Sparse Matrix Multiplication in C++

C++Server Side ProgrammingProgramming

Suppose we have two matrices A and B, we have to find the result of AB. We may assume that A's column number is equal to B's row number.

So, if the input is like [[1,0,0],[-1,0,3]] [[7,0,0],[0,0,0],[0,0,1]],

100
-103


700
000
001

then the output will be [[7,0,0],[-7,0,3]]

700
-703

To solve this, we will follow these steps −

  • r1 := size of A, r2 := size of B

  • c1 := size of A[0], c2 := size of B[0]

  • Define one 2D array ret of order r1 x c2

  • Define an array sparseA[r1] of pairs

  • for initialize i := 0, when i < r1, update (increase i by 1), do −

    • for initialize j := 0, when j < c1, update (increase j by 1), do −

      • if A[i, j] is not equal to 0, then −

        • insert { j, A[i, j] } at the end of sparseA[i]

  • for initialize i := 0, when i < r1, update (increase i by 1), do −

    • for initialize j := 0, when j < size of sparseA[i], update (increase j by 1), do −

      • for initialize k := 0, when k < c2, update (increase k by 1), do −

        • x := first element of sparseA[i, j]

        • if B[x, k] is not equal to 0, then −

          • ret[i, k] := ret[i, k] + second element of sparseA[i, j] * B[x, k]

  • return ret

Example 

Let us see the following implementation to get better understanding −

class Solution {
public:
   vector<vector<int<> multiply(vector<vector<int<>& A, vector<vector<int<>& B) {
      int r1 = A.size();
      int r2 = B.size();
      int c1 = A[0].size();
      int c2 = B[0].size();
      vector < vector <int< > ret(r1, vector <int< (c2));
      vector < pair <int, int> > sparseA[r1];
      for(int i = 0; i < r1; i++){
         for(int j = 0; j < c1; j++){
            if(A[i][j] != 0)sparseA[i].push_back({j, A[i][j]});
         }
      }
      for(int i = 0; i < r1; i++){
         for(int j = 0; j < sparseA[i].size(); j++){
            for(int k = 0; k < c2; k++){
               int x = sparseA[i][j].first;
               if(B[x][k] != 0){
                  ret[i][k] += sparseA[i][j].second * B[x][k];
               }
            }
         }
      }
      return ret;
   }
};

Input

{{1,0,0},{-1,0,3}},{{7,0,0},{0,0,0},{0,0,1}}

Output

[[7, 0, 0, ],[-7, 0, 3, ],]
raja
Published on 18-Nov-2020 16:33:17
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