Question

List all possible parenthesization of the product ABCDE of 5 matrices.

Answer 1

Solution:

Given,

=>Number of matrices = 5

=>Matrices are A, B, C, D and E

Explanation:

Total number of paranthesization:

=>Total number of paranthesization with n matrices = 1/n*{(2n-2)!/((n-1)!)^2}

=>Total number of paranthesization with 5 matrices = (2*5-2)!/((5-1)!)^2

=>Total number of paranthesization with 5 matrices = (8!/(4!^2))/5

=>Total number of paranthesization with 5 matrices = 40320/(576*5)

=>Total number of paranthesization with 5 matrices = 14

Finding paranthesization:

1: A(B((CD)E))

2: A(B(C(DE)))

3: A((BC)(DE))

4: A((B(CD))E)

5: A(((BC)D)E)

6: (AB)((CD)E)

7: (AB)(C(DE))

8: (A(BC))(DE)

9: ((AB)C)(DE)

10: (A(B(CD))E

11: (A(BC)D))E

12: ((AB)(CD))E

13: ((A(BC))D)E

14: (((AB)C)D)E

I have explained each and every part with the help of statements attached to the answer above.