Polyforms and their tilings, packings and tesselations is a Restricted Group with 85 members. Yahoo Groups Triangles with Mac Mahon's pieces Esser Message 1 of 14 , Apr 14, 2002 ----------------------- In his "Mathematic Magic Show" Gardner wrote about Mac Mahon's colored triangles and asked for solutions to fill a triangle with a complete set. For n=1 it's no problem but with the next possible n=24 we get 4624 pieces. With some modifications of my progam I got a solution for this case. Peter Roel Huisman Apr 14, 2002 ----------------------- ...n=24 we get 4624 pieces. With some modifications of my progam I got a solution for this case. Hang on a sec..let me check if it's a real solution... Nah, maybe not. Nice picture Peter! sterten@aol.com Apr 14, 2002 ----------------------- Peter: >File: smalltriangle.gif (91621 bytes) > >In his "Mathematic Magic Show" Gardner wrote about Mac Mahon's when was it ? >colored triangles and asked for solutions to fill a triangle >with a complete set. For n=1 it's no problem but with the next >possible n=24 we get 4624 pieces. With some modifications of >my progam I got a solution for this case. > >Peter 4624=C(24,3)*2+C(24,2)*2+24 so these are one-sided pieces. By looking at the picture I can't recognize 24 colors , but I did notice, that the corners look dark and the middle light. So I guess, you managed to reduce the colors by placing all pieces with a certain subset of colors first and then solving the reduced-colors-puzzle with an as-circular-as-possible shape for the endgame ? if I made no mistake ,there is one other possible triangle: 48 colors and two-sided pieces... Guenter Esser Apr 15, 2002 ----------------------- Hide message history ----- Original Message ----- From: To: Sent: Sunday, April 14, 2002 7:10 PM Subject: Re: [polyforms] Triangles with Mac Mahon's pieces > Peter: > > >File: smalltriangle.gif (91621 bytes) > > > >In his "Mathematic Magic Show" Gardner wrote about Mac Mahon's > Guenter: > when was it ? The book was printed in German in 1981 and Gardner cited Philpott's articles in JRM vol4 from 1971 and vol5 from 1972 > > >colored triangles and asked for solutions to fill a triangle > >with a complete set. For n=1 it's no problem but with the next > >possible n=24 we get 4624 pieces. With some modifications of > >my progam I got a solution for this case. > > > > 4624=C(24,3)*2+C(24,2)*2+24 > so these are one-sided pieces. Yes, the pieces are one-sided. > > By looking at the picture I can't recognize 24 colors , Perhaps you would like to zoom in? but I did > notice, that the corners look dark and the middle light. > So I guess, you managed to reduce the colors by placing all > pieces with a certain subset of colors first and then > solving the reduced-colors-puzzle with an as-circular-as-possible > shape for the endgame ? > first stage: all colors from 1 to 16 second stage: at least two colors form 1 to 16 third stage: one or two colors form 1 to 16 endgame: all colors from 17 to 24 and some pieces left from the third stage > if I made no mistake ,there is one other possible triangle: > 48 colors and two-sided pieces... > > With the formula (n^3+2n+3n^2)/6 for two-sided pieces we get 19600=140^2 pieces with 48 colors. But the number of edges with one color is 19600*3/48=1225. Therefore a uniform border cannot be realized. There may be some solutions for n>5000 due to the above mentioned articles. Peter Esser Apr 17, 2002 ----------------------- Hide message history ----- Original Message ----- From: To: Sent: Sunday, April 14, 2002 7:10 PM Subject: Re: [polyforms] Triangles with Mac Mahon's pieces > Peter: > > >File: smalltriangle.gif (91621 bytes) > > > >In his "Mathematic Magic Show" Gardner wrote about Mac Mahon's > Guenter: > when was it ? The book was printed in German in 1981 and Gardner cited Philpott's articles in JRM vol4 from 1971 and vol5 from 1972 > > >colored triangles and asked for solutions to fill a triangle > >with a complete set. For n=1 it's no problem but with the next > >possible n=24 we get 4624 pieces. With some modifications of > >my progam I got a solution for this case. > > > > 