Re: Un grand classique, mais toujour aussi long à résoudre

J'avais aussi commencé par une fonction récursive, mais elle ne dépassait la limite de recursion, je l'ai remplacé par une simple boucle while.

inputs = open(f'input_d10.txt').read()
grid=[]
for y,l in enumerate(inputs.strip().split('\n')):
 if 'S' in l:
 start = l.find('S'), y
 grid.append([*l])
H,W = len(grid)-1, len(grid[0])-1
pipes = {
 '|': ((0,1),(0,-1)), 
 '-': ((1,0),(-1,0)),
 'L': ((0,-1),(1,0)),
 'J': ((0,-1),(-1,0)),
 '7': ((-1,0),(0,1)),
 'F': ((1,0),(0,1)),
 '.': ((0,0),(0,0))
}
dirs = {s:{i:o, o:i} for s,(i,o) in pipes.items()}
def loop(dx,dy,x,y):
 l, path = 0, []
 while 1:
 if l and (x,y) == start:
 return l, path
 path.append((x,y))
 nx, ny = x+dx, y+dy
 if nx<0>ny or nx>W or ny>H:
 return None
 d = [(ndx,ndy) for dxy in map(dirs.get,grid[ny][nx]) 
 for (ndx,ndy) in dxy if (nx+ndx,ny+ndy)==(x,y)]
 if not d:
 return None
 ndx, ndy = dirs[grid[ny][nx]][d[0]]
 dx, dy, x, y, l = ndx, ndy, nx, ny, l+1
ll, path = 0, []
s = ''
for p in pipes:
 x,y = start
 dx, dy = pipes[p][0]
 grid[y][x] = p
 if lp:=loop(dx,dy,x,y):
 if lp[0]>ll:
 ll, path = lp
 s = p
part_1 = ll // 2
print(f"{part_1=}")
part_2 = 0
x,y = path[0]
grid[y][x] = s 
path = set(path)
prev_l = '.' * (W+1)
for y,l in enumerate(grid):
 inside = 0
 for x,c in enumerate(l):
 if ((x,y) in path and (c=='|' or 
 (c in 'JL' and prev_l[x] in '|7F'))):
 inside ^= 1
 elif (x,y) not in path:
 part_2 += inside
 prev_l = l
print(f"{part_2=}")