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.
- The A vector is of size NNZ and it stores the values of the non-zero elements of the matrix. The values appear in the order of traversing the matrix row-by-row
- The IA vector is of size m+1 stores the cumulative number of non-zero elements upto ( not including) the i-th row. It is defined by the recursive relation :
- IA[0] = 0
- IA[i] = IA[i-1] + no of non-zero elements in the (i-1) th row of the Matrix
- The JA vector stores the column index of each element in the A vector. Thus it is of size NNZ as well.
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:
// CPP program to find sparse matrix rep- // resentation using CSR #include <algorithm> #include <iostream> #include <vector> using namespace std;
typedef std::vector< int > vi;
typedef vector<vector< int > > matrix;
// Utility Function to print a Matrix void printMatrix( const matrix& M)
{ int m = M.size();
int n = M[0].size();
for ( int i = 0; i < m; i++) {
for ( int j = 0; j < n; j++)
cout << M[i][j] << " " ;
cout << endl;
}
} // Utility Function to print A, IA, JA vectors // with some decoration. void printVector( const vi& V, char * msg)
{ cout << msg << "[ " ;
for_each(V.begin(), V.end(), []( int a) {
cout << a << " " ;
});
cout << "]" << endl;
} // Generate the three vectors A, IA, JA void sparesify( const matrix& M)
{ int m = M.size();
int n = M[0].size(), i, j;
vi A;
vi IA = { 0 }; // IA matrix has N+1 rows
vi JA;
int NNZ = 0;
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
if (M[i][j] != 0) {
A.push_back(M[i][j]);
JA.push_back(j);
// Count Number of Non Zero
// Elements in row i
NNZ++;
}
}
IA.push_back(NNZ);
}
printMatrix(M);
printVector(A, ( char *) "A = " );
printVector(IA, ( char *) "IA = " );
printVector(JA, ( char *) "JA = " );
} // Driver code int main()
{ matrix M = {
{ 0, 0, 0, 0, 1 },
{ 5, 8, 0, 0, 0 },
{ 0, 0, 3, 0, 0 },
{ 0, 6, 0, 0, 1 },
};
sparesify(M);
return 0;
} |
import java.util.*;
// Java program to find sparse matrix // resentation using CSR public class GFG {
// Utility Function to print a Matrix
private static void printMatrix( int [][] M)
{
int m = M.length;
int n = (M.length == 0 ? 0 : M[ 0 ].length);
for ( int i = 0 ; i < m; i++) {
for ( int j = 0 ; j < n; j++) {
System.out.print(M[i][j] + " " );
}
System.out.println();
}
}
// Utility Function to print A, IA, JA vectors
// with some decoration.
private static void printVector(ArrayList<Integer> V,
String msg)
{
System.out.print(msg + "[ " );
for (var a : V) {
System.out.print(a + " " );
}
System.out.println( "]" );
}
// Generate the three vectors A, IA, JA
private static void sparesify( int [][] M)
{
int m = M.length;
int n = (M.length == 0 ? 0 : M[ 0 ].length), i, j;
ArrayList<Integer> A = new ArrayList<Integer>();
ArrayList<Integer> IA
= new ArrayList<Integer>(); // IA matrix has N+1
// rows
ArrayList<Integer> JA = new ArrayList<Integer>();
int NNZ = 0 ;
for (i = 0 ; i < m; i++) {
for (j = 0 ; j < n; j++) {
if (M[i][j] != 0 ) {
A.add(M[i][j]);
JA.add(j);
// Count Number of Non Zero
// Elements in row i
NNZ++;
}
}
IA.add(NNZ);
}
printMatrix(M);
printVector(A, "A = " );
printVector(IA, "IA = " );
printVector(JA, "JA = " );
}
// Driver code
public static void main(String[] args)
{
int [][] M = { { 0 , 0 , 0 , 0 , 1 },
{ 5 , 8 , 0 , 0 , 0 },
{ 0 , 0 , 3 , 0 , 0 },
{ 0 , 6 , 0 , 0 , 1 } };
// Function call
sparesify(M);
}
} // This code is contributed by Aarti_Rathi |
// C# program to find sparse matrix // resentation using CSR using System;
using System.Collections.Generic;
class GFG {
// Utility Function to print a Matrix
static void printMatrix( int [, ] M)
{
int m = M.GetLength(0);
int n = M.GetLength(1);
for ( int i = 0; i < m; i++) {
for ( int j = 0; j < n; j++)
Console.Write(M[i, j] + " " );
Console.WriteLine();
}
}
// Utility Function to print A, IA, JA vectors
// with some decoration.
