material added 25 February 2002
The date on the front of wolframscience.com is true. A
New Kind of Science is at the printer. New sample pages
are available for your perusal, and Stephen picked out a sequence
of new pages that will be shown one per week until the book is in
stores. The pages are not drafts, these are final versions. Page 43
is one of them, showing some of the history of patterns. A 70,000 year
old lattice engraving was discovered too late to be considered
for page 43.
698896 is a palindromic square number with an even number of digits. The next one is 637832238736. More terms in the series can be found at the Encyclopedia of Integer Sequences. A general discussion can be found at worldofnumbers.com. Toshi "Junk" Kato is intrigued by this sequence, and conjectures that there are an infinite number of base10 palindromic square numbers with an even number of digits. He will pay $50 to the first who can prove him right or wrong. (His money is probably pretty safe.) If you can make headway on this, let me know. Vaguely related are the squares using three digits. In word palindromes, 2002 has just been published, a 2002 word palindromic story.
Toby Gottfried has written a java applet for my Prime Heart puzzle. I tried solving it myself  not too hard at all. The New Yorker has written a nice article about various puzzlers. Stuart Anderson let me know about a very interesting paper on Squared Squares by Ian Gambini. It's in french, but has many interesting diagrams:
Juha Hyvönen sent two crossnumber puzzles. In the first, all entries across and down are square numbers. In the second, all entries across and down are triangular numbers. Curiously, both contain all ten digits. Can you find the unique solutions? Send answers.
material added 19 February 2002
When Andrea Gilbert presented her father's Maze of Life idea, I coined the termed "intelligent cell" as a modest contribution. Many others have contributed much more. Dr William Paulsen has been exploring how an intelligent cell can ride a glider creatively, and has found a sequence (744 steps from Ystart) that creates a glider gun! It's damn neat, watching an intelligent cell manipulate it's universe. (To watch it, undo all, then click through redo.) Andrea has also added new Plank maze puzzles at her site (clickmazes.com).
Serhiy Grabarchuk of Puzzles.com recently answered a question about folding puzzles. He was so comprehensive, I felt obligated to quote his paperfolding list.
There is a group of people competing for who can compute a million digits of Pi the fastest on an average PC. Stu's Pi Page lists the various links and records. His program can produce a meg of Pi in less than 9 seconds.
The Vegas Scrabble tournament was written up in the Village Voice.
Junk Kato has made a sliding block puzzle for Edward Hordern.
material added 9 February 2002
Ivars Peterson has written an article about Algebraic Hearts. I happen to have a small puzzle in it concerning the two digit primes, and a heart. If you can solve it, tell me  What is the number of love?
I've updated quite a few previous puzzles to link to a solvers list and solutions.
Erich Friedman: Here are a set of fifth powers with an odd property: A=27^5=14348907, B=39^5=90224199, C=1239^5=2919823044470199. C is composed of the same digits as A and B together. What is the smallest set of fourth powers with this property?
material added 5 February 2002
Siam News has published a $100, hundred digit contest by Nick Trefethen of Oxford University. There are lots of interesting problems here. If you think you can solve one of these, and would like to be a part of a Mathpuzzle team, let me know. I'll hook you up with a small group.
The Zen of Magic Squares, Circles, and Stars by Clifford Pickover. During the 2000 National Puzzler's League convention, octogenarian member Twisto shared with us some of the magnificent constructions of The Great Fubine, a man who turned to magic squares after losing all his money in the stock market crash of 1929. Twisto told me about how Fubine was mostly ignored by mathematicians of the time  but he thought the material was deserving of attention. I mentioned this to Cliff. He followed up big time: first by visiting Twisto at his home, and then by publishing it. Cliff uncovered all sorts of things while researching this book. Good stuff  if you like magic squares, this is a book you need. And, if you don't ... well, there's lots of magic squares in here. See if you can find it at your bookstore, at the very least. Another recent book on the same topic is Magic Square Lexicon: Illustrated, by Harvey D. Heinz and John R. Hendricks.
