Create matrix_chain.py

Blooverh · Blooverh · commit 67cbe42ebc4b · 2023-09-06T23:11:20.000-05:00
Mtraix chain multiplication, finding the min number of multiplication in a chain
diff --git a/TextProcessingChpt/matrix_chain.py b/TextProcessingChpt/matrix_chain.py
@@ -0,0 +1,32 @@
+
+# from typing import List
+# import math
+
+def matrix_chain(d):
+    """ d is a list of n+1 numbers such that size of kth matrix
+    id d[k] by d[k+1]
+    
+    Return a n by n table such that N[i][j] represents the min number
+    of multiplications needed to compute the product of Ai through Aj inclusive"""
+
+    n = len(d) - 1  # number of matrices
+    N = [[0] * n for i in range(n)]  # initilialize n by n result to 0
+
+    for b in range(1, n): #number of products in subchain
+        for i in range(n-b): # start of subchain
+            j = i + b
+            N[i][j] = min(N[i][k] + N[k+1][j] + d[i] * d[k+1] * d[j+1] for k in range(i, j))
+    
+    return N #or N[0][n-1]
+# A leetcode solution
+# def minScoreTriangulation( A: List[int]) -> int:
+#     SP, L = [[0]*50 for _ in range(50)], len(A)
+#     for i in range(2,L):
+#         for j in range(L-i):
+#             s, e, SP[s][e] = j, j + i, math.inf
+#             for k in range(s+1,e): SP[s][e] = min(SP[s][e], A[s]*A[k]*A[e] + SP[s][k] + SP[k][e])
+#     return SP[0][L-1]
+
+array = [2,3,4]
+print(matrix_chain(array))
+# print(minScoreTriangulation(array))
