221 lines
7.3 KiB
Python
221 lines
7.3 KiB
Python
G = []
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for line in open("input"):
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words = line.split()
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valve = words[1]
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rate = int(''.join(x for x in words[4] if x.isdigit()))
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edges = [x.strip(", ") for x in words[9:]]
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G.append((valve, rate, edges))
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#print(G)
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import sys, os; sys.path.append(os.path.join(os.path.dirname(__file__), "../lib"))
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import astar
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def solve():
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G.sort(key=lambda x: (-x[1],x[0]))
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B = {v: 1<<i for i,(v,_,_) in enumerate(G)} # bitmasks
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E = {B[v]: [B[e] for e in edges] for v,_,edges in G} # E[b] = edges of b
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R = {B[v]: r for v,r,_ in G if r} # R[b] -> rate
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AA = B['AA']
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all_closed = sum(R.keys())
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all_open = 0
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# TODO: memoize this
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def pressure(bits):
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pressure = 0
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for v,r in R.items():
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if bits&v:
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pressure += r
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return pressure
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assert pressure(all_open) == 0
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max_pressure = pressure(all_closed)
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# A* search minimizes costs
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# it can't maxmize anything
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# so we'll borrow an idea from https://github.com/morgoth1145/advent-of-code/blob/2bf7c157e37b3e0a65deedc6c88e42297d813d1d/2022/16/solution.py
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# and instead say that the cost of moving from one node to the next
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# is equal to the potential pressure we could have released from the closed pipes
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# or, in other words, we'll keep track of how much pressure builds up in
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# closed pipes instead of how much pressure is released from open pipes
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# let's shink the graph by finding the shortest path between
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# every pair of rooms (floyd-warshall), and then use that to build a
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# weighted graph which only has paths from any room to rooms with a valve.
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#
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# not only does this make our search space smaller,
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# it also helps by making it so that the cost changes on every step
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# (since opening a valve is the only thing that actually changes the pressure)
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# giving A* a much clearer signal about which paths are worth exploring
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dist = {}
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for v in E:
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dist[v,v] = 0
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for e in E[v]:
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dist[v,e] = 1
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dist[e,v] = 1
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for t in E:
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for u in E:
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for v in E:
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if (u,t) in dist and (t,v) in dist:
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dist[u,v] = min(dist.get((u,v),999999), dist[u,t] + dist[t,v])
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W = {} # weighted edges
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for u in E:
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W[u] = []
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for v in R:
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if (u,v) in dist:
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W[u].append((v, dist[u,v]))
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print(W[AA])
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# our heuristic cost has to be <= the actual cost of getting to the goal
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#
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# here's a simple one:
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# we know it takes at least 1 minute to open a valve,
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# and at least another minute to walk to the valve
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# so we can assign at least cost 2*pressure(closed_valves) to this node
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# (unless there is only 1 minute left)
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#
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# we can keep doing that until there are no valves left to open
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# or there is no time left.
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#
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# note that the nodes with the largest flow rate are assigned
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# the lowest position in the bitmap, so clearing the bits from
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# low to high will always give us the optimal order
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def heuristic(node):
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v, minutes, closed = node
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# assume we can open a valve every 2 minutes
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# how much would that cost?
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c = 0
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while closed and minutes > 0:
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c += pressure(closed) * min(minutes,2)
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closed &= (closed-1)
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minutes -= 2
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return c
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def is_goal(node):
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v, minutes, closed = node
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return minutes == 0 or closed == all_open
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info = {}
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def neighbors(node):
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v, minutes, closed = node
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if minutes not in info:
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print(info)
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info[minutes] = 0
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info[minutes] += 1
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if minutes <= 0:
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pass
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elif closed == all_open:
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yield 0, (v, 0, closed)
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else:
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# move to a closed valve (or maybe stay in
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# the same spot) and open it
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can_move = False
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for e, dist in W[v]:
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t = dist + 1
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if e&closed and t <= minutes:
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can_move = True
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c = pressure(closed)*t
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yield c, (e, minutes-t, closed&~e)
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# wait til the end
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if can_move == False:
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yield pressure(closed)*minutes, (v, 0, closed)
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minutes = 30
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start = (AA, minutes, all_closed)
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c, _, path = astar.search(start, is_goal, neighbors, heuristic)
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print(c)
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print(max_pressure*minutes - c)
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def heuristic2(node):
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if is_goal2(node):
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return 0
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v1, v2, min1, min2, closed = node
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# if the players are out of sync, assume the other player
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# will close one valve when they catch up
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if min1 != min2:
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closed &= closed - 1
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# assume we can open a valve every minute remaining
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# how much would that cost?
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c = 0
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pr = pressure(closed)
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m = min(min1, min2)
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while closed and m > 0:
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c += pr
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tmp = closed
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closed &= (closed-1)
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pr -= R[tmp-closed]
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m -= 1
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return c
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def worst2(node):
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_, _, min1, min2, closed = node
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return min(min1,min2) * pressure(closed)
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def is_goal2(node):
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_, _, min1, min2, closed = node
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return min1 == 0 and min2 == 0 or closed == all_open
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def neighbors2(node):
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v1, v2, min1, min2, closed = node
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if min(min1,min2) not in info:
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print(info)
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info.setdefault(min1, 0)
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info.setdefault(min2, 0)
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info[min1] += 1
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info[min2] += 1
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if min1 <= 0 and min2 <= 0:
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pass
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elif closed == all_open:
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yield 0, (v1, v2, 0, 0, closed)
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else:
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moved = False
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# either player can move
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# but we can't open a valve that would take less time to open
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# than the other player has already used. we've already paid
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# the cost for _not_ opening those valves and we can't
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# retroactively change that (no time travel)
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# if both players are in the same spot and have the same amount of
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# time remaining, then only let one of them move in order to break
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# symmetries (this can only happen in the start state)
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# TODO: are there more symmetries we can break?
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# move to a closed valve and open it
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discount1 = max(min1-min2, 0)
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for e, dist in W[v1]:
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t = dist + 1
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if e&closed and discount1 <= t <= min1:
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moved = True
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c = pressure(closed)*(t-discount1)
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yield c, (e, v2, min1-t, min2, closed&~e)
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if (v1, min1) != (v2, min2):
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discount2 = max(min2-min1, 0)
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for e, dist in W[v2]:
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t = dist + 1
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if e&closed and discount2 <= t <= min2:
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moved = True
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c = pressure(closed)*(t-discount2)
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yield c, (v1, e, min1, min2-t, closed&~e)
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# are there no moves left?
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# then wait out the timer
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if not moved:
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yield pressure(closed)*min(min1,min2), (v1, v2, 0, 0, closed)
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minutes = 26
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start2 = (AA, AA, minutes, minutes, all_closed)
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info.clear()
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c, _, path = astar.search(start2, is_goal2, neighbors2, heuristic2, worst=worst2)
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print(c)
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print(max_pressure*minutes - c)
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solve()
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