This actually looks like a job that the double entry accounting concept could help with.

Your transactions could be structured as bookkeeping entries thus:

```
Alice Bill Charles Balance
Alice -> Bill $10 10 10- 0 0
Bill -> Alice $1 9 9- 0 0
Bill -> Charles $5 9 4- 5- 0
Charles -> Alice $5 4 4- 0 0
```

And there you have it. At each transaction, you credit one ledger account and debit another so that the balance is always zero. At at the end, you simply work out the minimal number transactions to be applied to each account to return it to zero.

For this simple case, it’s a simple $4 transfer from Bill to Alice. What you need to do is to reduce at least one account (but preferably two) to zero for every transaction added. Let’s say you had the more complicated:

```
Alice Bill Charles Balance
Alice -> Bill $10 10 10- 0 0
Bill -> Alice $1 9 9- 0 0
Bill -> Charles $5 9 4- 5- 0
Charles -> Alice $5 4 4- 0 0
Charles -> Bill $1 4 5- 1 0
```

Then the transactions needed would be:

```
Bill -> Alice $4 0 1- 1 0
Bill -> Charles $1 0 0 0 0
```

Unfortunately, there are some states where this simple greedy strategy does not generate the best solution (kudos to `j_random_hacker`

for pointing this out). One example is:

```
Alan Bill Chas Doug Edie Fred Bal
Bill->Alan $5 5- 5 0 0 0 0 0
Bill->Chas $20 5- 25 20- 0 0 0 0
Doug->Edie $2 5- 25 20- 2 2- 0 0
Doug->Fred $1 5- 25 20- 3 2- 1- 0
```

Clearly, this could be reversed in four moves (since four moves is all it took to get there) but, if you choose your first move unwisely `(Edie->Bill $2)`

, five is the minimum you’ll get away with.

You can solve this *particular* problem with the following rules:

- (1) if you can wipe out two balances, do it.
- (2) otherwise if you can wipe out one balance and set yourself up to wipe out two in the next move, do it.
- (3) otherwise, wipe out any one balance.

That would result in the following sequence:

- (a) [1] not applicable, [2] can be achieved with
`Alan->Bill $5`

. - (b) [1] can be done with
`Chas->Bill $20`

. - (c) and (d), similar reasoning with Doug, Edie and Fred, for four total moves.

However, that works simply because of the small number of possibilities. As the number of people rises and the group inter-relations becomes more complex, you’ll most likely need an exhaustive search to find the minimum number of moves required (basically the rules 1, 2 and 3 above but expanded to handle more depth).

I think that is what will be required to give you the smallest number of transactions in all circumstances. However, it may be that that’s not required for the *best* answer (best, in this case, meaning maximum “bang per buck”). It may be that even the basic 1/2/3 rule set will give you a good-enough answer for your purposes.