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GH-116554: Relax list.sort()'s notion of "descending" runs (#116578)
* GH-116554: Relax list.sort()'s notion of "descending" run Rewrote `count_run()` so that sub-runs of equal elements no longer end a descending run. Both ascending and descending runs can have arbitrarily many sub-runs of arbitrarily many equal elements now. This is tricky, because we only use ``<`` comparisons, so checking for equality doesn't come "for free". Surprisingly, it turned out there's a very cheap (one comparison) way to determine whether an ascending run consisted of all-equal elements. That sealed the deal. In addition, after a descending run is reversed in-place, we now go on to see whether it can be extended by an ascending run that just happens to be adjacent. This succeeds in finding at least one additional element to append about half the time, and so appears to more than repay its cost (the savings come from getting to skip a binary search, when a short run is artificially forced to length MIINRUN later, for each new element `count_run()` can add to the initial run). While these have been in the back of my mind for years, a question on StackOverflow pushed it to action: https://stackoverflow.com/questions/78108792/ They were wondering why it took about 4x longer to sort a list like: [999_999, 999_999, ..., 2, 2, 1, 1, 0, 0] than "similar" lists. Of course that runs very much faster after this patch. Co-authored-by: Alex Waygood <Alex.Waygood@Gmail.com> Co-authored-by: Pieter Eendebak <pieter.eendebak@gmail.com>
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@ -128,6 +128,27 @@ class TestBase(unittest.TestCase):
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x = [e for e, i in augmented] # a stable sort of s
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check("stability", x, s)
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def test_small_stability(self):
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from itertools import product
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from operator import itemgetter
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# Exhaustively test stability across all lists of small lengths
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# and only a few distinct elements.
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# This can provoke edge cases that randomization is unlikely to find.
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# But it can grow very expensive quickly, so don't overdo it.
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NELTS = 3
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MAXSIZE = 9
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pick0 = itemgetter(0)
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for length in range(MAXSIZE + 1):
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# There are NELTS ** length distinct lists.
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for t in product(range(NELTS), repeat=length):
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xs = list(zip(t, range(length)))
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# Stability forced by index in each element.
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forced = sorted(xs)
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# Use key= to hide the index from compares.
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native = sorted(xs, key=pick0)
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self.assertEqual(forced, native)
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#==============================================================================
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class TestBugs(unittest.TestCase):
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@ -0,0 +1 @@
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``list.sort()`` now exploits more cases of partial ordering, particularly those with long descending runs with sub-runs of equal values. Those are recognized as single runs now (previously, each block of repeated values caused a new run to be created).
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@ -1618,10 +1618,11 @@ struct s_MergeState {
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/* binarysort is the best method for sorting small arrays: it does
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few compares, but can do data movement quadratic in the number of
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elements.
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[lo, hi) is a contiguous slice of a list, and is sorted via
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[lo.keys, hi) is a contiguous slice of a list of keys, and is sorted via
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binary insertion. This sort is stable.
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On entry, must have lo <= start <= hi, and that [lo, start) is already
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sorted (pass start == lo if you don't know!).
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On entry, must have lo.keys <= start <= hi, and that
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[lo.keys, start) is already sorted (pass start == lo.keys if you don't
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know!).
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If islt() complains return -1, else 0.
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Even in case of error, the output slice will be some permutation of
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the input (nothing is lost or duplicated).
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@ -1634,7 +1635,7 @@ binarysort(MergeState *ms, sortslice lo, PyObject **hi, PyObject **start)
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PyObject *pivot;
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assert(lo.keys <= start && start <= hi);
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/* assert [lo, start) is sorted */
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/* assert [lo.keys, start) is sorted */
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if (lo.keys == start)
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++start;
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for (; start < hi; ++start) {
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@ -1643,9 +1644,9 @@ binarysort(MergeState *ms, sortslice lo, PyObject **hi, PyObject **start)
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r = start;
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pivot = *r;
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/* Invariants:
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* pivot >= all in [lo, l).
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* pivot >= all in [lo.keys, l).
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* pivot < all in [r, start).
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* The second is vacuously true at the start.
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* These are vacuously true at the start.
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*/
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assert(l < r);
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do {
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@ -1656,7 +1657,7 @@ binarysort(MergeState *ms, sortslice lo, PyObject **hi, PyObject **start)
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l = p+1;
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} while (l < r);
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assert(l == r);
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/* The invariants still hold, so pivot >= all in [lo, l) and
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/* The invariants still hold, so pivot >= all in [lo.keys, l) and
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pivot < all in [l, start), so pivot belongs at l. Note
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that if there are elements equal to pivot, l points to the
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first slot after them -- that's why this sort is stable.
