find largest submatrix algorithm

Question

I have an N*N matrix (N=2 to 10000) of numbers that may range from 0 to 1000. How can I find the largest (rectangular) submatrix that consists of the same number?

Exampl

Answer 1

This is a work in progress

I thought about this problem and I think I may have a O(w*h) algorithm.

The idea goes like this:

• for any (i,j) compute the highest number of cells with the same value in the column j starting from (i,j). Store this values as heights[i][j].
• create an empty vector of sub matrix (a lifo)
• for all row: i
  • for all column: j
    • pop all sub matrix whose height > heights[i][j]. Because the submatrix with height > heights[i][j] cannot continue on this cell
    • push a submatrix defined by (i,j,heights[i][j]) where j is the farest coordinate where we can fit a submatrix of height: heights[i][j]
    • update the current max sub matrix

The tricky part is in the inner loop. I use something similar to the max subwindow algorithm to ensure it is O(1) on average for each cell.

I will try to formulate a proof but in the meantime here is the code.

#include 
#include 
#include 
#include 
#include 

typedef std::vector   row_t;
typedef std::vector matrix_t;

std::size_t height(matrix_t const& M) { return M.size(); }
std::size_t width (matrix_t const& M) { return M.size() ? M[0].size() : 0u; }

std::ostream& operator<<(std::ostream& out, matrix_t const& M) {
  for(unsigned i=0; i(out, ", "));
    out << std::endl;
  }
  return out;
}

struct sub_matrix_t {
  int i, j, h, w;
  sub_matrix_t(): i(0),j(0),h(0),w(1) {}
  sub_matrix_t(int i_,int j_,int h_,int w_):i(i_),j(j_),h(h_),w(w_) {}
  bool operator<(sub_matrix_t const& rhs) const { return (w*h)<(rhs.w*rhs.h); }
};


// Pop all sub_matrix from the vector keeping only those with an height
// inferior to the passed height.
// Compute the max sub matrix while removing sub matrix with height > h
void pop_sub_m(std::vector& subs,
           int i, int j, int h, sub_matrix_t& max_m) {

  sub_matrix_t sub_m(i, j, h, 1);

  while(subs.size() && subs.back().h >= h) {
    sub_m = subs.back();
    subs.pop_back();
    sub_m.w = j-sub_m.j;
    max_m = std::max(max_m, sub_m);
  }

  // Now sub_m.{i,j} is updated to the farest coordinates where there is a
  // submatrix with heights >= h

  // If we don't cut the current height (because we changed value) update
  // the current max submatrix
  if(h > 0) {
    sub_m.h = h;
    sub_m.w = j-sub_m.j+1;
    max_m = std::max(max_m, sub_m);
    subs.push_back(sub_m);
  }
}

void push_sub_m(std::vector& subs,
        int i, int j, int h, sub_matrix_t& max_m) {
  if(subs.empty() || subs.back().h < h)
    subs.emplace_back(i, j, h, 1);
}

void solve(matrix_t const& M, sub_matrix_t& max_m) {
  // Initialize answer suitable for an empty matrix
  max_m = sub_matrix_t();
  if(height(M) == 0 || width(M) == 0) return;

  // 1) Compute the heights of columns of the same values
  matrix_t heights(height(M), row_t(width(M), 1));
  for(unsigned i=height(M)-1; i>0; --i)
    for(unsigned j=0; j subs;
  for(int i=height(M)-1; i>=0; --i) {
    push_sub_m(subs, i, 0, heights[i][0], max_m);
    for(unsigned j=1; j