Martin Gardner frequently debunks myths and explains scams. One that has annoyed me lately is known as The Spanish Prisoner scam. Fabulous wealth is promised, for nothing. I've gotten three different letters in the past two weeks alone. It turns out that these are far nastier than I first believed. After all, I thought it was just a scam. The United State Secret Service explains the full horror. (I have forwarded all of these letters to the Secret Service, for their perusal.) Hoaxbusters is another good page explaining email hoaxes and scams, so is Snopes.
The Puzzling World of Barry R Clarke has been steadily growing for awhile now, and is well worth regular visits. At Puzzles.com, a monthly interactive logic puzzle has been added. Mario Velucchi has started "the ultimate Knight's Tour page of Links". Rawbrains.com has weekly patternbased problems.
Dave Tuller sent me this: Given an arbitrary polynomial with rational coefficients f(x), show that you can always find another polynomial with rational coefficients g(x) so that f(x)g(x) is a linear combination of prime powers of x. (a x^2 + b x^3 + c x^5 + d x^7 + ..., with rational {a,b,c,d,...}, and all powers of x prime) For example, 1+x works since x^2(1+x)=x^2+x^3. Send Answer.
Erich Friedman's 124 square challenge ... which I offered $10 for ... has been solved by Erich Friedman. I'll write up solvers and more things on Friday.
material added 28 January 2002
This week's entry requires a bit of work for some, but the results are well worth it. Many of the best mathematical files are in a ps.gz format. This first paragraph is aimed at Windows users that don't already have access to files of this type. First, you'll need to get AFPL Ghostscript 7.03 (Instructions,Download). Once you've installed that, you can get Ghostview (Instructions,Download). If you're on a Linux or Unix system, you likely have both already.
Now that you're equipped to look at the files, Don Knuth has released the first part of the long awaited Book 4. It is two parts: Generating all ntuples and Generating all permutations. These are fantastic articles, well worth perusing. One piece in these I'd like to fix involves the Chinese Rings puzzle. Somewhere(1, 2,3, 4), I read that Cardano had described these in one of his books. UIUC just happened to have an original copy of that particular 15th century book, and the chinese rings weren't in it. I'd like to find an original source, for Knuth's book. Can anyone help? This would imply an early understanding of an important math concept, by means of a puzzle.
I'd also like to list other great ps.gz documents. Feel free to send me suggestions.
Erich Friedman passed along a crossword by Berend Jan van der Zwaag. The solution is unique. Answer. This puzzle was solved by Michael J D Dufour, Juha Hyvönen, Juha Saukkola, Rick Shepherd, Kaijian Xu, Peter Exterkate, John Gowland, Jeff Smith, Mark Michell, jkmclean, Derrick Schneider, Chris Lusby Taylor, Dick Saunders Jr., Kaushik.Laskar, and Sudipta Das.
material added 20 January 2002
Helmut Postl did an exploration of the Tetrods, and
found that three identical shapes like those below could be made.
The holes are in the same place, too  where are they? He found a mirror symmetric solution for the
Iamodds. Working on my 1.9 * fourthroot(9.1) = 3.30000007 problem,
he noticed: 19 * fourthroot(9.1) :: 33, 19^44 * 9.1 :: 33^4, 19^4 *
91 :: 33^4 * 10, 11859211 :: 11859210.
Shyam
Sunder Gupta has a great new website on number recreations. An
example from his site: find three consecutive abundant numbers.
Erich Friedman: It is well known that 1^2 + 2^2 + ... + 24^2 = 70^2. Can we tile a 140x140 square with 4 copies of each square of size 1 through 24? I'll offer $10 for an answer to this.
Nick Baxter has started a 1000 Play Thinks Errata page. On item 150, many trees can all be touching each other, and thus be equidistant.
jkmclean, Nick Baxter, and Joseph DeVincentis all solved the Harbour puzzle.
