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Frequently Asked Questions (FAQS);faqs.468 I invented some new "scramble by rotation" puzzles myself. My favourite creation is the Twisty Torus. It is a torus puzzle with segments (which slide around 360 degrees) with multiple rings around the circumference. The computer puzzle simulation library I'm forming will be described in depth in DOTC #4 (The Domain of the Cube Newsletter). So if you have any interesting computer puzzle programs please email me and tell me all about them! Also to the people interested in obtaining a subscription to DOTC, who are outside of Canada (which it seems is just about all of you!) please don't send U.S. or non-Canadian stamps (yeah, I know I said Self-Addressed Stamped Envelope before). Instead send me an international money order in Canadian funds for $6. I'll send you the first 4 issues (issue #4 is almost finished). Mark Longridge Address: 259 Thornton Rd N, Oshawa Ontario Canada, L1J 6T2 Email: mark.longridge@canrem.com One other thing, the six bucks is not for me to make any money. This is only to cover the cost of producing it and mailing it. I'm just trying to spread the word about DOTC and to encourage other mechanical puzzle lovers to share their ideas, books, programs and puzzles. Most of the programs I've written and/or collected are shareware for C64, Amiga and IBM. I have source for all my programs (all in C or Basic) and I am thinking of providing a disk with the 4th issue of DOTC. If the response is favourable I will continue to provide disks with DOTC. -- Mark Longridge <mark.longridge@canrem.com> writes: It may interest people to know that in the latest issue of "Cubism For Fun" % (# 28 that I just received yesterday) there is an article by Herbert Kociemba from Darmstadt. He describes a program that solves the cube. He states that until now he has found no configuration that required more than 21 turns to solve. He gives a 20 move manoeuvre to get at the "all edges flipped/ all corners twisted" position: DF^2U'B^2R^2B^2R^2LB'D'FD^2FB^2UF'RLU^2F' or in Varga's parlance: dofitabiribirilobadafodifobitofarolotifa Other things #28 contains are an analysis of Square 1, an article about triangular tilings by Martin Gardner, and a number of articles about other puzzles. -- % CFF is a newsletter published by the Dutch Cubusts Club NKC. Secretary: Anneke Treep Postbus 8295 6710 AG Ede The Netherlands Membership fee for 1992 is DFL 20 (about$ 11). -- -- dik t. winter <dik@cwi.nl> References: E. C. Turner & K. F. Gold, "Rubik's Groups", American Mathematical Monthly, vol. 92 (1985), pp. 617-629. Cubelike Puzzles - What Are They and How Do You Solve Them? J.A. Eidswick A.M.M. March, 1986 Rubik's Revenge: The Group Theoretical Solution Mogens Esrom Larsen A.M.M. June-July, 1985 The Group of the Hungarian Magic Cube Chris Rowley Proceedings of the First Western Austrialian Conference on Algebra, 1982 Rubik's Cubic Compendium Erno Rubik, Tamas Varga, et al (Ed by David Singmaster) Oxford University Press, 1987 (Some chapters on mathematics of the cube.) David Singmaster, _Notes on Rubik's `Magic Cube'_ "Winning Ways" by Berlekamp, Elwyn R. Conway, John H. Guy, Richard K. Volume two, pages 760-768, 808, 809 ==> games/rubiks.magic.p <== How do you solve Rubik's Magic? ==> games/rubiks.magic.s <== The solution is in a 3x3 grid with a corner missing. +---+---+---+ +---+---+---+---+ | 3 | 5 | 7 | | 1 | 3 | 5 | 7 | +---+---+---+ +---+---+---+---+ | 1 | 6 | 8 | | 2 | 4 | 6 | 8 | +---+---+---+ +---+---+---+---+ | 2 | 4 | Original Shape +---+---+ To get the 2x4 "standard" shape into this shape, follow this: 1. Lie it flat in front of you (4 going across). 2. Flip the pair (1,2) up and over on top of (3,4). 3. Flip the ONE square (2) up and over (1). [Note: if step 3 won't go, start over, but flip the entire original shape over (exposing the back).] 4. Flip the pair (2,4) up and over on top of (5,6). 5. Flip the pair (1,2) up and toward you on top of (blank,4). 6. Flip the ONE square (2) up and left on top of (1). 