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Matrix Chain Multiplication ( A O ( N ^ 2 ) Solution ) | Matrix Mi has dimension p [ i - 1 ] x p [ i ] for i = 1. . n.
For simplicity of the program , one extra row and one extra column are allocated in dp [ ] [ ] . 0 th row and 0 th column of dp [ ] [ ] are not used.
cost is zero when multiplying one matrix ..
Simply following above recursive formula ..
Driver code"
Solution in Python:
assistant
#Here is code to matrix Chain Multiplication ( A O ( N ^ 2 ) Solution ) and the other steps required.
 def MatrixChainOrder ( p , n ) :
	dp = [ [ 0 for i in range ( n ) ] for i in range ( n ) ]
	for i in range ( 1 , n ) :
		dp [ i ] [ i ] = 0
	for L in range ( 1 , n - 1 ) :
		for i in range ( n - L ) :
			dp [ i ] [ i + L ] = min ( dp [ i + 1 ] [ i + L ] + p [ i - 1 ] * p [ i ] * p [ i + L ] , dp [ i ] [ i + L - 1 ] + p [ i - 1 ] * p [ i + L - 1 ] * p [ i + L ] )
	return dp [ 1 ] [ n - 1 ]
arr = [ 10 , 20 , 30 , 40 , 30 ]
size = len ( arr )
print ( " Minimum ▁ number ▁ of ▁ multiplications ▁ is " , MatrixChainOrder ( arr , size ) )