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@ -1671,7 +1672,7 @@ binarysort(MergeState *ms, sortslice lo, PyObject **hi, PyObject **start)
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p = start + offset;
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pivot = *p;
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l += offset;
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for (p = start + offset; p > l; --p)
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for ( ; p > l; --p)
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*p = *(p-1);
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*l = pivot;
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}
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@ -1682,56 +1683,115 @@ binarysort(MergeState *ms, sortslice lo, PyObject **hi, PyObject **start)
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return -1;
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}
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static void
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sortslice_reverse(sortslice *s, Py_ssize_t n)
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{
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reverse_slice(s->keys, &s->keys[n]);
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if (s->values != NULL)
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reverse_slice(s->values, &s->values[n]);
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}
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/*
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Return the length of the run beginning at lo, in the slice [lo, hi). lo < hi
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is required on entry. "A run" is the longest ascending sequence, with
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lo[0] <= lo[1] <= lo[2] <= ...
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or the longest descending sequence, with
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lo[0] > lo[1] > lo[2] > ...
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Boolean *descending is set to 0 in the former case, or to 1 in the latter.
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For its intended use in a stable mergesort, the strictness of the defn of
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"descending" is needed so that the caller can safely reverse a descending
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sequence without violating stability (strict > ensures there are no equal
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elements to get out of order).
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Return the length of the run beginning at slo->keys, spanning no more than
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nremaining elements. The run beginning there may be ascending or descending,
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but the function permutes it in place, if needed, so that it's always ascending
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upon return.
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Returns -1 in case of error.
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*/
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static Py_ssize_t
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count_run(MergeState *ms, PyObject **lo, PyObject **hi, int *descending)
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count_run(MergeState *ms, sortslice *slo, Py_ssize_t nremaining)
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{
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Py_ssize_t k;
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Py_ssize_t k; /* used by IFLT macro expansion */
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Py_ssize_t n;
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PyObject ** const lo = slo->keys;
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assert(lo < hi);
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*descending = 0;
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++lo;
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if (lo == hi)
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return 1;
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/* In general, as things go on we've established that the slice starts
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with a monotone run of n elements, starting at lo. */
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n = 2;
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IFLT(*lo, *(lo-1)) {
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*descending = 1;
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for (lo = lo+1; lo < hi; ++lo, ++n) {
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IFLT(*lo, *(lo-1))
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;
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else
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/* We're n elements into the slice, and the most recent neq+1 elments are
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* all equal. This reverses them in-place, and resets neq for reuse.
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*/
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#define REVERSE_LAST_NEQ \
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if (neq) { \
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sortslice slice = *slo; \
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++neq; \
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sortslice_advance(&slice, n - neq); \
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sortslice_reverse(&slice, neq); \
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neq = 0; \
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}
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/* Sticking to only __lt__ compares is confusing and error-prone. But in
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* this routine, almost all uses of IFLT can be captured by tiny macros
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* giving mnemonic names to the intent. Note that inline functions don't
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* work for this (IFLT expands to code including `goto fail`).
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*/
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#define IF_NEXT_LARGER IFLT(lo[n-1], lo[n])
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#define IF_NEXT_SMALLER IFLT(lo[n], lo[n-1])
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assert(nremaining);
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/* try ascending run first */
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for (n = 1; n < nremaining; ++n) {
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IF_NEXT_SMALLER
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break;
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}
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if (n == nremaining)
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return n;
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/* lo[n] is strictly less */
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/* If n is 1 now, then the first compare established it's a descending
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* run, so fall through to the descending case. But if n > 1, there are
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* n elements in an ascending run terminated by the strictly less lo[n].
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* If the first key < lo[n-1], *somewhere* along the way the sequence
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* increased, so we're done (there is no descending run).
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* Else first key >= lo[n-1], which implies that the entire ascending run
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* consists of equal elements. In that case, this is a descending run,
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* and we reverse the all-equal prefix in-place.
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*/
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if (n > 1) {
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IFLT(lo[0], lo[n-1])
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return n;
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sortslice_reverse(slo, n);
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}
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++n; /* in all cases it's been established that lo[n] has been resolved */
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/* Finish descending run. All-squal subruns are reversed in-place on the
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* fly. Their original order will be restored at the end by the whole-slice
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* reversal.
