I got my spiffy solid solids from Peda.com, and decided to roll each of the odder solids 100 times. Out of 100 rolls, the triangles came up 17 times on both the icosidecahedron and the cuboctahedron. On the rhombitruncated cuboctahedron, squares came up 5 times, hexagons 22 times, and decagons 73 times. If any students would like to get a set and roll them a lot as a class project, I'd be happy to post results. Are there rolling methods that can influence the odds?

Andrew Fenwick did a lovely job with Things that Roll. He's made his applet available. I much enjoyed playing around with it, especially the generated graphs. As might be expected, the simplest diagrams gave the most complicated graphs. One little design of mine seemed quite elegant, so I present it here. There are two ways to put a 1x2x3 slab at the center of a 5x7 grid. By a series of rolls, get from one to the other. Can you solve my Plus Maze? I also liked a 1x3 chasm maze I came up with. Based an idea by Adrian Fisher, the log you're rolling can span either of the two pits, but its ends can't land in them. Can you solve my Chasm maze? Answers.

Only Claudio Baiocchi found the intended solution for Erich Friedman's triangle puzzle. Matt Elder, Dave Hanson, Max Hille, Hugh Everett, Hagen von Eitzen, Jaap Scherphuis, and Nathan Stohler all found solutions involving infinite or overlapping triangles.

Joseph DeVincentis, Matt Elder, Dharmashankar Subramanian, Mike Gerfin, Andrew Marshall, Jim Reed, Robert Patterson and myself all found a slightly complicated method for making 100 with four 7s and one 1. Only Jeff Vermette found 177-77 = 100.

Wei-Hua Huang (now at Google) suggested Mathematical Curves with Lego.

From Erich Friedman: Try to find a collection of equal triangles with the property that every point that is the vertex of a triangle is the vertex of exactly 3 triangles. Is there a configuration that allows 4 vertices? (we don't think so). Also, he's made a new series of Number Puzzles.

Cihan Altay has posted the interesting answers to the Optimization Challenges.

Steve Finch has moved his wonderful Mathematical Constants site. Worth a look, if you haven't seen it. He would also like to see an improvement on the solutions of Shortest Paths when Lost problems.

If you've ever wanted a solid platonic solids in aluminum, with a cuboctahedron, icosidodecahedron, and rhombitruncated cuboctahedron , then take the link to peda.com. They have spiffy solid solids there. Their programs are nice, too, especially if you're on a tight budget -- you can try them for free. If you have a bit more, I'd recommend Mathematical Explorer.

Do you know what the second hardest substance is? While doing some research on boron nitride and boron carbide, I learned about the new substance BAM (Boron, Aluminum, Magnesium). Here is the story. In short, assistant scientist Bruce Cook was investigating thermoelectric properties of materials, when he found that he couldn't cut his sample with a diamond saw. That caught his attention. I want some BAM myself. Let me go on record by stating I want a BAM ring.

Anna Perez passed along the puzzle of how to make 100 using four 7's and one 1, using +,-,x, /. It has a valid solution, with no tricks. Rawal Vijay Dinesh passed along a different puzzle: Arrange 0 to 9 in a circular fashion such that the difference between any adjacent two numbers is either 3, or 4 or 5. I accidently solved the problem using only 1 to 9. That puzzle has 6 solutions, one of them special. Can you find it? Toss in 0, and the problem has no solution. Can you prove that? Solvers and solution.

Peter Esser:

Eric von Laudermann:

I visit your site often, but this is the first time I've ever written to you. I couldn't resist the squares with intersecting lines, and I solved a bunch of the more interesting ones. The one with 8 variables has a particularly beautiful symmetry. Bitmaps are attached. [Ed - I'll redraw the whole set of solutions, smaller]

At one point, you had a link to a page of 12 puzzles from the author of the book "which Way Did the Bicycle Go?" Is there a solution page for these? It took me forever to prove that it's impossible to tile the 8x8x8 cube (minus two corners) with 1x1x3 blocks. [Ed - that would be stanwagon.com. Solution]

Here's a lovely ambigram puzzle by Carlos Penedo. Can you figure out what it is? If not, here is the Answer.

