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Computationally solving "Easy Cube" and "Soma Cube" games / puzzles

Posted 10 years ago

Easy Cube is a reflexion game from Japan. It consists of a set of 7 pieces which are "tetris"-like (each piece is a simple geometrical shape made from 3,4 or 5 cubes), and a collection of 48 problems, where the goal is to use the pieces to fill a given shape (in 2d, or in 3d for the hardest problems).

The game box enter image description here

The 7 pieces enter image description here

2 examples of problems enter image description here

It is pleasant to play, with the difficulty slowly increasing whit the hardest problems. I decided to try to solve it using Mathematica. I did not use any advanced mathematics for this (which anyway I would not know :-)), and I wanted a method general enough so it could solve all the problems without any specific adpatation. I went for a purely random method, where the program places a first piece at random, then try to put a second piece, etc. When it fails to add a piece after a fixed number of attemps, it simply starts the whole process again. So this program does not learn anything... It just tries many many different possibilities, until it finds a solution ! Using Wolfram language, the code to make it and to vizualize to result was quite simple and effective (I show the code details at the end of this post for people interested). In particular, going from the 2d case to the 3d case needed very small changes only.

The two followings animations show the program in action for a 2d problem and a 3d problem enter image description here

enter image description here

The number of attempts needed to reach a solution is of course random, and depends on the difficulty of the problem. Typically, it takes between a few hundreds and a few thousands iterations, which takes between a few seconds to 1-2 minutes. Here a sample of 4 solutions found for a 2d problem : Sample solutions from a 2d problem

And 2 solutions from a 3d problem : Sample solutions from a 2d problem

Finally, I considered it if was possible to use this method to find all solutions for a given problem (the number of different solutions is an info given on the description of each problem). In principle, because the method is random, it is not well suited for this (one can never be sure that all solutions have been found). However, since the method is quite fast, I decided to try anyway. For a given problem, I computed 10000 solutions. Here is a plot of the number of attempts needed for each solution, and a histogram of this: Number of attempts for 10000 solutions

On average, finding a solution to this problem requested 2882 attempts...

From the 10000 solutions, one wants only different solutions. This is easily done with the DeleteDuplicates command:

SolgridQ26ListMReduced = 
 DeleteDuplicates@SolgridQ26ListM; Length@SolgridQ26ListMReduced

362

which gives 362 different solutions. However, since this problem geometry has a simple central symmetry, one also want to get rid of "symmetric copies" of the solutions. Defining the symmetry operation as the GsymGrid function, this is again done with a DeleteDuplicates command :

SolgridQ26ListMReducedSym = 
 DeleteDuplicates[
  SolgridQ26ListMReduced, #2 == 
    GSymGrid[#1] &]; Length@SolgridQ26ListMReducedSym

188

which gives 188 unique solutions. This is precisely the number which is given in the problem description ! So the method was able to find all solutions. Here is plot with the 188 solutions: all solutions


Details of the code

Definition of the pieces (2d case). Each piece is a simple list of the points make the piece. When then construct, for each piece, a list of copies of it with all possible orientations.