ListPlot[Table[ ( PrimePi[(n+1)^2]PrimePi[n^2] )  Floor[n/( Log[n+1]) Sqrt[n]] , {n,1,100000}]];
Is there always a prime between two square numbers? This is a famous unsolved math problem, so I took a quick look at it. In Mathematica, PrimePi[x] gives the number of primes less than or equal to x. The part in blue thus counts the number of primes between n^2 and (n+1)^2. The part in green is a lower bound I concocted. Up to n=100000, at least, it works pretty well.
material added 14 January 2002
M Oskar van Deventer has made a made a wonderful challenge he calls the Harbour puzzle. (I can't seem to get the last few blocks into the ship.)
Another Henley Mob creation is Andrea Gilbert's Cups and Peas puzzles. Marvelous.
Place 3 queens and 2 rooks on a chessboard so that every square is attacked. The solution is unique. You can see the solution to this and many similar questions at www.mround.pwp.blueyonder.co.uk/puzzles/minattak/ , which has a solver.
A lot of people have recently recommended 1000 Play Thinks by Ivan Moscovich to me: Nick Baxter, Will Shortz, and Ken Duisenberg. I also liked it. I spotted one error: In Playthink 150, Ivan shows how 3 trees can be equidistant (in a triangle) and asks if that is the highest number of equidistant trees. His answer is wrong. I would recommend looking in a bookstore for it, first. I found many copies in every bookstore I tried, and this is a huge book. So, save on your shipping costs. Any corrections you find should be emailed to Nick Baxter at nick@baxterweb.com.
Another marvelous book is Puzzlers' Tribute, a hardbound synopsis of Gathering for Gardner 4. Almost every famous recreational mathematician I know has a short, interesting paper inside it. An online version of the previous book is available for free at the site. I can hardly wait to see the next book in this series. Sadly, this book doesn't seem to have caught the attention of bookstore chains, and it really deserves it.
At some point during the week, I wondered if a grid could be divided into pairs of points where every distance was distinct. It turns out that this is possible on a 10x10 grid. This makes a marvelous little solitaire exercise. I found the answer below by hand. The two 5's are separated by the distance Sqrt[5], the 13's by Sqrt[13], and so on. Every possible distance in a 10x10 grid is represented.
The Association of Game and
Puzzle Collectors has a great site.
material added 6&7 January 20022002
For the puzzle of the week I'll defer to the NPR puzzle of the week. Quoting Will Shortz: "This is a timely challenge from listener Ed Pegg, Jr. of Champaign, Illinois. He points out that 2002 is an unusual number because it's a palindromic number (it reads forward and backward the same). Now double that: 20022002. The challenge is to find three different palindromic numbers that can be multiplied together to get the palindromic number 20,022,002. Each of these numbers must have at least 2 digits. What numbers are they?" You can enter the challenge at the National Public Radio site.
I discovered the above while concocting a different
problem. I've seen two palindromic years so far: 1991 and 2002. If you
multiply those together, you get 3985982, which isn't palindromic.
Few consecutive palindromic number sets have a palindromic product.
101 × 111 × 121 = 1356531 is one example. Find the next
triple of palindromic numbers with a palindromic product. Answer.
Much of the above I found using The Mathematical Explorer, and its Divisor function. If you'd like a side journey into a surprising series of math facts, try these: Euler constant (Gamma) and Harmonic Number.
The above plot shows Power[Apply[Times, ContinuedFraction[k,n]],1/n], for various k. I picked p, 2p, g, Log[2], 2^(1/3), 3^(1/3). As n increases, the geometric mean of all of these numbers converge to the Khinchin constant (blue line). That's pretty neat.
As something of an ultimate "oops", Dr. Robert Smith looked at the wellaccepted "Earth eventually gets engulfed by expanding Sun" argument, and noticed that noone had bothered to account for Sun mass being converted to energy. It turns out that the Earth will not suffer a firey doom within a red giant inferno! Huzzah! As the sun loses mass, the Earth will spiral out to a orbit beyond the reach of the eventual red giant, 7 billions years from now. Here is the report. Just goes to show that it never hurts to check a calculation. I might be celebrating too soon, though  our galaxy will collide with another in about 3 billion years.
