How to find the largest rectangular submatrix












3












$begingroup$


I have a sparse non-symmetric binary matrix with a block structure. The dimensions of the matrix are thousands long x less than one hundred wide.



How do I identify the largest non-contiguous rectangular sub-matrix consisting only of 1-entries?



edit: The pattern is either very sparse, or there are very few large submatrices, however the blocks are likely to overlap.



In the simplified example below, it would be elements {2,4} towards {6,6}.




$mat=left( begin{array}{ccc}
1&1&1&1&0&0\
1&1&1&1&1&1\
0&0&0&1&1&1\
1&1&0&1&1&1\
1&1&1&1&1&1\
1&1&1&1&1&1\
end{array} right)$







matrix graphs-and-networks regions







share|improve this question











$endgroup$












  • $begingroup$
    How is this related to graphs and networks? Might there be an additional structure behind the matrix? For example, if mat is an adjacency matrix of a graph, you might be looking for a maximal complete subgraph or a clique. Szabolcs' package "IGraphM`" has tools for that...
    $endgroup$
    – Henrik Schumacher
    2 hours ago








  • 1




    $begingroup$
    You still want to solve the problem for general binary matrices? (It is not unlikely that his discrete optimization problem is a very hard...)
    $endgroup$
    – Henrik Schumacher
    2 hours ago






  • 1




    $begingroup$
    If you want non-contiguous too, then this is indeed the clique problem, which is NP-complete, and there's not going to be a simpler solution. Use FindClique, then filter for blocks that also have 1s on the diagonal.
    $endgroup$
    – Szabolcs
    2 hours ago








  • 1




    $begingroup$
    Before FindClique, remove every row/column that has a 0 on the diagonal.
    $endgroup$
    – Szabolcs
    2 hours ago






  • 1




    $begingroup$
    I added the size (thousands long, less than a hundred wide and that I am looking for non-contiguous solutions.
    $endgroup$
    – Sander
    1 hour ago
















3












$begingroup$


I have a sparse non-symmetric binary matrix with a block structure. The dimensions of the matrix are thousands long x less than one hundred wide.



How do I identify the largest non-contiguous rectangular sub-matrix consisting only of 1-entries?



edit: The pattern is either very sparse, or there are very few large submatrices, however the blocks are likely to overlap.



In the simplified example below, it would be elements {2,4} towards {6,6}.




$mat=left( begin{array}{ccc}
1&1&1&1&0&0\
1&1&1&1&1&1\
0&0&0&1&1&1\
1&1&0&1&1&1\
1&1&1&1&1&1\
1&1&1&1&1&1\
end{array} right)$







matrix graphs-and-networks regions







share|improve this question











$endgroup$












  • $begingroup$
    How is this related to graphs and networks? Might there be an additional structure behind the matrix? For example, if mat is an adjacency matrix of a graph, you might be looking for a maximal complete subgraph or a clique. Szabolcs' package "IGraphM`" has tools for that...
    $endgroup$
    – Henrik Schumacher
    2 hours ago








  • 1




    $begingroup$
    You still want to solve the problem for general binary matrices? (It is not unlikely that his discrete optimization problem is a very hard...)
    $endgroup$
    – Henrik Schumacher
    2 hours ago






  • 1




    $begingroup$
    If you want non-contiguous too, then this is indeed the clique problem, which is NP-complete, and there's not going to be a simpler solution. Use FindClique, then filter for blocks that also have 1s on the diagonal.
    $endgroup$
    – Szabolcs
    2 hours ago








  • 1




    $begingroup$
    Before FindClique, remove every row/column that has a 0 on the diagonal.
    $endgroup$
    – Szabolcs
    2 hours ago






  • 1




    $begingroup$
    I added the size (thousands long, less than a hundred wide and that I am looking for non-contiguous solutions.
    $endgroup$
    – Sander
    1 hour ago














3












3








3





$begingroup$


I have a sparse non-symmetric binary matrix with a block structure. The dimensions of the matrix are thousands long x less than one hundred wide.



How do I identify the largest non-contiguous rectangular sub-matrix consisting only of 1-entries?



edit: The pattern is either very sparse, or there are very few large submatrices, however the blocks are likely to overlap.



In the simplified example below, it would be elements {2,4} towards {6,6}.




$mat=left( begin{array}{ccc}
1&1&1&1&0&0\
1&1&1&1&1&1\
0&0&0&1&1&1\
1&1&0&1&1&1\
1&1&1&1&1&1\
1&1&1&1&1&1\
end{array} right)$







matrix graphs-and-networks regions







share|improve this question











$endgroup$




I have a sparse non-symmetric binary matrix with a block structure. The dimensions of the matrix are thousands long x less than one hundred wide.