Pieces Definitions
PTriangle = {{0, 0}, {0, 1}, {1, 0}};
PSquare = {{0, 0}, {0, 1}, {1, 0}, {1, 1}};
PLine = {{0, 0}, {0, 1}, {0, 2}};
PL = {{0, 0}, {0, 1}, {1, 0}, {2, 0}};
PPodium = {{0, 0}, {0, 1}, {-1, 0}, {1, 0}};
PSnake = {{0, 0}, {0, 1}, {1, 0}, {-1, 1}};
PBulky = {{0, 0}, {1, 0}, {0, 1}, {1, 1}, {-1, 1}};
(* Rotation by angle \[Pi]/2 matrix*)
Mrot = {{0, 1}, {-1, 0}};
(* Reflection by y-axis matrix *)
Mref = {{-1, 0}, {0, 1}};
MUnit = {{1, 0}, {0, 1}};
(*Operators which perform a given number of rotation, and of \
reflection*)
RotatePiece[piece_, n_] := Nest[Mrot.# &, #, n] & /@ piece
ReflectPiece[piece_, n_] := 
 If[n == 0, (MUnit.#) & /@ piece, (Mref.#) & /@ piece]
Each piece with all the possible orientations
PTriangleAll = {PTriangle, RotatePiece[PTriangle, 1], 
   RotatePiece[PTriangle, 2], RotatePiece[PTriangle, 3]};
PSquareAll = {PSquare};
PLineAll = {PLine, RotatePiece[PLine, 1]};
PLAll = Module[{PLr = ReflectPiece[PL, 1]}, {PL, RotatePiece[PL, 1], 
    RotatePiece[PL, 2], RotatePiece[PL, 3], PLr, RotatePiece[PLr, 1], 
    RotatePiece[PLr, 2], RotatePiece[PLr, 3]}];
PPodiumAll = {PPodium, RotatePiece[PPodium, 1], 
   RotatePiece[PPodium, 2], RotatePiece[PPodium, 3]};
PSnakeAll = {PSnake, RotatePiece[PSnake, 1], ReflectPiece[PSnake, 1], 
   RotatePiece[ReflectPiece[PSnake, 1], 1]};
PBulkyAll = 
  Module[{PBulkyr = ReflectPiece[PBulky, 1]}, {PBulky, 
    RotatePiece[PBulky, 1], RotatePiece[PBulky, 2], 
    RotatePiece[PBulky, 3], PBulkyr, RotatePiece[PBulkyr, 1], 
    RotatePiece[PBulkyr, 2], RotatePiece[PBulkyr, 3]}];
tPieces = {PBulkyAll, PLAll, PPodiumAll, PSnakeAll, PSquareAll, 
   PLineAll, PTriangleAll};
tPiecesNames = {"Bu", "Ll", "Po", "Sn", "Sq", "Li", "Tr"};

Functions which place the pieces

Piece placement functions
PutPiece[piece_, piecename_, grid_] := 
 Module[{IniPt, IniPtxy, Pts, Vcheck, PtsinGrid, Pos, i, gridx}, 
  gridx = grid;
  IniPt = RandomChoice[Position[gridx, 0]];
  Pts = Map[IniPt + # &, RandomChoice[piece]];
  PtsinGrid = gridx[[Sequence @@ #]] & /@ Pts;
  Vcheck = 
   If[PtsinGrid ==  Table[0, Length[PtsinGrid]], "success", "failure"];
  If[Vcheck == "success", 
   Do[gridx[[Sequence @@ Pts[[i]]]] = piecename, {i, 
     Length[Pts]}]; {"success", gridx}, {"failure", gridx}]
  ]
AddOnePiece[gridin_, piece_, piecename_, nmax_] := 
 Module[{cc = "failure", nn = 0, aa}, 
  While[(cc != "success") && (nn < nmax), 
   aa = PutPiece[piece, piecename, gridin]; cc = aa[[1]]; 
   nn = nn + 1]; aa]

Grid definition, and plotting functions The initial grid is a ensemble of 0, embedded in a series of 1, forming a matrix. When a piece is placed the 0's it occupies are replaced by the name of the piece

General Grid Definitions
TheGrid0 = Table[1, {i, -6, 6, 1}, {j, -6, 6, 1}];
GridDefine[coord_] :=  
 Module[{gridout}, gridout = TheGrid0; 
  Do[gridout[[Sequence @@ (coord[[i]] + {7, 7}) ]] = 0, {i, 
    Length[coord]}]; gridout]
DrawPieceRectangle[piecename_, grid_, color_] := {color, 
  Map[Rectangle[{#[[1]] - 0.5 - 7, #[[2]] - 0.5 - 7}, {#[[1]] + 0.5 - 
       7, #[[2]] + 0.5 - 7}, RoundingRadius -> 0.] &, 
   Position[grid, piecename]]}
mycolors = ColorData[24];
PlotGrid[grid_, plotrange_] := 
 Graphics[{{White, Rectangle[{-6, -6}, {6, 6}]},
   DrawPieceRectangle[0, grid, White], 
   DrawPieceRectangle[1, grid, Black], 
   DrawPieceRectangle[tPiecesNames[[#]], grid, 
      mycolors[If[# != 6, #, # + 4]]] & /@ Range[7]}, 
  PlotRange -> plotrange]