Many answers for the Constellation puzzle are here. An online spelling bee (English) is at the MSNBC site. I've added much more to the Pancakes page.
material added 2 January 2002
Scott Kim did a lovely article on Knot Puzzles in the latest Discover magazine. He mentions the Picture Hanging problem.
Xort[A_List] := Fold[Join[Reverse[Take[##]],Drop[##]]&, A, Flatten[Position[BitXor@@@Partition[Append[A,1],2,1],1]]] is an interesting method for sorting pancakes of two different sizes. In the picture, a green line indicates samesize/noflop, and a red line indicates different/flop. A computer uses Xor to tell if two bits or the same. This can be called the Pancake Sort, or perhaps just Xort.
Pancake Xorting
In other words, a spatula goes down a stack of pancakes, and flops the pancakes above it every time the sizes differ. When the spatula reaches the last pancake, the stack is flopped so that larger pancakes are on the bottom. This is a reversible sorting technique. By saving the BitXor list, the original state can be reconstructed. Larger integers can be sorted by considering each bit in a series. Can the BitXor lists be combined in any interesting way to sort Pancake stacks in an optimized way? Send answer. Many people responded to Eric's pancake problem from last week. Turns out this was at Sloane's Integer Sequences, but there was a typo in the sequence there. jkmclean pointed out the interesting fact that the sequence could increase by no more than 2.
Peter Esser found a lovely solution for a particular type of polyform:
Eric Iverson: 2002 has the property that the product of the nonzero digits equals their sum. What was the last year in which this happened, and the next?
Erich Friedman: I couldn't let the year end without making a new years puzzle. Start with the annulus of sqaures shown below. Place 4 horizontal dominoes (each covering two adjacent squares) so that the remaining squares can be tiled with dominoes in exactly 2002 ways. I think there's only one solution (up to symmetry). Have a happy new year.
Several people sent answers to Theo Gray's Spieker Center Table challenge. Only Dick Saunders Jr. and I solved Serhiy Grabarchuk's Constellation puzzle (he runs Puzzles.com). It's very nice, so I'm repeating it, hoping more people will try to solve it. Answers.
material added 27 December 2001
Eric Weisstein and I chatted a bit about the Pancake sorting problem. If you have a spatula, and 7 pancakes of size 17 on a plate, what is the minimum number of flops you'll need to get the pancakes sorted from smallest to largest? Eric took a brute force look at the problem. 2 pancakes  1 flop. 3 pancakes  3 flops {1,3,2}. 4 pancakes  4 flops {2, 4, 1, 3}. 5 pancakes  5 flops {1, 3, 2, 5, 4}. 6 pancakes  7 flops. There are only two "worst case" stacks: {4, 6, 2, 5, 1, 3} and {5, 3, 6, 1, 4, 2}. Below is the method for sorting out the first stack. Can you you sort the second stack with 7 flops? Answer. The worst case for 7 pancakes is unknown, so far as I know. Can someone analyze this? The sequence so far is 1 3 4 5 7, if you can extend this, let me know.
Eric's worst 6 pancake stack needs 7 spatula flops to sort.
I've recently been studying the possibilities of Domino Poker. No real sane reason  it just struck me as something that needed to be investigated, and it hasn't left me alone since. I love the feel of a domino set. I'd absolutely love to have a mini double9 set. Suppose I draw 5 dominoes at random. I might get something like , or 66, 62, 44, 51, 45. Looking at the distribution of numbers, I've gotten 2 singles, a double, and 2 triples. Or {1~1~2~3~3}. There are 28!/23!/5! = 98280 possible hands with the 28 double6 dominoes. Here is an analysis of all possible distributions: 105={2~4~4}, 105={3~3~4}, 105={1~1~1~1~6}, 420={1~1~3~5}, 420={1~1~4~4}, 420={1~2~2~5}, 420={1~1~1~1~1~1~4}, 462={1~1~1~1~1~5}, 798={2~2~2~2~2}, 840={1~3~3~3}, 840={2~2~2~4}, 1680={2~2~3~3}, 1680={1~1~1~2~5}, 3360={1~2~3~4}, 4200={1~1~1~3~4}, 4305={1~1~1~1~3~3}, 4620={1~1~1~1~1~2~3}, 5250={1~1~1~1~2~2~2}, 6090={1~1~1~1~2~4}, 7980={1~1~2~2~2~2}, 8190={1~1~2~2~4}, 10920={1~2~2~2~3}, 11970={1~1~2~3~3}, 23100={1~1~1~2~2~3}. The rarest hands have only 3 different numbers or 6 of a kind. My random hand was the second most common. There are many other hands to consider. For example, there are straights and other things. is a closed chain where all the connecting sums have 8 pips. Many different letters can be formed, like A or O.
