![]() There are two cases by which we can solve this multiplication: (M2x M3)+M4, M2+(M3 x M4)Īs 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.After solving both cases we choose the case in which minimum output is there.Īs 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.There are two cases by which we can solve this multiplication: ( M1 x M2) + M3, M1+ (M2x M3).Now the third diagonal will be solved out in the same way. After that second diagonal is sorted out and we get all the values corresponded to it.We initialize the diagonal element with equal i,j value with '0'.We have to sort out all the combination but the minimum output combination is taken into consideration. In Dynamic Programming, initialization of every method done by '0'.So we initialize it by '0'.It will sort out diagonally. On the basis of sequence, we make a formula, for M i -> p as column and p as row. Here P0 to P5 are Position and M1 to M5 are matrix of size (pi to pi-1) We compute the optimal solution for the product of 2 matrices. Let us proceed with working away from the diagonal. ![]() Example Problem of Matrix Chain MultiplicationĮxample-1 : We are given the sequence.
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