2
$\begingroup$

Thanks to your help to solve previous problems, I have been able to advance in the learning of mathematica. It seems that this project to make a solver for the 8-puzzle has surpassed my abilities in mathematica, since I have not been able to do a function that calculates the amount of linear conflicts given a board, it is very important to keep in mind that it should not be Count the zero to make this calculation. Searching on the web to reach this web page, where the explanation given by the author helped me understand that it is a linear conflict, here is a screenshot, so that you also know what it is.

part one

part two

Here is what I have been able to do, very much to my dismay, if it takes zero into account to say how many conflicts there are on a board.

sameRow[i_, j_] := Quotient[i-1, 3] == Quotient[j-1, 3];
 adjacentColumn[i_, j_] := Abs[i-j] == 1;
 sameColumn[i_, j_] := Mod[i-1, 3] == Mod[j-1, 3];
 adjacentRow[i_, j_] := Abs[i-j] == 3;
 directos[board_] := Module[{v = Flatten[board]/.0->9},
    Total[Table[If[v[[v[[i]]]] == i &&
         (sameRow[i, v[[i]]] && adjacentColumn[i, v[[i]]] ||
          sameColumn[i, v[[i]]] && adjacentRow[i, v[[i]]]), 1, 0], {i, 1, 9}]]/2];

Testit with the board above, we have

directos [{[4,2,5},{1,0,6},{3,8,7}}]

But if we try it with the next one, we'll notice that it takes zero into account, which should not be.

directos[{{1,2,3},{4,5,0},{7,8,6}}]

Do you know any way to achieve this goal? It is important to advance in the construction of the solver I am planning to do. Thanks in advance for your help, without which I can not move forward.

$\endgroup$
2
  • 1
    $\begingroup$ some typos to fix: Module[{v = Flatten[board] /. 0 -> 9}, Total[....]] $\endgroup$ – kglr Jul 17 '18 at 23:04
  • $\begingroup$ @kglr Thank you for your observation, and modify them according to your suggestion $\endgroup$ – bullitohappy Jul 18 '18 at 1:05
1
$\begingroup$

Add the condition that i and v[[i]] cannot be 9 (the value of the blank tile after the replacement 0 -> 9):

ClearAll[directos]
directos[board_] := Module[{v = Flatten[board] /. 0 -> 9}, 
   Total[Table[Boole[v[[i]] != 9 &&  i != 9 && 
    (v[[v[[i]]]] == i && (sameRow[i, v[[i]]] && adjacentColumn[i, v[[i]]] || 
    sameColumn[i, v[[i]]] && adjacentRow[i, v[[i]]]))], {i, 1, 9}]] / 2];

The examples in OP:

directos[{{4, 2, 5}, {1, 0, 6}, {3, 8, 7}}]

1

directos[{{1, 2, 3}, {4, 5, 0}, {7, 8, 6}}]

0

$\endgroup$
2
  • $\begingroup$ Thank you very much for your answer, I expected more answers to know what improvements I could make, I'm going ahead to know more about MMA. $\endgroup$ – bullitohappy Jul 19 '18 at 19:02
  • $\begingroup$ @bullitohappy, my pleasure. Thank you for the accept. $\endgroup$ – kglr Jul 19 '18 at 21:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.