# Blanche Dissections

Posted 4 years ago
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 Recently I found out about Mondrian Art Puzzles, and decided to take a look. The idea is to divide a unit-sided square into differently sized unit-sided rectangles so that the area difference between the largest and smallest rectangles are minimized. One good first step is to find possible sets of rectangles. Here's code to do that. size = 10; bb =Table[#[],{#[]}]&/@Select[{Quiet[Times@@#[]],Length[#]}&/@Table[Select[Union[Sort/@Transpose[{Divisors[n],Reverse@Divisors[n]}]],Last[#]0&]; Grid[Take[{Last[#]-First[#],Length[#],#}&/@SortBy[Union[Select[Flatten[Table[Union[Select[With[{sub=Flatten[Take[bb,{a,b}]]},Subsets[sub,{Length[sub]-3,Length[sub]}]],Total@#==size^2&]], {a,1,Length[bb]-3},{b,a+3,Length[bb]}],2],Length[#]>0&]],Last[#]-First[#]&],15]] The next step is to find a tiling with those rectangles. I actually used Burr Tools for that step, but the techniques at the Easy Cube discussion might let Mathematica solve them. I beat the existing best solutions for everything above 10. Here's the best I found for 10x10 to 17x17. Turn's out these were inspired by Blanche's Dissection -- Divide a square into 7 differently sized rectangles with the same area. I wondered if Mathematica could solve this problem. Select[{a,b,c,d,e,f,g,h}/.RootReduce[Solve[ {e(b+c+d) == a(e+f) == g(a+b) == c(f+g) == b f == d(f+g+h) ==h(a+b+c),a+b+c+d==e+f+g+h==1} ]],Min[#]>0&][] {1/15 (8-Sqrt),1/15 (1+Sqrt),1/15 (-1+Sqrt),1/15 (7-Sqrt), 1/14 (7-Sqrt),1/42 (-5+5 Sqrt),5/21,1/21 (8-Sqrt)} The code for that is easy! I couldn't find any dissections online for a 8-rectangle Blanche-type dissection of a square, so I pondered if Mathematica would work again. sol =Select[{a,b,c,d,e,f,g,h,i}/.RootReduce[ Solve[{(a+b)(f+g)==(c+d+e)(f)==c g==d(g+ h)==(g+h+i)e==h (b+c)==(a)(i+h)==i(b+c+d), a+b+c+d+e==f+g+h+i==1}]],Min[#]>0&][]; xx = RootReduce[FoldList[Plus,0,Take[sol,5]]]; yy = RootReduce[FoldList[Plus,0,Drop[sol,5]]]; Graphics[{EdgeForm[Black],White, Rectangle[{xx[],yy[]},{xx[],yy[]}],Rectangle[{xx[],yy[]},{xx[],yy[]}],Rectangle[{xx[],yy[]},{xx[],yy[]}],Rectangle[{xx[],yy[]},{xx[],yy[]}],Rectangle[{xx[],yy[]},{xx[],yy[]}],Rectangle[{xx[],yy[]},{xx[],yy[]}],Rectangle[{xx[],yy[]},{xx[],yy[]}],Rectangle[{xx[],yy[]},{xx[],yy[]}]}] So yes, it works. Can we find a lot more? Can we find a dissection where all the side lengths are integers? Possibly. A network of rectangles that may lead to a solution is equivalent to a polyhedral graph. The original Blanche corresponds to the 5-wheel graph, and the new one here corresponds to Johnson 12. We have data on thousands of polyhedral graphs, so I just need to make some code that turns graphs into rectangle layouts. To be continued.... Answer
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Posted 4 years ago - another post of yours has been selected for the Staff Picks group, congratulations !We are happy to see you at the top of the "Featured Contributor" board. Thank you for your wonderful contributions, and please keep them coming! Answer
Posted 4 years ago
 I needed some rectangled rectangles, and I had a bunch over at Mondrian Puzzles and Mondrian Four-Coloring. Those had limited sets, so I went and got more Bouwcamp codes from Catalogues of Simple Perfect Squared Rectangles.I put all of that into an attached notebook. Here are 16 ways to dissect a square into noncongruent equal area rectangles: What I really need is code that turns a graph into an Interval Visibility Graph. The code I have there goes the other direction. Part of what I need is at Canonical Polyhedra, The basic idea is to take one of those 3D polyhedral graphs. Then, from the midpoint of an edge, project the other edges and vertices onto a 2D surface and delete the chosen edge. Convert the vertices into intervals in an interval visibility graph. Repeat that for every edge of the original graph. Repeat all that for lots of polyhedral graphs. It'll take me a few hours I don't have at the moment to program that.This is also a mathoverflow question Tiling a square with rectangles. To be continued. Attachments: Answer
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