How do I identify the largest non-contiguous rectangular sub-matrix consisting only of 1-entries?



edit: The pattern is either very sparse, or there are very few large submatrices, however the blocks are likely to overlap.



In the simplified example below, it would be elements {2,4} towards {6,6}.




$mat=left( begin{array}{ccc}
1&1&1&1&0&0\
1&1&1&1&1&1\
0&0&0&1&1&1\
1&1&0&1&1&1\
1&1&1&1&1&1\
1&1&1&1&1&1\
end{array} right)$







matrix graphs-and-networks regions




matrix graphs-and-networks regions






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 14 mins ago







Sander

















asked 2 hours ago









SanderSander

1,109512




1,109512












  • $begingroup$
    How is this related to graphs and networks? Might there be an additional structure behind the matrix? For example, if mat is an adjacency matrix of a graph, you might be looking for a maximal complete subgraph or a clique. Szabolcs' package "IGraphM`" has tools for that...
    $endgroup$
    – Henrik Schumacher
    2 hours ago








  • 1




    $begingroup$
    You still want to solve the problem for general binary matrices? (It is not unlikely that his discrete optimization problem is a very hard...)
    $endgroup$
    – Henrik Schumacher
    2 hours ago






  • 1




    $begingroup$
    If you want non-contiguous too, then this is indeed the clique problem, which is NP-complete, and there's not going to be a simpler solution. Use FindClique, then filter for blocks that also have 1s on the diagonal.
    $endgroup$
    – Szabolcs
    2 hours ago








  • 1




    $begingroup$
    Before FindClique, remove every row/column that has a 0 on the diagonal.
    $endgroup$
    – Szabolcs
    2 hours ago






  • 1




    $begingroup$
    I added the size (thousands long, less than a hundred wide and that I am looking for non-contiguous solutions.
    $endgroup$
    – Sander
    1 hour ago


















  • $begingroup$
    How is this related to graphs and networks? Might there be an additional structure behind the matrix? For example, if mat is an adjacency matrix of a graph, you might be looking for a maximal complete subgraph or a clique. Szabolcs' package "IGraphM`" has tools for that...
    $endgroup$
    – Henrik Schumacher
    2 hours ago








  • 1




    $begingroup$
    You still want to solve the problem for general binary matrices? (It is not unlikely that his discrete optimization problem is a very hard...)
    $endgroup$
    – Henrik Schumacher
    2 hours ago






  • 1




    $begingroup$
    If you want non-contiguous too, then this is indeed the clique problem, which is NP-complete, and there's not going to be a simpler solution. Use FindClique, then filter for blocks that also have 1s on the diagonal.
    $endgroup$
    – Szabolcs
    2 hours ago








  • 1




    $begingroup$
    Before FindClique, remove every row/column that has a 0 on the diagonal.
    $endgroup$
    – Szabolcs
    2 hours ago






  • 1




    $begingroup$
    I added the size (thousands long, less than a hundred wide and that I am looking for non-contiguous solutions.
    $endgroup$
    – Sander
    1 hour ago
















$begingroup$
How is this related to graphs and networks? Might there be an additional structure behind the matrix? For example, if mat is an adjacency matrix of a graph, you might be looking for a maximal complete subgraph or a clique. Szabolcs' package "IGraphM`" has tools for that...
$endgroup$
– Henrik Schumacher
2 hours ago






$begingroup$
How is this related to graphs and networks? Might there be an additional structure behind the matrix? For example, if mat is an adjacency matrix of a graph, you might be looking for a maximal complete subgraph or a clique. Szabolcs' package "IGraphM`" has tools for that...
$endgroup$
– Henrik Schumacher
2 hours ago






1




1




$begingroup$
You still want to solve the problem for general binary matrices? (It is not unlikely that his discrete optimization problem is a very hard...)
$endgroup$
– Henrik Schumacher
2 hours ago




$begingroup$
You still want to solve the problem for general binary matrices? (It is not unlikely that his discrete optimization problem is a very hard...)
$endgroup$
– Henrik Schumacher
2 hours ago




1




1




$begingroup$
If you want non-contiguous too, then this is indeed the clique problem, which is NP-complete, and there's not going to be a simpler solution. Use FindClique, then filter for blocks that also have 1s on the diagonal.
$endgroup$
– Szabolcs
2 hours ago






