adventofcode2023/day17/sol.py

70 lines
2.0 KiB
Python
Raw Normal View History

import astar
def read_input(f):
map = []
for line in f:
map.append([int(x) for x in line.strip()])
return map
def solve(map,part=1):
start = (0,0,(0,0),0)
goal = (len(map)-1,len(map[-1])-1)
def is_goal(x):
return x[:2] == goal
2023-12-22 09:16:15 +00:00
def heuristic(state):
# manhattan distance. this is correct because the cost
# at every point is >= 1 and we can only move one square
# at a time NSEW.
if part == 2 and state[3] < 4:
y,x,d,count = state
# account for forced moves away from the goal
if d[0] < 0 or d[1] < 0:
return (4-count)*2 + abs(y-goal[0]) + abs(x-goal[1])
return abs(state[0]-goal[0]) + abs(state[1]-goal[1])
def neighbors(state):
y,x,dir,count = state
next = []
def go(dx,dy):
if (-dx,-dy) == dir:
2023-12-22 09:16:15 +00:00
# no turning back
return
newx = x + dx
newy = y + dy
if (dx,dy) == dir:
newcount = count+1
else:
newcount = 1
if part == 1:
if newcount > 3:
return
if part == 2:
if newcount > 10:
return
if newcount <= count < 4:
return
if (0 <= newy < len(map)) and (0 <= newx < len(map[newy])):
next.append((map[newy][newx], (newy, newx, (dx,dy), newcount)))
go(-1,0)
go(+1,0)
go(0,-1)
go(0,+1)
return next
2023-12-22 09:16:15 +00:00
# turns out it's actually faster to not use the heuristic, so don't
cost, node, path = astar.search(start, is_goal, neighbors)
print(cost)
print([(x,y) for (y,x,_,_) in path])
print(''.join(dir2s(d) for (_,_,d,_) in path))
def dir2s(d):
if d == (0,1): return "v"
if d == (0,-1): return "^"
if d == (1,0): return ">"
if d == (-1,0): return "<"
return "?"
import sys
map = read_input(sys.stdin)
solve(map, part=1)
solve(map, part=2)