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JavaScript program to check idempotent matrix

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Release: 2023-09-01 20:09:16
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JavaScript 程序检查幂等矩阵

Impotent matrix is ​​a square matrix with the same number of rows and columns, when we multiply the matrix with itself, the result will be equal to the same matrix. We are going to get a matrix and we have to determine if it is idempotent.

Mathematically

If a matrix ix M is given, then M is an idempotent matrix and should follow the following properties -

M*M = M
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Matrix multiplication

Multiplying one matrix by another matrix produces another matrix, and if the given matrix is ​​a square matrix of N*N, the resulting matrix will also have the same dimensions (N*N).

Each index (i, j) of the result matrix of the multiplication of two matrices A and B is the sum of the multiplications of the j-th column of matrix A and the i-th column of matrix B.

enter

Mat = [ [1, 0],
	    [0, 1]]
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Output

Yes, the given matrix is an Idempotent matrix.
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illustrate

For the cell (0,0): We have to multiply [1,0] with [1,0] => 1* 1 + 0 * 0 = 1. 
For the cell (0,1): We have to multiply [0,1] with [1,0] => 0* 1 + 0 * 1 = 0. 
For the cell (1,0): We have to multiply [1,0] with [0,1] => 1* 0 + 0 * 1 = 0.
For the cell (1,1): We have to multiply [0,1] with [0,1] => 0* 0 + 1 * 1 = 1.  
So, the final matrix we will get is: [ [1, 0], [0, 1]]
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method

We have seen the examples and methods of multiplying two matrices, now let us look at the steps to implement the code to find whether a given matrix is ​​idempotent.

  • First, we will create a function that will take a single parameter, which will be the matrix to be found, whether it is idempotent or not.

  • We will get the length of the matrix and use it to iterate through each cell of the matrix via a for loop.

  • In each index or cell we will get the value present in the current cell of the answer matrix using the above steps.

  • We will use a for loop to iterate through the current column and row and get their multiplicative sum.

  • If the current sum is equal to the current index value, we will move to the next value, otherwise false will be returned.

  • Based on the return value, print the statement whether the current matrix is ​​idempotent.

Example

// function to check if the current matrix is an Idempotent matrix or not
function check(mat){

   // getting the size of the given matrix. 
   var n = mat.length;    
   
   // traversing over the given matrix 
   for(var i = 0;i < n; i++){
      for(var j = 0; j<n; j++){
      
         // for the current cell calculating the value present in the resultant matrix 
         
         // variable to store the current cell value 
         var cur = 0;
         for(var k = 0; k<n; k++){
            cur += mat[k][j]*mat[i][k];
         }
         
         // if the current value is not equal to value we got for this cell then return false 
         if(cur != mat[i][j]){
            return false;
         }
      }
   }
   
   // returing true as all the values matched 
   return true;
}

// defining the matrix 
var mat = [[2, -2, -4],
           [-1, 3, 4 ],
           [1, -2, -3]]
console.log("The given matrix is: ")
console.log(mat);

// calling the function to check if the current matrix is idempotent matrix or not 
if(check(mat)){
   console.log("The given matrix is idempotent matrix")
}
else {
   console.log("The given matrix is not idempotent matrix")
}
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Time and space complexity

The time complexity of the above code is O(N^3), where N is the number of rows of the given matrix. For each cell, we must multiply the current column by the current row to produce a factor, or N, for a total of N^N cells.

The space complexity of the above code is O(1) because we are not using any extra space to store the matrix.

in conclusion

In this tutorial, we implemented a JavaScript program to check whether a given matrix is ​​idempotent. An idempotent matrix is ​​a square matrix with the same number of rows and columns, and when we multiply the matrix with itself, the result will be equal to the same matrix. We implemented the code in O(N^3) time complexity and worked in O(1) space complexity.

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source:tutorialspoint.com
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