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.

更新于 2026年9月28日已验证的示例：4

Each person's share

小数位数：2；取最近值，等距时取偶数末位

Total spent: $195.50
Payments to settle up: 3
People: 4
Who pays whom: Jo pays Alex $48.88; Priya pays Alex $18.88; Sam pays Alex $3.38

4 people spent $195.50, so each owes $48.88. 3 payments settle everything: Jo pays Alex $48.88; Priya pays Alex $18.88; Sam pays Alex $3.38.

Paid minus share

Payments to settle up 行数：3

从	到	Amount
Jo	Alex	$48.88
Priya	Alex	$18.88
Sam	Alex	$3.38

计算方法 S

Total and equal share
$\frac{195.50}{4} = 48.875$
Balance: paid minus share
Alex 120 − 48.875 = 71.125; Sam 45.5 − 48.875 = −3.375; Priya 30 − 48.875 = −18.875; Jo 0 − 48.875 = −48.875
Fewest payments
4 people are owed or owe money. Grouping them into sets whose balances cancel exactly, each set of g people settles in g − 1 payments: 3 in total.
Amounts are rounded to the cent for display, so a payment can differ from the exact balance by less than a cent.

关于Group expense splitter for trips and parties

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.

计算示例

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

核验来源：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

核验来源：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.

核验来源：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

核验来源：Hand calculation: 120/3 = 40; Alex +20, Sam −20, Kim 0

常见问题

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.

“Group expense splitter for trips and parties”有多准确？

准确性取决于输入值和方法的假设。十进制运算使用50位有效数字，但估算、数值方法和源数据的精度可能较低；显示时的舍入并不能消除这些限制。已按独立来源核验的计算示例：4。例如，“Three payers and one person who paid nothing”根据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进行核验。

这种方法出自哪里？

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

关于此计算器

\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})

来源

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

已对照来源验证

此计算器包含 4 个已解示例，答案来自独立来源。这些示例会在测试套件中运行，你也可以在此运行验证。