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68 lines
1.8 KiB
Python
68 lines
1.8 KiB
Python
import sys
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"""
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Dynamic Programming
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Implementation of Matrix Chain Multiplication
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Time Complexity: O(n^3)
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Space Complexity: O(n^2)
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Reference: https://en.wikipedia.org/wiki/Matrix_chain_multiplication
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"""
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def matrix_chain_order(array: list[int]) -> tuple[list[list[int]], list[list[int]]]:
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"""
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>>> matrix_chain_order([10, 30, 5])
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([[0, 0, 0], [0, 0, 1500], [0, 0, 0]], [[0, 0, 0], [0, 0, 1], [0, 0, 0]])
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"""
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n = len(array)
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matrix = [[0 for _ in range(n)] for _ in range(n)]
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sol = [[0 for _ in range(n)] for _ in range(n)]
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for chain_length in range(2, n):
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for a in range(1, n - chain_length + 1):
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b = a + chain_length - 1
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matrix[a][b] = sys.maxsize
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for c in range(a, b):
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cost = (
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matrix[a][c] + matrix[c + 1][b] + array[a - 1] * array[c] * array[b]
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)
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if cost < matrix[a][b]:
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matrix[a][b] = cost
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sol[a][b] = c
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return matrix, sol
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def print_optimal_solution(optimal_solution: list[list[int]], i: int, j: int):
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"""
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Print order of matrix with Ai as Matrix.
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"""
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if i == j:
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print("A" + str(i), end=" ")
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else:
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print("(", end=" ")
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print_optimal_solution(optimal_solution, i, optimal_solution[i][j])
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print_optimal_solution(optimal_solution, optimal_solution[i][j] + 1, j)
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print(")", end=" ")
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def main():
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"""
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Size of matrix created from array [30, 35, 15, 5, 10, 20, 25] will be:
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30*35 35*15 15*5 5*10 10*20 20*25
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"""
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array = [30, 35, 15, 5, 10, 20, 25]
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n = len(array)
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matrix, optimal_solution = matrix_chain_order(array)
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print("No. of Operation required: " + str(matrix[1][n - 1]))
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print_optimal_solution(optimal_solution, 1, n - 1)
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if __name__ == "__main__":
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main()
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