4624=C(24,3)*2+C(24,2)*2+24 > so these are one-sided pieces. Yes, the pieces are one-sided. > > By looking at the picture I can't recognize 24 colors , Perhaps you would like to zoom in? but I did > notice, that the corners look dark and the middle light. > So I guess, you managed to reduce the colors by placing all > pieces with a certain subset of colors first and then > solving the reduced-colors-puzzle with an as-circular-as-possible > shape for the endgame ? > first stage: all colors from 1 to 16 second stage: at least two colors form 1 to 16 third stage: one or two colors form 1 to 16 endgame: all colors from 17 to 24 and some pieces left from the third stage > if I made no mistake ,there is one other possible triangle: > 48 colors and two-sided pieces... > > With the formula (n^3+2n+3n^2)/6 for two-sided pieces we get 19600=140^2 pieces with 48 colors. But the number of edges with one color is 19600*3/48=1225. Therefore a uniform border cannot be realized. There may be some solutions for n>5000 due to the above mentioned articles. Peter I sent this message on monday but I didn't get it back. Sorry, if you get it once more. Peter sterten@aol.com Apr 17, 2002 ----------------------- >> Peter: >> >> >File: smalltriangle.gif (91621 bytes) >> > >> >In his "Mathematic Magic Show" Gardner wrote about Mac Mahon's > >> Guenter: >> when was it ? > >The book was printed in German in 1981 and Gardner cited Philpott's >articles in JRM vol4 from 1971 and vol5 from 1972 >> >> >colored triangles and asked for solutions to fill a triangle >> >with a complete set. For n=1 it's no problem but with the next >> >possible n=24 we get 4624 pieces. With some modifications of >> >my progam I got a solution for this case. >> > >> >> 4624=C(24,3)*2+C(24,2)*2+24 >> so these are one-sided pieces. > >Yes, the pieces are one-sided. >> >> By looking at the picture I can't recognize 24 colors , > >Perhaps you would like to zoom in? > > but I did >> notice, that the corners look dark and the middle light. >> So I guess, you managed to reduce the colors by placing all >> pieces with a certain subset of colors first and then >> solving the reduced-colors-puzzle with an as-circular-as-possible >> shape for the endgame ? >> >first stage: all colors from 1 to 16 >second stage: at least two colors form 1 to 16 >third stage: one or two colors form 1 to 16 >endgame: all colors from 17 to 24 and some pieces left from the third stage > >> if I made no mistake ,there is one other possible triangle: >> 48 colors and two-sided pieces... >> >> >With the formula (n^3+2n+3n^2)/6 for two-sided pieces we get 19600=140^2 >pieces with 48 colors. But the number of edges with one color is >19600*3/48=1225. yes >Therefore a uniform border cannot be realized. There may the borderlength is 3*140=420 , which is much less than 1225. What am I missing ? Maybe 3 colors for the border ,one for each side , were also acceptable. >be some solutions for n>5000 due to the above mentioned articles. > >Peter > >I sent this message on monday but I didn't get it back. Sorry, if you get it >once more. Yes, I'd got it, thanks. Maybe you expected an answer, sorry then. What you wrote was somehow expected and the way how I would have done it. But you can never know in advance how well such thinks work in practice.How difficult was it ? Where can we expect the limit for this kind of puzzles ? Can we construct a puzzle of this sort which we expect has solutions but these will be too hard to find ? Does Gardner also mention 3d-puzzles with cubes or tetraedal disphenoids (or however it was called, the "tetras" which can build pyramides) ? Guenter sterten@aol.com Apr 17, 2002 ----------------------- >> By looking at the picture I can't recognize 24 colors , > >Perhaps you would like to zoom in? can you post it in ASCII with letters instead of colors ? { so I can check easily by computer, whether you cheated ;-) } A A B A C D A A E A C A D C E F G H . . . . . . . . Esser Apr 18, 2002 ----------------------- me: > >> > >With the formula (n^3+2n+3n^2)/6 for two-sided pieces we get 19600=140^2 > >pieces with 48 colors. But the