The final solving functions

Full solving function
Get the solution, and at each step provides a plot named p, which can be dynamically visualized using a cell with Dynamic[p]
SolveGrid[InitGrid_, plotrange_] := 
 Module[{imax, nattempt, grid, i, vc}, imax = 0; nattempt = 0; 
  While[imax < 8, grid[1] = InitGrid; i = 1; vc = "success"; 
   While[(i < 8) && (vc == \!\(\*
TagBox[
StyleBox["\"\<success\>\"",
ShowSpecialCharacters->False,
ShowStringCharacters->True,
NumberMarks->True],
FullForm]\)),
     {vc, grid[i + 1]} =  
     AddOnePiece[grid[i], tPieces[[i]], tPiecesNames[[i]], 30]; 
    p = PlotGrid[grid[i + 1], plotrange]; 
    If[vc == "success" || i < 7, i++; imax = i, i++]]; nattempt++]; 
  Print[nattempt]; PlotGrid[grid[8], plotrange]; grid[8]]
Get ony the solution, without final plot, nor dynamical vizualisation
SolveGridBare[InitGrid_] := 
 Module[{imax, nattempt, grid, i, vc}, imax = 0; nattempt = 0; 
  While[imax < 8, grid[1] = InitGrid; i = 1; vc = "success"; 
   While[(i < 8) && (vc == \!\(\*
TagBox[
StyleBox["\"\<success\>\"",
ShowSpecialCharacters->False,
ShowStringCharacters->True,
NumberMarks->True],
FullForm]\)),
     {vc, grid[i + 1]} =  
     AddOnePiece[grid[i], tPieces[[i]], tPiecesNames[[i]], 30]; 
    If[vc == "success" || i < 7, i++; imax = i, i++]]; nattempt++]; 
  Print[nattempt]; grid[8]]
SolveGridBare2[InitGrid_] := 
 Module[{imax, nattempt, grid, i, vc}, imax = 0; nattempt = 0; 
  While[imax < 8, grid[1] = InitGrid; i = 1; vc = "success"; 
   While[(i < 8) && (vc == \!\(\*
TagBox[
StyleBox["\"\<success\>\"",
ShowSpecialCharacters->False,
ShowStringCharacters->True,
NumberMarks->True],
FullForm]\)),
     {vc, grid[i + 1]} =  
     AddOnePiece[grid[i], tPieces[[i]], tPiecesNames[[i]], 30]; 
    If[vc == "success" || i < 7, i++; imax = i, i++]]; 
   nattempt++]; {grid[8], nattempt}]

Example of use

coordQ35 = 
  Join[Table[{i, 2}, {i, -2, 3}], Table[{i, -2}, {i, -2, 3}], 
   Table[{i, 1}, {i, 0, 3}], Table[{i, 0}, {i, 0, 3}], 
   Table[{i, -1}, {i, 0, 3}], Table[{-2, j}, {j, -1, 1}]];

TheGridQ35 = GridDefine[coordQ35];
SolgridQ35ex1 = SolveGridBare[TheGridQ35];
PlotGrid[SolgridQ35ex1, {{-3.5, 4.5}, {-3.5, 3.5}}]

A dynamic view of the process (similar to the animations shown above) is obtained by running a cell "Dynamic[p]" before calling SolveGrid[TheGridQ35,{{-3.5, 4.5}, {-3.5, 3.5}}]

Pieces definition for the 3 case (here all the possible orientations are obtained by applying randomly rotations and reflection on the initial piece)