Double9 distributions are as follows: 360={2~4~4}, 360={3~3~4}, 945={1~1~1~1~1~1~1~1~1~1}, 1260={1~1~1~1~6}, 2520={1~1~3~5}, 2520={1~1~4~4}, 2520={1~2~2~5}, 5040={1~3~3~3}, 5040={2~2~2~4}, 9576={2~2~2~2~2}, 10080={2~2~3~3}, 13860={1~1~1~1~1~5}, 20160={1~2~3~4}, 20160={1~1~1~2~5}, 47250={1~1~1~1~1~1~1~1~2}, 50400={1~1~1~3~4}, 50400={1~1~1~1~1~1~4}, 75600={1~1~1~1~1~1~1~3}, 98280={1~1~2~2~4}, 129150={1~1~1~1~3~3}, 131040={1~2~2~2~3}, 143640={1~1~2~3~3}, 182700={1~1~1~1~2~4}, 239400={1~1~2~2~2~2}, 359100={1~1~1~1~1~1~2~2}, 554400={1~1~1~1~1~2~3}, 630000={1~1~1~1~2~2~2}, 693000={1~1~1~2~2~3}. I prefer this set of pieces, and I'll try to have a complete writeup on Domino Poker next week.
Roel Huisman, who runs an excellent polyform site, handsolved the Tetrods problem from my last update. Noone has solved the Iamodds problem, yet  and to be fair, I don't know if it has one.
Robert Abbott: I sent Martin Gardner the latest Hoyle book by Philip Morehead, and my latest booklet on Eleusis. Martin wrote back to congratulate me and also said, "Incidently, my column on Eleusis is reprinted in The Colossal Book of Mathematics, just published by Norton. It contains fifty of what I considered my most significant columns." Robert has updated his Things That Roll page with a new 2x2 block maze by Adrian Fisher.
material added 17 December 2001
Theo Gray, a Mathematica frontend developer, recently built a table. And not just any table  it's a fullblown 345 Spieker Circle triangle table, with inlays. He kept a photo diary of his effort. Eric Weisstein has added this to his Spieker Circle description. Eric and I both helped a tiny bit, mainly by pointing to Clark Kimberling's excellent Encyclopedia of Triangle Centers, and by perusing CK's Triangle Book (one of my favorites). Theo came up with the built in puzzle, which is my puzzle of the week. ABC is a 345 triangle, with midpoints DEF. Prove that the incenter of triangle ABC is on the incircle of DEF, as shown below. Answer.
Theo Gray's Triangle Table
Robert Abbott's Eleusis is now in Hoyles. Congrats! He writes about it. Bob Kraus has added a CrossNumber Confusion puzzle to his page. Marko Brenko has made some very interesting "constant protection" Chess Mazes. I had no idea that a King+Knight could make such a tricky maze.
I absolutely adored issue 56 of Cubism For Fun. Every page of it. In particular, I enjoyed a writeup about Alex Randolf's "Tetrods". He took the 5 tetrominoes, and drilled a single hole in each in all possible ways to make 12 pieces. Then he tossed in a 1x1 square with a hole. The total area of these pieces is 49. It is quite easy to make a 7x7 square. Trickier is to make a 7x7 square with 4fold symmetry. If the 1x1 piece is left out, can 3 identical shapes be made? 4? Here's the same sort of idea with Iamonds ... I suppose I'll call these Iamodds.
The 16 Iamodds. Can the holes be symmetrical? Answer.
