Example of Matrix Chain Multiplication

Example: We are given the sequence {4, 10, 3, 12, 20, and 7}. The matrices have size 4 x 10, 10 x 3, 3 x 12, 12 x 20, 20 x 7. We need to compute M [i,j], 0 ≤ i, j≤ 5. We know M [i, i] = 0 for all i.

Let us proceed with working away from the diagonal. We compute the optimal solution for the product of 2 matrices.

Here P₀ to P₅ are Position and M₁ to M₅ are matrix of size (p_i to p_i-1)

On the basis of sequence, we make a formula

In Dynamic Programming, initialization of every method done by '0'.So we initialize it by '0'.It will sort out diagonally.

We have to sort out all the combination but the minimum output combination is taken into consideration.

Calculation of Product of 2 matrices:
1. m (1,2) = m1  x m2
           = 4 x 10 x  10 x 3
           = 4 x 10 x 3 = 120
		   
2. m (2, 3) = m2 x m3
            = 10 x 3  x  3 x 12
            = 10 x 3 x 12 = 360
			
3. m (3, 4) = m3 x m4 
            = 3 x 12  x  12 x 20
            = 3 x 12 x 20 = 720
			
4. m (4,5) = m4 x m5
           = 12 x 20  x  20 x 7
           = 12 x 20 x 7 = 1680

• We initialize the diagonal element with equal i,j value with '0'.
• After that second diagonal is sorted out and we get all the values corresponded to it

Now the third diagonal will be solved out in the same way.

Now product of 3 matrices:

M [1, 3] = M1 M2 M3

1. There are two cases by which we can solve this multiplication: ( M₁ x M₂) + M₃, M₁+ (M₂x M₃)
2. After solving both cases we choose the case in which minimum output is there.

M [1, 3] =264

As Comparing both output 264 is minimum in both cases so we insert 264 in table and ( M₁ x M₂) + M₃ this combination is chosen for the output making.

M [2, 4] = M2 M3 M4

1. There are two cases by which we can solve this multiplication: (M₂x M₃)+M₄, M₂+(M₃ x M₄)
2. After solving both cases we choose the case in which minimum output is there.

M [2, 4] = 1320

As Comparing both output 1320 is minimum in both cases so we insert 1320 in table and M₂+(M₃ x M₄) this combination is chosen for the output making.

M [3, 5] = M3  M4  M5

1. There are two cases by which we can solve this multiplication: ( M₃ x M₄) + M₅, M₃+ ( M₄xM₅)
2. After solving both cases we choose the case in which minimum output is there.

M [3, 5] = 1140

As Comparing both output 1140 is minimum in both cases so we insert 1140 in table and ( M₃ x M₄) + M₅this combination is chosen for the output making.

Now Product of 4 matrices:

M [1, 4] = M1  M2 M3 M4

There are three cases by which we can solve this multiplication:

1. ( M₁ x M₂ x M₃) M₄
2. M₁ x(M₂ x M₃ x M₄)
3. (M₁ xM₂) x ( M₃ x M₄)

After solving these cases we choose the case in which minimum output is there

M [1, 4] =1080

As comparing the output of different cases then '1080' is minimum output, so we insert 1080 in the table and (M₁ xM₂) x (M₃ x M₄) combination is taken out in output making,

M [2, 5] = M2 M3 M4 M5

There are three cases by which we can solve this multiplication:

1. (M₂ x M₃ x M₄)x M₅
2. M₂ x( M₃ x M₄ x M₅)
3. (M₂ x M₃)x ( M₄ x M₅)

After solving these cases we choose the case in which minimum output is there

M [2, 5] = 1350

As comparing the output of different cases then '1350' is minimum output, so we insert 1350 in the table and M₂ x( M₃ x M₄ xM₅)combination is taken out in output making.

Now Product of 5 matrices:

M [1, 5] = M1  M2 M3 M4 M5

There are five cases by which we can solve this multiplication:

1. (M₁ x M₂ xM₃ x M₄ )x M₅
2. M₁ x( M₂ xM₃ x M₄ xM₅)
3. (M₁ x M₂ xM₃)x M₄ xM₅
4. M₁ x M₂x(M₃ x M₄ xM₅)

After solving these cases we choose the case in which minimum output is there

M [1, 5] = 1344

As comparing the output of different cases then '1344' is minimum output, so we insert 1344 in the table and M₁ x M₂ x(M₃ x M₄ x M₅)combination is taken out in output making.

Final Output is:

Step 3: Computing Optimal Costs: let us assume that matrix A_i has dimension p_i-1x p_i for i=1, 2, 3....n. The input is a sequence (p₀,p₁,......p_n) where length [p] = n+1. The procedure uses an auxiliary table m [1....n, 1.....n] for storing m [i, j] costs an auxiliary table s [1.....n, 1.....n] that record which index of k achieved the optimal costs in computing m [i, j].

The algorithm first computes m [i, j] ← 0 for i=1, 2, 3.....n, the minimum costs for the chain of length 1.