Pieces Definitions
PTriangle = {{0, 0, 0}, {0, 1, 0}, {1, 0, 0}};
PSquare = {{0, 0, 0}, {0, 1, 0}, {1, 0, 0}, {1, 1, 0}};
PLine = {{0, -1, 0}, {0, 0, 0}, {0, 1, 0}};
PL = {{0, 0, 0}, {0, 1, 0}, {1, 0, 0}, {2, 0, 0}};
PPodium = {{0, 0, 0}, {0, 1, 0}, {-1, 0, 0}, {1, 0, 0}};
PSnake = {{0, 0, 0}, {0, 1, 0}, {1, 0, 0}, {-1, 1, 0}};
PBulky = {{0, 0, 0}, {1, 0, 0}, {0, 1, 0}, {1, 1, 0}, {-1, 1, 0}};
(* Rotation by angle \[Pi]/2 matrix*)
Mrotz = {{0, 1, 0}, {-1, 0, 0}, {0, 0, 1}};
Mrotx = {{1, 0, 0}, {0, 0, 1}, {0, -1, 0}};
Mrotx = {{0, 0, -1}, {0, 1, 0}, {1, 0, 0}};
(* Reflection  matrix *)
Mrefx = {{-1, 0, 0}, {0, 1, 0}, {0, 0, 1}};
Mrefy = {{1, 0, 0}, {0, -1, 0}, {0, 0, 1}};
Mrefz = {{1, 0, 0}, {0, 1, 0}, {0, 0, -1}};
MUnit = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}};
(*Operators which perform a given number of rotation, and of \
reflection*)
RotatePiecex[piece_, n_] := Nest[Mrotx.# &, #, n] & /@ piece
RotatePiecey[piece_, n_] := Nest[Mrotx.# &, #, n] & /@ piece
RotatePiecez[piece_, n_] := Nest[Mrotz.# &, #, n] & /@ piece
ReflectPiecex[piece_, n_] := 
 If[n == 0, (MUnit.#) & /@ piece, (Mrefx.#) & /@ piece]
ReflectPiecey[piece_, n_] := 
 If[n == 0, (MUnit.#) & /@ piece, (Mrefy.#) & /@ piece]
ReflectPiecez[piece_, n_] := 
 If[n == 0, (MUnit.#) & /@ piece, (Mrefz.#) & /@ piece]
Each piece with all the possible orientations
MakeAllPieces[piece_] := 
 DeleteDuplicates@
  Table[Module[{Rota, Rotb, Rotc, Refa, Refb, Refc, na, nb, nc, ma, 
     mb, mc},
    Rota = RandomChoice[{RotatePiecex, RotatePiecey, RotatePiecez}];
    Rotb = RandomChoice[{RotatePiecex, RotatePiecey, RotatePiecez}]; 
    Rotc = RandomChoice[{RotatePiecex, RotatePiecey, RotatePiecez}];
    Refa = RandomChoice[{ReflectPiecex, ReflectPiecey, ReflectPiecez}];
    Refb = RandomChoice[{ReflectPiecex, ReflectPiecey, ReflectPiecez}];
    Refc = RandomChoice[{ReflectPiecex, ReflectPiecey, ReflectPiecez}];
    na = RandomInteger[{0, 3}]; nb = RandomInteger[{0, 3}]; 
    nc = RandomInteger[{0, 3}];
    ma = RandomInteger[{0, 1}]; mb = RandomInteger[{0, 1}]; 
    mc = RandomInteger[{0, 1}];
    Refc[Rotc[Refb[Rotb[ Refa[Rota[piece, na], ma], nb], mb], nc], 
     mc]], {i, 1, 2000}]
PTriangleAll = MakeAllPieces[PTriangle]; Length@PTriangleAll
24
PSquareAll = MakeAllPieces[PSquare]; Length@PSquareAll
24
PLineAll = MakeAllPieces[PLine]; Length@PLineAll
6
PLAll = MakeAllPieces[PL]; Length@PLAll
24
PPodiumAll = MakeAllPieces[PPodium]; Length@PPodiumAll
24
PSnakeAll = MakeAllPieces[PSnake]; Length@PSnakeAll
24
PBulkyAll = MakeAllPieces[PBulky]; Length@PBulkyAll
24
tPieces = {PBulkyAll, PLAll, PPodiumAll, PSnakeAll, PSquareAll, 
   PLineAll, PTriangleAll};
tPiecesNames = {"Bu", "Ll", "Po", "Sn", "Sq", "Li", "Tr"};