number of edges with one color is > >19600*3/48=1225. > guenter: > yes > > >Therefore a uniform border cannot be realized. > > the borderlength is 3*140=420 , which is much less than 1225. > What am I missing ? > 1225 is odd. The left edges must match and the number must be even. > Maybe 3 colors for the border ,one for each side , were also acceptable. > Same argument as above. > What you wrote was somehow expected and the way how I would have done it. > But you can never know in advance how well such thinks work in > practice.How difficult was it ? It didn't take much time (< one hour) to finish the first three stages, but then I got stuck. At last I changed the program so that I could manually remove some pieces of the third stage. After inspecting about 10 solutions for the first three stages the endgame needs about four hours on a 166 Mhz. > > Where can we expect the limit for this kind of puzzles ? For the onesided triangles there is only a finite number of solutions for (n^3+2n+3n^2)/6 =m^2 according to Philpott's article. ;-) Yes, it's a good question, but I haven't got an idea. If I compare the constructions of the irregular hexagon with 1376 pieces and the triangle with 4624 pieces, the main difference was a more complicated planning of the stages. > Can we construct a puzzle of this sort which we expect > has solutions but these will be too hard to find ? > > Does Gardner also mention 3d-puzzles with cubes or > tetraedal disphenoids (or however it was called, the "tetras" which > can build pyramides) ? > > Yes, he mentioned cubes. http://mitglied.lycos.de/jkoeller/ has something on Mac Mahon's cubes on his site. Peter Esser Apr 18, 2002 ----------------------- Hide message history ----- Original Message ----- From: To: Sent: Thursday, April 18, 2002 5:25 AM Subject: Re: [polyforms] Triangles with Mac Mahon's pieces > >> By looking at the picture I can't recognize 24 colors , > > > >Perhaps you would like to zoom in? > > can you post it in ASCII with letters instead of colors ? > { so I can check easily by computer, whether you cheated ;-) } > > > > A A > B > A C D A > A E > A C A D C E > F G H > . . . . . . . . > If I read your example correctly, you only show the triangles with top up. The other triangles can be derived from this representation. In my representation only the triangles with top down are shown. Here is the size 68 triangle in ASCII: Peter sterten@aol.com Apr 19, 2002 ----------------------- hi Peter, >1225 is odd. The left edges must match and the number must be even. ahh,yes. So start with any predefined border you like. >It didn't take much time (< one hour) to finish the first three stages, but >then I got stuck. At last I changed the program so that I could manually >remove some pieces of the third stage. >After inspecting about 10 solutions for the first three stages the endgame >needs about four hours on a 166 Mhz. a real human-computer cooperation ! Can you estimate the number of solutions ? >> Where can we expect the limit for this kind of puzzles ? >For the onesided triangles there is only a finite number of solutions for >(n^3+2n+3n^2)/6 =m^2 according to Philpott's article. ;-) but they are ridiculously large , as I understood your previous message >Yes, it's a good question, but I haven't got an idea. If I compare the >constructions of the irregular hexagon with 1376 pieces and the triangle >with 4624 pieces, the main difference was a more complicated planning of the >stages. which hexagon ? McMahon-like-colored hexagons to tile a planar shape ? I'm too lazy to derive the corresponding formula for the number of pieces. But for the solvability of the puzzle it doesn't matter a lot whether the set is complete is not , or if we even allow some repeated pieces. >> Can we construct a puzzle of this sort which we expect >> has solutions but these will be too hard to find ? I mean, by adding pieces or adding colors or adding dimensions or adding sides (or removing) can we construct puzzles of ever increasing difficulty and decreasing number of solutions per time unit ? I'm not sure about the optimum number of pieces,colors,edges. We