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- Unit 3 Exercise 15.3-6 Discussion Question

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By Danny Elliott in the group COMP372 Design and Analysis of Algorithms August 6, 2017 - 2:57pm Comments (2)

Proof that finding the best sequence of exchanges for currencies exhibits optimal substructure when commission C_{k} = 0 for all k trades.

Let U_{1}, U_{2}, ..., U_{n} be a collection of currencies. We can represent all possible sequences from U_{1} to U_{n} as a directed graph where each node is a currency and is connected to each other node in the graph. Each directed edge represents a currency conversion value.

This problem is somewhat similar to the Unweighted Shortest Path problem only in this case, rather than hitting the fewest nodes from the start to end points, we wish to maximize the resulting value when exchanging from U_{1} along the sequence (path) to U_{n}

For simplicity, assume nodes are only visited once and that U_{1} ≠ U_{n}

Any path P from U_{1} -->_{P} U_{n} can be broken into sub-paths U_{1} -->_{P1} w -->_{P2} U_{n} (where node w can equal U_{1,} U_{n,} or some other currency in the collection). The number of edges in P equals the sum of edges in P1 and P2.

We claim that if P is the optimal path from U_{1} to U_{n} (that is, the sequence will offer the highest return of currency exchange) then P1 must be the optimal path from U_{1} to w and P2 must be the optimal path from w to U_{n}

By the cut and paste argument, if there was a more optimal path P1’ from U_{1} to w, then we could cut out P1 and past in P1’ to give U_{1} -->_{P1’} w -->_{P2} U_{n} as a more optimal path than P. However, this contradicts P’s optimality. Likewise, if there was a more optimal path P2’ from w to U_{n} , then we could cut out P2 and past in P2’ to give U_{1} -->_{P1} w -->_{P2’} U_{n} as a more optimal path than P. However, this contradicts P’s optimality.

This proves that finding the best sequence of exchanges for currencies exhibits optimal substructure when commission C_{k} = 0 for all k trades.

When C_{k} are arbitrary values for all k trades, finding the best sequence of exchanges no longer has optimal substructure. This is because the sub-problem of finding the best return along the path is not independent from the problem of minimizing commissions based on the number of trades. What was once an optimal path when C_{k} = 0 could no longer be optimal when C_{k} is some other value.

This is for course discussion.

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## Comments

I'd say that the main reason why there is no guarantee that the sub-problems may give optimal solutions, because the commission rates are arbitrary values so as long as that is the case then the problem won't exhibit optimal substructure property.

Great proof of the optimal substructure. It is fun to also think of counterexamples for when the commission rates are nonzero!