Grid definition and visualisation for the 3d case

General Grid Definitions
TheGrid0 = Table[1, {i, -6, 6, 1}, {j, -6, 6, 1}, {k, -6, 6, 1}];
GridDefine[coord_] :=  
 Module[{gridout}, gridout = TheGrid0; 
  Do[gridout[[Sequence @@ (coord[[i]] + {7, 7, 7}) ]] = 0, {i, 
    Length[coord]}]; gridout]
DrawPieceCube[piecename_, grid_, color_] := {Glow[color], 
  Opacity[0.75], Specularity[0], EdgeForm[Thick], 
  Map[Cuboid[{#[[1]] - 0.5 - 7, #[[2]] - 0.5 - 7, #[[3]] - 0.5 - 
       7}] &, Position[grid, piecename]]}
mycolors = ColorData[24];
PlotGrid3D[grid_, plotrange_] := 
 Graphics3D[{(*{White,Cuboid[{-6,-6,-6},{6,6,6}]},*)

   DrawPieceCube[0, grid, White],(*DrawPieceCube[1,grid,Black],*)
   DrawPieceCube[tPiecesNames[[#]], grid, 
      mycolors[If[# != 6, #, # + 4]]] & /@ Range[7]}, 
  PlotRange -> plotrange, Lighting -> None, Boxed -> False]
PlotGrid3D[grid_, plotrange_, pov_] := 
 Graphics3D[{(*{White,Cuboid[{-6,-6,-6},{6,6,6}]},*)

   DrawPieceCube[0, grid, White],(*DrawPieceCube[1,grid,Black],*)
   DrawPieceCube[tPiecesNames[[#]], grid, 
      mycolors[If[# != 6, #, # + 4]]] & /@ Range[7]}, 
  PlotRange -> plotrange, Lighting -> None, Boxed -> False, 
  ViewPoint -> pov]
15 Replies
Attachments:
POSTED BY: Sjoerd de Vries

Wow, this is great. Your code succeeds at being very efficient while being reasonnably simple and short. I did not know the ListCorrelate[] command, but it is very effective here.

Also, randomly trying the pieces is not really efficient.

Yes, I agree with this, it ts the most basic way, and the least efficient

For instance, randomly combining rotation matrices and hoping you'll eventually get all necessary rotations just doesn't feel good

I disagree with this. While it effectively uses RandomChoice[], in practice there is no randomness in the results for all pieces orientations (thanks to the large number of tries), and it is fast enough (and it needs to be done only once for a given set of pieces). I think it is a matter of taste, but for me it was a lazy way to get all possible orientations without the risk of forgetting some combination of transformations. :-)

Thank you for the notebook - many things to learn !

Glad you liked it. I hope you have version 10, as I used a few of its features.

POSTED BY: Sjoerd de Vries

@Thibaut Jonckheere Is it also possible to make the program learn about the game? Say, rather trying brute force all the solutions it can intelligently decide some better moves?

POSTED BY: Vijay Sharma

Well, I don't know. It would be nice to mix this basic solving with machine learning, to improve it by eliminating some tries which are obviously wrong. For example, one could imagine that the machine could learn that it is bad to leave an empty space between two neighboring pieces that cannot be filled by any piece (this is something a human does naturally when trying to solve the puzzle). On a more elaborate level the machine could learn some typical combinations of 2 or 3 pieces which gives useful shapes. But as I have no experience with machine learning, I don't see what would be the way to train the machine. If any expert has some idea about it...

I have attached the notebook version of the code (SOMA set of pieces) for convenience. It contains all the code (the pieces names are still "incorrect", but the pieces themselves are ok), and a few examples.

Attachments:

The "Easy cube" seems to be a simplified version of the "Soma cube". So, I wonder how easy would be to adapt the code given in this discussion for "Soma cube" constructs. E.g.

SOMA Figures A426450

POSTED BY: Anton Antonov

Indeed, it is nearly the same problem ! Thank you for letting me know about SOMA.

The only difference is in the pieces set, which is here:

The SOMA pieces

The 4 first pieces are also present in Easy Cube, the last three are different (the total volume of the 7 pieces is the same: 27 "unit cubes", which means the two games share identical problems).

