## 25^{th} Friday Fun Session – 7^{th} Jul 2017

Given a set of currencies and some exchange rates among them, we want to find if there exists an arbitrage; meaning to exploit the discrepancies in the exchange rates and transform one unit of a certain currency to more than one unit of the same, thus making a profit.

This is the solution to JLTi Code Jam – Jun 2017 problem.

### Let us walk through an example

Let us take the example as mentioned here. We can start with 1 USD, convert that to SGD (1.380 SGD), then convert that to MYR (1.380 * 3.080 MYR), then convert to INR (1.380 * 3.080 * 15.120 INR), then convert to GBP (1.380 * 3.080 * 15.120 * 0.012 GBP), then convert that back to USD (1.380 * 3.080 * 15.120 * 0.012 * 1.30 = 1.0025503488 USD).

We end up with more than 1 USD. That means, we have an arbitrage in this set of exchange rates. The profit making cycle here is: USD -> SGD -> MYR -> INR -> GBP. And since it is a cycle, we can start from any currency within it. For example, SGD -> MYR -> INR -> GBP -> USD also represents the same cycle.

### The transformation

In general, if we have to make a profit, the respective rates in the cycle, when multiplied, should give more than 1, as we have seen in the above example.

### Negative cycle in Bellman-Ford

After some simple transformation of the profit making condition, we see, if we take *negative of log rate*, and use that as the edge cost/distance, then finding profit making cycle is equivalent to finding negative cycle in the corresponding graph. And we can do so using Bellman-Ford algorithm.

To be precise, each of the currencies would be considered as a vertex. If there exists an exchange rate *r* between two currencies then there would be a directed edge between the corresponding vertices, and *–log r* would be the associated cost/distance of that edge.

### Source of Bellman-Ford

The next question comes: using which vertex as source shall we run the Bellman-Ford? Let us see the below graph.

Suppose, we have a single profit making cycle here: GBP-> AUD -> CAD. In that case, if we start with USD as source vertex, we will never detect this cycle.

### Add extra currency as source

To solve this problem, we need to add an extra currency, and then create edges from it to all the existing currencies with cost 0. Now using this extra vertex (EXT) as source we have to run Bellman-Ford and that would ensure that we can detect a cycle, if there exist one.

GitHub: Manipulating Money Exchange