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The minimum number of scalar multiplication required, for parenthesization of a matrix-chain product whose sequence of dimensions for four matrices is $< 5,10,3,12,5> $ is

  1. $630$  
  2. $580$ 
  3. $480$  
  4. $405$
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could anyone plz explain this method more clearly?
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Option D Is correct
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5 Answers

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4) 405 scalar multiplication is required.

 

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helpfull solution thankyou
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The question is incomplete

The given dimensions are <5, 10, 3, 12, 5>

now we have 4 matrices

1st matrix size 5X10

2nd matrix size 10X3

3rd matrix 3X12

4th matrix 12X 5

let the matrices be A ,B,C and D since we have to parenthesize the matrix chain following cases are possible

Case 1:

(A*B)*(C*D)

Case 2:

(A*(B*C))*D

case 3:

A*(B*(C*D))

case 4:

A*((B*C)*D)

case 5:

((A*B)*C)*D

 

we have to evaluate each case find the case with minimum scalar multiplications

Case 1:

first we multiply first and second matrices so total number of scalar multiplications are 5*10*3=150

the resultant matrix size is 5X3 let it be T1

then by multiplying C and D matrices total number of scalar multiplications possible are 3*12*5=180

the resultant matrix size is 3X5 let it be T2

finally by multiplying T1 and T2  the total number of scalar multiplications possible are 5*3*5=75

hence total number of scalar multiplications possible are 150+180+75=405

similarly evaluate all the cases and the case with minimum number of scalar multiplications will be the answer
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option D 405

((5x10 . 10x3). (3x12 . 12x5))

5x10x3 + 3x12x5 +5x3x5=405

2 Comments

how to do you calculate  to get 405.. could you please explain...? thanks
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Caption

 

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The question is incomplete

The given dimensions are <5, 10, 3, 12, 5>

now we have 4 matrices

1st matrix size 5X10

2nd matrix size 10X3

3rd matrix 3X12

4th matrix 12X 5

let the matrices be A ,B,C and D since we have to parenthesize the matrix chain following cases are possible

Case 1:

(A*B)*(C*D)

Case 2:

(A*(B*C))*D

case 3:

A*(B*(C*D))

case 4:

A*((B*C)*D)

case 5:

((A*B)*C)*D

 

we have to evaluate each case find the case with minimum scalar multiplications

Case 1:

first we multiply first and second matrices so total number of scalar multiplications are 5*10*3=150

the resultant matrix size is 5X3 let it be T1

then by multiplying C and D matrices total number of scalar multiplications possible are 3*12*5=180

the resultant matrix size is 3X5 let it be T2

finally by multiplying T1 and T2  the total number of scalar multiplications possible are 5*3*5=75

hence total number of scalar multiplications possible are 150+180+75=405

similarly evaluate all the cases and the case with minimum number of scalar multiplications will be the answer
Answer:

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