day 21 cleanup
parent
c5d4f796e0
commit
a940dff830
279
day21/sol.py
279
day21/sol.py
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@ -1,40 +1,58 @@
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import sys
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import time
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import numpy
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map = [x.strip() for x in sys.stdin]
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if len(map) < 80:
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def read_map(input):
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map = [x.strip() for x in input]
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if len(map) < 80:
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print(map)
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Y = len(map)
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X = len(map[0])
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assert all(len(row) == X for row in map)
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Y = len(map)
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X = len(map[0])
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assert all(len(row) == X for row in map)
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assert X == Y
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grid = {}
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#start = None
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#for i, row in enumerate(map):
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# for j,c in enumerate(row):
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# if c in "#.":
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# pass
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# elif c == "S":
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# start = (i,j)
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# else:
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# raise Exception("invalid char %s at (%s,%s)" % (c, i, j))
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# grid[i,j] = d
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#print(grid)
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return map
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def draw(mask, fill):
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return
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if len(fill) < 50:
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for i, row in enumerate(fill):
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print("".join(".O#!"[f + 2*mask[i,j]] for j,f in enumerate(row)))
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print()
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def new():
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return numpy.zeros((Y,X+1), dtype='uint8')
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def solve(grid):
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def solve(map):
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Y = len(map)
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X = len(map[0])
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fill = numpy.zeros((Y,X+1), dtype='uint8')
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mask = numpy.zeros((Y,X+1), dtype='uint8')
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for i, row in enumerate(map):
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for j, c in enumerate(row):
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if c == '#':
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mask[i,j] = 1
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if c == 'S':
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start = (i,j)
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# add a border on the side of the map (to prevent wraparound)
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mask[:, -1] = 1
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# start at the start
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fill[start] = 1
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for i in range(64):
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fill = step(mask, fill)
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#draw(mask, fill)
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if i == 6-1: # sample
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print(fill.sum())
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print("part1 = ", fill.sum())
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def solve2(map):
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Y = len(map)
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X = len(map[0])
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fill = numpy.zeros((Y,X), dtype='uint8')
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mask = numpy.zeros((Y,X), dtype='uint8')
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for i, row in enumerate(map):
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@ -44,152 +62,96 @@ def solve(grid):
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if c == 'S':
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start = (i,j)
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S = 11
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if X <= 11:
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S = 15
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fill = numpy.c_[numpy.tile(fill,(S,S)), numpy.zeros(Y*S, dtype='uint8')]
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mask = numpy.c_[numpy.tile(mask,(S,S)), numpy.zeros(Y*S, dtype='uint8')]
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cache = {} # bitmap->k
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maps = {} # k->bitmap
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cycles = {} # k->[k']
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print(fill.shape, mask.shape)
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target_steps = 4064
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fill[start] = 1
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values, period, d2 = simulate(fill, mask, max_steps=500)
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fill[S//2*Y + start[0], S//2*X + start[1]] = 1
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values, d2 = sequence(fill, mask, target_steps, X)
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for n in 6,10,50,100,500,1000, 5000, 26501365:
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print(n,extrapolate(n, values, X, d2))
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return
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print(n, extrapolate(n, values, period, d2))
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# find the tiles reachable in n steps from each possible start
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# position
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reachable = {}
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breach = {}
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center = slice(S//2*Y, (S//2+1)*Y), slice(S//2*X, (S//2+1)*X)
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left = slice(S//2*Y, (S//2+1)*Y), slice((S//2-1)*X, S//2*X)
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right = slice(S//2*Y, (S//2+1)*Y), slice((S//2+1)*X, (S//2+2)*X)
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up = slice((S//2-1)*Y, (S//2 )*Y), slice(S//2*X, (S//2+1)*X)
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down = slice((S//2+1)*Y, (S//2+2)*Y), slice(S//2*X, (S//2+1)*X)
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dirs = [left,right,up,down]
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for i in range(Y):
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for j in range(X):
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if i in (0,Y-1) or j in (0,X-1) or (i,j) == start:
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fill = numpy.zeros((Y*S,X*S+1), dtype='uint8')
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fill[S//2*Y + i, S//2*X + j] = 1
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reachable[i,j] = [fill]
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found = [0,0,0,0]
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while True:
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s0 = fill
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s1 = step(mask, s0)
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s2 = step(mask, s1)
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fill = s2
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n1 = len(reachable[i,j])
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n2 = len(reachable[i,j])+1
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reachable[i,j].append(s1)
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reachable[i,j].append(s2)
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for d in range(4):
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if not found[d]:
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if s1[dirs[d]].any():
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found[d] = n1
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elif s2[dirs[d]].any():
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found[d] = n2
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if (s0 == s2)[center].all():
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break
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breach[i,j] = found
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print(i,j,len(reachable[i,j]), found)
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#draw(mask, fill)
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def simulate(fill, mask, max_steps):
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Y,X = mask.shape
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assert Y == X, "need a square matrix"
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period = X
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tiles = 1
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for (i,j),fills in reachable.items():
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fills = reachable[i,j]
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when = breach[i,j]
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for d in range(4):
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if when[d]:
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f = fills[when[d]][dirs[d]]
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num_breach_points = sum(f.ravel())
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assert num_breach_points > 0
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print(i,j,d, num_breach_points == 1, num_breach_points)
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if num_breach_points > 1:
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draw(mask, fills[when[d]])
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return
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# maximum number of steps for a tile to become completely reachable
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M = max(len(r) for r in reachable.values())
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states = {}
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active = {}
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reachable[start]
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states = [((0,0),[(0,start)])]
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for iters in range(target_steps):
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assert fill.any()
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fill = step(mask, fill)
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fill = step(mask, fill)
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#print("\033[2J") # clear screen
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#draw(mask, fill)
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for u in range(S):
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for v in range(S):
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small = fill[Y*u:Y*(u+1), X*v:X*(v+1)]
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#if (u,v) == (1,1): print(small)
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b = small.tobytes()
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super[u,v] = cache.setdefault(b, len(cache))
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print("========")
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print(super)
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print(flush=True)
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#time.sleep(.1)
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# look for symmetries
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def look():
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for u in range(S):
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for v in range(u,S):
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if super[u,v] != super[v,u]:
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return False
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return True
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#print(fill.tobytes())
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def sequence(fill, mask, target_steps, period):
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original_mask = mask
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prev = [0]
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c1, d1 = 0, 0
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c2, d2 = 0, 0
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prev2 = []
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for i in range(target_steps):
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diffs = []
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for i in range(max_steps):
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# expand the map if necessary
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if fill[0].any() or fill[-1].any() or fill[:,0].any() or fill[:,-1].any():
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draw(mask,fill)
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print("expanding...")
