What algorithm to use to determine minimum number of actions required to get the system to "Zero" state?

Question

This is kind of more generic question, isn't language-specific. More about idea and algorithm to use.

The system is as follows:

It registers small loans between groups of friends. Alice and Bill are going to lunch, Bill's card isn't working, so Alice pays for his meal, $10.
The next day Bill and Charles meet each other on a railway station, Chales has no money for ticket, so Bill buys him one, for $5. Later that day Alice borrows $5 from Charles and $1 from Bill to buy her friend a gift.

Now, assuming they all registered that transactions in the system, it looks like this:

Alice -> Bill $10
Bill -> Alice $1
Bill -> Charles $5
Charles -> Alice $5

So, now, only thing that needs to be done is Bill giving Alice $4 (he gave her $1 and Charlie transferred his $5 to Alice alredy) and they're at the initial state.

If we scale that to many diffrent people, having multiple transaction, what would be the best algorithm to get as little transactions as possible?

Answer 1

If you take states as nodes of graph then you will be able to use shortest path algorithm to know the answer.

Answer 2

You should be able to solve this in O(n) by first determining how much each person owes and is owed. Transfer the debts of anyone who owes less than he is owed to his debtors (thus turning that person into an end point). Repeat until you can't transfer any more debts.

Answer 3

Well, the first step is to totally ignore the transactions. Just add them up. All you actually need to know is the net amount of debt a person owes/owns.

You could very easily find transactions by then creating a crazy flow graph and finding max flow. Then a min cut...

Some partial elaboration: There is a source node, a sink node, and a node for each person. There will be an edge between every pair of nodes except no edge between source node and sink node. Edges between people have infinite capacity in both directions. Edges coming from source node to person have capacity equal to the net debt of the person (0 if they have no net debt). Edges going from person node to sink node have capacity equal to the net amount of money that person is owed (0 if they have no net owed).

Apply max flow and/or min cut will give you a set of transfers. The actual flow amount will be how much money will be transfered.

Answer 4

Only if someone owes more than 2 people, whom also owe to the same set, can you reduce the number of transactions from the simple set.

That is, the simple set is just find each balance and repay it. That's no more than N! transactions.

If A owes B and C, and some subset of B C owe each other, so B owes C, then instead of: A -> B, A -> C (3 transactions). You'd use: A -> B, B -> C (2 transactions).

So in other words you are building a directed graph and you want to trim vertices on order to maximize path length and minimize total edges.

Sorry, I don't have an algorithm for you.

Answer 5

This actually looks like a job for Superman, oops, I mean double entry accounting.

Your transactions should 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

Update:

There are some states where a 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 (in terms of "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.

I'm reminded of people complaining that calculating PI with the formula 355/113 isn't accurate. In fact, it has an error of about one part in 125 million so that the calculation of the circumference of a circle as big as the whole Earth would only be out by about the length of your average car.

Back-of-the-envelope proof follows for those of a more disbelieving nature:

  2 x PI x R (for R = 6300km)       2 x (355/113) x R
= 2 x PI x 6,300,000m             = 2 x (355/113) x 6,300,000m
= 39,584,067.44m                  = 39,584,070.8m

Answer 6

Intuitively, this sounds like an NP-complete problem (it reduces to a problem very like bin packing), however the following algorithm (a modified form of bin packing) should be pretty good (no time for a proof, sorry).

1. Net out everyone's positions, i.e. from your example above:

   Alice = $4 Bill = $-4 Charles = $0

2. Sort all net creditors from highest to lowest, and all debtors from lowest to highest, then match by iterating over the lists.

3. At some point you might need to split a person's debts to net everything out - here it is probably best to split into the biggest chunks possible (i.e. into the bins with the most remaining space first).

This will take something like O(n log n) (again, proper proof needed).

See the Partition Problem and Bin Packing for more information (the former is a very similar problem, and if you limit yourself to fixed precision transactions, then it is equivalent - proof needed of course).

Answer 7

I found a practical solution which I implemented in Excel:

• find out who ows the most

• let that person pay the complete amount he ows to the one who should get the most

• that makes the first person zero

• repeat this proces taken into account that one of (n-1) persons has a changed amount

It should result in a maximum of (n-1) transfers and the nice thing about it is that no one is doing more than one payment. Take the (modified) example of jrandomhacker:

a=-5 b=25 c=-20 d=3 e=-2 f=-1 (sum should be zero!)

1. c -> b 20. result: a=-5 b=5 c=0 d=3 e=-2 f=-1

2. a -> b 5 result: a=0 b=0 c=0 d=3 e=-2 f=-1

3. e -> d 2 result a=0 b=0 c=0 d=1 e=0 f=-1

4. f -> d 1

Now, everyone is satisfied and no one is bothered by making two or more payments. As you can see, it is possible that one person recieves more than one payment (person d in this example).