7. Flip the pair (2,4) up and toward you. Your puzzle won't be completely solved, but this is how to get the shape. Notice that 3,5,6,7,8 don't move. ==> games/scrabble.p <== What are some exceptional scrabble games? ==> games/scrabble.s <== The shortest scrabble game: The Scrabble Players News, Vol. XI No. 49, June 1983, contributed by Kyle Corbin of Raleigh, NC: [J] J U S S O X [X]U which can be done in 4 moves, JUS, SOX, [J]US, and [X]U. In SPN Vol. XI, No. 52, December 1983, Alan Frank presented what he claimed is the shortest game where no blanks are used, also four moves: C WUD CUKES DEY S This was followed in SPN, Vol. XII No. 54, April 1984, by Terry Davis of Glasgow, KY: V V O[X] [X]U, which is three moves. He noted that the use of two blanks prevents such plays as VOLVOX. Unfortunately, it doesn't prevent SONOVOX. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Record for the highest scrabble score in a single turn (in a legal position): According to the Scrabble Players Newspaper (since renamed to Scrabble Players News) issue 44, p13, the highest score for one turn yet discovered, using the Official Scrabble Players Dictionary, 1st ed. (the 2nd edition is now in use in club and tournament play) and the Websters 9th New Collegiate Dictionary, was the following: d i s e q u i l i b r a t e D . . . . . . . e . . . . . . e . . . . . . . e . . . . . o m r a d i o a u t o g r a p(h)Y . . . . . . . . . . . w a s T . . . . . . . . . . b e . . h . . . . . . . . . . a . . g o . . . c o n j u n c t i v a L . . . . . . . . . . . . . n o . . . . . . . f i n i k i n G . . . . . . . a . . . (l) e i . . . . . . . d . s p e l t Z . . . . . . w e . . . . . . e . . . . . . r . . . . . . o r m e t h o x y f l u r a n e S for 1682 points. According to the May 1986 issue of GAMES, the highest known score achievable in one turn is 1,962 points. The word is BENZOXYCAMPHORS formed across the three triple-word scores on the bottom of the board. Apparently it was discovered by Darryl Francis, Ron Jerome, and Jeff Grant. As for other Scrabble trivia, the highest-scoring first move based on the Official Scrabble Players Dictionary is 120 points, with the words JUKEBOX, QUIZZED, SQUEEZE, or ZYMURGY. If Funk & Wagnall's New Standard Dictionary is used then ZYXOMMA, worth 130 points, can be formed. The highest-scoring game, based on Webster's Second and Third and on the Oxford English Dictionary, was devised by Ron Jerome and Ralph Beaman and totalled 4,142 points for the two players. The highest-scoring words in the game were BENZOXYCAMPHORS, VELVETEEN, and JACKPUDDINGHOOD. The following example of a SCRABBLE game produced a score of 2448 for one player and 1175 for the final word. It is taken from _Beyond Language_ (1967) by Dmitri Borgman (pp. 217-218). He credits this solution to Mrs. Josefa H. Byrne of San Francisco and implies that all words can be found in _Webster's Second Edition_. The two large words (multiplied by 27 as they span 3 triple word scores) are ZOOPSYCHOLOGIST (a psychologist who treats animals rather than humans) and PREJUDICATENESS (the condition or state of being decided beforehand). The asterisks (*) represent the blank tiles. (Please excuse any typo's). Board Player1 Player2 Z O O P S Y C H O L O G I S T ABILITY 76 ERI, YE 9 O N H A U R O W MAN, MI 10 EN 2 * R I B R O V E I FEN, FUN 14 MANIA 7 L T I K E G TABU 12 RIB 6 O L NEXT 11 AM 4 G I AX 9 END 6 I T IT, TIKE 10 LURE 6 * Y E LEND, LOGIC*AL 79 OO*LOGICAL 8 A R FUND, JUD 27 ATE, MA 7 L E N D M I ROVE 14 LO 2 E A Q DARE, DE 13 ES, ES, RE 6 W A X F E N U RE, ROW 14 IRE, IS, SO 7 E T A B U I A DARED, QUAD 22 ON 4 E N A M D A R E D WAX, WEE 27 WIG 9 P R E J U D I C A T E N E S S CHIT, HA 14 ON 2 PREJUDICATENESS, AN, MANIAC, QUADS, WEEP 911 OOP 8 ZOOPSYCHOLOGIST, HABILITY, TWIG, ZOOLOGICAL 1175 -------------------------------------- Total: 2438 93 F, N, V, T in loser's hand: +10 -10 -------------------------------------- Final Score: 2448 83 --------------------------------------------------------------------------- It is possible to form the following 14 7-letter OSPD words from the tiles: HUMANLY FATUOUS AMAZING EERIEST ROOFING TOILERS QUIXOTE JEWELRY CAPABLE PREVIEW BIDDERS HACKING OVATION DONATED ==> games/square-1.p <== Does anyone have any hints on how to solve the Square-1 puzzle? Xref: bloom-picayune.mit.edu rec.puzzles:18145 news.answers:3076 Newsgroups: rec.puzzles,news.answers Path: