0
$\begingroup$

Problem: find a 7X7 matrix of distinct cubed integers with identical row and column sums.

Well, I have the following code (that came from @kglr from my previous question):

ClearAll[findMat]
findMat = 
  Module[{csums = ConstantArray[#, 3], 
     mats = Select[DuplicateFreeQ[Join @@ #] &]@
       Subsets[Select[DuplicateFreeQ]@
         IntegerPartitions[#, {3}, Range[Ceiling[#^(1/2)]]^2], {3}]}, 
    Join @@ ((Select[Total[#] == csums &]@
          Tuples[{{#[[1]]}, Permutations[#[[2]]], 
            Permutations[#[[3]]]}]) & /@ mats)] &;
solsB = ParallelTable[findMat[n] /. {} -> Nothing, {n, 0, 100000}];
{Total[#[[1, 1]]]} & /@ solsB

And this code solves the described problem for a 3X3 matrix of distinct squared integers with identical row and column sums.

How can I edit my code, such that this problem can be solved for a 7X7 matrix of distinct cubed integers with identical row and column sums?

$\endgroup$
  • $\begingroup$ Crossposted here. $\endgroup$ – Rohit Namjoshi Apr 20 at 19:42
  • $\begingroup$ @Sumit I tried that and I also add Permutations[#[[4]]],Permutations[#[[5]]],Permutations[#[[6]]],Permutations[#[[7]]] to the Tuples-function, but it took a very very long time to complete the calculation. $\endgroup$ – Jan Apr 21 at 13:23

Your Answer

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

Browse other questions tagged or ask your own question.