# Group expense splitter for trips and parties

> Expense splitter: enter who paid what to get each person's equal share and who owes whom, settled with the fewest possible payments.

Interactive version: https://www.calcopenly.com/everyday/trip-expense-splitter
Subject: Everyday calculators

Each person's share is the total spent divided by the number of people. A balance is what someone paid minus that share: a positive balance is owed money and a negative one owes it. The calculator then splits the balances into as many groups as possible whose balances cancel out exactly, because a group of g people can always settle in g − 1 payments.

With the defaults, Alex, Sam and Priya paid 195.50 between them and Jo paid nothing, so each share is 48.875. Three payments settle everything: Jo pays Alex 48.88, Priya pays Alex 18.88 and Sam pays Alex 3.38.

Finding the fewest payments is NP-hard in general (Verhoeff, 2004), so the exact search runs when up to 16 people have a non-zero balance; larger groups fall back to paying the largest creditor from the largest debtor. Payments are shown to the cent, so each can differ from the exact balance by less than a cent.

## Inputs

- **Who paid what**: One payment per line: “Alex paid 120 for food” or “Alex, 120, food”.
- **Everyone sharing the costs**: Separate names with commas. Include people who paid nothing; leave blank to split among the payers only.

## Results

- Each person's share — main result
- Total spent
- Payments to settle up
- People
- Who pays whom

## Formula

$$
\text{share} = \frac{\sum \text{paid}}{n},\qquad \text{balance}_i = \text{paid}_i - \text{share},\qquad \text{payments} = k - \max(\text{zero-sum groups})
$$

## Worked examples

### Three payers and one person who paid nothing

- Who paid what: Alex paid 120 for groceries / Sam paid 45.50 for fuel / Priya paid 30 for snacks
- Everyone sharing the costs: Alex, Sam, Priya, Jo
- **Total spent: 195.50**
- **Each person's share: 48.88**
- **Payments to settle up: 3**
- **Who pays whom: Jo pays Alex $48.88; Priya pays Alex $18.88; Sam pays Alex $3.38**
- Checked against: Python decimal: 195.50/4 = 48.875; balances −48.875, −18.875, −3.375 rounded half-even to cents; 3 payments by brute-force subset search

### A case where matching the largest amounts first needs 4 payments, not 3

- Who paid what: A paid 14 / B paid 13 / C paid 12 / D paid 5 / E paid 6
- **Each person's share: 10.00**
- **Payments to settle up: 3**
- Checked against: Python brute-force partition into zero-sum groups {A, E}, {B, C, D}: 5 − 2 = 3; largest-first greedy gives 4

### Everyone paid the same

- Who paid what: A, 10 / B, 10 / C, 10
- **Each person's share: 10.00**
- **Payments to settle up: 0**
- **Who pays whom: Everyone is even — no payments needed.**
- Checked against: Definition: every balance is 0

### A payer missing from the list is added

- Who paid what: Alex paid 60 / Sam paid 20 / Kim paid 40
- Everyone sharing the costs: Alex, Sam
- **People: 3**
- **Each person's share: 40.00**
- **Payments to settle up: 1**
- **Who pays whom: Sam pays Alex $20.00**
- Checked against: Hand calculation: 120/3 = 40; Alex +20, Sam −20, Kim 0

## Questions

### How do you split expenses in a group?

Add up everything spent, divide by the number of people to get the equal share, then subtract that share from what each person paid. For 195.50 spent across four people the share is 48.875: Alex, who paid 120, is owed 71.125, and Jo, who paid nothing, owes 48.875. Everyone with a negative balance pays someone with a positive one.

### What is the fewest number of payments needed to settle up?

At most one fewer than the number of people with a non-zero balance, and fewer when the balances split into groups that cancel out, since each group of g people needs g − 1 payments. Tom Verhoeff's 2004 paper shows that finding the minimum is at least as hard as the subset-sum problem, so an exact answer needs a search.

### Why can paying off the biggest debts first need extra payments?

It can miss groups that cancel out. With balances of +4, +3, +2, −5 and −4 (people A to E in the worked example), matching the largest amounts first takes 4 payments. Splitting them into {+4, −4} and {+3, +2, −5} settles the first pair in 1 payment and the trio in 2, so 3 in total.

### What if some costs are shared by only part of the group?

This splitter shares every cost equally among everyone listed. For a cost that only some people share, such as two people's train tickets, run it separately with just those people, then add the payments from both runs. The combined list settles every debt, though it may use one or two more payments than a single optimised plan.

### How accurate is the group expense splitter?

Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 4 worked examples whose answers come from independent sources; for example, “Three payers and one person who paid nothing” is checked against Python decimal: 195.50/4 = 48.875; balances −48.875, −18.875, −3.375 rounded half-even to cents; 3 payments by brute-force subset search.

### Where does the method come from?

Verhoeff, T. (2004). Settling multiple debts efficiently: an invitation to computing science. Informatics in Education 3(1).

## Sources

- [Verhoeff, T. (2004). Settling multiple debts efficiently: an invitation to computing science. Informatics in Education 3(1)](https://www.win.tue.nl/~wstomv/publications/settling-debts.pdf)