bloom-picayune.mit.edu!enterpoop.mit.edu!snorkelwacker.mit.edu!usc!wupost!spool.mu.edu!uunet!questrel!chris From: uunet!questrel!chris (Chris Cole) Subject: rec.puzzles FAQ, part 10 of 15 Message-ID: <puzzles-faq-10_717034101@questrel.com> Followup-To: rec.puzzles Summary: This posting contains a list of Frequently Asked Questions (and their answers). It should be read by anyone who wishes to post to the rec.puzzles newsgroup. Sender: chris@questrel.com (Chris Cole) Reply-To: uunet!questrel!faql-comment Organization: Questrel, Inc. References: <puzzles-faq-1_717034101@questrel.com> Date: Mon, 21 Sep 1992 00:09:31 GMT Approved: news-answers-request@MIT.Edu Expires: Sat, 3 Apr 1993 00:08:21 GMT Lines: 1487 Archive-name: puzzles-faq/part10 Last-modified: 1992/09/20 Version: 3 ==> games/square-1.s <== SHAPES 1. There are 29 different shapes for a side, counting reflections: 1 with 6 corners, 0 edges 3 with 5 corners, 2 edges 10 with 4 corners, 4 edges 10 with 3 corners, 6 edges 5 with 2 corners, 8 edges 2. Naturally, a surplus of corners on one side must be compensated by a deficit of corners on the other side. Thus there are 1*5 + 3*10 + C(10,2) = 5 + 30 + 55 = 90 distinct combinations of shapes, not counting the middle layer. 3. You can reach two squares from any other shape in at most 7 transforms, where a transform consists of (1) optionally twisting the top, (2) optionally twisting the bottom, and (3) flipping. 4. Each transform toggles the middle layer between Square and Kite, so you may need 8 transforms to reach a perfect cube. 5. The shapes with 4 corners and 4 edges on each side fall into four mutually separated classes. Side shapes can be assigned values: 0: Square, Mushroom, and Shield; 1: Left Fist and Left Paw; 2: Scallop, Kite, and Barrel; 3. Right Fist and Right Paw. The top and bottom's sum or difference, depending on how you look at them, is a constant. Notice that the side shapes with bilateral symmetry are those with even values. 6. To change this constant, and in particular to make it zero, you must attain a position that does not have 4 corners and 4 edges on each side. Almost any such position will do, but returning to 4 corners and 4 edges with the right constant is left to your ingenuity. 7. If the top and bottom are Squares but the middle is a Kite, just flip with the top and bottom 30deg out of phase and you will get a cube. COLORS 1. I do not know the most efficient way to restore the colors. What follows is my own suboptimal method. All flips keep the yellow stripe steady and flip the blue stripe. 2. You can permute the corners without changing the edges, so first get the edges right, then the corners. 3. This transformation sends the right top edge to the bottom and the left bottom edge to the top, leaving the other edges on the same side as they started: Twist top 30deg cl, flip, twist top 30deg ccl, twist bottom 150deg cl, flip, twist bottom 30deg cl, twist top 120deg cl, flip, twist top 30deg ccl, twist bottom 150deg cl, flip, twist bottom 30deg cl. Cl and ccl are defined looking directly at the face. With this transformation you can eventually get all the white edges on top. 4. Check the parity of the edge sequence on each side. If either is wrong, you need to fix it. Sorry -- I don't know how! (See any standard reference on combinatorics for an explanation of parity.) 5. The following transformation cyclically permutes ccl all the top edges but the right one and cl all the bottom edges but the left one. Apply the transformation in 3., and turn the whole cube 180deg. Repeat. This is a useful transformation, though not a cure-all. 6. Varying the transformation in 3. with other twists will produce other results. 7. The following transformation changes a cube into a Comet and Star: Flip to get Kite and Kite. Twist top and bottom cl 90deg and flip to get Barrel and Barrel. Twist top cl 30 and bottom cl 60 and flip to get Scallop and Scallop. Twist top cl 60 and bottom cl 120 and flip to get Comet and Star. The virtue of the Star is that it contains only corners, so that you can permute the corners without altering the edges. 8. To reach a Lemon and Star instead, replace the final bottom cl 120 with a bottom cl 60. In both these transformation the Star is on the bottom. 