static void printVector(List< int > V, string msg)
{
Console.Write(msg + "[ " );
foreach ( var a in V) { Console.Write(a + " " ); }
Console.WriteLine( "]" );
}
// Generate the three vectors A, IA, JA
static void sparesify( int [, ] M)
{
int m = M.GetLength(0);
int n = M.GetLength(1), i, j;
List< int > A = new List< int >();
List< int > IA
= new List< int >(); // IA matrix has N+1 rows
List< int > JA = new List< int >();
int NNZ = 0;
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
if (M[i, j] != 0) {
A.Add(M[i, j]);
JA.Add(j);
// Count Number of Non Zero
// Elements in row i
NNZ++;
}
}
IA.Add(NNZ);
}
printMatrix(M);
printVector(A, "A = " );
printVector(IA, "IA = " );
printVector(JA, "JA = " );
}
// Driver code
public static void Main()
{
int [, ] M = {
{ 0, 0, 0, 0, 1 },
{ 5, 8, 0, 0, 0 },
{ 0, 0, 3, 0, 0 },
{ 0, 6, 0, 0, 1 },
};
// Function call
sparesify(M);
}
} // This code is contributed by Aarti_Rathi |
# Python 3 program to find sparse matrix # resentation using CSR class GFG :
# Utility Function to print a Matrix
@staticmethod
def printMatrix( M) :
m = len (M)
n = ( 0 if len (M) = = 0 else len (M[ 0 ]))
i = 0
while (i < m) :
j = 0
while (j < n) :
print ( str (M[i][j]) + " " , end = "")
j + = 1
print ()
i + = 1
# Utility Function to print A, IA, JA vectors
# with some decoration.
@staticmethod
def printVector( V, msg) :
print (msg + "[ " , end = "")
for a in V :
print ( str (a) + " " , end = "")
print ( "]" )
# Generate the three vectors A, IA, JA
@staticmethod
def sparesify( M) :
m = len (M)
n = ( 0 if len (M) = = 0 else len (M[ 0 ]))
i = 0
j = 0
A = []
IA = []
# IA matrix has N+1
# rows
JA = []
NNZ = 0
i = 0
while (i < m) :
j = 0
while (j < n) :
if (M[i][j] ! = 0 ) :
A.append(M[i][j])
JA.append(j)
# Count Number of Non Zero
# Elements in row i
NNZ + = 1
j + = 1
IA.append(NNZ)
i + = 1
GFG.printMatrix(M)
GFG.printVector(A, "A = " )
GFG.printVector(IA, "IA = " )
GFG.printVector(JA, "JA = " )
# Driver code
@staticmethod
def main( args) :
M = [[ 0 , 0 , 0 , 0 , 1 ], [ 5 , 8 , 0 , 0 , 0 ], [ 0 , 0 , 3 , 0 , 0 ], [ 0 , 6 , 0 , 0 , 1 ]]
# Function call
GFG.sparesify(M)
if __name__ = = "__main__" :
GFG.main([])
|
// JS program to find sparse matrix // resentation using CSR class GFG { // Utility Function to print a Matrix
printMatrix( M)
{
let m = M.length
let n = (m == 0) ? 0 : M[0].length
let i = 0
while (i < m)
{
let j = 0
while (j < n)
{
process.stdout.write((M[i][j]) + " " )
j += 1
}
console.log()
i += 1
}
}
// Utility Function to print A, IA, JA vectors
// with some decoration.
printVector( V, msg)
{
process.stdout.write(msg + "[ " )
process.stdout.write(V.join( " " ) )
console.log( "]" )
}
// Generate the three vectors A, IA, JA
sparesify( M)
{
let m = M.length
let n = (m == 0) ? 0 : M[0].length
let j = 0
let A = []
let IA = []
// IA matrix has N+1
// rows
let JA = []
let NNZ = 0
let i = 0
while (i < m)
{
j = 0
while (j < n)
{
if (M[i][j] != 0)
{
A.push(M[i][j])
JA.push(j)
// Count Number of Non Zero
// Elements in row i
NNZ += 1
}
j += 1
}
IA.push(NNZ)
i += 1
}
g.printMatrix(M)
g.printVector(A, "A = " )
g.printVector(IA, "IA = " )
g.printVector(JA, "JA = " )
}
} // Driver code let M = [[0, 0, 0, 0, 1], [5, 8, 0, 0, 0], [0, 0, 3, 0, 0], [0, 6, 0, 0, 1]] // Function call let g = new GFG()
g.sparesify(M) // This code is contributed by phasing17.
|
Output
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
- The sparsity of the matrix = ( Total No of Elements – Number of Non Zero Elements) / ( Total No of Elements) or (1 – NNZ/mn ) or ( 1 – size(A)/mn ) .
- The direct array based representation required memory 3 * NNZ while CSR requires ( 2*NNZ + m + 1) memory.
- CSR matrices are memory efficient as long as
. - Similar to CSR there exits CSC which stands for Compressed Sparse Columns. It is the column analogue for CSR.
- The ‘New’ Yale format further compresses the A and JA vectors into 1 vector.