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*/
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Py_ssize_t neq = 0;
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for ( ; n < nremaining; ++n) {
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IF_NEXT_SMALLER {
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/* This ends the most recent run of equal elments, but still in
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* the "descending" direction.
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*/
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REVERSE_LAST_NEQ
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}
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else {
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IF_NEXT_LARGER /* descending run is over */
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break;
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else /* not x < y and not y < x implies x == y */
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++neq;
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}
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}
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else {
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for (lo = lo+1; lo < hi; ++lo, ++n) {
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IFLT(*lo, *(lo-1))
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break;
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}
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REVERSE_LAST_NEQ
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sortslice_reverse(slo, n); /* transform to ascending run */
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/* And after reversing, it's possible this can be extended by a
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* naturally increasing suffix; e.g., [3, 2, 3, 4, 1] makes an
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* ascending run from the first 4 elements.
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*/
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for ( ; n < nremaining; ++n) {
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IF_NEXT_SMALLER
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break;
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}
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return n;
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fail:
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return -1;
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#undef REVERSE_LAST_NEQ
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#undef IF_NEXT_SMALLER
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#undef IF_NEXT_LARGER
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}
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/*
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@ -2449,14 +2509,6 @@ merge_compute_minrun(Py_ssize_t n)
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return n + r;
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}
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static void
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reverse_sortslice(sortslice *s, Py_ssize_t n)
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{
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reverse_slice(s->keys, &s->keys[n]);
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if (s->values != NULL)
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reverse_slice(s->values, &s->values[n]);
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}
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/* Here we define custom comparison functions to optimize for the cases one commonly
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* encounters in practice: homogeneous lists, often of one of the basic types. */
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@ -2824,15 +2876,12 @@ list_sort_impl(PyListObject *self, PyObject *keyfunc, int reverse)
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*/
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minrun = merge_compute_minrun(nremaining);
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do {
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int descending;
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Py_ssize_t n;
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/* Identify next run. */
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n = count_run(&ms, lo.keys, lo.keys + nremaining, &descending);
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n = count_run(&ms, &lo, nremaining);
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if (n < 0)
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goto fail;
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if (descending)
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reverse_sortslice(&lo, n);
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/* If short, extend to min(minrun, nremaining). */
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if (n < minrun) {
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const Py_ssize_t force = nremaining <= minrun ?
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@ -212,24 +212,43 @@ A detailed description of timsort follows.
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Runs
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----
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count_run() returns the # of elements in the next run. A run is either
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"ascending", which means non-decreasing:
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count_run() returns the # of elements in the next run, and, if it's a
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descending run, reverses it in-place. A run is either "ascending", which
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means non-decreasing:
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a0 <= a1 <= a2 <= ...
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or "descending", which means strictly decreasing:
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or "descending", which means non-increasing:
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a0 > a1 > a2 > ...
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a0 >= a1 >= a2 >= ...
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Note that a run is always at least 2 long, unless we start at the array's
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last element.
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last element. If all elements in the array are equal, it can be viewed as
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both ascending and descending. Upon return, the run count_run() identifies
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is always ascending.
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The definition of descending is strict, because the main routine reverses
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a descending run in-place, transforming a descending run into an ascending
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run. Reversal is done via the obvious fast "swap elements starting at each
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end, and converge at the middle" method, and that can violate stability if
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the slice contains any equal elements. Using a strict definition of
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descending ensures that a descending run contains distinct elements.
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Reversal is done via the obvious fast "swap elements starting at each
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end, and converge at the middle" method. That can violate stability if
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the slice contains any equal elements. For that reason, for a long time
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the code used strict inequality (">" rather than ">=") in its definition
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of descending.
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Removing that restriction required some complication: when processing a
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descending run, all-equal sub-runs of elements are reversed in-place, on the
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fly. Their original relative order is restored "by magic" via the final
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"reverse the entire run" step.
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This makes processing descending runs a little more costly. We only use
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`__lt__` comparisons, so that `x == y` has to be deduced from
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`not x < y and not y < x`. But so long as a run remains strictly decreasing,
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only one of those compares needs to be done per loop iteration. So the primsry
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extra cost is paid only when there are equal elements, and they get some
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compensating benefit by not needing to end the descending run.
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There's one more trick added since the original: after reversing a descending
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run, it's possible that it can be extended by an adjacent ascending run. For
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example, given [3, 2, 1, 3, 4, 5, 0], the 3-element descending prefix is
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reversed in-place, and then extended by [3, 4, 5].
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If an array is random, it's very unlikely we'll see long runs. If a natural
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run contains less than minrun elements (see next section), the main loop
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