I mentioned I would mention some Palm OS programs I like. In Math Recreations, I like Turmite by Ken Krugler, CA Explorer by Andrew Brault, Igraph by Justin Manus, P.Fract by Rainer Persicke, Sort by Peter Novotnik. In general Science, I like Particle Data Book by Douglas Lowder, Chem Table by Robert Eng, Planetarium by Andreas Hofer, Gene by Y Kanai, Pocket C by Jeremy Dewey, Plua by Marcio Migueletto de Andrade, and pDraft by Harry Konstas. For Games, I like Adventure by Markus Bosshard, Boulder Palm by Wojciech Martusewicz, Tactorus by Thomas Gamet, Pocket Rogue by Takebayashi Tomoaki, Yahdice by Mirek Wojtowicz, and AIGO by A Iizuka. For puzzles, I like 3dBlockout by Kadon Enterprises, Collisions 5000 by Graham Rowbottom, Vexed by James McCombe, Twizzle by Richard Hocking, SlitherLink and Nurikabe by Hobbit's Room, and Exodus by Craig McMahon. Have I missed anything really cool?
material added 11 December 2001
Puzzle of the Week: Small decimals have just one digit on each side of the dot. 3.3 is a small decimal. There is one small decimal with a curious property. If you multiply its reverse and its fourth root, you get a number remarkably close to 3.3 . So, for example, if I start with 1.3, hit the square root button twice, and multiply by 3.1, I get 3.31. That's kind of close, but another small decimal is remarkably closer. What is it? Many people sent me the Answer. Eric Weisstein has a page devoted to Almost Integers.
The big new Mersenne prime from 5 December, and Richard Schroeppel's posting of DN Lehmer's classic factoring story, "Hunting Big Game in the Theory of Numbers" inspired me to examine primes. I wondered what the lowest prime factor of (10^(10^7))! + 1 might be. Such a prime would have at least ten million digits. Mathematica beckoned to me to do a little study of the lowest prime factors of various factorials +1, so I took a look via Table[{n, FactorInteger[(n1)! + 1, FactorComplete > False][[1, 1]]}, {n, 1, 100}]. I did a bit of clean up on the data by entering simpler forms for factorial primes like 77!+1, and by substituting xx for numbers Mathematica couldn't factor quickly. Record factorial primes discussed here.
In the data below, note how the smallest factor and n are the same whenever n is prime. This is known as Wilson's Theorem. Perusing the data, I noticed a different sequence: {4,7}, {6,11}, {10,19}, {22,43}, {24,47}, and so on. Wilson basically asked about p((p1)!+1), where "" indicates divisibility. Does p divide into (p1)!+1, in other words. (26 is true, since 6/2 is an integer.)
{{1, 2}, {2, 2}, {3, 3}, {4, 7}, {5, 5}, {6, 11}, {7, 7}, {8, 71}, {9, 61}, {10, 19}, {11, 11}, {12, 11! + 1}, {13, 13}, {14, 83}, {15, 23}, {16, 59}, {17, 17}, {18, 661}, {19, 19}, {20, 71}, {21, 20639383}, {22, 43}, {23, 23}, {24, 47}, {25, 811}, {26, 401}, {27, 1697}, {28, 27! + 1}, {29, 29}, {30, 14557}, {31, 31}, {32, 257}, {33, 2281}, {34, 67}, {35, 67411}, {36, 137}, {37, 37}, {38, 37! + 1}, {39, 14029308060317546154181}, {40, 79}, {41, 41}, {42, 41! + 1}, {43, 43}, {44, 59}, {45, 694763}, {46, 293}, {47, 47}, {48, 168070783}, {49, 12893}, {50, 1021}, {51, 149}, {52, 71}, {53, 53}, {54, 9293}, {55, 12318573951317236818169524329}, {56, 79}, {57, 55921523591}, {58, 6323}, {59, 59}, {60, 16567}, {61, 61}, {62, 71}, {63, 6379}, {64, 71}, {65, 233}, {66, 131}, {67, 67}, {68, 309013}, {69, 4139}, {70, 83}, {71, 71}, {72, 6653}, {73, 73}, {74, 73! + 1}, {75, 449}, {76, 293}, {77, 123259}, {78, 77! + 1}, {79, 79}, {80, 7158703}, {81, 672937}, {82, 163}, {83, 83}, {84, xx}, {85, xx}, {86, 383}, {87, 109}, {88, 4663}, {89,89}, {90, 179}, {91, xx}, {92, 317}, {93, xx}, {94, 271}, {95, 486353434961}, {96, 191}, {97, 97}, {98, 251}, {99, xx}, {100, 199}}  This data is in the form {n, smallest factor of ((n1)! + 1)}.
