new A* optimization: worst-case pruning

i'm not sure if this has been described in the literature exactly but it
is similar to existing branch-and-bound techniques.

it doesn't speed up the search directly (the number of visited nodes is
the same) but it does cut the queue length down by an order of magnitude
and shave a couple seconds off of day 16, presumably due to reduced memory
pressure or something.

day16/sol.py

@@ -152,6 +152,10 @@ def solve():
             m -= 1
         return c
 
+    def worst2(node):
+        _, _, min1, min2, closed = node
+        return min(min1,min2) * pressure(closed)
+
     def is_goal2(node):
         _, _, min1, min2, closed = node
         return min1 == 0 and min2 == 0 or closed == all_open
@@ -209,7 +213,7 @@ def solve():
     minutes = 26
     start2 = (AA, AA, minutes, minutes, all_closed)
     info.clear()
-    c, _, path = astar.search(start2, is_goal2, neighbors2, heuristic2)
+    c, _, path = astar.search(start2, is_goal2, neighbors2, heuristic2, worst=worst2)
     print(c)
     print(max_pressure*minutes - c)
 

lib/astar.py

@@ -1,6 +1,6 @@
 from heapq import heappush, heappop
 
-def search(start, is_goal, neighbors, heuristic=None):
+def search(start, is_goal, neighbors, heuristic=None, worst=None):
     if heuristic == None:
         def heuristic(x):
             return 0
@@ -16,6 +16,8 @@ def search(start, is_goal, neighbors, heuristic=None):
         i += 1
         heappush(q, (heuristic(s), i, s, None))
     done = {}
+    if worst:
+        best_worst = min(worst(s) for s in start)
     while q:
         z, _, this, prev = heappop(q)
         if this in done:
@@ -41,8 +43,15 @@ def search(start, is_goal, neighbors, heuristic=None):
             c = cost_so_far + c
             if n not in done or c < done[n][0]:
                 h = heuristic(n)
-                if h is None:
-                    continue
+                if worst:
+                    if c+h > best_worst:
+                        # if the best possible cost for this node
+                        # is worse than the lowest worst-case cost we've seen
+                        # then don't even bother exploring it
+                        continue
+                    w = worst(n)
+                    if c+w < best_worst:
+                        best_worst = c+w
                 i += 1
                 heappush(q, (c+h, i, n, this))