Theo Gray has built a Periodic Table. A full photo history is available. I helped him a bit, and hope to write up some of my elemental discoveries soon. As an example, take a look at this molybdenum ingot.

The Catalan conjecture, that 8 and 9 are the only consectutive powers, has been proven. This has been expected for awhile, as mentioned by Ivars Peterson. There, he mentions "Preda Mihailescu of the Swiss Federal Institute of Technology in Zurich has proved a theorem that is likely to lead to a solution of Catalan's conjecture." The theorem bounded any possible answers to the conjecture within a computer searchable space, and it seems Mihailescu has finished the search.

Mathematician Bill Tutte has died. Among his many accomplishments, he disproved Tait's conjecture. I talked about this and about him on 19 November. In 2001, one of his conjectures was proven. Snark Theorem: Any Snark has a Petersen Graph minor. (Conjectured by Tutte, called Snarks by Gardner, proven by Robertson, Sanders, Seymour and Thomas.)

NPL puzzler Kray will run Intercoastal Altercations, a team puzzle solving event, on May 11th. See his site for details, if you would like to put a team together.

"That I make examinación has adapted to the left the reduction of you, the ' causes that I go to the layers of scherblockes. Nothing is applicable. And nothing to stop approximately the suspension arrives." -- A Beatles lyric put through The Babelizer. Can you identify it? Answer.

While on a plane flight, I tried to solve the Petersen Graph in Pathos. I wasn't successful, but I did notice a nice two-crossing configuration with straight lines in a 4x5 grid. I wondered how small of a grid it would fit on, and managed to get it onto a 3x4 grid. In fact, I put it on a 3x3 with an added dot on the top. One of the puzzles of the week is draw the edges of the Petersen graph using the vertices of the central figure below. Send Answer. Next to it is a small grid representation of a cube.

Ramprasath Lakshminarasimhan solved the Petersen Graph. A's winning move is the green path. I received wonderful analyses of the cube, K5, and K3-3. I include there the findings of Richard Hoshino, who has been studying a similar game where the graph can never be disconnected. The cube, K5, K3-3, and Petersen are all unsolved in that variant. Send Analyses.

Yoshuki Kotani presented a very nice puzzle at G4G5. I forwarded it to Will Shortz, and he used it as the NPR puzzle of the week, with Dr. Kotani's permission. It's solvable entirely with logic.

The MIT Mystery Hunt has been updated with a new year of puzzles.

Patrick Hamlyn has succeeded in placing the 84 unholey hexakites into 7 hexagons.

Pathos is a game invented and investigated by Frank Harary and Desh Ranjan. Of all the things I saw at Gathering for Gardner, this was my favorite. There are two players, A and B, and a graph G.Aalways moves first. A begins by erasing the edges of any path in G. Then B erases the edges of any path that remains, and so forth. Whenever an isolated node appears, it is erased. The player who erases the last edge wins.

Legal (blue) and Illegal (red) moves in Pathos.

The above image shows some legal (blue) and illegal (red) moves in Pathos. The blue moves are all valid paths. (It is fine for the graph to become unconnected following a move.) The first red move is illegal because it is a cycle, not a path. The second and third red moves contain a cycle, so they are not paths. The fourth red figure has disconnected moves, so it isn't a path. The final red is a tree, not a path. All of the red moves would be illegal.

Graphs grouped into first and second player wins.

The above image shows some connected graphs that are known wins. The first block contains graphs that are wins for A, the first player. The second block contains graphs which are wins for B, the second player. Any path is a win for A. Any cycle is a win for B. Any game with two identical graphs is a win for B (mirror strategy). The first blue move, above, is a winning move for A. Much remains unknown about Pathos. Who wins K5, the connected graph of 5 nodes? Who wins on the Petersen graph? Who wins on the cube? Answers.

For my puzzle of the week, I've determined that A wins on K3-3, the graph below. What is a winning move for A? Answers. (Note that 4-1-6-3-2 is a valid move, since a path may cross over itself. But then B would play 1-2-5-4.)

Find a winning Pathos move for the first player.

Andrei Khodulyov has solved many problems involving rigid polygons. You can see the results at Erich Friedman's Math Magic. Nick Baxter is preparing the World Puzzle Championship qualifier. If you want to hone your skills for it, pick up World-Class Puzzles from the World Puzzle Championship. I talk a bit about these particular puzzles here.