Adapting the code is very easy: only three of the piece definitions need to be changed, and the rest of the code can be used as it is. Here is for example two solutions of problem 426 (as shown in your picture) with SOMA pieces:

Solution for A426 enter image description here

The difficulty of these problems is probably higher than the Easy Cubes one. For the above solution of problem A426, I made two runs, using respectively ~88000 and 75000 iterations. This probably means that there is only a very small number of different solutions, and finding the solution is more time-consuming (typically between 10 minutes to a small multiple of it ?!). So the code given here should work for all the SOMA problems, but using a more advanced method (e.g. the dancing links algorithm by D. Knuth mentionned by Sander Huisman) would be a good idea if one needs all solutions of all problems... But anyway it is nice to see that a very naive method, quickly progammed, can give solutions.


Adapted code: definition of the pieces for SOMA problems (I did not make the effort to change the pieces names... the new pieces replace the definition of PLine, PSquare and PBulky, all the rest is unchanged)

Pieces Definitions
In[2]:= PTriangle = {{0, 0, 0}, {0, 1, 0}, {1, 0, 0}};
In[3]:= PSquare = {{0, 0, 0}, {0, 0, 1}, {1, 0, 0}, {1, -1, 0}};
In[4]:= PLine = {{0, 0, 0}, {1, 0, 0}, {0, 1, 0}, {0, 1, 1}};
In[5]:= PL = {{0, 0, 0}, {0, 1, 0}, {1, 0, 0}, {2, 0, 0}};
In[6]:= PPodium = {{0, 0, 0}, {0, 1, 0}, {-1, 0, 0}, {1, 0, 0}};
In[7]:= PSnake = {{0, 0, 0}, {0, 1, 0}, {1, 0, 0}, {-1, 1, 0}};
In[8]:= PBulky = {{0, 0, 0}, {1, 0, 0}, {0, -1, 0}, {0, 0, 1}};
In[9]:= (* Rotation by angle \[Pi]/2 matrix*)
In[10]:= Mrotz = {{0, 1, 0}, {-1, 0, 0}, {0, 0, 1}};
In[11]:= Mrotx = {{1, 0, 0}, {0, 0, 1}, {0, -1, 0}};
In[12]:= Mrotx = {{0, 0, -1}, {0, 1, 0}, {1, 0, 0}};
In[13]:= (* Reflection  matrix *)
In[14]:= Mrefx = {{-1, 0, 0}, {0, 1, 0}, {0, 0, 1}};
In[15]:= Mrefy = {{1, 0, 0}, {0, -1, 0}, {0, 0, 1}};
In[16]:= Mrefz = {{1, 0, 0}, {0, 1, 0}, {0, 0, -1}};
In[17]:= MUnit = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}};
In[18]:= (*Operators which peform a given number of rotation, and of \
reflection*)
In[19]:= RotatePiecex[piece_, n_] := Nest[Mrotx.# &, #, n] & /@ piece
In[20]:= RotatePiecey[piece_, n_] := Nest[Mrotx.# &, #, n] & /@ piece
In[21]:= RotatePiecez[piece_, n_] := Nest[Mrotz.# &, #, n] & /@ piece
In[22]:= ReflectPiecex[piece_, n_] := 
 If[n == 0, (MUnit.#) & /@ piece, (Mrefx.#) & /@ piece]
In[23]:= ReflectPiecey[piece_, n_] := 
 If[n == 0, (MUnit.#) & /@ piece, (Mrefy.#) & /@ piece]
In[24]:= ReflectPiecez[piece_, n_] := 
 If[n == 0, (MUnit.#) & /@ piece, (Mrefz.#) & /@ piece]
Each piece with all the possible orientations
In[25]:= MakeAllPieces[piece_] := 
 DeleteDuplicates@
  Table[Module[{Rota, Rotb, Rotc, Refa, Refb, Refc, na, nb, nc, ma, 
     mb, mc},
    Rota = RandomChoice[{RotatePiecex, RotatePiecey, RotatePiecez}];
    Rotb = RandomChoice[{RotatePiecex, RotatePiecey, RotatePiecez}]; 
    Rotc = RandomChoice[{RotatePiecex, RotatePiecey, RotatePiecez}];
    Refa = RandomChoice[{ReflectPiecex, ReflectPiecey, ReflectPiecez}];
    Refb = RandomChoice[{ReflectPiecex, ReflectPiecey, ReflectPiecez}];
    Refc = RandomChoice[{ReflectPiecex, ReflectPiecey, ReflectPiecez}];
    na = RandomInteger[{0, 3}]; nb = RandomInteger[{0, 3}]; 
    nc = RandomInteger[{0, 3}];
    ma = RandomInteger[{0, 1}]; mb = RandomInteger[{0, 1}]; 
    mc = RandomInteger[{0, 1}];
    Refc[Rotc[Refb[Rotb[ Refa[Rota[piece, na], ma], nb], mb], nc], 
     mc]], {i, 1, 2000}]
In[26]:= PTriangleAll = MakeAllPieces[PTriangle]; Length@PTriangleAll
Out[26]= 24
In[27]:= PSquareAll = MakeAllPieces[PSquare]; Length@PSquareAll
Out[27]= 48
In[28]:= PLineAll = MakeAllPieces[PLine]; Length@PLineAll
Out[28]= 48
In[29]:= PLAll = MakeAllPieces[PL]; Length@PLAll
Out[29]= 24
In[30]:= PPodiumAll = MakeAllPieces[PPodium]; Length@PPodiumAll
Out[30]= 24
In[31]:= PSnakeAll = MakeAllPieces[PSnake]; Length@PSnakeAll
Out[31]= 24
In[32]:= PBulkyAll = MakeAllPieces[PBulky]; Length@PBulkyAll
Out[32]= 48
In[33]:= tPieces = {PBulkyAll, PLAll, PPodiumAll, PSnakeAll, PSquareAll, 
   PLineAll, PTriangleAll};
In[34]:= tPiecesNames = {"Bu", "Ll", "Po", "Sn", "Sq", "Li", "Tr"};