9. The following transformation cyclically permutes all but the bottom left rear. It sends the top left front to the bottom, and the bottom left front to the top. Go to Comet and Star. Twist star cl 60. Go to Lemon and Star -- you need not return all the way to the cube, but do it if you're unsure of yourself by following 7 backwards. Twist star cl 60. Return to cube by following 8 backwards. With this transformation you should be able to get all the white corners on top. 10. Check the parity of the corner sequences on both sides. If the bottom parity is wrong, here's how to fix it: Go to Lemon and Star. The colors on the Star will run WWGWWG. Twist it 180 and return to cube. 11. If the top parity is wrong, do the same thing, except that when you go from Scallop and Scallop to Lemon and Star, twist the top and bottom ccl instead of cl. The colors on the Star should now run GGWGGW. 12. Once the parity is right on both sides, the basic method is to go to Comet and Star, twist the star 120 cl (it will be WGWGWG), return to cube, twist one or both sides, go to Comet and Star, undo the star twist, return to cube, undo the side twists. With no side twists, this does nothing. If you twist the top, you will permute the top corners. If you twist the bottom, you will permute the bottom corners. Eventually you will get both the top and the bottom right. Don't forget to undo the side twists -- you need to have the edges in the right places. Happy twisting.... -- Col. G. L. Sicherman gls@windmill.att.COM ==> games/think.and.jump.p <== THINK & JUMP: FIRST THINK, THEN JUMP UNTIL YOU ARE LEFT WITH ONE PEG! O - O O - O / \ / \ / \ / \ O---O---O---O---O BOARD DESCRIPTION: To the right is a model of \ / \ / \ / \ / the Think & Jump board. The O---O---O---O---O---O O's represent holes which / \ / \ / \ / \ / \ / \ contain pegs. O---O---O---O---O---O---O \ / \ / \ / \ / \ / \ / O---O---O---O---O---O DIRECTIONS: To play this brain teaser, you begin / \ / \ / \ / \ by removing the center peg. Then, O---O---O---O---O moving any direction in the grid, \ / \ / \ / \ / jump over one peg at a time, O - O O - O removing the jumped peg - until only one peg is left. It's harder then it looks. But it's more fun than you can imagine. SKILL CHART: 10 pegs left - getting better 5 pegs left - true talent 1 peg left - you're a genius Manufactured by Pressman Toy Corporation, NY, NY. ==> games/think.and.jump.s <== Three-color the board in the obvious way. The initial configuration has 12 of each color, and each jump changes the parity of all three colors. Thus, it is impossible to achieve any position where the colors do not have the same parity; in particular, (1,0,0). If you remove the requirement that the initially-empty cell must be at the center, the game becomes solvable. The demonstration is left as an exercise. Karl Heuer rutgers!harvard!ima!haddock!karl karl@haddock.ima.isc.com Here is one way of reducing Think & Jump to two pegs. Long simplifies Balsley's scintillating snowflake solution: 1 U-S A - B C - D 2 H-U / \ / \ / \ / \ 3 V-T E---F---G---H---I 4 S-H \ / \ / \ / \ / 5 D-M J---K---L---M---N---O 6 F-S / \ / \ / \ / \ / \ / \ 7 Q-F P---Q---R---S---T---U---V 8 A-L \ / \ / \ / \ / \ / \ / 9 S-Q W---X---Y---Z---a---b 10 P-R / \ / \ / \ / \ 11 Z-N c---d---e---f---g 12 Y-K \ / \ / \ / \ / 13 h-Y h - i j - k 14 k-Z The board should now be in the snowflake pattern, i.e. look like o - * * - o / \ / \ / \ / \ *---o---*---o---* \ / \ / \ / \ / *---*---*---*---*---* / \ / \ / \ / \ / \ / \ o---o---o---o---o---o---o \ / \ / \ / \ / \ / \ / *---*---*---*---*---* / \ / \ / \ / \ *---o---*---o---* \ / \ / \ / \ / o - * * - o where o is empty and * is a peg. The top and bottom can now be reduced to single pegs individually. For example, we could continue 15 g-T 16 Y-a 17 i-Z 18 T-e 19 j-Y 20 b-Z 21 c-R 22 Z-X 23 W-Y 24 R-e which finishes the bottom. The top can be done in a similar manner. -- Chris Long ==> games/tictactoe.p <== In random tic-tac-toe, what is the probability that the first mover wins? ==> games/tictactoe.s <== Count cases. First assume that the game goes on even after a win. (Later figure out who won if each player gets a row of three.) Then there are 9!/5!4! possible final boards, of which 8*6!