The National Public Radio puzzle of the week is by me.

The Gathering for Gardner was lots of fun, and I have much to report on. Quickies: You can look at some of the pictures I showed at my talk. Dan Murray has started a page which includes my Dice paper. Erich Friedman's Math Magic this month is about L packings in rectangles. (L's are concatenations of squares in the shape of an L.) He's offering \$10 to the first square packing of consecutive L's, L's of width 1, 2, 3, ... n. Cihan Altay is starting a PQRST Solving competition. Peter Esser is offering a solver for MacMahon's colored triangles. Wei-Hwa Huang has solutions for the Five problem posted, and he solved the Six problem.

Currently, the worst known convex packer in 2D space is the smoothed octagon. The packing density is about .902, compared with the .906 packing density of the circle. Ulam conjectured that the sphere was the worst packing object in 3D space. Was he right? I don't think so. Can someone calculate the best lattice packing density of the smoothed octagon when rotated on and axis of symmetry?

Elwyn Berlekamp will award at least one prize of \$1000 (Canadian) for the best Clobber composition. A "free" clobber position is also fine... you don't necessarily have to start with a checkerboard pattern.

ANNOUNCEMENT of CLOBBER PROBLEM COMPOSITION CONTEST

CLOBBER is a new combinatorial game invented last summer in Halifax by Michael Albert, J. P. Grossman, and Richard Nowakowski. The first competitive CLOBBER tournament was held at Dagstuhl, Germany, in February 2002.

Clobber is played by two players, white and black, on a rectangular nxm checkerboard. In the initial position, all squares are occupied by a stone, with white stones on the white squares and black stones on the black squares. A player moves by picking up one of his/her stones and "clobbering" an opponent's stone on an adjacent square (horizontally or vertically). The clobbered stone is removed from the board and replaced by the stone that was moved. The game ends when one player, on his/her turn, is unable to move, and then that player loses.

The game typically decomposes into many smaller positions. Values of some of these positions have been cataloged , and tools for further study of clobber and other games are available.

This note announces a \$1,000 (Canadian) PRIZE FOR THE BEST CLOBBER PROBLEM COMPOSITION(S).

At least one winner will be announced at the following meeting:
Third International Conference on Computers and Games
Email: cg2002@cs.ualberta.ca
URL: http://www.cs.ualberta.ca/~cg2002

If many outstanding problems are submitted, there will be more than one prize. The winning problems and their solutions are intended to be published, with commentary by the judge(s).

A composed problem must specify a position and which player is to move next. Each submission must also be accompanied by a solution proposed by the composer. Although the solution might include a modest number of computer-generated values, figures, or tables, it must be intelligible to humans who have no access to any machines. The solution should indicate how to play against all plausible opposing strategies. Ideally, the proof that the solution is correct should also require no machine assistance.

Difficult-yet-elegant problems which utilize combinatorial game concepts such as atomic weights and/or thermography are especially welcome (e.g., see the Childish Hackenbush "Lollipops" problem in Chapter 8 of Wining Ways). It is very desirable (but not required) that a problem have a history going back to the conventional starting position. The board size must not exceed 10 x 10. Smaller board sizes are more desirable unless they entail excessive compromises with the primary goals of difficulty and elegance.

Any entry, including solution, should not exceed three 8.5 x 11 inch pages. To be eligible for a prize, an entry should be submitted to
berlek@math.berkeley.edu and must be received before 11:59 PM, PDT, on July 20, 2002.

material added 31 March 2002 (Happy Easter)

Happy Easter! (A cross died, for you). Solutions by Juha Saukkola and Nick Baxter. Jim Boyce proves 44 maximal.

Next weekend, I'll be going to the Gathering for Gardner convention. Wei-Hwa Huang will present the following puzzle there:

"This year's theme is Five. Find a set S of lattice points on a 2-dimensional grid, such that there is a special distance D where each point in S is exactly length D away from exactly 5 other points in the set. What is the smallest possible distance D? What is the smallest size for S? Wei-Hwa's answer will be given at G4G5, and will be posted on mathpuzzle.com soon after."

If the theme had been four, instead, the above solution would do. Each point is at a distance of Sqrt[5] from exactly 4 other points. The lattice Wei-Hwa considered is a square lattice, but I'd be very interested in seeing a solution with a distance of 7 on a triangular lattice, since I couldn't solve it. (The 3-5-7 triangle has a 120 degree angle). Answer.

