day 16 python alternate solution (astar)

day16/search.py

@@ -0,0 +1,97 @@
+G = []
+for line in open("input"):
+    words = line.split()
+    valve = words[1]
+    rate = int(''.join(x for x in words[4] if x.isdigit()))
+    edges = [x.strip(", ") for x in words[9:]]
+    G.append((valve, rate, edges))
+
+#print(G)
+
+import sys, os; sys.path.append(os.path.join(os.path.dirname(__file__), "../lib"))
+import astar
+
+def search():
+    V = sorted(v for v,_,_ in G) # vertices
+    B = {v: 1<<i for i, v in enumerate(V)} # bitmasks
+    E = {B[v]: [B[e] for e in edges] for v,_,edges in G} # edge[b] -> b
+    R = {B[v]: r for v,r,_ in G} # rewards: R[b] = reward
+
+    all_closed = sum(B.values())
+    all_open = 0
+
+    minutes = 30
+    start = (B['AA'], minutes, all_closed)
+
+    # A* search minimizes costs
+    # it can't maxmize anything
+    # so we'll borrow an idea from https://github.com/morgoth1145/advent-of-code/blob/2bf7c157e37b3e0a65deedc6c88e42297d813d1d/2022/16/solution.py
+    # and instead say that the cost of moving from one node to the next
+    # is equal to the potential pressure we could have released from the closed pipes
+    # or, in other words, we'll keep track of how much pressure builds up in
+    # closed pipes instead of how much pressure is released from open pipes
+
+    # let's shink the graph by finding the shortest path between
+    # every pair of rooms (floyd-warshall), and then building a graph which only has
+    # paths from the starting room to rooms with a valve
+    # and from any room with a valve to any other room with a valve
+
+    # our heuristic cost has to be <= the actual cost of getting to the goal
+    # here's a simple one:
+    # we know it takes at least 1 minute to open a valve,
+    # and at least another minute to walk to the valve
+    # so we can assign at least cost 2*pressure(closed_valves) to this node
+    # (unless there is only 1 minute left)
+    def heuristic(node):
+        v, minutes, closed = node
+        if closed == all_open:
+            m = minutes
+        else:
+            m = min(minutes,1)
+        return m*pressure(closed)
+
+    def pressure(bits):
+        pressure = 0
+        for v,r in R.items():
+            if bits&v:
+                pressure += r
+        return pressure
+
+    def is_goal(n):
+        v, minutes, closed = n
+        return minutes == 0
+
+    info = {}
+    def neighbors(n):
+        v, minutes, closed = n
+        if minutes not in info:
+            print(info)
+            info[minutes] = 0
+        info[minutes] += 1
+        if minutes <= 0:
+            pass
+        elif closed == all_open:
+            c = pressure(closed) * minutes
+            yield c, (v, 0, closed)
+        else:
+            c = pressure(closed)
+            if v&closed and R[v]:
+                yield c, (v, minutes-1, closed & ~v)
+            for e in E[v]:
+                yield c, (e, minutes-1, closed)
+
+
+    c, _, path = astar.search(start, is_goal, neighbors, heuristic)
+    print(c)
+    print(pressure(all_closed)*30 - c)
+
+    #maxpair = []
+    #def pairs():
+    #    O = sorted(best.keys())
+    #    for i in range(len(O)):
+    #        for j in range(i,len(O)):
+    #            if not O[i] & O[j]:
+    #                yield(best[O[i]]+best[O[j]])
+    #print(max(pairs()))
+
+search()

lib/astar.py

@@ -1,10 +1,14 @@
 from heapq import heappush, heappop
 
-def search(start, goal, neighbors, heuristic=None):
+def search(start, is_goal, neighbors, heuristic=None):
     if heuristic == None:
         def heuristic(x):
             return 0
     # TODO: callable goal
+    if not callable(is_goal):
+        goal = is_goal
+        def is_goal(this):
+            return this == goal
     if not isinstance(start, list):
         start = [start]
     i = 0 # tiebreaker
@@ -22,7 +26,7 @@ def search(start, goal, neighbors, heuristic=None):
 
         done[this] = (cost_so_far, prev)
 
-        if this == goal:
+        if is_goal(this):
             print("astar: visited", len(done), "nodes")
             # reconsruct the path
             n = this