I've been fascinated by elements recently. My favorite, now, is Cesium --The Femme Fatale of all elements -- as beautiful as it is deadly. Curiously, when Uranium fissions, it doesn't break up into equal pieces, it tends to break up into Cesium and Strontium. For a spectacular picture of all the elements, see http://private.addcom.de/pniok/ .

John Gowland: A couple of years ago you sent me a list of primes whose products were 5-digit numbers which could be considered as a straight in Poker. You challenged me to construct a crossnumber with the information. As with all crossnumbers, you have to restrict the conditions or you have too much information. I looked at the products of two primes only and realised that combinations of 12345 or 45678, for instance, would always be divisible by 3. I then looked at 34567 and found my condition. As ever, even if you don't need it, I still enjoyed constructing it!

34567.(by John Gowland) In this crossnumber, p and q are primes whose product is some arrangement of 34567. This product is clued as a 3 digit number r consisting of 3 consecutive digits. For example, 101 x 647 = 65347; the latter would be entered as 653, 534, or 347. Across lights are denoted by capitals and down lights are denoted by small letters. There are no zeroes in the completed diagram. Hint grid.

Samantha Levin: I wrote a logic puzzle which I think your readers might enjoy. It was inspired by one that was part of the 2002 MIT Mystery Hunt (that puzzle is at http://www.mit.edu/~puzzle/02/red/B/Puzzle.html).

1. In "20 Questions," the magic word would fall under the category "Vegetable."
2. In "20 Questions," the magic word would fall under the category "Mineral."
3. In "20 Questions," the magic word would fall under the category "Animal."
4. Exactly one-third of the true statements fall between the first statement and this statement, inclusive.
5. The statement listed four statements prior to this one is true.
6. At most one of the statements numbered with a power of two is false.
7. Exactly half of the statements numbered with a multiple of 7 are true.
8. At most half of the true statements fall between the first statement and this statement, inclusive, and if the magic word has an even number of letters then it is exactly half.
9. The magic word can be obtained by writing down the numbers of all the true statements, assigning a letter to each using the pattern A=1, B=2, etc., and anagramming the result.
10. This statement and the next have different truth values.
11. At least half the statements are true.
12. None of the statements numbered with a multiple of 4 are true.
13. This statement would be just as true as it is now if it were replaced with "Every statement that begins with 'The magic word' is true."
14. Of this statement and the preceding two, an odd number are true.
15. Statements 6 and 7 are either both true or both false.
16. Exactly one of the final two statements is true.
17. There is exactly one vowel in the magic word.
18. The magic word is a commonly used English word of at least 6 letters, worth an even number of Scrabble points; and if its letters were put in alphabetical order, no more than three of them would be consecutive.
19. Statements 4 and 5 are either both true or both false.
20. Exactly half the statements whose numbers yield a remainder of two when divided by six are true.

What is the magic word? Answer.

One problem I created for the WPC, based on an idea by Scott Kim, is the quintessential intelligentsia qualifying test final analysis. To solve it, place letters into the grid below so that a chess king can spell out each word and phrase: quintessential, intelligentsia, qualify test, final analysis. The king may revisit a square in the course of spelling out a word or phrase, but may not stay in the same square twice in a row. One letter has been added to help you. Can you solve it? Answer and Solvers. If you like this puzzle, Scott Kim put a slew of easier challenges in the current Games magazine. If you don't have Games in your country (it's an american games and puzzles magazine), then tell me what magazines you do have in your country. I'd like to make a list of them.

