Question

4. Suppose you had n matrices with dimensions: ai xbi ,a2 x b2. . . . ,a,, X bn. Your goal is to determine, given two integers s and t, whether it is possible to multiply a sequence from the list of given matrices together, in any order and possibly not using all of the matrices, to end up with a matrix with dimensions s × t. For example, if the list of matrix dimensions is A : 3 × 5, B : 5 × 7,C : 7x9, D : 9 × 5, E : 9 x 3, and F : 7 x 5 we can construct a 9 × 9 matrix as D E A B C. (4 points for correct algorithm description, 3 for correctness proof, and 3 for efficiency and time analysis.).

Please show work clearly. Thanks

Answer 1

Problem is simple and interesting. Given the sequence of n matrices and we want to whether it is possible to obtain matrix of dimension say s x t. In order to solve this problem, we will create a directed graph G=(V,E) where number of vertices in V will be equal to number of distinct integers in set S = {a₁, b₁, a₂, b₂, .....,a_n,b_n} . Thus let there be m distinct values in the given set S and hence we have vertices V={v1, v2,...,vm } which are distinct integers in set S.

Now we will create a directed edge (u,v) if and only if there is some matrix in the given matrix set with the dimension u x v.

Since there are n matrices, so we will get n edges.

Claim 1 :- It is possible to obtain matrix of dimension s x t if and only if there exist a directed path from vertex s to vertex t.

Proof :- If there exist path from vertex s to vertex t, this means by following the set of edges in the path we will be able to reach vertex t and equivalently by taking the product of matrices corresponding to the edges in path from s to t, we will be able to get matrix of dimension s x t. Similarly if it is possible to obtain matrix of dimension s x t, then by taking the edges corresponding to the matrices in multiplication , we will be able to obtain path from s to t.

Hence finding matrix of dimension s x t is equivalent to finding path from s to t in graph G = (V,E ) and path from s to t can be found using algorithm like Breadth First Search with starting vertex s. If there exist directed path from s to t then vertex t will be in BFS tree with root vertex s.

Since |E|=n, and |V|<= 2n, so creating graph G=(V,E) will take O(n) time and doing BFS will also take O(n) time. And entire task will be done in O(n) time.

Please comment for any clarification.