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C# Program for Matrix Chain Multiplication | DP-8

Last Updated : 09 Nov, 2023
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Write a C# program for a given dimension of a sequence of matrices in an array arr[], where the dimension of the ith matrix is (arr[i-1] * arr[i]), the task is to find the most efficient way to multiply these matrices together such that the total number of element multiplications is minimum.

Examples:

Input: arr[] = {40, 20, 30, 10, 30}
Output: 26000
Explanation: There are 4 matrices of dimensions 40×20, 20×30, 30×10, 10×30.
Let the input 4 matrices be A, B, C, is, and D.
The minimum number of multiplications is obtained by putting parenthesis in the following way (A(BC))D.
The minimum is 20*30*10 + 40*20*10 + 40*10*30

Input: arr[] = {1, 2, 3, 4, 3}
Output: 30
Explanation: There are 4 matrices of dimensions 1×2, 2×3, 3×4, 4×3.
Let the input 4 matrices be A, B, C, and D.
The minimum number of multiplications is obtained by putting parenthesis in the following way ((AB)C)D.
The minimum number is 1*2*3 + 1*3*4 + 1*4*3 = 30

Input: arr[] = {10, 20, 30}
Output: 6000
Explanation: There are only two matrices of dimensions 10×20 and 20×30.
So there is only one way to multiply the matrices, cost of which is 10*20*30

C# Program for Matrix Chain Multiplication using Recursion:

Two matrices of size m*n and n*p when multiplied, they generate a matrix of size m*p and the number of multiplications performed are m*n*p.

Now, for a given chain of N matrices, the first partition can be done in N-1 ways. For example, sequence of matrices A, B, C and D can be grouped as (A)(BCD), (AB)(CD) or (ABC)(D) in these 3 ways. 

So a range [i, j] can be broken into two groups like {[i, i+1], [i+1, j]}, {[i, i+2], [i+2, j]}, . . . , {[i, j-1], [j-1, j]}.

  • Each of the groups can be further partitioned into smaller groups and we can find the total required multiplications by solving for each of the groups.
  • The minimum number of multiplications among all the first partitions is the required answer.

Step-by-step approach:

  • Create a recursive function that takes i and j as parameters that determines the range of a group.
    • Iterate from k = i to j to partition the given range into two groups.
    • Call the recursive function for these groups.
    • Return the minimum value among all the partitions as the required minimum number of multiplications to multiply all the matrices of this group.
  • The minimum value returned for the range 0 to N-1 is the required answer.

Below is the implementation of the above approach.

C#




// C# code to implement the
// matrix chain multiplication using recursion
 
using System;
 
class GFG {
 
    // Matrix Ai has dimension p[i-1] x p[i]
    // for i = 1..n
    static int MatrixChainOrder(int[] p, int i, int j)
    {
 
        if (i == j)
            return 0;
 
        int min = int.MaxValue;
 
        // Place parenthesis at different places
        // between first and last matrix,
        // recursively calculate count of multiplications
        // for each parenthesis placement
        // and return the minimum count
        for (int k = i; k < j; k++)
        {
            int count = MatrixChainOrder(p, i, k)
                        + MatrixChainOrder(p, k + 1, j)
                        + p[i - 1] * p[k] * p[j];
 
            if (count < min)
                min = count;
        }
 
        // Return minimum count
        return min;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = new int[] { 1, 2, 3, 4, 3 };
        int N = arr.Length;
 
        // Function call
        Console.Write(
            "Minimum number of multiplications is "
            + MatrixChainOrder(arr, 1, N - 1));
    }
}
 
// This code is contributed by Sam007.


Output

Minimum number of multiplications is 30

The time complexity of the solution is exponential
Auxiliary Space: O(1)

C# Program for  Matrix Chain Multiplication using Dynamic Programming (Memoization):

Step-by-step approach:

  • Build a matrix dp[][] of size N*N for memoization purposes.
  • Use the same recursive call as done in the above approach:
    • When we find a range (i, j) for which the value is already calculated, return the minimum value for that range (i.e., dp[i][j]).
    • Otherwise, perform the recursive calls as mentioned earlier.
  • The value stored at dp[0][N-1] is the required answer.

