The coin change problem is a well studied problem in Computer Science, and is a popular example given for teaching students Dynamic Programming. The problem is simple – given an amount and a set of coins, what is the minimum number of coins that can be used to pay that amount?
So, for example, if we have coins for 1,2,5,10,20,50,100 (like we do now in India), the easiest way to pay Rs. 11 is by using two coins – 10 and 1. If you have to pay Rs. 16, you can break it up as 10+5+1 and pay it using three coins.
The problem with the traditional formulation of the coin change problem is that it doesn’t involve “change” – the payer is not allowed to take back coins from the payee. So, for example, if you’ve to pay Rs. 99, you need to use 6 coins (50+20+20+5+2+2). On the other hand, if change is allowed, Rs. 99 can be paid using just 2 coins – pay Rs. 100 and get back Re. 1.
So how do you determine the way to pay using fewest coins when change is allowed? In other words, what happens to the coin change problems when negative coins can be used? (Paying 100 and getting back 1 is the same as paying 100 and (-1) ) .
Unfortunately, dynamic programming doesn’t work in this case, since we cannot process in a linear order. For example, the optimal way to pay 9 rupees when negatives are allowed is to break it up as (+10,-1), and calculating from 0 onwards (as we do in the DP) is not efficient.
For this reason, I’ve used an implementation of Dijkstra’s algorithm to determine the minimum number of coins to be used to pay any amount when cash back is allowed. Each amount is a node in the graph, with an edge between two amounts if the difference in amounts can be paid using a single coin. So there is an edge between 1 and 11 because the difference (10) can be paid using a single coin. Since cash back is allowed, the graph need not be directed.
So all we need to do to determine the way to pay each amount most optimally is to run Dijkstra’s algorithm starting from 0. The breadth first search has complexity $latex O(M^2 n)$ where is the maximum amount we want to pay, while is the number of coins.
I’ve implemented this algorithm using R, and the code can be found here. I’ve also used the algorithm to compute the number of coins to be used to pay all numbers between 1 and 10000 under different scenarios, and the results of that can be found here.
You can feel free to use this algorithm or code or results in any of your work, but make sure you provide appropriate credit!
PS: I’ve used “coin” here in a generic sense, in that it can mean “note” as well.