Theodore Gray has made a 3x6 foot version of his creation available on E-bay. I'll risk being a periodic periodic table table poster poster by mentioning the auction again, later. You can see the table and the poster auction here. In an effort to fill out the remaining elements, Theo and I are starting a "Wanted Elements of the Week." This week, we want to locate Boron, Scandium, and Vanadium. If you can give us visible samples of any of these elements, or sell them to us at a reasonable price, please let me know. We'd even take compounds of these elements, in a pinch. I did manage to track down a stamp-sized piece of Boron foil for \$2040.00, but that price seems a tad high. If elements and particles intrigue you, then you owe it to yourself to pick up the best free tiny blue thing aroung, the Particle Data Book. A new version of it will be coming out soon, so send in your address so you can get it free.

Erich Friedman noted square numbers 69696, 56722567225, and 95540955409. He wondered if there were more.
Denis Borris found a few with 19 digits (root on right), using a simple program:
1 10803324 1 10803324 1 (1052631579), 1 22009616 1 22009616 1 (1104579631),
4 31208078 4 31208078 4 (2076555028), 4 43213296 4 43213296 4 (2105263158),
4 88038464 4 88038464 4 (2209159262), 9 02430788 9 02430788 9 (3004048583),
9 97229916 9 97229916 9 (3157894737)
Richard Heylen found a method to generate an infinite number of these
129040592478166186761489901502889757103604704754068821088679921443542038411^2 =
16651474507116159856813341545933208272942509227175391166243089978499940492\
166514745071161598568133415459332082729425092271753911662430899784999404921
What is the method? As a hint, Richard says "As you might guess, the whole thing revolves around factorisations of 10^n+1. Because the number of factors in 10^n+1 generally increases as n increases, we get more and more of these numbers of a given length." (If you figure it out his method, please write me.)

Richard also sent a few other numbers that can be squared, as samples:
204314987546823982381066977511198358076663273062132807686418481254054869292^2
230762992688093026681489642356714998431787626040608497448167915008115529705^2
300245405658393315181104571316264157230554105835946632285612904286075581227^2
140594059405940594059405940594059405940594059405940594059405940594059405941^2

He also notes "50251406405844705^2= 2525203845765370 2525203845765370 25 has the digits going through the cycle twice plus two digits rather than the normal twice plus 1 but these are not terribly easy to find. I don't think one can find a square where the digits go through the cycle three times and a bit but I can't see a proof. You might be able to prove that there is a finite number of them. I won't spill the beans about the method until others have had a go and, more importantly, until I have tidied up and tied up all the loose ends."

Last week, my Windows system (I also have a Linux system) started crashing pretty regularly, and was slowing down a lot. I found the information about startup applications very, very useful. Now, my system is nice and speedy again (by year 2000 standards).

In a bit of preparation for the World Puzzle Championship qualifier (register by the 13th if you want to compete), Erich Friedman sent me a few puzzles. In the first, Spiral galaxies, you must divide the square into rotationally symmetric polyominoes. Just three of the pentominoes are rotationally symmetric.

Spoiler warning: If you click here, you'll see the answer to the puzzle above. You'll also see Erich's write-up on how Spiral Galaxy puzzles are NP-complete, by finding NOT, AND, and OR gates, signal duplications, wire crossings, and wire shifts.

Similarly, Minesweeper is NP-complete. If you have Mirek's Cellebration, WireWorld is NP-complete. I don't know of an easier way to lay out boolean circuits. Here is the .mcl version of the picture below.

Erich Friedman: I did a computer search for squares whose digits repeat in blocks but with at least two periods. the only such squares less than 100 million are 69696, 56722567225, and 95540955409. are these the only ones? Ed: Are they? Seems like an interesting question. He also tells me that Joseph DeVincentis solved the L packing challenge.

I've fled from Verio (with \$100/Gig bandwidth overage fees). I'm now on Yahoo, and it seems okay so far. Another hosting service that looks even better, costwise, is tera-byte.com.

