Dynamic Programming Implementation of matrix Chain Multiplication Time Complexity: On3 Space Complexity: On2 Finds optimal order to multiply matrices array  int Print order of matrix with Ai as matrix Print the solution optimalsolution  int i  int j  int Testing for matrixchainordering Size of matrix created from above array will be 3035 3515 155 510 1020 2025
INF = float("inf")


def matrix_chain_order(array):
    """Finds optimal order to multiply matrices

    array -- int[]
    """
    n = len(array)
    matrix = [[0 for x in range(n)] for x in range(n)]
    sol = [[0 for x in range(n)] for x in range(n)]
    for chain_length in range(2, n):
        for a in range(1, n-chain_length+1):
            b = a+chain_length-1

            matrix[a][b] = INF
            for c in range(a, b):
                cost = matrix[a][c] + matrix[c+1][b] + array[a-1]*array[c]*array[b]
                if cost < matrix[a][b]:
                    matrix[a][b] = cost
                    sol[a][b] = c
    return matrix, sol
# Print order of matrix with Ai as matrix

def print_optimal_solution(optimal_solution,i,j):
    """Print the solution

    optimal_solution -- int[][]
    i -- int[]
    j -- int[]
    """
    if i==j:
        print("A" + str(i),end = " ")
    else:
        print("(", end=" ")
        print_optimal_solution(optimal_solution, i, optimal_solution[i][j])
        print_optimal_solution(optimal_solution, optimal_solution[i][j]+1, j)
        print(")", end=" ")


def main():
    """
    Testing for matrix_chain_ordering
    """
    array=[30,35,15,5,10,20,25]
    length=len(array)
    #Size of matrix created from above array will be
    # 30*35 35*15 15*5 5*10 10*20 20*25
    matrix, optimal_solution = matrix_chain_order(array)

    print("No. of Operation required: "+str((matrix[1][length-1])))
    print_optimal_solution(optimal_solution,1,length-1)

if __name__ == '__main__':
    main()