$begingroup$
If you want non-contiguous too, then this is indeed the clique problem, which is NP-complete, and there's not going to be a simpler solution. Use FindClique, then filter for blocks that also have 1s on the diagonal.
$endgroup$
– Szabolcs
2 hours ago






1




1




$begingroup$
Before FindClique, remove every row/column that has a 0 on the diagonal.
$endgroup$
– Szabolcs
2 hours ago




$begingroup$
Before FindClique, remove every row/column that has a 0 on the diagonal.
$endgroup$
– Szabolcs
2 hours ago




1




1




$begingroup$
I added the size (thousands long, less than a hundred wide and that I am looking for non-contiguous solutions.
$endgroup$
– Sander
1 hour ago




$begingroup$
I added the size (thousands long, less than a hundred wide and that I am looking for non-contiguous solutions.
$endgroup$
– Sander
1 hour ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

A brute force approach:



mat = {{1, 1, 1, 1, 0, 0}, {1, 1, 1, 1, 1, 1}, {0, 0, 0, 1, 1, 1}, {1, 1, 0, 1, 1, 1}, 

   {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}};

pairs = Transpose /@ MaximalBy[DeleteDuplicates[CoordinateBounds /@ 

   Subsets[SparseArray[mat]["NonzeroPositions"], {2}]], 

     Min[#] Total[#, 2] &@mat[[## & @@ Span @@@ #]] &]



{{{2, 4}, {6, 6}}}







share|improve this answer









$endgroup$





















    1












    $begingroup$

    Update: This answer is not correct but for referencing, because the Subsets does not give all the possible slices of the matrix.



    mat = {{1, 1, 1, 1, 0, 0}, {1, 1, 1, 1, 1, 1}, {0, 0, 0, 1, 1, 1}, {1,
    
    1, 0, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}};
    

    
mat//MatrixForm//TeXForm



    $left(
    begin{array}{cccccc}
    1 & 1 & 1 & 1 & 0 & 0 \
    1 & 1 & 1 & 1 & 1 & 1 \
    0 & 0 & 0 & 1 & 1 & 1 \
    1 & 1 & 0 & 1 & 1 & 1 \
    1 & 1 & 1 & 1 & 1 & 1 \
    1 & 1 & 1 & 1 & 1 & 1 \
    end{array}
    right)$




    lst = Subsets@mat;
    

    
result = DeleteDuplicates@(MatrixForm /@ Select[lst, DeleteDuplicates@Flatten@# == {1} &]) // Sort;
    
result // TeXForm



    $
    left{left(
    begin{array}{cccccc}
    1 & 1 & 1 & 1 & 1 & 1 \
    end{array}
    right),left(
    begin{array}{cccccc}
    1 & 1 & 1 & 1 & 1 & 1 \
    1 & 1 & 1 & 1 & 1 & 1 \
    end{array}
    right),left(
    begin{array}{cccccc}
    1 & 1 & 1 & 1 & 1 & 1 \
    1 & 1 & 1 & 1 & 1 & 1 \
    1 & 1 & 1 & 1 & 1 & 1 \
    end{array}
    right)right}
    $




    Is this ok? I'm not sure to apply Transpose to the last one of the result.






    share|improve this answer











    $endgroup$













    • $begingroup$
      Thanks Jerry, there seem to be two issues: 1 the largest sub-matrix in the example is 5x3, your answer results in (after transpose) a 6x3 sub-matrix; Also, I am concerned the Subset will explode beyond memory capacity once we work with large matrices? 2. I would like to recover the coordinates of where the sub-matrix is residing.
      $endgroup$
      – Sander
      50 mins ago












    • $begingroup$
      @Sander Here is a similar question in other language, see geeksforgeeks.org/…
      $endgroup$
      – Jerry
      43 mins ago











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    A brute force approach:



    mat = {{1, 1, 1, 1, 0, 0}, {1, 1, 1, 1, 1, 1}, {0, 0, 0, 1, 1, 1}, {1, 1, 0, 1, 1, 1}, 
    
   {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}};
    
pairs = Transpose /@ MaximalBy[DeleteDuplicates[CoordinateBounds /@ 
    
   Subsets[SparseArray[mat]["NonzeroPositions"], {2}]], 
    
     Min[#] Total[#, 2] &@mat[[## & @@ Span @@@ #]] &]



    {{{2, 4}, {6, 6}}}







    share|improve this answer









    $endgroup$


















      3












      $begingroup$

      A brute force approach:



      mat = {{1, 1, 1, 1, 0, 0}, {1, 1, 1, 1, 1, 1}, {0, 0, 0, 1, 1, 1}, {1, 1, 0, 1, 1, 1}, 
      