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tiles += 2
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mask = numpy.tile(original_mask, (tiles,tiles))
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fill = numpy.pad(fill, [(Y,Y), (X,X)])
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# take one more step and count the number of reachable squares
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fill = step(mask, fill)
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#fill = step(mask, fill)
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n = int(fill.sum())
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#if len(prev) > period:
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# c1, d1 = d1, n - prev[-period]
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# c2, d2 = d2, d1 - c1
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# prev2.append(d2)
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# look for patterns
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# although the number of squares is somewhat unpreditable from step
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# to step (due to the maze-like structure of the mask), the fact that
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# the mask repeats in tiles (and the fact that there are unobstructed
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# pathways in the orthogonal and diagonal directions) means that, in
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# the long run, the flood fill will even out and -importantly- it should
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# have some recognizable pattern every N steps (where N is the period
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# of the tiling - 11 in the sample, 131 in the input). this is because
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# we enter new tiles every N steps, and although it's hard to predict
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# exactly how many of the squares in each tile we'll have visited,
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# it should be the same number in every tile (or rather, there should
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# be some small set of repeated tile-states, and we should be able to
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# predict how many of each tile-state there will be).
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#
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# SO the first step is to get the difference between the current
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# number of reachable squares and the number N steps ago.
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#
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# d1(i) = f(i) - f(i - N)
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#
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# we then have to discover some pattern in that sequence.
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# we know the number of reachable squares will grow roughly as the
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# square of the number of steps (because the map is 2D) so
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# we should be looking for a quadratic relation.
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# we can use second-order differences (see day 9) to do that.
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# d1 is already a first-order difference, so take the difference
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# between d1s to get a second-order difference. if our assumption is
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# correct, then d1 should be a linear sequence and d2 should be a
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# constant.
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#
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# d2(i) = d1(i) - d1(i-N)
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# = (f(i) - f(i-N)) - (f(i-N) - f(i-N-N))
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#
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#
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# there may be some unstability at the beginning of the simulation
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# so we need to wait until the d2 values for every step in the period
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# all agree.
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#
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# once it settles down, this gives us a set of N (11, 131, whatever)
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# equations we can use to predict the number of squares after any
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# future number of steps
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#
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d2 = 0
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if len(prev) >= 2*period:
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# find the second differece
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d2 = (n - prev[-period]) - (prev[-period] - prev[-2*period])
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prev2.append(d2)
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diffs.append(d2)
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prev.append(n)
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print(i,n,d1,d2,sep="\t",flush=True)
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if len(prev2) > period and len(set(prev2[-period:])) == 1:
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print(i, n, d2, sep="\t", flush=True)
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if len(diffs) > period and all_same_value(diffs[-period:]):
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print("gotcha!")
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return prev, prev2[-1]
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break
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if len(prev2) > period*2 and prev2[-period*2:-period] == prev2[-period:]:
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print("gotcha!")
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break
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if len(prev2) > period*3 and prev2[-period*3:-period] == prev2[-period*2:]:
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print("gotcha!")
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break
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if fill[0].any() or fill[-1].any() or fill[:,0].any() or fill[:,-1].any():
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draw(mask,fill)
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break
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return
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return prev, period, diffs[-1]
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assert False, "failed to find a stable pattern"
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return prev, period, None
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def all_same_value(list):
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if len(list) < 1:
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return False
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x = list[0]
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return all(x == y for y in list)
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def extrapolate(n, values, period, d2):
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if n < len(values):
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@ -207,8 +169,6 @@ def extrapolate(n, values, period, d2):
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def step(mask, old):
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# flood fill
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# note that the provided map has a 1-tile border of empty spaces
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# which we will use to our advantage
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#draw(fill)
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fill = numpy.zeros(old.shape, dtype='uint8')
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y,x = fill.shape
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@ -217,7 +177,6 @@ def step(mask, old):
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f = numpy.roll(f, 1) | numpy.roll(f, -1)
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if i > 0: f |= (old[i-1] == 1)
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if i < y-1: f |= (old[i+1] == 1)
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f[-1] = False
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f &= (mask[i] == 0)
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#print(old, f)
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if f.any():
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@ -227,4 +186,6 @@ def step(mask, old):
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return fill
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solve(grid)
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map = read_map(sys.stdin)
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solve(map)
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solve2(map)
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