Edward Brisse has made available a large number of formulae for Triangle Centers. Another grand list is at The Encyclopedia of Triangle Centers. On a recent long trip, I took Clark Kimberling's book with me, Triangle Centers and Central Triangles. It's one of my favorite books. I've been pondering a method for examining each triangle center in a polygonal way. Let A B C D E be the vertices of a pentagon (this can be changed to your favorite n-gon). Take one of the many triangle centers, and apply it to A B C, then B C D, then C D E, then D E A, and then E A B. You now have a new pentagon. Repeat the above process. Four things can happen: 1. Convergence to a point, 2. Divergence, 3. Cyclic behavior, 4. Chaotic behavior. Which Triangle Centers are also convergent Polygon centers? Perhaps I'll have some clues next week.

David Bush wonders what happens in a pentagonal version of Hex. I like his name for it: Pex. I still like Gonnect quite a lot.

There is a trio of very nice Java sites for the Riemann Zeta Function: Plotter, Wave, Partials. I'm trying to figure out an easy way to plot the final false convergents of the Zeta function. In the pictures below, the Blue Square is in the spiral of the final convergent. The previous spirals are all false convergents.

See the spectacular Periodic Table table at theodoregray.com. We've added a lot of things to it since I premiered it here.

This page is hosted on Yahoo. Winners of my Tetrods contest are Adam Dewbery, Helmut Postl, Brendan Owen, Juha Hyvönen, Stanislaw Jedrus, Roel Huisman, Carlos Gil, Dario Uri, Patrick Hamlyn, and Sterten Guenterbrink.

Tom Turrittin sent me the following observations on squares. Someday, I'll try to clean this up and study it more closely. In the meantime, Send an Analysis if you get far with the unsolved cases.

Site disruption. My former service provider, Verio, charges \$100 per gig of bandwidth in overage fees. The going rate for bandwidth from more reputable companies is \$2 per gig of bandwidth. I had an overage in May, and had to suffer a rather severe shakedown by Verio. At the moment, Verio has triple billed me for June. I'm moving to a different provider. If you run a website, and your provider doesn't advertise reasonable rates for bandwidth overages (Earthlink and Verio both charge \$100 per gig), you should consider finding a more reputable provider immediately. I hope to have everything in place with Yahoo as my provider by 10 June, and an update.

The qualifying round for the World Puzzle Championship will be at http://wpc.puzzles.com this year. You must sign up by June 13th.

SIAM 100-Digit Challenge solutions are available at Stan Wagon's page. At the bottom of Worm Pattern Inventory, there is a list of 11 simple programs whose outcome is unknown. Eric Solomon has made a java program of gravity mazes. And here's nice page about Soma. I've added another page on Pathos ... I'll try to summarize it all soon.

Ron Zeno-- I've created the attached small file of examples of sorting networks for 2 to 16 inputs with the best-known number of comparisons and delay-time. You mentioned that you were interested in creating something like this yourself, so I thought you'd be like a copy. The file is designed to be displayed at 2x or 3x magnification. The networks for n > 8 are all my own creations, though they are structurally identical to those in Knuth.

Aron Fay -- For the petersen graph in a 3x4 grid, I solved the puzzle, and included a graphic for it. Bjorn Moreau -- Attached you'll find a graphical representation of my answer. I also attach a solution to the dodecagon.

Alexandre Owen -- Here are petersen graphs on four different arrangements of vertices. A crude metric for the difficulty of this sort of puzzle is the mean number of vertices that each vertex cannot "see". From left to right, these puzzles have difficulties of 2.4, 2.4, 3.2, and 1.6.