can also color the vertices or the centers or use colored n-iamonds instead of just triangles etc. >> Does Gardner also mention 3d-puzzles with cubes or >> tetraedal disphenoids (or however it was called, the "tetras" which >> can build pyramides) ? >> >> >Yes, he mentioned cubes. http://mitglied.lycos.de/jkoeller/ has something >on Mac Mahon's cubes on his site. yes , thanks. Lots of other tilings too. I haven't already read all of them.. I displayed 24 pictures of the ASCII-file , one for each color. (solution is correct, Brendan !) No pattern was found, only that you spared all pieces with colors 17-24 for the center. Maybe it's better to spare only those which don't contain previous colors. So, e.g. exhaust all pieces which contain a color < 13 first , using as few sides with color >12 as possible , and then you are left with a real 12-color subpuzzle. Partition the colors again, if possible etc. Guenter Esser Apr 21, 2002 ----------------------- Hide message history ----- Original Message ----- From: To: Sent: Friday, April 19, 2002 1:34 PM Subject: Re: [polyforms] Triangles with Mac Mahon's pieces hi Guenter, > >1225 is odd. The left edges must match and the number must be even. > > ahh,yes. So start with any predefined border you like. > > May be 16 triangels of size 35 and each with three differently colored edges are possible. Can a subset be selected so that the triangles are permutations of each other? > > >> Where can we expect the limit for this kind of puzzles ? > >For the onesided triangles there is only a finite number of solutions for > >(n^3+2n+3n^2)/6 =m^2 according to Philpott's article. ;-) > > but they are ridiculously large , as I understood your previous message > Yes, I think so, too. > >Yes, it's a good question, but I haven't got an idea. If I compare the > >constructions of the irregular hexagon with 1376 pieces and the triangle > >with 4624 pieces, the main difference was a more complicated planning of the > >stages. > > which hexagon ? A hexagon constructed from triangles. I mentioned it two weeks ago, but didn't post the gif. It's incuded now. > But for the solvability of the puzzle it doesn't matter a lot > whether the set is complete is not , or if we even allow some > repeated pieces. You are right, but I like complete sets. > > >> Can we construct a puzzle of this sort which we expect > >> has solutions but these will be too hard to find ? > > I mean, by adding pieces or adding colors or adding dimensions > or adding sides (or removing) can we construct puzzles of ever > increasing difficulty and decreasing number of solutions > per time unit ? > I'm not sure about the optimum number of pieces,colors,edges. What's the highest number of hexagons with uniform border in a given figure? This question seems to be hard to solve. > > We can also color the vertices or the centers or use colored n-iamonds > instead of just triangles etc. Now I try to do it with isosceles triangles as mentioned in my message from 16.02.2002. > > >> Does Gardner also mention 3d-puzzles with cubes or > >> tetraedal disphenoids (or however it was called, the "tetras" which > >> can build pyramides) ? > > I displayed 24 pictures of the ASCII-file , one for each color. > (solution is correct, Brendan !) Thank you for checking the solution. > Maybe it's better to spare only those which don't contain > previous colors. So, e.g. exhaust all pieces which contain > a color < 13 first , using as few sides with color >12 as possible , > and then you are left with a real 12-color subpuzzle. > Partition the colors again, if possible etc. > In this case the main problem is the transition to the last stage, because the common boundary must only have colors >12 , which you would like to spare. Peter sterten@aol.com Apr 22, 2002 ----------------------- Peter: >> >1225 is odd. The left edges must match and the number must be even. >> >> ahh,yes. So start with any predefined border you like. >> >> >May be 16 triangels of size 35 and each with three differently colored edges >are possible. >Can a subset be selected so that the triangles are permutations of