I decided to ask about (4n1)((2n1)!+1) and (4n1)((2n1)!1). Here are the n under 1000 that qualify for each.
{2, 3, 5, 11, 12, 17, 20, 26, 32, 33, 41, 45, 48, 50, 57, 66, 87, 92, 96, 105, 108, 111, 120, 123, 126, 131, 141, 143, 150, 152, 155, 158, 171, 173, 182, 185, 186, 197, 206, 210, 216, 222, 237, 248, 255, 263, 272, 273, 281, 288, 297, 323, 330, 342, 356, 363, 372, 375, 378, 393, 395, 396, 416, 417, 431, 437, 446, 456, 467, 468, 477, 483, 488, 501, 510, 516, 528, 533, 536, 551, 572, 578, 587, 593, 596, 606, 612, 626, 633, 638, 645, 648, 665, 671, 678, 680, 692, 701, 705, 720, 722, 735, 741, 750, 753, 770, 771, 780, 791, 792, 798, 815, 827, 830, 831, 840, 843, 848, 852, 878, 882, 887, 890, 896, 906, 911, 915, 918, 923, 956, 963, 966, 987, 992} {1, 6, 8, 15, 18, 21, 27, 35, 38, 42, 53, 56, 60, 63, 68, 71, 77, 78, 83, 90, 95, 110, 116, 117, 122, 125, 137, 147, 161, 162, 165, 180, 188, 203, 207, 215, 221, 227, 228, 230, 242, 243, 246, 258, 260, 266, 276, 291, 293, 306, 308, 315, 320, 321, 326, 327, 332, 350, 357, 360, 362, 365, 368, 371, 381, 383, 386, 390, 392, 402, 405, 407, 425, 440, 447, 453, 458, 462, 470, 495, 497, 500, 503, 507, 521, 522, 525, 545, 552, 560, 561, 563, 567, 585, 588, 600, 603, 615, 617, 635, 636, 662, 666, 668, 672, 675, 677, 683, 698, 711, 713, 726, 732, 743, 755, 756, 767, 797, 801, 813, 818, 825, 833, 836, 837, 866, 867, 873, 875, 885, 893, 902, 908, 930, 932, 935, 942, 945, 951, 962, 977, 978, 980, 981, 983, 986}
Here is a plot of the gaps between the numbers on the first list. It turns out this is an aspect of Wilson's Corollary. The original Wilson Theorem is that p((p1)!+1) for any prime number p. A lesser known result is that p((p2)!1) for any prime number p. What I rediscovered is that for any prime of the form 4k1 divides ((2k1)!)^2 1. This implies (4k1)((2k1)!+1) or (4k1)((2k1)!1), if (4k1) is prime. Which one? I don't know. I'll pay $5 to the first person that can send me a formula for it. I made a plot of the progressive +1 and 1 terms for the 4k1 primes and it's pretty chaotic. I also noticed that for primes of the form 4k+1, (4k+1)(((2k1)!)^2 + 4).
I thought the above might lead to a good puzzle, but I managed to solve it. I learned something in the process.
Junk Kato: Small puzzle for mathpuzzle challenger. Here is a sample of 2piece Sliding Block Puzzle in 3x3 board. What is largest diameter of 2piece Sliding Block Puzzle in 6x6 board? (Send Answer) Taniguchi's Programs might help.