Below is the implementation of the above approach:

C#




// C# program using memoization
using System;
class GFG
{
     
    static int[,] dp = new int[100, 100];
 
// Function for matrix chain multiplication
static int matrixChainMemoised(int[] p, int i, int j)
{
    if (i == j)
    {
    return 0;
    }
    if (dp[i, j] != -1)
    {
    return dp[i, j];
    }
    dp[i, j] = Int32.MaxValue;
    for (int k = i; k < j; k++)
    {
    dp[i, j] = Math.Min(
        dp[i, j], matrixChainMemoised(p, i, k)
        + matrixChainMemoised(p, k + 1, j)
        + p[i - 1] * p[k] * p[j]);
    }
    return dp[i,j];
}
 
static int MatrixChainOrder(int[] p, int n)
{
    int i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
}
 
// Driver code
static void Main()
{
    int[] arr = { 1, 2, 3, 4 };
    int n = arr.Length;
     
    for(int i = 0; i < 100; i++)
    {
        for(int j = 0; j < 100; j++)
        {
            dp[i, j] = -1;
        }
    }
 
    Console.WriteLine("Minimum number of multiplications is " +
                    MatrixChainOrder(arr, n));
}
}
 
// This code is contributed by divyeshrabadiya07.


Output

Minimum number of multiplications is 18

Time Complexity: O(N3 )
Auxiliary Space: O(N2) ignoring recursion stack space

C# Program for Matrix Chain Multiplication using Dynamic Programming (Tabulation):

In iterative approach, we initially need to find the number of multiplications required to multiply two adjacent matrices. We can use these values to find the minimum multiplication required for matrices in a range of length 3 and further use those values for ranges with higher lengths. 
Build on the answer in this manner till the range becomes [0, N-1].

Step-by-step approach:

  • Iterate from l = 2 to N-1 which denotes the length of the range:
    • Iterate from i = 0 to N-1:
      • Find the right end of the range (j) having l matrices.
      • Iterate from k = i+1 to j which denotes the point of partition.
        • Multiply the matrices in range (i, k) and (k, j).
        • This will create two matrices with dimensions arr[i-1]*arr[k] and arr[k]*arr[j].
        • The number of multiplications to be performed to multiply these two matrices (say X) are arr[i-1]*arr[k]*arr[j].
        • The total number of multiplications is dp[i][k]+ dp[k+1][j] + X.
  • The value stored at dp[1][N-1] is the required answer.

Below is the implementation of the above approach:

C#




// Dynamic Programming C# implementation of
// Matrix Chain Multiplication.
// See the Cormen book for details of the
// following algorithm
using System;
 
class GFG
{
 
    // Matrix Ai has dimension p[i-1] x p[i]
    // for i = 1..n
    static int MatrixChainOrder(int[] p, int n)
    {
 
        /* For simplicity of the program, one
        extra row and one extra column are
        allocated in m[][]. 0th row and 0th
        column of m[][] are not used */
        int[, ] m = new int[n, n];
 
        int i, j, k, L, q;
 
        /* m[i, j] = Minimum number of scalar
        multiplications needed
        to compute the matrix A[i]A[i+1]...A[j]
        = A[i..j] where dimension of A[i] is
        p[i-1] x p[i] */
 
        // cost is zero when multiplying
        // one matrix.
        for (i = 1; i < n; i++)
            m[i, i] = 0;
 
        // L is chain length.
        for (L = 2; L < n; L++)
        {
            for (i = 1; i < n - L + 1; i++)
            {
                j = i + L - 1;
                if (j == n)
                    continue;
                m[i, j] = int.MaxValue;
                for (k = i; k <= j - 1; k++)
                {
                    // q = cost/scalar multiplications
                    q = m[i, k] + m[k + 1, j]
                        + p[i - 1] * p[k] * p[j];
                    if (q < m[i, j])
                        m[i, j] = q;
                }
            }
        }
 
        return m[1, n - 1];
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = new int[] { 1, 2, 3, 4 };
        int size = arr.Length;
 
        Console.Write("Minimum number of "
                    + "multiplications is "
                    + MatrixChainOrder(arr, size));
    }
}
 
// This code is contributed by Sam007


Output

Minimum number of multiplications is 18

Time Complexity: O(N3 )
Auxiliary Space: O(N2)

Please refer complete article on Matrix Chain Multiplication | DP-8 for more details!



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