Javascript Program for Diagonally Dominant Matrix
In mathematics, a square matrix is said to be diagonally dominant if for every row of the matrix, the magnitude of the diagonal entry in a row is larger than or equal to the sum of the magnitudes of all the other (non-diagonal) entries in that row. More precisely, the matrix A is diagonally dominant if
For example, The matrix
is diagonally dominant because
|a11| ? |a12| + |a13| since |+3| ? |-2| + |+1|
|a22| ? |a21| + |a23| since |-3| ? |+1| + |+2|
|a33| ? |a31| + |a32| since |+4| ? |-1| + |+2|
Given a matrix A of n rows and n columns. The task is to check whether matrix A is diagonally dominant or not.
Examples :
Input : A = { { 3, -2, 1 },
{ 1, -3, 2 },
{ -1, 2, 4 } };
Output : YES
Given matrix is diagonally dominant
because absolute value of every diagonal
element is more than sum of absolute values
of corresponding row.
Input : A = { { -2, 2, 1 },
{ 1, 3, 2 },
{ 1, -2, 0 } };
Output : NO
The idea is to run a loop from i = 0 to n-1 for the number of rows and for each row, run a loop j = 0 to n-1 find the sum of non-diagonal element i.e i != j. And check if diagonal element is greater than or equal to sum. If for any row, it is false, then return false or print “No”. Else print “YES”.
Javascript
<script>
function isDDM(m, n)
{
for (let i = 0; i < n; i++)
{
let sum = 0;
for (let j = 0; j < n; j++)
sum += Math.abs(m[i][j]);
sum -= Math.abs(m[i][i]);
if (Math.abs(m[i][i]) < sum)
return false ;
}
return true ;
}
let n = 3;
let m = [[ 3, -2, 1 ],
[ 1, -3, 2 ],
[ -1, 2, 4 ]];
if (isDDM(m, n))
document.write( "YES" ) ;
else
document.write( "NO" );
</script>
|
Output :
YES
Time Complexity: O(N2)
Auxiliary Space: O(1)
Please refer complete article on Diagonally Dominant Matrix for more details!
Last Updated :
12 Jan, 2023
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