<Start>
<Finish>
# # # # # # # # # #
# A # # # #
#
# B B # > # B B #
# # # # #
A #
# # # # # # # # # # 5 moves
Lloyd King has started a new puzzle site with some nice ideas. Andrea Gilbert has started analyzing Denki Block puzzles. These are kind of like sliding magnet puzzles. Andrea has found a very difficult puzzle in a 4x4 grid.
Zome Systems has updated their site. The Big Bag of Parts is a great deal.
In response to my message from 5 December, Ron Zeno took a look at sorting 10 elements, and found an improvement over the best known network (Waksman, 1969), which had length 29, delay 9. Zeno found a lot of length 29, delay 8 networks. Ron Zeno: "Waksman did the real work, I just analyzed the solution. There are literally over a million equivalent networks: 1. Any set of 5 disjoint pairs works for the first 5 comparators (945 possibilities), 2. Any 8 of the 10 elements can be used for the next 4 comparators (45 possibilities, and there are 3 or so arrangements of the 8 chosen that will work), 3. After that, there are at least 8 variations of the remaining network = 945*45*3*8 = 1,020,600. Lots of solutions in a vast solution space. (about 9,495^8 different networks of delay 8. That makes the chance of randomly finding one is about 1 in 6.6 * 10^26)." Here is Zeno's list of pairs: {{1, 2}, {3, 4}, {5, 6}, {7, 8}, {9, 10}, {1, 3}, {2, 4}, {5, 7}, {6, 8}, {1, 9}, {2, 7}, {8, 10}, {1, 5}, {3, 8}, {4, 10}, {6, 9}, {2, 6}, {3, 5}, {4, 9}, {7, 8}, {2, 3}, {4, 7}, {5, 6}, {8, 9}, {3, 5}, {4, 6}, {7, 8}, {4, 5}, {6, 7}}. Ron noticed that another 8delay network can be gotten just by starting with Waksman's original sorter and shifting the last operation back two positions.Ron "It's too trivial a change to not have been found by Waksman." Ron also pointed me to a published paper for 9 elements, I. Parberry, "A Computer Assisted Optimal Depth Lower Bound for NineInput Sorting Networks", Mathematical Systems Theory, Vol. 24, pp. 101116, 1991. A second paper by Parberry is Parberry: "On the Computational Complexity of Optimal Sorting Network Verification". Proceedings of The Conference on Parallel Architectures and Languages Europe, SpringerVerlag Lecture Notes in Computer Science, Vol. 506, pp. 252269, June 1991.


Tom Thomas, Roosevelt University: "Bob Kraus suggested that I contact you about a search that I am doing with a high school math teacher. We are looking for logic problems that come from different cultures, e.g., from Asia, Africa that have been translated into English. We are trying to emphasize the international interest in logic and felt that providing high school students with a sampler of logic puzzles that come from different parts of the world would help to make this point. Do you know of any resource that we can consult that would include such information?" If can help here, please send an email to Tom Thomas.
Colin Backhurst, Daniel Reeves, and Ramprasath Lakshminarasimhan (who sent a wonderful analysis) solved Bob Kraus's 8 question challenge.
Don Woods: "I write with some embarrassment this time. I noticed that you had posted some answers to my 20 Questions puzzle, and when I looked at them they didn't match the answer I had in mind, yet (unlike some I'd seen so far) they did appear to be correct. After some effort I found the bug in the program I'd written that "proved" the uniqueness of my intended solution. Oops! Sheer bad luck that the bug happened to be such that the one solution it did report was the one I expected, instead of some other. Sigh.With some effort, I've managed to swap a few questions around and tweak a couple answers here and there, and believe I have "uncooked" the puzzle. I enclose it here." If you can solve 20 Q, Mark 2, write me.
I just so happen to be playing Don's classic game Adventure on my Visor Edge (which just had a $50 drop in price). I've been programming in Lua and C on the thing. I'll try to have a rundown on neat Palm OS software in my next update.
material added 5 December 2001
2^13466917  1 is the latest, greatest mersenne prime, discovered by michael cameron, a 20year old canadian. BBC Story.