fix(combinations-replacement): return an empty list for a non-positive group size

combinationsReplacement recurses on x.slice(i, x.length), which at i = 0 is
as long as the input, so the only exit left is the k === 1 base case. For
k <= 0 the counter starts below that case and moves away from it, and the
recursion overflows the stack. The sibling combinations recurses on
x.slice(i + 1, x.length), runs out of input and returns [] for the same
arguments.

Guard the non-positive group size up front so both functions agree.
This commit is contained in:
dualfroz committed 2026-09-03 14:04:54 +02:00
1 parent 1aa682fd34
commit 24fe35e78e
3 files changed
+38

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@@ -0,0 +1,5 @@
---
"simple-statistics": patch
---
Fix `combinationsReplacement` overflowing the stack for a non-positive group size. `combinations` recurses on `x.slice(i + 1, x.length)`, so it always runs out of input and returns `[]` for `k <= 0`. `combinationsReplacement` recurses on `x.slice(i, x.length)`, which at `i = 0` is as long as `x`, so with `k` counting down past its `k === 1` base case the recursion never terminates: `combinationsReplacement([1, 2, 3], 0)` and `combinationsReplacement([1, 2, 3], -1)` both threw `RangeError: Maximum call stack size exceeded`. Both now return `[]`, matching the sibling function.
+10
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@@ -13,6 +13,16 @@
function combinationsReplacement(x, k) { function combinationsReplacement(x, k) {
const combinationList = []; const combinationList = [];
// There is no group of zero or fewer objects to draw, so there is nothing
// to enumerate. This case has to be caught up front: unlike `combinations`,
// which recurses on `x.slice(i + 1, x.length)` and therefore always runs
// out of input, the recursion below re-uses `x.slice(i, x.length)`, which
// at `i = 0` is just as long as `x`. With `k` never reaching 1, that
// recursion never terminates and overflows the stack.
if (k <= 0) {
return combinationList;
}
for (let i = 0; i < x.length; i++) { for (let i = 0; i < x.length; i++) {
if (k === 1) { if (k === 1) {
// If we're requested to find only one element, we don't need // If we're requested to find only one element, we don't need
+23
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@@ -15,4 +15,27 @@ describe("combinations", function () {
[2, 2] [2, 2]
]); ]);
}); });
it("returns no combinations when choosing zero at a time", function () {
assert.deepEqual(
ss.combinationsReplacement(Object.freeze([1, 2, 3]), 0),
[]
);
});
it("returns no combinations for a negative group size", function () {
assert.deepEqual(
ss.combinationsReplacement(Object.freeze([1, 2, 3]), -1),
[]
);
});
it("agrees with combinations on non-positive group sizes", function () {
const x = Object.freeze([1, 2, 3]);
assert.deepEqual(
ss.combinationsReplacement(x, 0),
ss.combinations(x, 0)
);
assert.deepEqual(
ss.combinationsReplacement(x, -1),
ss.combinations(x, -1)
);
});
}); });