At the Gathering for Gardner 5, my puzzle exchagne was Alex Randolph's TETRODS, because I really liked the puzzles given by Franz-Josef Schulte in his article about them in Cubism For Fun. There is a full set of 30 puzzles in DIE DREI by Franjos. I gave several extra sets to the Cubism For Fun group, and one of them got back to Franz-Josef, who sent me some extra puzzles. Basically, rearrange the 13 pieces in the first figure to get each of the following 3 figures. Each is a tetromino with a hole, and the pieces may be flipped over. There is also a single square with a hole. The first puzzle has a 5x4 blank area. As a fourth puzzle, find a 7x7 square with a 7x3 blank area. I'll send a copy of Walter Hoppe's laser-cut rendition of Tetrods to the ten best sets of solutions I receive by 11 PM, CST, 3 Jun 2002. Send your mailing address and solutions.

An interview with puzzlemaster Scott Kim that is now online at http://www.puzzlementonline.com/scottkim.htm .

Michael Rios sent me a logic problem: The Fourth Annual E R Emmet Verity Competition was recently held in my hometown. Five contestants were judged on their ability to tell the truth under a variety of circumstances and were rated first (highest) through fifth (lowest). Afterwards, each of the five contestants made two comments:
Alf:     Bert wasn't first.
Charlie finished lower than Duggie.
Bert:    I was second.
Alf finished lower than Edmund.
Charlie: Alf finished higher than of Duggie.
Bert was fourth.
Duggie:  Edmund was second.
Charlie wasn't last.
Edmund:  Bert wasn't first.
I was third.
I noted with interest that the person who was ranked first told two true statements, the person ranked last told two false statement, and the other
three contestants told one true and one false statement. Can you determine the order of the five contestants? Answer and Solvers.

Sales of A New Kind of Science have exceeded all expectations. There is quite a lot of new material available at wolframscience.com. One sample page to look at in particular is 908, with a graph based on a million points of the Ulam sequence. Does the 13.5 n result hold for the next million terms?

The impact of Bill Tutte can be seen seen in recent articles on his life (1, 2, 3, 4). I'm intrigued by the series of events: 1. A Nazi officer made some minor correction to a lengthy directive, and had it recoded and resent. 2. In order to at least look busy, Bill copied the two intercepted Nazi codes onto graph paper, and looked for patterns. 3. After four months of intense analysis, he spotted several patterns, enough to determine that the Nazi FISH machine used sprockets with 41 and 31 teeth. 4. The techniques he developed needed a large computational device, so COLOSSUS was built, the world's first electronic computer. 5. Bill's techniques, with the assistance of COLOSSUS, led to the routine cracking of Nazi messages. 6. The now-possible deduction of Nazi dispositions and intentions allowed for Allied invasion of Normandy in June 1944.

On 14 May, A New Kind of Science will be in stores. In 1981, a young Stephen Wolfram was interested in a class of simple programs, and decided to look at all 256 possibilities. Several of these simple programs led to surprising results. Since then, Stephen developed Mathematica, then used it to exhaustively analyze trillions of simple programs of many varieties. The book teems with thousands of intriguing results, many connected directly with real-world phenomena. Here is a small sample from page 833... he looked for 3-color cellular automata that doubled the width of an initial pattern. Out of 7625597484987 possible rules, 4277 of them satisfied the requirement. Here are six of them:

From page 833 of A New Kind of Science (© 2002 Stephen Wolfram, LLC)

One item of note is the book's 2400 dpi printing process. To see it, find NKS in a bookstore and page through it.

I've learned quite a bit about the elements in the process of helping with the Periodic Table. Here's an interesting story about Building a Breeder Reactor at home. Of the elements i've recently gotten, Indium, Tungsten, and Bismuth are my favorites. Gallium was a big disappointment.

Matthew Prins pointed this out to me, at Sport Illustrated. "The USFSA has come up with a less radical plan [to revamp
the judging of figure skating], which it will present to the ISU Congress. Under its proposal, the 6.0 system stays in place, except the high and low marks are thrown out. The median mark of the remaining seven would be displayed as the skater's score." Let's think about this. There are nine scores, they ditch the high and low, and they take the median of the other seven? Couldn't they just save themselves a step?