   {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}};
      
pairs = Transpose /@ MaximalBy[DeleteDuplicates[CoordinateBounds /@ 
      
   Subsets[SparseArray[mat]["NonzeroPositions"], {2}]], 
      
     Min[#] Total[#, 2] &@mat[[## & @@ Span @@@ #]] &]



      {{{2, 4}, {6, 6}}}







      share|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        A brute force approach:



        mat = {{1, 1, 1, 1, 0, 0}, {1, 1, 1, 1, 1, 1}, {0, 0, 0, 1, 1, 1}, {1, 1, 0, 1, 1, 1}, 
        
   {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}};
        
pairs = Transpose /@ MaximalBy[DeleteDuplicates[CoordinateBounds /@ 
        
   Subsets[SparseArray[mat]["NonzeroPositions"], {2}]], 
        
     Min[#] Total[#, 2] &@mat[[## & @@ Span @@@ #]] &]



        {{{2, 4}, {6, 6}}}







        share|improve this answer









        $endgroup$



        A brute force approach:



        mat = {{1, 1, 1, 1, 0, 0}, {1, 1, 1, 1, 1, 1}, {0, 0, 0, 1, 1, 1}, {1, 1, 0, 1, 1, 1}, 
        
   {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}};
        
pairs = Transpose /@ MaximalBy[DeleteDuplicates[CoordinateBounds /@ 
        
   Subsets[SparseArray[mat]["NonzeroPositions"], {2}]], 
        
     Min[#] Total[#, 2] &@mat[[## & @@ Span @@@ #]] &]



        {{{2, 4}, {6, 6}}}








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        kglrkglr

        180k9200413




        180k9200413























            1












            $begingroup$

            Update: This answer is not correct but for referencing, because the Subsets does not give all the possible slices of the matrix.



            mat = {{1, 1, 1, 1, 0, 0}, {1, 1, 1, 1, 1, 1}, {0, 0, 0, 1, 1, 1}, {1,
            
    1, 0, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}};
            

            
mat//MatrixForm//TeXForm



            $left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 0 & 0 \
            1 & 1 & 1 & 1 & 1 & 1 \
            0 & 0 & 0 & 1 & 1 & 1 \
            1 & 1 & 0 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right)$




            lst = Subsets@mat;
            

            
result = DeleteDuplicates@(MatrixForm /@ Select[lst, DeleteDuplicates@Flatten@# == {1} &]) // Sort;
            
result // TeXForm



            $
            left{left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right),left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right),left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right)right}
            $




            Is this ok? I'm not sure to apply Transpose to the last one of the result.






            share|improve this answer











            $endgroup$













            • $begingroup$
              Thanks Jerry, there seem to be two issues: 1 the largest sub-matrix in the example is 5x3, your answer results in (after transpose) a 6x3 sub-matrix; Also, I am concerned the Subset will explode beyond memory capacity once we work with large matrices? 2. I would like to recover the coordinates of where the sub-matrix is residing.
              $endgroup$
              – Sander
              50 mins ago












            • $begingroup$
              @Sander Here is a similar question in other language, see geeksforgeeks.org/…
              $endgroup$
              – Jerry
              43 mins ago
















            1












            $begingroup$

            Update: This answer is not correct but for referencing, because the Subsets does not give all the possible slices of the matrix.



            mat = {{1, 1, 1, 1, 0, 0}, {1, 1, 1, 1, 1, 1}, {0, 0, 0, 1, 1, 1}, {1,
            
    1, 0, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}};
            

            
mat//MatrixForm//TeXForm



            $left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 0 & 0 \
            1 & 1 & 1 & 1 & 1 & 1 \
            0 & 0 & 0 & 1 & 1 & 1 \
            1 & 1 & 0 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right)$




            lst = Subsets@mat;
            

            
result = DeleteDuplicates@(MatrixForm /@ Select[lst, DeleteDuplicates@Flatten@# == {1} &]) // Sort;
            
result // TeXForm



            $
            left{left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right),left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right),left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right)right}
            $




            Is this ok? I'm not sure to apply Transpose to the last one of the result.






            share|improve this answer











            $endgroup$













            • $begingroup$
              Thanks Jerry, there seem to be two issues: 1 the largest sub-matrix in the example is 5x3, your answer results in (after transpose) a 6x3 sub-matrix; Also, I am concerned the Subset will explode beyond memory capacity once we work with large matrices? 2. I would like to recover the coordinates of where the sub-matrix is residing.
              $endgroup$
              – Sander
              50 mins ago