It is possible to arrange observers so that each observer cannot see 4 others (solution by Aron Fay). Can this be extended to 5?

I took a look through puzzles that few have solved, and one find by Guenter Stertenbrink is particularly nice. You might recall an earlier puzzle where a 17*17 square could be dissected into 7 differently sized dominoes and trominoes. Guenter solved that, and sent me this extension: "My Computer found a 20*20 with 11 pieces , and the 17*17 can also be done with 9 pieces."

Last week, I wondered about the relation between Loch and Khinchin. The above plots X for p, 2p, g, Log[2], 2^(1/3), and 3^(1/3). It kind of answers my question. It isn't a proof, though, so my \$5 challenge remains open.

Jerry Slocum will present some of his puzzles on the Martha Stewart Living Television program on the CBS network on April Fool's Day (Monday April 1, 2002).

Find a cubic die, and some double-6 dominoes. Imagine that the dominoes are 1x2 in size and foldable, and that the die is 3x3x3 in size. Toss out the 0-0 domino, and completely cover the face of the die with the 27 others. Pretty easy. Now, give yourself 1 point for every domino square that is on the matching die square. The most points I was able to get was 40 points. I'm thinking 42 points might be possible, with 7 matching pips on each cube face. Samantha Levin, Roger Phillips, Nathan Stohler, and Clinton Weaver all beat my score of 40. Jim Boyce, Juha Saukkola and Nick Baxter found solutions for the maximum score of 44.

Darrel C Jones sent me an alphametric: MEOWTH + JAMES = JESSIE. Substitute 0-9 for the letters. Juha Saukkola, Federico Kereki, Clinton Weaver, Roger Phillips, Jared Marks, Nathan Stohler, Ishihama Yoshiaki, Samantha Levin, Brian Trial, Peter Exterkate, and Joseph DeVincentis solved it.

Three Nice Pentomino Coloring Problems by Alexandre Owen Muñiz is well worth a look. So is the Guide to Appearances of Mathematics and Mathematicians on The Simpsons. Junk Kato let me know where to find more Japanese Temple Geometry problems.

A G Clarke, Al Stanger, Dane Brooke, Chris Lusby.Taylor, clark cooper, Juha Hyvönen, neil harris, David Yamanishi, Joseph DeVincentis, Jorge Montero, John Gowland, Jeremy Galvagni, Heinrich Hemme, Graeme McRae, George Tolley, Derek Bosch, David J Bush, Darrel C Jones, Clinton Weaver, Kaushik Laskar, Kent Cooper, Jared Marks, Mark Michell, Larry Rosenhein, Nick Baxter, Ole Poul Pedersen, Peter Exterkate, Nathan Stohler, Roger Phillips, Seth Kleinerman, jkmclean, Lyman Hurd, Jeff Vermette, Mark J. Tilford, Matt Coury, Jesse L Locust, and Michael Dufour solved Federico Kereki's chessborad puzzle.

Pi continued fraction = {3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, 1, 84, 2, 1, 1, 15, 3, 13, 1, 4}

Sequence A001203 has a story behind it, as told by Hans Havermann (sent to the Sequence Fan list 3/14):

"In late 1984, I complained to Martin Gardner, then mathematical-games columnist for Scientific American, about the paucity of simple continued fraction terms for the number pi. C.D. Olds' "Continued Fractions" had whetted my 1960s teenage-appetite with only the first 23 numbers of the sequence. Mr. Gardner let me know that a Bill Gosper had calculated many thousands of terms and was kind enough to supply me with Mr. Gosper's address.

Incredibly, after I sent Mr. Gosper my inquiry, he mailed me a 205-page Xerox of his (19 August 1977) 204103-term computer calculation. I still have it. Bill went on to calculate 17001303 terms in 1985, a feat that (if I remember correctly) was mentioned at the time in "Science News".

By 1995, I was able to calculate only a measly 10000 terms of the number using Mathematica on my first Macintosh. Lack of adequate memory was the major impediment to progress. That year I looked up Bill Gosper's e-mail and asked for an electronic version of his calculations. He replied: "I just lost a disk with 17000000. When I get a new one, we'll see how well my old backup tapes work."

