Improve hash method for bitmapsets: some examination of actual outputs

shows that adding a circular shift between words greatly improves the
distribution of hash outputs.

1 files changed, 20 insertions, 6 deletions

diff --git a/src/backend/nodes/bitmapset.c b/src/backend/nodes/bitmapset.c
index 5d8327c108..5285293c72 100644
--- a/src/backend/nodes/bitmapset.c
+++ b/src/backend/nodes/bitmapset.c
@@ -769,22 +769,36 @@ bms_first_member(Bitmapset *a)
  *
  * Note: we must ensure that any two bitmapsets that are bms_equal() will
  * hash to the same value; in practice this means that trailing all-zero
- * words cannot affect the result.  Longitudinal XOR provides a reasonable
- * hash value that has this property.
+ * words cannot affect the result.  The circular-shift-and-XOR hash method
+ * used here has this property, so long as we work from back to front.
+ *
+ * Note: you might wonder why we bother with the circular shift; at first
+ * glance a straight longitudinal XOR seems as good and much simpler.  The
+ * reason is empirical: this gives a better distribution of hash values on
+ * the bitmapsets actually generated by the planner.  A common way to have
+ * multiword bitmapsets is "a JOIN b JOIN c JOIN d ...", which gives rise
+ * to rangetables in which base tables and JOIN nodes alternate; so
+ * bitmapsets of base table RT indexes tend to use only odd-numbered or only
+ * even-numbered bits.  A straight longitudinal XOR would preserve this
+ * property, leading to a much smaller set of possible outputs than if
+ * we include a shift.
  */
 uint32
 bms_hash_value(const Bitmapset *a)
 {
 	bitmapword	result = 0;
-	int			nwords;
 	int			wordnum;
 
-	if (a == NULL)
+	if (a == NULL || a->nwords <= 0)
 		return 0;				/* All empty sets hash to 0 */
-	nwords = a->nwords;
-	for (wordnum = 0; wordnum < nwords; wordnum++)
+	for (wordnum = a->nwords; --wordnum > 0; )
 	{
 		result ^= a->words[wordnum];
+		if (result & ((bitmapword) 1 << (BITS_PER_BITMAPWORD - 1)))
+			result = (result << 1) | 1;
+		else
+			result = (result << 1);
 	}
+	result ^= a->words[0];
 	return (uint32) result;
 }