            • $begingroup$
              @Sander Here is a similar question in other language, see geeksforgeeks.org/…
              $endgroup$
              – Jerry
              43 mins ago














            1












            1








            1





            $begingroup$

            Update: This answer is not correct but for referencing, because the Subsets does not give all the possible slices of the matrix.



            mat = {{1, 1, 1, 1, 0, 0}, {1, 1, 1, 1, 1, 1}, {0, 0, 0, 1, 1, 1}, {1,
            
    1, 0, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}};
            

            
mat//MatrixForm//TeXForm



            $left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 0 & 0 \
            1 & 1 & 1 & 1 & 1 & 1 \
            0 & 0 & 0 & 1 & 1 & 1 \
            1 & 1 & 0 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right)$




            lst = Subsets@mat;
            

            
result = DeleteDuplicates@(MatrixForm /@ Select[lst, DeleteDuplicates@Flatten@# == {1} &]) // Sort;
            
result // TeXForm



            $
            left{left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right),left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right),left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right)right}
            $




            Is this ok? I'm not sure to apply Transpose to the last one of the result.






            share|improve this answer











            $endgroup$



            Update: This answer is not correct but for referencing, because the Subsets does not give all the possible slices of the matrix.



            mat = {{1, 1, 1, 1, 0, 0}, {1, 1, 1, 1, 1, 1}, {0, 0, 0, 1, 1, 1}, {1,
            
    1, 0, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}};
            

            
mat//MatrixForm//TeXForm



            $left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 0 & 0 \
            1 & 1 & 1 & 1 & 1 & 1 \
            0 & 0 & 0 & 1 & 1 & 1 \
            1 & 1 & 0 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right)$




            lst = Subsets@mat;
            

            
result = DeleteDuplicates@(MatrixForm /@ Select[lst, DeleteDuplicates@Flatten@# == {1} &]) // Sort;
            
result // TeXForm



            $
            left{left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right),left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right),left(
            begin{array}{cccccc}
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            1 & 1 & 1 & 1 & 1 & 1 \
            end{array}
            right)right}
            $




            Is this ok? I'm not sure to apply Transpose to the last one of the result.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 1 hour ago

























            answered 1 hour ago









            JerryJerry

            1,252112




            1,252112












            • $begingroup$
              Thanks Jerry, there seem to be two issues: 1 the largest sub-matrix in the example is 5x3, your answer results in (after transpose) a 6x3 sub-matrix; Also, I am concerned the Subset will explode beyond memory capacity once we work with large matrices? 2. I would like to recover the coordinates of where the sub-matrix is residing.
              $endgroup$
              – Sander
              50 mins ago












            • $begingroup$
              @Sander Here is a similar question in other language, see geeksforgeeks.org/…
              $endgroup$
              – Jerry
              43 mins ago


















            • $begingroup$
              Thanks Jerry, there seem to be two issues: 1 the largest sub-matrix in the example is 5x3, your answer results in (after transpose) a 6x3 sub-matrix; Also, I am concerned the Subset will explode beyond memory capacity once we work with large matrices? 2. I would like to recover the coordinates of where the sub-matrix is residing.
              $endgroup$
              – Sander
              50 mins ago












            • $begingroup$
              @Sander Here is a similar question in other language, see geeksforgeeks.org/…
              $endgroup$
              – Jerry
              43 mins ago
















            $begingroup$
            Thanks Jerry, there seem to be two issues: 1 the largest sub-matrix in the example is 5x3, your answer results in (after transpose) a 6x3 sub-matrix; Also, I am concerned the Subset will explode beyond memory capacity once we work with large matrices? 2. I would like to recover the coordinates of where the sub-matrix is residing.
            $endgroup$
            – Sander
            50 mins ago






            $begingroup$
            Thanks Jerry, there seem to be two issues: 1 the largest sub-matrix in the example is 5x3, your answer results in (after transpose) a 6x3 sub-matrix; Also, I am concerned the Subset will explode beyond memory capacity once we work with large matrices? 2. I would like to recover the coordinates of where the sub-matrix is residing.
            $endgroup$
            – Sander
            50 mins ago














            $begingroup$
            @Sander Here is a similar question in other language, see geeksforgeeks.org/…
            $endgroup$
            – Jerry
            43 mins ago




            $begingroup$
            @Sander Here is a similar question in other language, see geeksforgeeks.org/…
            $endgroup$
            – Jerry
            43 mins ago


















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