/2!4! - 2*6*4!/0!4! - 3*3*4!/0!4! - 1 = 98 have a row of three Xs. The first term is 8 rows times (6 choose 2) ways to put down the remaining 2 Xs. The second term is the number of ways X can have a diagonal row plus a horizontal or vertical row. The third term is the number of ways X can have a vertical and a horizontal row, and the 4th term is the number of ways X can have two diagonal rows. All the two-row configurations must be subtracted to avoid double-counting. There are 8*6!/1!5! = 48 ways O can get a row. There is no double- counting problem since only 4 Os are on the final board. There are 6*2*3!/2!1! = 36 ways that both players can have a row. (6 possible rows for X, each leaving 2 possible rows for O and (3 choose 2) ways to arrange the remaining row.) These cases need further consideration. There are 98 - 36 = 62 ways X can have a row but not O. There are 48 - 36 = 12 ways O can have a row but not X. There are 126 - 36 - 62 - 12 = 16 ways the game can be a tie. Now consider the 36 configurations in which each player has a row. Each such can be achieved in 5!4! = 2880 orders. There are 3*4!4! = 1728 ways that X's last move completes his row. In these cases O wins. There are 2*3*3!3! = 216 ways that Xs fourth move completes his row and Os row is already done in three moves. In these cases O also wins. Altogether, O wins 1728 + 216 = 1944 out of 2880 times in each of these 36 configurations. X wins the other 936 out of 2880. Altogether, the probability of X winning is ( 62 + 36*(936/2880) ) / 126. win: 737 / 1260 ( 0.5849206... ) lose: 121 / 420 ( 0.2880952... ) draw: 8 / 63 ( 0.1269841... ) 1000000 games: won 584865, lost 288240, tied 126895 Instead, how about just methodically having the program play every possible game, tallying up who wins? Wonderful idea, especially since there are only 9! ~ 1/3 million possible games. Of course some are identical because they end in fewer than 8 moves. It is clear that these should be counted multiple times since they are more probable than games that go longer. The result: 362880 games: won 212256, lost 104544, tied 46080 #include <stdio.h> int board[9]; int N, move, won, lost, tied; int perm[9] = { 0, 1, 2, 3, 4, 5, 6, 7, 8 }; int rows[8][3] = { { 0, 1, 2 }, { 3, 4, 5 }, { 6, 7, 8 }, { 0, 3, 6 }, { 1, 4, 7 }, { 2, 5, 8 }, { 0, 4, 8 }, { 2, 4, 6 } }; main() { do { bzero((char *)board, sizeof board); for ( move=0; move<9; move++ ) { board[perm[move]] = (move&1) ? 4 : 1; if ( move >= 4 && over() ) break; } if ( move == 9 ) tied++; #ifdef DEBUG printf("%1d%1d%1d\n%1d%1d%1d w %d, l %d, t %d\n%1d%1d%1d\n\n", board[0], board[1], board[2], board[3], board[4], board[5], won, lost, tied, board[6], board[7], board[8]); #endif N++; } while ( nextperm(perm, 9) ); printf("%d games: won %d, lost %d, tied %d\n", N, won, lost, tied); exit(0); } int s; int *row; over() { for ( row=rows[0]; row<rows[8]; row+=3 ) { s = board[row[0]] + board[row[1]] + board[row[2]]; if ( s == 3 ) return ++won; if ( s == 12 ) return ++lost; } return 0; } nextperm(c, n) int c[], n; { int i = n-2, j=n-1, t; while ( i >= 0 && c[i] >= c[i+1] ) i--; if ( i < 0 ) return 0; while ( c[j] <= c[i] ) j--; t = c[i]; c[i] = c[j]; c[j] = t; i++; j = n-1; while ( i < j ) { t = c[i]; c[i] = c[j]; c[j] = t; i++; j--; } return 1; } ==> geometry/K3,3.p <== Can three houses be connected to three utilities without the pipes crossing? _______ _______ _______ | oil | |water| | gas | |_____| |_____| |_____| _______ _______ _______ |HOUSE| |HOUSE| |HOUSE| | one | | two | |three| ==> geometry/K3,3.s <== The problem you describe is to draw a bipartite graph of 3 nodes connected in all ways to 3 nodes, all embedded in the plane. The graph is called K3,3. A famous theorem of Kuratowsky says that all graphs can be embedded in the plane, EXCEPT those containing K3,3 or K5 (the complete graph on 5 vertices, i.e., the graph with 5 nodes and 10 edges) as a subgraph. So your problem is a minimal example of a graph that cannot be embedded in the plane. The proofs that K5 and K3,3 are non-planar are really quite easy, and only depend on Euler's Theorem that F-E+V=2 for a planar graph. For K3,3 V is 6 and E is 9, so F would have to be 5. But each face has at least 4 edges, so E >= (F*4)/2 = 10, contradiction. For