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If most of the elements in the matrix are zero then the matrix is called a sparse matrix. It is wasteful to store the zero elements in the matrix since they do not affect the results of our computation. This is why we implement these matrices in more efficient representations than the standard 2D Array. Using more efficient representations we can cut down space and time complexities of operations significantly.
We have discussed at 4 different representations in following articles :
In this article, we will discuss another representation of the Sparse Matrix which is commonly referred as the Yale Format.
The CSR (Compressed Sparse Row) or the Yale Format is similar to the Array Representation (discussed in Set 1) of Sparse Matrix. We represent a matrix M (m * n), by three 1-D arrays or vectors called as A, IA, JA. Let NNZ denote the number of non-zero elements in M and note that 0-based indexing is used.
To find the no of non-zero elements in say row i, we perform IA[i+1] - IA[i]. Notice how this representation is different to the array based implementation where the second vector stores the row indices of non-zero elements.
The following examples show how these matrixes are represented.
Examples:
Input : 0 0 0 0 5 8 0 0 0 0 3 0 0 6 0 0 Solution: When the matrix is read row by row, the A vector is [ 5 8 3 6] The JA vector stores column indices of elements in A hence, JA = [ 0 1 2 1]. IA[0] = 0. IA[1] = IA[0] + no of non-zero elements in row 0 i.e 0 + 0 = 0. Similarly, IA[2] = IA[1] + 2 = 2 IA[3] = IA[2] + 1 = 3 IA[4] = IA[3]+1 = 4 Therefore IA = [0 0 2 3 4] The trick is remember that IA[i] stores NNZ upto and not-including i row. Input : 10 20 0 0 0 0 0 30 0 4 0 0 0 0 50 60 70 0 0 0 0 0 0 80 Output : A = [10 20 30 4 50 60 70 80], IA = [0 2 4 7 8] JA = [0 1 1 3 2 3 4 5]
Algorithm:
SPARSIFY (MATRIX) Step 1: Set M to number of rows in MATRIX Step 2: Set N to number of columns in MATRIX Step 3: I = 0, NNZ = 0. Declare A, JA, and IA. Set IA[0] to 0 Step 4: for I = 0 ... N-1 Step 5: for J = 0 ... N-1 Step 5: If MATRIX [I][J] is not zero Add MATRIX[I][J] to A Add J to JA NNZ = NNZ + 1 [End of IF] Step 6: [ End of J loop ] Add NNZ to IA [ End of I loop ] Step 7: Print vectors A, IA, JA Step 8: END
Implementation:
0 0 0 0 1 5 8 0 0 0 0 0 3 0 0 0 6 0 0 1 A = [ 1 5 8 3 6 1 ] IA = [ 0 1 3 4 6 ] JA = [ 4 0 1 2 1 4 ]
Time Complexity : O(n x m)
Auxiliary Space: O(n + m)
Notes