151
151
if b_ancestors.has_key(revision):
152
152
a_intersection.append((a_distance, a_order, revision))
153
153
b_intersection.append((b_ancestors[revision][1], a_order, revision))
157
157
a_closest = __get_closest(a_intersection)
160
160
b_closest = __get_closest(b_intersection)
161
161
assert len(b_closest) != 0
164
164
if a_closest[0] in b_closest:
166
166
elif b_closest[0] in a_closest: