6:15 6:18 - - Z6/ 6:05 6:08 6:14 6:20 6:27 6:33 6:42 6:45 - - Z6/ 6:33 6:36 6:42 6:48 6:55 7:01 7:10 7:13 - - Z6 6:55 6:59 7:05 7:11 7:18 7:24 7:33 7:40 7:43 7:50 Z6 7:26 7:30 7:36 7:42 7:49 7:55 8:04 8:11 8:14 8:21 Z6 7:50 7:53 7:58 8:03 8:09 8:14 8:20 8:27 8:30 8:37 Z6 8:17 8:20 8:25 8:30 8:36 8:41 8:47 8:54 8:57 9:04 Z6 8:47 8:50 8:55 9:00 9:06 9:11 9:17 9:24 9:27 9:34 Z6 9:17 9:20 9:25 9:30 9:36 9:41 9:47 9:54 9:57 10:04 Z6/ 9:47 9:50 9:55 10:00 10:06 10:11 10:17 10:24 - - Page 7 of 7 ^L --- Article 978982 of rec.arts.sf.fandom: From: Dan Hoey Newsgroups: rec.arts.sf.fandom Subject: Re: This newsgroup was touted in a Denvention newsletter Date: Fri, 15 Aug 2008 21:56:18 -0400 Organization: Naval Research Laboratory, Washington, DC Keith F. Lynch wrote: > Dan Hoey wrote: >> Keith F. Lynch wrote: >>> You certainly can search and destroy a \324 character (a lowercase >>> a with a hat over it) or anything else in Emacs. What makes you >>> think otherwise? >> Like I said, your mileage will almost certainly vary. In Emacs >> running in a shell window in a MacOSX Terminal.app session, the >> '\324's are different from the 'a circumflex'es, and you can't >> query-replace the former. > What version of Emacs is that? Here on Panix I'm using: > GNU Emacs 21.3.1 (i386-unknown-netbsdelf1.5.2) of 2003-04-08 This is the GNU Emacs 22.1.1 (mac-apple-darwin, Carbon Version 1.6.0) of 2008-05-28 that came with MacOSX 10.5. > What does it say when you put the cursor on the character(s) and > press control-X then equal-sign? I get: > Char: â (0342, 226, 0xe2, file 0xE2) It says: Char: \303 (195, #o303, #xc3, part of display "\303\242"->"\342") This is with the cursor on the \ in "Jan van\342M-^@M-^Yt Ent" in the output of pdftotext -enc UTF=8 -layout newsletter9.pdf -|cat -v The "\342" is displayed in red, and the cursor won't go under any other part of the \342. Dan --- Newsgroups: rec.puzzles From: Dan Hoey Date: Fri, 15 Aug 2008 22:06:33 -0400 Subject: Re: A puzzling bedside story Bill wrote: > I am still curious about the meaning of "badges"? Probably a fetid mustelid, though it could be a mephitid of genus Mydaus. Save the Asiatic stink badges! Dan Hoey haoyuep@aol.com --- Newsgroups: rec.arts.sf.fandom From: Dan Hoey Date: Fri, 15 Aug 2008 22:18:17 -0400 Subject: Re: This newsgroup was touted in a Denvention newsletter Keith F. Lynch wrote: > Dan Hoey wrote: [...] >> pdftotext -enc UTF=8 -layout newsletter9.pdf -|cat -v [...] > When I tried the above pdftotext command, exactly as you had it, I got: > Error: Couldn't find unicodeMap file for the 'UTF=8' encoding > Error: Couldn't get text encoding Sorry, that's a typo. the command is pdftotext -enc UTF-8 -layout newsletter9.pdf -|cat -v Dan --- Newsgroups: rec.puzzles From: Dan Hoey Date: Mon, 18 Aug 2008 05:49:39 -0400 Subject: Re: Jack Doing Whole Job heich1 wrote: > Here is a problem by J.A.H. Hunter. > Jack and Jill had been given a huge basket of peanuts to shell. Jack > started alone, and worked steadily for two-thirds of the time which > Jill would have taken to do the whole job by herself; after > that, however, Jack stopped work. Jill now took her lazy brother's > place, and finished the job alone. If the two children had worked > together. from start to finish. the whole job would have taken 12 > minutes less time, and Jack would have done exactly the same amount > of shelling. > How long would it have taken Jack to do the whole job himself? Here is a solution that is perhaps more direct than Ilan's but a little more explanatory than willem's. Jack does the same amount of shelling in both cases, so the remainder left to Jill is the same in both cases. So Jack's shift is the same length of time in both cases, and so is Jill's. But when working together, their shifts are the same, so all shifts are equal. The two shifts worked consecutively are twelve minutes longer than the shift worked together, so a shift is twelve minutes long. We are given that Jill does two-thirds of the work in a shift, so Jack does one-third in a shift. Therefore Jack working alone would take three shifts, or 36 minutes. Suppose the fourth sentence had read, "Then the two children were given another basket of the same size which they shelled together, completing the two baskets in a total of 36 minutes, yet Jack did exactly the same amount of shelling on both both baskets." Would the puzzle have been at all more difficult or easier? Are puzzles that happen to mention their solution more or less interesting? As to the question of why Jack is called "lazy" if he does the same amount of work in both cases, I earlier misremembered this puzzle as asking for the amount of time Jill would take to do the whole job, in which case Jack would have been shiftless. That not being the case, we may observe that compared to Jill's speedy peanut-shelling, Jack's work is half-fast. But the real reason is the same reason that Jack is named "Jack" and is Jill's brother and that they are shelling peanuts instead of Tbilisi. Merely corroborative detail, intended to give artistic verisimilitude to an otherwise bald and unconvincing narrative. [Gur Zvxnqb, npg 2] Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Mon, 18 Aug 2008 12:53:23 -0400 Subject: Re: Partition a square into rectangles Dan Hoey wrote: > Risto Lankinen wrote: >> no.glam wrote: >>> Hello, >>> Prove/disprove that a 3000X3000 square can be partitioned into 5X9 >>> rectangles (the rectangles can be mixed, either 5X9 or 7X9). > and Risto disproves it. David J.C. MacKay proves a startling generalization at http://www.inference.phy.cam.ac.uk/mackay/rectangles/ using a graph-theoretical argument. The result can be stated as: Let G be any additive subgroup of the real numbers. If a rectangle can be partitioned into a finite number of rectangular tiles, where each tile has at least one side-length in G, then the whole rectangle has at least one side in G. The result is not a much of a generalization if G is a subgroup of the rationals, since with a finite number of rectangles we can find a common denominator and convert it to the integer problem. But MacKay's argument works if if G is the algebraic numbers--quite a different situation. Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Mon, 18 Aug 2008 20:39:24 -0400 Subject: Re: Hound Catches Rabbit Steve Fry wrote: > Peter Heichelheim writes: >> ... I am using this method to find out how to solve the >> problem. >> Peter Heichelheim > I don't believe you. > Many of usl told you the correct answer for the Hound catching > the Rabbit, and that Hunter's answer was wrong. You just refuse > to believe it. Hunter composed the puzzle, so obviously he intended his answer to be correct. Either he intends that the hound can catch the rabbit by shortening its last leap, or he intends that the hound can deviate from a straight line in chasing the rabbit, so as to catch the rabbit with leaps of exactly nine feet. In either case, the hound catches the rabbit fifty feet from the burrow, as Hunter intended. The problem is not merely that Peter refuses to believe your answer, it is that there are better answers than yours to believe, and they have the advantage of agreeing with the composer's intended answer to the puzzle. Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Tue, 19 Aug 2008 19:14:07 -0400 Subject: Re: Hound Catches Rabbit Steve Fry wrote: > "Dan Hoey" wrote: >> The problem is not merely that Peter refuses to believe your >> answer, it is that there are better answers than yours to believe, >> and they have the advantage of agreeing with the composer's >> intended answer to the puzzle. > You are wrong. You are not convincing. Alexander Pope wrote, in "An Essay on Criticism": 'Tis with our Judgments as our Watches, none Go just alike, yet each believes his own. Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Thu, 21 Aug 2008 19:33:58 -0400 Subject: Re: Enigma 1501 - Five sets at Wimbledon James Dow Allen wrote: > Remark that in a fifth set going to extra points > (score of 8-6, or 9-7, etc.) the server wins an odd > number of games. To rule out 4, 6, or 8 in the 6-4 set, you need the slightly stronger result that if the point spread is 2, then the server wins an odd number of games. Proof: If the point spread is 2, the score is N+1 to N-1, and the number of games is 2N, with each player serving N times. If the winner wins K of his serves, then he must win N+1-K of the loser's serves. The loser then wins N-(N+1-K) = K-1 of his serves, and the total number of server wins is 2K-1, QED. This can be generalized to show that in any sport consisting of games with alternating servers, the server wins an odd number of games if the point spread is 2 mod 4, and an even number of games if the point spread is 0 mod 4. The parity for odd point spreads depends on whether the first server wins. Dan Hoey haoyuep at aol.com --- Article 979890 of rec.arts.sf.fandom: From: Dan Hoey Newsgroups: rec.arts.sf.fandom Subject: Re: Teeny things to put in spaceships when you're bored... Date: Sun, 24 Aug 2008 10:35:40 -0400 Organization: Aioe.org NNTP Server Marcus L. Rowland wrote: > Dare I ask why there are air conditioner boxes with fans etc. on a > structure in space? Presumably one of those "joke" things you described earlier, but thanks for pointing it out--I had missed it. Way cool. I wonder if having some laundry on a line out to dry might work. Oddly enough, that may be the easiest way to dry laundry in space, but you might not want to waste the water. Dan Hoey haoyuep at aol.com --- Article 979893 of rec.arts.sf.fandom: From: Dan Hoey Newsgroups: rec.arts.sf.fandom Subject: Re: Teeny things to put in spaceships when you're bored... Date: Sun, 24 Aug 2008 10:58:41 -0400 Organization: Aioe.org NNTP Server Karl Johanson wrote: > I added some spilled milk and a sticky note to the window in this cover > image. It's in space, but not really a space ship per se, but rather a > cable vehicle. (The second monitor from the right originally had the > UPC code.) Glad you had it there, because I missed the cleanup scene at first. Does a squeegee wiper really make sense in 0G? Was the milk spilled, or was it thrown by Marcia? The asteroid gravity lab is neat. I was going to suggest that Asteroid 46610 would be a more interesting number, but it seems someone got there first ( http://en.wikipedia.org/wiki/46610_Besixdouze ). Dan Hoey haoyuep at aol.com --- Article 980278 of rec.arts.sf.fandom: From: Dan Hoey Newsgroups: rec.arts.sf.fandom Subject: Re: Teeny things to put in spaceships when you're bored... Date: Tue, 26 Aug 2008 12:42:42 -0400 Organization: Naval Research Laboratory, Washington, DC Marcus L. Rowland wrote: > William December Starr writes >> Marcus L. Rowland said: >>> http://ffutures.livejournal.com/489602.html#cutid1 >> Are all those orange dots with black curlicues on them scattered >> throughout the interior of the ship fire extinguishers? >> -- wds > Yes - I know you can vent the ship to space, but sometimes a squirt of > foam will do the job faster and easier... Which reminds me--spittoons! Every space cowboy needs a spittoon now and then. Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Wed, 27 Aug 2008 16:38:46 -0400 Subject: Re: (again ?) Skolem-like puzzle sequence Eric A. wrote: > We are looking for an 18-digit integer (like > 946131483695200285) where we have "d" digits > between two d's (here: one digit between two > 1's, zero digit between the two 0's, nine > digits between the two 9's, etc.) > Those 18 digits form thus 9 pairs of digits -- > the 9 pairs being different one from another. > Now cut this integer into chunks so to make > a finite monotonically increasing sequence > (like this one, for instance, among others): > 9,46,131,483,695,200285 > .... and we sum the terms: > 9+46+131+483+695+200285 = 201649 > Question: > Find the integer which, properly chunked, will > give the smallest possible sum. > --- > Example #2: > 849362432856900151 could give: > 8+49+362+432+856+900151 = 901858 > .... but if we move the first digit "8" to the > end (which is still sound), 849362432856900151 > becomes 493624328569001518 > .... which, properly chunked, will give: > 4+9+362+432+856+900+1518 = 4080 > This is my personal best so far. s p o i l e r s p a c e s p o i l e r s p a c e s p o i l e r s p a c e s p o i l e r s p a c e s p o i l e r s p a c e s p o i l e r s p a c e The best is 456784151637283200, which can be chunked into 4+5+6+7+8+41+51+63+72+83+200 = 540. > P.S. > a 20-digit integer such as described is impossible > for parity reasons. I'll be delighted to receive > the full b-file list of all such 18-digit integers > (if someone would be kind enough to compute it). Done. There are 10216 such 18-digit strings, or 9060 such integers if we prohibit leading zeroes. List or Java program available upon e-mailed request. Dan Hoey haoyuep at aol.com --- Newsgroups: sci.math, rec.puzzles From: Dan Hoey Date: Wed, 27 Aug 2008 18:31:24 -0400 Subject: Re: Range of a two-variable polynomial tchow@lsa.umich.edu wrote: > The crux of the matter was this: Let f be a polynomial function from > R^2 to R, i.e., a polynomial in two real variables x and y with real > coefficients. Is it possible for the range of f to be the open interval > (0, infinity)? Perhaps some traction can be attained by considering the polynomials f_theta(r) = f(r cos theta, r sin theta). Each is a polynomial in r that attains its minimum at solutions of f_theta'(r)=0, and the degrees of the f_theta are bounded by deg f. Perhaps the set {(r,theta) : f_theta'(r)=0} is compact, or at least well-enough behaved that we can show that f attains its minimum there. But I have no proof. Dan Hoey haoyuep at aol.com --- Newsgroups: sci.math.research From: Dan Hoey Date: Tue, 2 Sep 2008 12:00:01 +0000 (UTC) Subject: Re: This Week's Finds in Mathematical Physics (Week 269) On finding the minimal-length set of boundaries partitioning R^2 into cells of equal area, John Baez wrote: > So what Hales had to do is rule out cells with curved edges. This > is harder than you might think. In fact, for clusters of finitely > many cells, the optimal shapes can be curved, even near the middle! > You can see pictures in the review by Frank Morgan above, or here: I looked at a smaller problem a few decades ago, that of the minimal- length set of boundaries partitioning a square into n cells of equal area. The boundaries were generally circular arcs meeting in in triples. It looked like a model based on cell pressure might simplify the search. I made very little progress on the problem, but I'd be interested in any further work. The problem of partitioning a disc would also be interesting. Dan Hoey haoyuep@aol.com --- Newsgroups: rec.puzzles, sci.math, rec.games.abstract From: Dan Hoey Date: Tue, 02 Sep 2008 17:59:58 -0400 Subject: Re: Line-Segments In Circle Game Leroy Quet wrote: > Here is a game of mine that is NOT based on an n-by-n grid... It is also the first of your games to which I suspect applicability of combinatorial game theory. > The game is for 2 players. > Draw a circle on a piece of paper. (Preferably the circle is drawn > with a compass, but this is not necessary. The circle should be drawn > carefully, however.) > Put a dot at the circle's center. > Players take turns drawing straight line-segments (preferably with a > straight-edge) within the circle as follows: Given the rules, the circle can be replaced by any simple curve of strictly positive curvature. That is, the curve would be the boundary of a convex shape and would include no straight segment. For the center we may use any point of the interior. > *A segment can go from the dot at the circle's center to the edge of > the circle. > *A segment can go from {an intersection of another segment and the > circle} to {another intersection of a segment and the circle}. I advise combining this into a single rule (R1) and making it explicit that these are the only permissible moves. It would make sense to call these two kinds of moves "radii" and "chords". We should probably have a simple name for those points on the circle to which radii have been drawn. For the time being, I'll call them "used points". > *A segment can pass through at most ONE line-segment that has already > been drawn. The fact that this rule (R2) applies only to previously-drawn line- segments makes the possible ending diagrams depend on order of moves. For instance, the move sequence 1. OA, 2. OB, 3. OC, 4. AC(crossing OB), 5. OD(crossing AC) could not be played with moves 4 and 5 interchanged, because AC could not be drawn crossing OB and OD. This suggests a variant ruleset in which (R2) is replaced by (R2') No segment may cross more than one other segment. which would prohibit move 5 above. > *A segment can not coincide with a previously drawn segment. (ie. > Two segments can coincide at at most one point.) > *A segment cannot be drawn from the center to the circle that is 180 > degrees away from the segment last drawn by the other player, if the > last segment was drawn from the center to the circle. (This prevents > a player from simply matching the other player's moves, and so > automatically winning.) Calling these (R3) and (R4), another alternate ruleset would replace (R4) with (R4') No two radii may be collinear. I suspect this ruleset would be more tractable, because it removes the entailment inherent in rule (R4). (Entailment is the restriction of moves depending on the immediately preceding move). It would also simplify (R3) to prohibit drawing a segment that had previously been drawn. While the rules appear to permit a continuum of positions, the description of a position can be simplified. Let A,B,C,... be symbols for the used points, except that we denote by x' a used point that is 180 degrees from a used point x. We also denote by (x') an unused point 180 degrees from a used point x. The actual positions of the points is then irrelevant; only the circular order of the symbols x, x', and (x') matters. In addition, we would record which segments xy (or xy' or x'y') had been drawn, and which arcs of the circle are traversed by two chords, thus prohibiting further radii. In order to enforce rule (R4), it is also be necessary to note the last move if it is a radius. If rule (R4') is used, we would not need to use symbols x' or (x'). Instead, we would only have to record which pair of neighboring used points are separated by more than 180 degrees, if any. The rule for drawing chords is simple: a "neighbor chord" is a chord between two used points that have no intervening used points. A "crossing chord" is a chord between two used points that have exactly one intervening used point. An undrawn neighbor chord may always be drawn. An undrawn crossing chord may be drawn if and only if the intervening point has no crossing chord. No other chords may be drawn. I'm adding rec.games.abstract to this reply, in hopes that someone may be able to attack this with the theory of impartial games. Perhaps entailment is not quite the bugbear I fear, and rule (R4) can be dealt with as well. Dan Hoey haoyuep at aol.com --- Newsgroups: sci.math From: Dan Hoey Date: Wed, 03 Sep 2008 16:52:31 -0400 Subject: Re: Question about polynomial roots tchow@lsa.umich.edu wrote: > Ray Vickson wrote: >> However, I am still wondering how one can show that arctan(4/3) or >> arctan(3/4) are not rational multiples of pi. Is there a theorem about this? > The general theorem uses the concept of an algebraic integer. A complex > number x is an algebraic integer if it satisfies some polynomial equation > of the form x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + ... + a_0 = 0, where > the coefficient of the leading term x^n is 1 and the a_i are integers. > Two basic facts about algebraic integers that you can find proved in > textbooks on abstract algebra are: > 1. The sum of two algebraic integers is an algebraic integer. > 2. The only algebraic integers that are also rational numbers are > the integers. > Any root of x^n - 1 = 0 for integer n is an algebraic integer, from the > definition of an algebraic integer. On the other hand, if (3/5) + (4/5)i > were an algebraic integer, then (3/5) - (4/5)i would also be an algebraic > integer (they would in fact satisfy the same equation with integer > coefficients), and so their sum would be. But their sum is 6/5, which > is a rational number and not an integer. > There is probably a simpler proof if all you care about is the number > (3/5) + (4/5)i, but in my opinion the "right" viewpoint is the above one. Your explanation is probably the best for anyone who has not studied field extensions. Once the reader knows about uniqueness of minimal polynomials, it's more concise to say (or prove, as you did) that every cyclotomic polynomial is monic, and that the minimal polynomial for t=(3+4i)/5 is 5z^2-6z+5 (which you get from (z-t)(z-t*)). Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Thu, 25 Sep 2008 19:33:44 -0400 Subject: Re: Enigma 1505 - Brief N-counter ken wrote: > On Sep 24, 1:20 pm, Chappy wrote: >> i.e. Ken's solution has three pieces that are the same shape. >> My solution has only 3 differently shaped pieces. My program found 203 solutions. My tired eyes found only one in which all six hexominoes are congruent, but I haven't taught the program to back me up on that. Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Fri, 26 Sep 2008 23:58:18 -0400 Subject: Re: Homework Richard Heathfield wrote: > David Burn said: >> The integers from 1 to 12 are randomly divided into three groups of >> four integers apiece. >> This process is repeated three times. >> If after this has happened, it can be said of any integer that it has >> appeared in the same group as any other integer on all three repetitions >> of the process, you will have failed. >> What is the chance that you will fail? [Monte Carlo code snipped] > Output: > Trials: 1000000 > Successes: 237784 > Failures: 762216 > p(fail) = 0.762216 The exact probability is 2825881/3705625, about .762592 . Dan haoyuep@aol.com --- Newsgroups: rec.puzzles, sci.math From: Dan Hoey Date: Tue, 30 Sep 2008 17:42:43 -0400 Subject: Re: Wind Around Solitaire Leroy Quet wrote: > Here is a 1-player game played on an n-by-n grid, where n is odd. (I > suggest an n of about 13 to 21 for beginners.) > The player places the numbers 1,2,3,...n^2 into the grid. > 1 goes in the center square. > Each number (2k+1) must go either below, above, left of, or right of > the square with (2k-1) in it. > Each number (2k) can go anywhere in the grid. > Numbers can only be placed in empty squares. > The player's score is the number of times the path of odd integers > goes completely around the center square clockwise before the player > is unable to place any more numbers in the grid. Why should we count only the odd integers in the path? What happens when one segment of the path passes through the center point? It should probably count as 1/2 of a circle, but is it clockwise or counterclockwise? Perhaps we should count how many times the path goes around a point P=((n+1+e)/2, (n+1+e^2)/2), where e is a positive number small enough that no line between two grid points contains P. I think e=1/n is small enough. Dan Hoey haoyuep@aol.com --- Newsgroups: rec.puzzles, sci.math From: Dan Hoey Date: Tue, 30 Sep 2008 20:20:40 -0400 Subject: Re: Wind Around Solitaire Leroy Quet wrote: > Dan Hoey wrote: >> Leroy Quet wrote: >>> Here is a 1-player game played on an n-by-n grid, where n is odd. >>> (I suggest an n of about 13 to 21 for beginners.) >>> The player places the numbers 1,2,3,...n^2 into the grid. >>> 1 goes in the center square. >>> Each number (2k+1) must go either below, above, left of, or right of >>> the square with (2k-1) in it. >>> Each number (2k) can go anywhere in the grid. >>> Numbers can only be placed in empty squares. >>> The player's score is the number of times the path of odd integers >>> goes completely around the center square clockwise before the player >>> is unable to place any more numbers in the grid. I misread the problem, and treated the integers as points instead of squares. You clarified: > The player's score is the number of times the path of odd integers > goes completely around the CENTER SQUARE clockwise before the player > is unable to place any more numbers in the grid. I still don't get it. If 2m-1 is below the center square and 2m+1 is above the center square, does that go halfway around clockwise or counter-clockwise? Or did you intend that the path consist of adjacent consecutive odd integers? That seems too easy a game to play, since you just fill in a dense spiral with odd integers, while placing even integers arbitrarily beyond the spiral. I'm sure I'm missing something. Can you supply the position of a game that is played far enough that the path goes twice around the center square? Dan --- Newsgroups: rec.puzzles, sci.math From: Dan Hoey Date: Wed, 01 Oct 2008 18:39:25 -0400 Subject: Re: Wind Around Solitaire Leroy Quet wrote: > Yes, the game as written is indeed too easy to play. > Rule update: The even integers must go in squares immediately adjacent > to already filled in squares. > I suspect that even with this rule update, the game is too easy to get > the maximum possible score. If we require that each even integer be placed immediately adjacent to the previous (odd) integer, that might make it difficult enough. Dan --- Newsgroups: sci.math.research From: Dan Hoey Date: 3 Oct 2008 07:25:33 -0400 Subject: Re: arbitrary factorization as a difference of cubes? Robert Israel wrote: > Along the same lines, > (a+b+c+d)^4 - (a+b+c-d)^4 - (a+b-c+d)^4 - (a-b+c+d)^4 - (-a+b+c+d)^4 > +(a+b-c-d)^4 + (a-b+c-d)^4 + (a-b-c+d)^4 = 192 a b c d Indeed, it appears that (#P)! \product_(p \in P) 2p =? Sum_(Q \subset P) (-1)^#Q ( Sum_(p \in P) PM(p,Q) )^#P, where PM(p,Q) = -p if p \in Q, p otherwise. There should be a name for this equation, but I don't know it. Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Sat, 04 Oct 2008 07:49:21 -0400 Subject: Re: Enigma 1505 - Brief N-counter Dan Hoey wrote: > ken wrote: >> On Sep 24, 1:20 pm, Chappy wrote: >>> i.e. Ken's solution has three pieces that are the same shape. >>> My solution has only 3 differently shaped pieces. > My program found 203 solutions. My tired eyes found only one in which > all six hexominoes are congruent, but I haven't taught the program to > back me up on that. Here's are a couple of the solutions. The one-shape solution is #52, but my favorite solution is #193, because it has two four-corners points. The whole list is about 32Kb, a little large for a Usenet post. ._._._._._._. Solution #52 | . |_._. . | 1 + 2 + 7 + 8 + 9 + 10= 37 |_._._._|_._| 3 + 4 + 5 + 6 + 11 + 12= 41 | . |_._. . | 13 + 14 + 19 + 20 + 21 + 22= 109 |_._._._|_._| 15 + 16 + 17 + 18 + 23 + 24= 113 | . ._._| . | 25 + 26 + 27 + 28 + 31 + 32= 169 |_._|_._._._| 29 + 30 + 33 + 34 + 35 + 36= 197 ._._._._._._. Solution #193 | | . | . ._| 1 + 7 + 13 + 14 + 19 + 25= 79 | |_. |_. | | 2 + 3 + 8 + 9 + 15 + 16= 53 | ._|_._|_| | 4 + 5 + 6 + 10 + 11 + 17= 53 | | |_. |_. | 12 + 18 + 23 + 24 + 30 + 36= 143 |_| . | . | | 20 + 26 + 27 + 31 + 32 + 33= 169 |_._._|_._|_| 21 + 22 + 28 + 29 + 34 + 35= 169 Dan Hoey haoyuep at aol.com --- Newsgroups: sci.math From: Dan Hoey Date: Sat, 04 Oct 2008 08:04:31 -0400 Subject: Re: tomic polynomial: CONJECTURE 3 quasi wrote: > On Thu, 02 Oct 2008 03:01:32 GMT, Gerry Myerson wrote: >> I have. What is a tomic polynomial? What is a tommy polynomial? > Presumably this: > Calling polynomials "tomic" based on their coefficients being among {-1,0,1} is presumably based on the misapprehension that cyclotomic polynomials possess this property. Can we say anything true and interesting about the possible coefficients of cyclotomic polynomials? Perhaps an upper bound U(n,m) on the number of coefficients of magnitude m in the nth cyclotomic polynomial? Dan Hoey haoyuep at aol.com --- Newsgroups: sci.math From: Dan Hoey Date: Sun, 05 Oct 2008 13:33:29 -0400 Subject: Re: SAT and NP-Completeness Yaakov Davis, quoting Joshua Cranmer in part, wrote: >> ...All I see is a specific case. and replied: > I meant a 'general case' *like* the one specified. Sorry for not > being clear. I'm sorry Yaakov isn't clear, too, but he hasn't done anything to be more clear. He hasn't said what cases would be 'like' the one he gave. Without making that explicit, there is no way of telling what general case he is talking about. If he wants to stop being sorry, he needs to /describe/ the general case with /description./ After all, Joshua actually wrote (in slightly larger part): >> What's the "general case"? All I see is a specific case. I'm sorry that Joshua's question was apparently not clear to Yaakov. Dan Hoey --- Newsgroups: sci.math From: Dan Hoey Date: Sun, 05 Oct 2008 14:58:16 -0400 Subject: Re: Bit-permutation function and Cantor Dust? mike3 wrote: > Consider the following function on the interval [0, 1]. > For x < 1: > 1. Obtain the binary expansion of the fractional part of x (infinitely > long: if it "terminates" then just append an infinite string of zeroes.) > 2. Divide this into bit pairs, so the first 2 digits becomes pair 1, > the next 2 are another pair, and so on. > 3. Reverse the bits in each pair. > 4. Take this to be the binary expansion of the fractional part of > f(x), which has integer part 0. [...] > The graph looks sort of like a "Cantor dust" (cartesian product of 2 > copies of a Cantor set) that has been skewed and stretched or rotated. > The dust looks more like "middle half" than "middle third", however. > (i.e. remove the middle 2 quarters of the segument when divided into > 4 quarters.) Let P be the set of (x,f(x)) in [0,1]x[0,1]. P is the union of P_1 = P/4, P_2 = (P+(2,1))/4, P_3 = (P+(1,2))/4, and P_4 = (P+(3,3))/4. where addition is interpreted as translation and division as scaling. Let T be the linear transform defined by T((x,y))=(2x-y, 2y-x). Then T(P) is the union of T(P_1) = T(P)/4, T(P_2) = (T(P)+(3,0))/4, T(P_3) = (T(P)+(0,3))/4, and T(P_4) = (T(P)+(3,3))/4, which is the middle-halves set you describe. Dan Hoey haoyuep at aol.com --- Newsgroups: sci.math From: Dan Hoey Date: Sun, 05 Oct 2008 21:40:18 -0400 Subject: Re: Bit-permutation function and Cantor Dust? mike3 wrote: > On Oct 5, 12:58 pm, Dan Hoey wrote: >> mike3 wrote: >> [...] >>> The graph looks sort of like a "Cantor dust" (cartesian product of 2 >>> copies of a Cantor set) that has been skewed and stretched or rotated. >>> The dust looks more like "middle half" than "middle third", however. >>> (i.e. remove the middle 2 quarters of the segument when divided into 4 >>> quarters.) >> Let P be the set of (x,f(x)) in [0,1]x[0,1]. P is the union of >> P_1 = P/4, >> P_2 = (P+(2,1))/4, >> P_3 = (P+(1,2))/4, and >> P_4 = (P+(3,3))/4. >> where addition is interpreted as translation and division as scaling. >> Let T be the linear transform defined by T((x,y))=(2x-y, 2y-x). >> Then T(P) is the union of >> T(P_1) = T(P)/4, >> T(P_2) = (T(P)+(3,0))/4, >> T(P_3) = (T(P)+(0,3))/4, and >> T(P_4) = (T(P)+(3,3))/4, >> which is the middle-halves set you describe. > Hmm. But how does the transform given relate to the function > described? The transform describes the inverse of the way in which the middle-halves set has been "skewed and stretched or rotated" to form the graph of the function. Dan Hoey haoyuep at aol.com --- Newsgroups: sci.math From: Dan Hoey Date: Mon, 06 Oct 2008 06:09:53 -0400 Subject: Re: Bit-permutation function and Cantor Dust? I wrote: > The transform describes the inverse of the way in which the > middle-halves set has been "skewed and stretched or rotated" > to form the graph of the function. Another way of looking at this is that the middle-halves set consists of those coefficient pairs (r,s) such that every digit in the base-4 expansion of r and s is either 0 or 3. Thus each digit position can be written as (3r_i, 3s_i) where r_i and s_i are in {0,1}. When mapped by the inverse transform U(r,s) = (2r/3 + s/3, 2s/3 + r/3), each digit position is mapped to x = 2r_i + s_i, y = 2s_i + r_i, which describes the operation of exchanging pairs of bits of x to form y. Dan Hoey haoyuep at aol.com --- Newsgroups: rec.games.abstract From: Dan Hoey Date: Sat, 15 Nov 2008 01:02:22 -0500 Subject: Re: The game similar to NIM alberto.hi8@gmail.com wrote: > I have to consider the following game: > all rules are the same like in classic 'nim' game except one rule - > the number of stones in heaps are nondecreasing (e.g. (1,2,2,6,7), > not (5,1,7)). After every move this rule must be keep. So player can > remove stones from heap X only and only if in heap to the left are > less piles and he can't remove more stones, than difference between > heap X and heap to the left. [...] I'm pretty sure this is called "the silver dollar game" (or maybe "the silver dollar game without the silver dollar" in _On Numbers and Games_. I'm pretty sure that a sequence of order-heaps h_1, h_2, ..., h_n acts like a Nim game with heaps related to differences of adjacent h_i , that is h_n - h_(n-1), h_(n-2) - h_(n-3), ..., h_(4-(n mod 2)) - h_(3-(n mod 2)), h_(2-(n mod 2)) - h_(1-(n mod 2)) where h_0=0. Basically, moving in order-heaps congruent to n (mod 2) is like taking stones from a Nim heap, and moving in order-heaps NOT congruent to n is like ADDING stones to a Nim heap, which can always be answered by reversion (taking the stones back out by moving in the next higher order-heap). Reversion is a little iffy when we deal in the Lost World of misère games, since reverting the ending moves loses. So there may be a difference in the misère Nim endgame. Dan Hoey haoyuep@aol.com --- Wikipedia Talk:Misère Perhaps the importance (now rated low) should be increased due to the links from Category:Combinatorial game theory cites. -Dan Hoey 09:48, 15 November 2008 (UTC) --- Newsgroups: rec.games.abstract From: Dan Hoey Date: Sun, 16 Nov 2008 16:16:22 -0500 Subject: Re: The game similar to NIM alberto.hi8@gmail.com wrote: > I have to consider the following game: > all rules are the same like in classic 'nim' game except one rule - > the number of stones in heaps are nondecreasing (e.g. (1,2,2,6,7), > not (5,1,7)). After every move this rule must be keep. So player can > remove stones from heap X only and only if in heap to the left are > less piles and he can't remove more stones, than difference between > heap X and heap to the left. [...] This is indeed "The Silver Dollar game, without the dollar" from J H Conway's "On Numbers and Games", except that he considers the game in which the heaps (which are modeled by coin positions) are strictly decreasing. As before, the largest heap and every second heap from there down is treated as a Nim heap whose size is the number of its options, and moves to the other heaps are reversible. The misère case also works exactly as in misère Nim. If the winner is forced to move in the ignored heaps, the best move is to remove one token, leaving the opponent with only one nonreversible move. Eventually, the nonreversible move will be the last move, losing the game. Dan Hoey haoyuep@aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Sun, 16 Nov 2008 20:34:22 -0500 Subject: Re: What power of 3 is closest to a whole number of millions? Laretal wrote: >>> There was a puzzle in the Sunday Times a couple of weeks ago that went >>> like this. What power of the number 3 is the closest to a whole >>> number of millions? Can you calculate it with logic and algebra alone >>> - i.e. not through trial and error? > 0 is not the answer given. That is too easy. It has to be a higher > power. The full text of the puzzle as it was sent to me is as > follows: > For a school project, Penny was investigating the powers of 3. She > wrote some down and explained to Paul that the first power was 3, > the second power was 9, the third power was 27, and so on. Soon the > numbers were in the millions and Paul asked how close to a whole > number of millions one of the high powers could be. Simply by some > clever logic and algebra Penny was soon able to answer the question. > She aslo found the lowest power of 3 that actually came that close. > Which power was it?eg 248th Phi(10^6)=400000, so 3^400000 = 1 mod 10^6 and 3^n hits +-1 mod 10^6 at an interval of n a factor of 400000. I worked down from PowerMod(3,400000,10^6) to PowerMod(3,50000,10^6)=1, then found that PowerMod(3,10000,10^6)=200001 and PowerMod(3,25000,10^6)=500001. So the answer is 50000. The Sunday Times may have a solution that uses cleverer logic and algebra, though, so I'll try another way. We start with 3^2 = 9, so 2|n. 3^4 and 3^10 are wrong mod 100, so we use 3^20 = 78401 (mod 10^6), so 20|n. 3^40 is wrong mod 1000, but 3^100=522001 (mod 10^6), so 100|n. 3^200 is wrong mod 10^4, but 3^500=610001 (mod 10^6), so 500|n. 3^1000 and 3^2500 are wrong mod 10^5, but 3^5000=100001 (mod 10^6), so 5000|n. Then 3^25000 and 3^10000 are wrong mod 10^6, so 3^50000=1 (mod 10^6) is the first solution. Dan Hoey haoyuep at aol.com --- Newsgroups: rec.puzzles From: Dan Hoey Date: Mon, 17 Nov 2008 10:37:39 -0500 Subject: Re: What power of 3 is closest to a whole number of millions? mike wrote: > Laretal says... [...] >> For a school project, Penny was investigating the powers of 3. She >> wrote some down and explained to Paul that the first power was 3, >> the second power was 9, the third power was 27, and so on. Soon the >> numbers were in the millions and Paul asked how close to a whole >> number of millions one of the high powers could be. Simply by some >> clever logic and algebra Penny was soon able to answer the question. >> She aslo found the lowest power of 3 that actually came that close. >> Which power was it?eg 248th > So presumably the puzzle is asking (none to clearly) either: > "For what integers m > 0 and n > 0 is 3^m-n*10^6 a minimum?" > or: > "For what integers m > 0 and n > 0 is 3^m/(n*10^6) closest to one?" > Any other possible interpretations? Only that the puzzle does not ask for n, and asking "how close to a whole number of millions" suggests absolute closeness, not relative closeness. So the question is "What integer m > 0 minimizes min {| 3^m-n*10^6 | : n integer }?" Mark Wooding has already answered this question, much more cleverly than I did. In case my linguistic argument is not convincing to you, note that there is no answer to your second interpretation; the form 3^m/(n*10^6) achieves values arbitrarily close to one. Dan haoyuep@aol.com --- Newsgroups: sprouts-theory From: Dan Hoey Date: Tue, 16 Dec 2008 10:16:54 -0500 Subject: Re: Further thoughts on Peltier-Purvis Jeff Peltier wrote: > Yes this is true and really amazing as 4spots has genus 1°²·· and 0L2 has > 2¹4.., how is that sum of the two has genus 3°²°·· ? 2^14 is not tame. You can't tell the genus of the sum from the genus of the addends unless all are tame. Dan --- Newsgroups: sprouts-theory From: Dan Hoey Date: Mon, 22 Dec 2008 14:41:26 -0500 Subject: Re: Focardi and Luccio DannyPurvis wrote: > In 2003 I wrote a critique of the analysis of 7+ by Focardi and > Luccio. My guess is that I made some big errors. My critique is at > http://www.geocities.com/chessdp/focardi.htm. Unfortunately, the > link within the article to the Focardi / Luccio analysis is broken. I saw your critique, and it seemed accurate. They arrived at the right answer, but one of the steps was wrong. Dan --- Newsgroups: sprouts-theory From: Dan Hoey Date: Mon, 22 Dec 2008 18:44:55 -0500 Subject: Re: Focardi and Luccio DannyPurvis wrote: > In 2003 I wrote a critique of the analysis of 7+ by Focardi and > Luccio. My guess is that I made some big errors. My critique is at > http://www.geocities.com/chessdp/focardi.htm. Unfortunately, the link > within the article to the Focardi / Luccio analysis is broken. Google found their paper at http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.21.212 The problem is on page 13, "Strategy for x_0 = 7", case 1. They claim that after 7+ 1(8)2 1(9)1[2,3] 4(10)9 the second player should play 9(11)9[5,6], ensuring two survivors in the region with spot 3 and three survivors outside that region. They claim the region with spot 3 is a slight variant of a region with two spots "but the strategy is exactly the same." However, in the two-spot game, two survivors can be guaranteed by the second player. In this game, the first player can play 2(12)2[3,8], forcing one survivor. Dan --- Newsgroups: sprouts-theory From: Dan Hoey Date: Tue, 23 Dec 2008 15:53:06 -0500 Subject: Re: Focardi and Luccio I wrote: > DannyPurvis wrote: >> In 2003 I wrote a critique of the analysis of 7+ by Focardi and >> Luccio. My guess is that I made some big errors. My critique is at >> http://www.geocities.com/chessdp/focardi.htm. Unfortunately, the >> link within the article to the Focardi / Luccio analysis is broken. > Google found their paper at > http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.21.212 > The problem is on page 13, "Strategy for x_0 = 7", case 1. They > claim that after 7+ 1(8)2 1(9)1[2,3] 4(10)9 the second player > should play 9(11)9[5,6], ensuring two survivors in the region > with spot 3 and three survivors outside that region. They claim > the region with spot 3 is a slight variant of a region with two > spots "but the strategy is exactly the same." However, in the > two-spot game, two survivors can be guaranteed by the second > player. In this game, the first player can play 2(12)2[3,8], > forcing one survivor. It might be useful to look at just what happens after 7+ 1(8)2 1(9)[2,3] 4(10)9. If I recall correctly, the second player can still win (by forcing an odd number of survivors). However, it may be that the second player cannot force a _particular_ number of survivors. I'd be interested in a simple example (ideally a starting position) in which a player can win, but cannot force the game to last a particular number of moves. This would be a counterexample to the observation that the game usually boils down to a battle over whether the number of survivors is N or N+1. Dan --- Newsgroups: sprouts-theory From: Dan Hoey Date: Wed, 24 Dec 2008 15:00:38 -0500 Subject: Re: Sprouts notation Yper Cube sent me a note requesting input on Sprouts notation suitable for a position library as in GLOP. With his permission, I'm responding to sprouts-theory generally. Yper Cube wrote: > I liked the parenthesis idea.It doesn't work in positions arousing from > torii and proj. planes surfaces but maybe it can still be used sometimes. > (Example: we can't use it in .abcabc. > but we can write .abacbc. > as .(b)(b). , can't we? I think this is a good idea when possible. > I would also prefer GLOP to use "{" to mark the start of a region and not > only to mark its ending and "." to mark for boundary starting as well. Like, > instead of: > 0.0.AB.}AB.}]! > use something like > {.0.0.AB.}{.AB.}]! > or maybe > [.0.0.AB.}{.AB.}] > In a recent sprouts theory post, I mentioned another possible > technique to copmact notation further. > Use of * or ^ when we have multiple identical boundaries. > Example. instead of 0.0.0.0.}], use: > 0*4.}] > (GLOP uses the above now) > or > [{(.0.)^4}] > or > [{(.0.)*4}] I strongly disagree with this. We should save our matching brackets for things that need to be grouped recursively, whereas boundaries within regions within lands are handled by hierarchy. Boundaries within a region should be _separated_ (as with "."), regions should be separated (say, with "/"), and lands should be separated (say, with "%"). The use of a trailing separator is superfluous and should be avoided, so the above would be 0.0.AB/AB . The "!" for end-of-position should only be needed if there's some other thing following the position. A trailing separator should be permitted on input, but ignored. For nonplanar regions, I suggest a region should be suffixed with +n for T^n or -n for P^n. Since the region will always be followed by "/", "%", "!", or end-of-string, there is no restriction on multiple digits for n. > Using parenthesis, we can extend this to include more complex > objects (than boundaries) > Like, instead of 0.0.0.0.2.2.2.2.2.AB.CD.EF.}AB.}CD.}EF.} ,use: > {(.0.)*4(.2.)*5(.AB.{.AB.})*3} This can be done in several ways without the leading or trailing separators. Perhaps the simplest is to always use a kind of brackets (say "<>"), so that <0.>4 means 0.0.0.0 <1.2A/>2 means 1.2A/1.2A <12>32 means 1212122 <(>3<(<)>2>3 means (((())())()) . The <...> is always followed by a single digit specifying the repeat. A position canonicalizer may not want to use all of these constructs, but they should be recognized on input. Your last example in which the parentheses imply a change of variables is somewhat problematic. I think it should be done with a different approach, which I call the "partial position" representation. I've talked about partial positions before--I think they will eventually appear in the database by themselves, along with simplifications such as the ^x:x.0 = ^x:x2 that has been observed before. A partial position is a land that has some unmatched pivots called "parameters". The parameters are listed in the front of the partial position in a sort of lambda notation, like ^xyz:1xy2z.2 . This refers to a part of a position whose pivots are to be matched up with another part of the position in a way to be determined. Copies of the partial position may be matched up in several different places. When there is only one parameter, the presence of that parameter outside the partial position denotes a pivot attached to a copy of the partial position. So ^x:xA/A1%xx.x means AB.C/AD/D1/BE/E1/CF/F1 . Partial positions can instantiate other partial positions: ^x:x((1))%^z:x.xz%zzx means ABC/D.EA/D((1))/E((1))/F.GB/F((1))/G((1))/C((1)) . With a multiple-parameter partial position instantiated more than once, we must specify which parameters go with which instantiation. In the simplest case, on a plane where the matching pivots appear in the same region (and therefore on a single boundary), the parameters parenthesize, so we can specify parameters as [x...y...z] without ambiguity. For instance, ^xyz:1xy2z.2%xy[z1[xyz]yx]z would mean ABF1GHJEDC/1AB2C.2/1DE2F.2/1GH2J.2 . In more complicated situations, we can specify multiple parameter sets for a partial position as in ^xyz^tuv:x(y(z))%xtuyvz for ADEBFC/A(B(C))/D(E(F)) . The primary parameter set can still be parenthesized if necessary, but the others must be unambiguously matched with each other. Finally, I would like a construct for specifying the join of two partial positions. I propose using "=xyz=tuv" , where the first "=" replaces the preceding "%" character. So ^xyzw:xyzw1=xyzw=xywz refers to the (nonplanar) position ABCD1/ABDC1 On alphabets: Glop introduced the idea of lower-case characters for pier spots, which can appear on multiple boundaries within a position. upper-case characters were reserved for pivots. The problem is that we sometimes run out of one kind or another. With the addition of parameter letters, the problem is worse, so I propose a more general approach. For readability, a program is encouraged to use ABC... for pivots, abc... for labeled pier spots, and xyz... for parameters so far as possible. The rules for recognizing which is which should be loosened so that any letter can be used for any of these purposes. A letter that appears in a lambda expression "^xyz:" is always a parameter, and those are the only parameters. For multiple parameters in brackets, each parameter must appear just once except within inner brackets. Other multiple parameter sets appearing on a boundary are assumed to match each other, so ^xy%xy.xy means AB.CD/AB/CD; we need ^xy^zw:xy%xz.yw to refer to AC.BD/AB/CD . No other unbracketed parameters of the set may appear on that boundary. Other sets of unbracketed parameters that appear within a region are assumed to match each other, so ^xy%Axy/A1xy means ABC/A1CD/BC/CD ; again we can't have other stray members of the set in the region. If none of these is possible, each set of parameters can appear only once in a land: ^xy^zw^tu:xz.yt/uz for AD.BE/FC/AB/CD/EF . Excluding the parameters, a letter that appears twice on a boundary is a pier spot, and may appear only twice on that boundary. Excluding these, a letter that appears on different boundaries in the same region may appear only twice in the region, and the two letters match each other. (This can only happen on nonplanar boards). After all of these are taken into account, remaining letters must appear exactly twice per land, and match each other. Just in case we end up with a need for more than 52 letters, I believe we should allow an sequence like &word; to be used for a letter, where "word" is composed entirely of letters. So the grammar we have is := | | "!" := ( "+" | "-" ) ":" := | "=" "=" | "%" := "^" ":" | "^" ":" := | "%" | "%" := | "/" | "/" := | ( "+" | "-" ) := | "." | "." := | := | "(" ")" | "[" "]" := | 1 | 2 | "&" ";" := | This grammar accidentally prohibits extended letters as parameters, but maybe that's a good thing. There are other requirements such as that parameter lists can't repeat a letter, but that is for symantic analysis. The gamelabel is in case someone needs to mark the position as normal or misere or specify an initial number of spots. I suppose I should allow the number or sign to be omitted. The use of "<>" for repetition is a metalanguage used for abbreviating this language. It is expanded textually first before applying this grammar. We still have a set of brackets "{}" that aren't used, but I find they tend to look too much like parentheses to be very useful. Other symbols such as @~#$*_\,;?'"` may turn out to have some use as well. Finally, I should note that GLOP databases need to have a version number at the top so that older versions can be recognized and converted to newer ones. Dan --- Newsgroups: sprouts-theory From: Dan Hoey Date: Mon, 29 Dec 2008 13:08:18 -0500 Subject: Re: Focardi and Luccio lapinot@tuxfamily.org wrote: > Hi Dan, > you should have a look to these positions (in Glop-notation) : > 2AB.}AB.}]2AB.}AB.}]! for a normal game > 1.2.2.}]2AB.}AB.}]! for a misere game. Thanks for the example. I don't believe Focardi and Luccio have addressed this sort of thing in their theory, while it is fundamental to the strategy of nimbers *2 or greater. Dan --- Newsgroups: sprouts-theory From: Dan Hoey Date: Wed, 31 Dec 2008 15:21:21 -0500 Subject: A lousy periodicity theorem In _Sprouts Game on Compact Surfaces_, Julien Lemoine and Simon Viennot proved that beyond a certain number (depending on the region), increasing the genus of an orientable region cannot affect a sprouts game, nor can increasing the genus of a non-orientable region by two affect the game. This reminds me of an observation I made that is provable in the same sort of way (unless I've overlooked something). A "louse" is a boundary consisting of a single degree-2 point that does not appear anywhere else; "2." in Glop notation. We can change a sprouts game by adding a louse to a region. The theorem is that beyond a certain number (depending on the region), adding two lice to a region does not affect the game. I'm pretty sure that the result can be strengthened to show that (in the presence of enough lice) adding one louse to a region will change the Sprague-Grundy value of the position by the nimber *1. This should hold in both normal and misère play. Dan Hoey ---