When Mathematica 4 came out in May 1999, I immediately ordered my upgrade. They had incorporated the ContinuedFraction function into the body of the program (it used to reside in an add-on package) and, on my first try, I was able to come up with 10 million terms (of pi) on my G3/300/384. Next I calculated 17 million. Finally, playing with assorted larger values, I managed to get 20 million terms without running out of memory. The calculation took 6 hours and resulted in a 62 MB output file. Simon Plouffe of Plouffe's Inverter believed the 20 million terms were a record and was kind enough to give them web-space: <http://pi.lacim.uqam.ca/piDATA/>

One year later, a RAM doubling allowed me to calculate 40 million terms. And, in October 2000, a 400 MHz iMac G3 equipped with a GigaByte of RAM churned out 53 million terms in under 20 hours.

Last month I became the proud owner of an 800 MHz flat-panel iMac G4, running OS X. Because of OS X's significantly different memory architecture, I was able to churn out 100 million terms with no more RAM than I had in my previous computer. Filled with bold expectations I tried for a billion, only to run into a Mathematica overflow-error. And my subsequent 200-million attempt shut down the Mathematica kernel because of a lack of memory. So there *was* a limit to what even OS X could do for me.

On 10 March, I finished a 62.5-hour run that generated 160 million terms. Mathematica did not have enough memory, alas, to input the entire file, so I was forced to use a text-editor (BBEdit Lite) to search and subsequently shape the file (I wanted 100 numbers per line).

Gosper's 878783625 [A033089] at position 11504931 [A033090] is still unsurpassed. 5983 is the smallest number yet to appear. There's still a gap to bridge between the 160 million attained and the 200 million no-go. In the meantime, I'm web-sharing the 180 million term files: <http://odo.ca/~haha/j/seq/cfpi/>.

Ed again. Neat story. That last link is well worth a look. Hans does a lot of exploring with the Khinchin Constant, one of the least understood constants (it is only known to 7350 places). My first question after reading this was "How many decimals does 160 million terms of Pi give us?" My question was answered by Eric Weisstein, who pointed me to Loch's Theorem. If you have m terms of Pi, and count n digits of decimal accuracy, then

So Hans has calculated about 160000000 * 1.0306408341 ~ 165 million decimal terms. Loch's theorem doesn't just apply to Pi, it applies to almost every number. The list of numbers where Loch and Khinchin don't apply includes rational numbers, quadratics, and powers of E. Is there a number that works for Khinchin, but not Loch? Is there a number that works for Loch, but not Khinchin? I'll conjecture {Khinchin numbers} = {Loch numbers}, and I'll pay \$5 for either a proof, or counterexample.

Federico Kereki sent me this: I cut myself a rectangle out of a chessboard. (All cuts were along the square lines, so the rectangle has integer sides.) Were I to tell you either its area, or its perimeter, or the length of its diagonal, you wouldn't be able to determine the dimensions of the rectangle. What's its size?

It's always good to check chessvariants.com and zillions-of-games.com every once in awhile. At the first, I noticed LOTR Chess puzzles. With a bit more searching, I found the new site for Go Variations. One thing not listed at Other Boards is the third dimension. Ishihama Yoshiaki has been studying 3-D Go on the lattice of a Diamond. It's being called (surprise) Diamond Go. I suggested that he add Graphite Go, and he did. I'm curious about how well a 2 layer hexagonal game will work. I've started a small game of Graphite with him on 7 hexagons, or 48 vertices, and it seems a bit better than 7x7 go, so far. (Incidently, take a look at his Genetic Programming page, especially if you have a Mac).

Curiously, the idea of hexagonal columns has recently been applied to drying cornstarch. This is a great double experiment. First, prepare the thixotropic mixture, water (1 part) and cornstarch (2 parts). (Websites about this at Kid Wizard,ChemInst,Argo). If you have it in a bowl, you can try a gentle toss, as if you were trying to splash it, and the suspension will stay in the bowl. Once done, pour it onto a shallow plate, and let it dry over a period of days. Once dry, put some cardboard on top, and flip the plate over. The cornstarch will dry to form hexagonal columns, much like thebasalt columns of Devil's Postpile or Giant's Causeway. It's caused due to the fact that hexagons are the most efficient form of stress relief. You can read about this at nature.com. Amazingly, the properties of drying cornstarch were only noticed a few years ago by Dr. Gerhard Müller, Eduardo Jagla, and Alberto Rojo.

Juha Hyvönen sent me the below solutions for fixed tetrominoes + pentomino. Clinton Weaver, Peter Esser, Andrew Clarke, Robert Henderson, and Roel Huisman sent solutions.

