ancestors: remove unnecessary handling of 'left'

If one of the initial nodes also is an ancestor then that most be the only
ancestor. There is no need for additional bookkeeping.

mercurial/ancestor.py

@@ -31,7 +31,7 b' def ancestors(pfunc, *orignodes):'
         poison = 1 << (i + 1)
 
         gca = set()
-        interesting = left = len(nodes)
+        interesting = len(nodes)
         nv = len(seen) - 1
         while nv >= 0 and interesting:
             v = nv
@@ -45,8 +45,6 b' def ancestors(pfunc, *orignodes):'
                     gca.add(v)
                     sv |= poison
                     if v in nodes:
-                        left -= 1
-                        if left <= 1:
                             # history is linear
                             return set([v])
             if sv < poison:

mercurial/parsers.c

@@ -1208,7 +1208,7 b' static PyObject *find_gca_candidates(ind'
 	const bitmask allseen = (1ull << revcount) - 1;
 	const bitmask poison = 1ull << revcount;
 	PyObject *gca = PyList_New(0);
-	int i, v, interesting, left;
+	int i, v, interesting;
 	int maxrev = -1;
 	long sp;
 	bitmask *seen;
@@ -1230,7 +1230,7 b' static PyObject *find_gca_candidates(ind'
 	for (i = 0; i < revcount; i++)
 		seen[revs[i]] = 1ull << i;
 
-	interesting = left = revcount;
+	interesting = revcount;
 
 	for (v = maxrev; v >= 0 && interesting; v--) {
 		long sv = seen[v];
@@ -1251,11 +1251,8 b' static PyObject *find_gca_candidates(ind'
 				}
 				sv |= poison;
 				for (i = 0; i < revcount; i++) {
-					if (revs[i] == v) {
-						if (--left <= 1)
+					if (revs[i] == v)
 							goto done;
-						break;
-					}
 				}
 			}
 		}