Example of use on problem A

Solve problem A426 
In[71]:= coordA426 = 
 Join[Table[{0, j, 0}, {j, -3, 2}], Table[{1, j, 0}, {j, -3, 2}], 
  Table[{-1, j, 0}, {j, -3, 2}],
  Table[{0, j, 1}, {j, -1, 2}], Table[{-1, j, 1}, {j, -1, 2}], 
  Table[{-1, j, 2}, {j, 2, 2}]]
Out[71]= {{0, -3, 0}, {0, -2, 0}, {0, -1, 0}, {0, 0, 0}, {0, 1, 0}, {0, 2, 
  0}, {1, -3, 0}, {1, -2, 0}, {1, -1, 0}, {1, 0, 0}, {1, 1, 0}, {1, 2,
   0}, {-1, -3, 0}, {-1, -2, 0}, {-1, -1, 0}, {-1, 0, 0}, {-1, 1, 
  0}, {-1, 2, 0}, {0, -1, 1}, {0, 0, 1}, {0, 1, 1}, {0, 2, 
  1}, {-1, -1, 1}, {-1, 0, 1}, {-1, 1, 1}, {-1, 2, 1}, {-1, 2, 2}}
In[72]:= TheGridA426 = GridDefine[coordA426];
In[73]:= Plus @@ Flatten[ 1 - TheGridA426]
Out[73]= 27

In[75]:= SolgridA426ex1 = SolveGridBare[TheGridA426];
During evaluation of In[75]:= 88530
In[78]:= PlotGrid3D[SolgridA426ex1, All]

Thanks, I will play with this!

POSTED BY: Anton Antonov

enter image description here - 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!

POSTED BY: EDITORIAL BOARD

Nicely done! I always wanted to program the dancing links algorithm that is used for 'exact cover' problems like this one! But your algorithm seems nice too. I'm gonna study it a bit!

POSTED BY: Sander Huisman

Thank you for the info ! I did not know the dancing links algorithm. Having a look at the wikipedia page about it, it seems very interesting (and way more powerful than my basic algorithm). I have matter to study here.

I've seen a c++ code dancing links code for sudoku, it looked really complex. I'm quite sure it is not easy ;)

POSTED BY: Sander Huisman
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