K5 V is 5 and E is 10, so F = 7. In this case each face has at least 3 edges, so E >= (F*3)/2 = 10.5, contradiction. The difficult part of Kuratowsky is the proof in the other direction! A quick, informal proof by contradiction without assuming Euler's Theorem: Using a map in which the houses are 1, 2, and 3 and the utilities are A, B, and C, there must be continuous lines that connect the buildings and divide the area into three sections bounded by the loops A-1-B-2-A, A-1-B-3-A, and A-2-B-3-A. (One of the areas is the infinite plane *around* whichever loop is the outer edge of the network.) C must be in one of these three areas; whichever area it is in, either 1, or 2, or 3, is *not* part of the loop that rings its area and hence is inaccessible to C. The usual quibble is to solve the puzzle by running one of the pipes underneath one of houses on its way to another house; the puzzle's instructions forbid crossing other *pipes*, but not crossing other *houses*. ==> geometry/bear.p <== If a hunter goes out his front door, goes 50 miles south, then goes 50 miles west, shoots a bear, goes 50 miles north and ends up in front of his house. What color was the bear? ==> geometry/bear.s <== The hunter's door is in one of two locations. One is a foot or so from the North Pole, facing north, such that his position in front of the door is precisely upon the North Pole. Since that's a ridiculous place to build a house and since bears do not roam within fifty miles of the pole, the bear is either imaginary or imported, and there is no telling what color it is. There is another place (actually a whole set) on earth from which one can go fifty miles south, fifty miles west, and fifty miles north and end up where one started. Consider the parallel of latitude close enough to the South Pole that the circumference of the earth at that latitude is 50/n miles, for some integer n. Take any point on that parallel of latitude and pick the point fifty miles north of it. Situate the hunter's front porch there. The hunter goes fifty miles south from his porch and is at a point we'll call A. He travels fifty miles west, going n times around the earth, and is at A again, where he shoots the bear. Fifty miles north from A he is back home. Since bears are not indigenous to the Antarctic, again the bear is either imaginary or imported and there is no telling what color it might be. ==> geometry/bisector.p <== If two angle bisectors of a triangle are equal, then the triangle is isosceles (more specifically, the sides opposite to the two angles being bisected are equal). ==> geometry/bisector.s <== The following proof is probably from Altshiller-Court's College Geometry, since that's where I first saw the problem. Let the triangle be ABC, with angle bisectors BE and CD. Let F be such that BEFD is a parallelagram. Let x = measure of angle CBE = angle DBE, y = measure of angle BCD = angle DCE, x' = measure of angle EFC, y' = measure of angle ECF. (You will probably want to draw a picture.) Suppose x > y. Consider the triangles EBC and DCB. Since BC = BC and BE = CD, we must have CE > BD. Now, since BD = EF, we have that CE > EF, so that x' > y'. Thus x+x' > y+y'. But, triangle FDC is isosceles, since DF = BE = DC, so x+x' = y+y', a contradiction. Similarly, we cannot have x < y. Therefore the base angles of ABC are equal, making ABC an isosceles triangle. QED ==> geometry/calendar.p <== Build a calendar from two sets of cubes. On the first set, spell the months with a letter on each face of three cubes. Use lowercase three-letter abbreviations for the names of all twelve months (e.g., "jan", "feb", "mar"). On the second set, number the days with a digit on each face of two cubes (e.g., "01", "02", etc.). ==> geometry/calendar.s <== First note that there are *nineteen* different letters in the month abbreviations (abcdef gjlmno prstuv y) so to get them all on the eighteen faces of 3 cubes, you know right away you're going to have to resort to trickery. So I wrote them all down and looked at which ones could be reversed to make another letter in the set. The only pair that jumped out at me was the d/p pair. Now I knew that it was at least feasible, as long as it wasn't necessary to duplicate any letters.