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Ivenika [448]
3 years ago
9

Create a method called fixArray(int[][] array, int row, int col, int value) that sets the [row][column] to the correct value. Th

en, call the fixArray method three times - once for each value change that you are supposed to make.
Computers and Technology
1 answer:
PolarNik [594]3 years ago
7 0

Answer:

  1. public class Main {
  2.    public static void main (String [] args) {
  3.        int[][] myArray = {{1,5,6}, {7, 9, 2}};
  4.        fixArray(myArray, 1, 2, 12);
  5.        System.out.println(myArray[1][2]);
  6.    }
  7.    private static void fixArray(int[][] array, int row, int col, int value){
  8.        array[row][col] = value;
  9.    }
  10. }

Explanation:

The solution code is written in Java.

Firstly, create the method fixArray with that takes four inputs, array, row, col and value (Line 11). Within the method body, use row and col as index to address a particular element from array and set the input value to it (Line 12).

Next, we test the method in the main program using a sample array (Line 4) and we try to change the row-1 and col-2 element from 2 to 12 (Line 5).

The print statement in Line 7 will display 12 in console.

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Suppose you wish to develop a matrix-multiplication algorithm that is asymptotically faster than Strassen’s algorithm. Your algo
Kazeer [188]

Answer:

The number of subproblems are given as T(n)=a*T(n/8)+\theta(n^2) while the value of T(n) to be less than S(n) is for 342.

Explanation:

The number of subproblems are given as

T(n)=a*T(n/8)+\theta(n^2)

Asymptotic running time for Strassen’s algorithm is S(n)=\theta(n^{log(7)})

Now, when a increases, number of subproblems determines the asymptotic running time of the problem and case 1 of master theorem applies. So, in worst case, asymptotic running time of the algorithm will be

T(n)=\theta(n^{logb(a)})=\theta(n^{log8(a)})=\theta(n^{log_{2}(a^{1/3})})

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