Mark's Math Pages has a very nice section on Japanese Temple problems.

I pondered a possible game this weekend, where cards or dominoes would represent the first 14 triangular numbers: 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, and 91 (D0 = 0). By summing three of these numbers, all but one target number from 1 to 172 can be represented (which one is missing?). I'd use 1-100 in a game. Gauss found a difficult proof that any integer was the sum of three triangulars. I wondered if I might spot a pattern in it all that would lead to an easier proof, but didn't. John Conway gave me some advice: "You might like to look in my little book, The Sensual Form, which discusses Legendre's 3-squares theorem (of which the three triangles theorem is the particular case 8n+3) in a way that will probably help you to understand what's going on. The proof hasn't really changed since the days of Gauss and Legendre, and my guess is that it will remain hard."

While trying to figure out a fair distribution of cards, I noticed that D7 (28) was essential. 29, 44, 50, 53, 74, 83, 119, and 239 all require D7 for the D+D+D sum. The first seven, and D10 and D12, are all essential. 36 (D8) seemed to be nonessential, and I recursively constructed a whole list of nonessential Ds. The list starts {8, 11, 16, 17, 23, 24, 29, 31, 36, 38, 41, 43, 45, 49, 50, 59, 60, 61, 62, 65}. Here is a list of how to solve D+D+D = X for the first thousand integers. Richard Schroeppel noted that every positive integer is the sum of four squares, and you don't need 49 (or a bunch of others).

1 0 0 0 0 0 1 1 1 1 1
0 1 0 0 0 0 0 1 2 2 1
0 0 1 0 0 0 1 0 1 2 2
0 0 0 1 0 0 2 1 0 1 2
0 0 0 0 1 0 2 2 1 0 1
0 0 0 0 0 1 1 2 2 1 0

This all got started when Jack Wert sent me a solution for finding an odd coin among 1092, in 7 weighings. I also pondered a game using the ternary Golay code, but that hasn't gotten anywhere either. As long as I'm rambling, let me point out the harpsichord piece La Gemissante by Jean François Dandrieu 1682-1738. It sounds like something from a Hayao Miyazaki movie. I also had fun with close approximations this weekend. e ~ 3 - (5/63)^(1/2), p ~ 4 - (105/166)^(1/3), g ~ 1/15 + (35/263)^(1/3), and (23/9)^5 ~ 109. Richard Guy tells me the last one is well known (to those who well know it), to be the best known example (due to Eric Reyssat) in connection with the abc conjecture.

For more approximations, Marc LeBrun noticed g ~ 3^(-1/2) and ln 2 ~ 1/sqrt(2+1/sqrt(151+1/sqrt(746))). Simon Plouffe sent log(13981) ~ 9.5454545454...., log(163) ~ 163/32, Zeta(3) ~ log(19)/sqrt(6), exp(689/(Pi*396) ) ~ 689 / 396, log(124)^3 ~ 112, log(5275)^2 ~ 75, and log(631)^3 ~ 268.

Alex Fink, Jeff Smith, and Rick Shepard solved Juha Hyvöne's crossnumber puzzles. Serhiy Grabarchuk recommends the 3D sliding puzzles by Paul Thomas. Ivan Skvarca has put together a page commemorating the International Day of Symmetry. The New York Times recently gave their readers a choice of two headlines: "Another Round of Layoffs Is Planned At First Boston" and "Another Round of Layoffs Are Planned At First Boston".

I also tried these: There are 54 tetromino configurations in 3-space. Can all of them be packed perfectly in a 6x6x6 cube? (I don't know) There are 19 tetromino configurations in 2-space. Can all of them, and a pentomino, be packed in a 9x9 square? Yes! Roel Huisman sent a lovely answer with the X pentomino at the center. Can you find it? Send Answer. Patrick Hamlyn found a nice solution for splitting one hexomino, and using the rest to spell out 2002 in a square. The puzzle idea is by Rodolfo Kurchan.

Dario Uri has tracked down the Chinese Rings puzzle to Luca Pacioli, around the year 1500, in the manuscript "De Viribus Quantitatis", which recorded the library of the University of Bologna at that time. Problem 107: "Do Cavare et Mettere una Strenghetta Salda in al Quanti Anelli Saldi, Difficil Caso" (Remove and put a little bar joined in some joined rings. Difficult case) He sent me a copy of the very pages which describes it.