each >other? you mean: once we made one of the 35-35-35s , we can get the 15 others by permuting the colors ? Each of the 16 triangle would have to contain 3 unicolored,141 double-colored and 1081 tri-colored pieces. I don't know whether it's possible nor can I prove that it's not. It should be a permutation of order 16. So in cycle notation : (1,2,...,16)(17,..,32)(33,..,48) would be a candidate. take pieces: 1-1-1,17-17-17,33-33-33 1-1-{2..48},17-17-{1..16,18..48},33-33-{1..32,34..48} 2- } ... } haven't already figured this out 48- } but suppose, we find such a decomposition, do you think you can solve such a 1225-pieces-triangle with 48 colors ? >>> irregular hexagon with 1376 pieces and the triangle with 4624 pieces, >>> the main difference was a more complicated planning of >>> the stages. >> >> which hexagon ? >A hexagon constructed from triangles. I mentioned it two weeks ago, but >didn't post the gif. It's incuded now. I see. I remember vaguely. With all these tilings here , I forget after some weeks what exactly has been tiled with which :-( You tiled a 39-8-8-39-8-8 hex with 1376 triangles. I was speculating you'd used "McMahon-hexagons". >> I mean, by adding pieces or adding colors or adding dimensions >> or adding sides (or removing) can we construct puzzles of ever >> increasing difficulty and decreasing number of solutions >> per time unit ? >> I'm not sure about the optimum number of pieces,colors,edges. > >What's the highest number of hexagons with uniform border in a given >figure? This question seems to be hard to solve. the shapes are hexagons ? are they regular , mutually congruent ? all 6 sides have the same color or only each side is unicolorous ? >> Maybe it's better to spare only those which don't contain >> previous colors. So, e.g. exhaust all pieces which contain >> a color < 13 first , using as few sides with color >12 as possible , >> and then you are left with a real 12-color subpuzzle. >> Partition the colors again, if possible etc. >> >In this case the main problem is the transition to the last stage, because >the common boundary must only have colors >12 , which you would like to >spare. sorry,I don't understand the problem another idea: make 1156 2-2-2 triangles from the 4624 pieces , with the 3 edge-pairs having same color. Then you could use these to recurse to the 34-34-34 subpuzzle. Then maybe to 17-17-17. Guenter Esser Apr 27, 2002 ----------------------- Hide message history ----- Original Message ----- From: To: Sent: Monday, April 22, 2002 3:30 PM Subject: Re: [polyforms] Triangles with Mac Mahon's pieces > >May be 16 triangels of size 35 and each with three differently colored edges > >are possible. > >Can a subset be selected so that the triangles are permutations of each > >other? > > you mean: once we made one of the 35-35-35s , we can get the 15 others > by permuting the colors ? Yes. > > but suppose, we find such a decomposition, I think I have found one: Colors 0..47; triangles 0..15. Let (a,b,c) one piece. We can use it for triangle n, if (a+b+c)*11 mod 16 =n and a div 16, b div 16 and c div 16 are not the same numbers. Otherwise we check a normalised piece (a1,b1,c1)=((a-n) mod 16, (b-n) mod 16,(c-n) mod 16) for the four special cases (1,1,14), (2,3,11), (5,5,6) or (7,10,15). If one of this cases is matched we must increment n by 1. do you think you can > solve such a 1225-pieces-triangle with 48 colors ? > Didn't try yet. > > Peter Esser Message 14 of 14 , May 2, 2002 ----------------------- Guenter: > I was speculating you'd used "McMahon-hexagons". > > > >> I mean, by adding pieces or adding colors or adding dimensions > >> or adding sides (or removing) can we construct puzzles of ever > >> increasing difficulty and decreasing number of solutions > >> per time unit ? I have tried McMahon hexagons now, three colors and onesided pieces. I finished the same construction as Miro Vichera for three different edge types (to be found on Andrew's site). It's not much difference between color matching and edge matching, but compared with the same number of square or triangle pieces, it's much more to do. Peter