Encore des maths, du Python et de la POO

Ma solution aurait pu être mieux optimisée pour la lisibilité et la redondance, mais elle fonctionne. Je définis une classe pour les nœuds, surcharge quelques opérateurs, prétraite quelques données, et trouve le ppcm en utilisant le module math de Python.

#!/bin/python3
from math import lcm
class Nodon:
 def __init__(self,name,left,right):
 self.name = name
 self.right = right
 self.left = left
 def __repr__(self):
 return self.name
 def __eq__(self,stringnode):
 if self.name == stringnode:
 return True
 return False
 def endswith(self,letter):
 if self.name[2] == letter:
 return True
 return False
 def go(self,direction):
 if direction == "R":
 return self.right
 if direction == "L":
 return self.left
 raise ValueError
def parse_node(puzzle):
 direction = puzzle[0]
 nodes = []
 for line in puzzle[1:]:
 if line == "":
 continue
 nodes.append(Nodon(line[:3],line[7:10],line[12:15]))
 return direction, nodes
def run(direction, nodes):
 now = nodes[nodes.index("AAA")]
 nb = 0
 while now != "ZZZ":
 for d in direction:
 now = now.go(d)
 nb += 1
 if now == "ZZZ":
 break
 return nb
def nodesendswith(nodes, l):
 for n in nodes:
 if n.endswith(l) == False:
 return False
 return True
def beforerun(nodes):
 for n in nodes:
 n.right = nodes[nodes.index(n.right)]
 n.left = nodes[nodes.index(n.left)]
 return nodes
def runboth(direction, nodes):
 now = []
 for n in nodes:
 if n.endswith("A"):
 now.append(n)
 nb = 0
 ends = []
 while len(now) != 0:
 for d in direction:
 nb += 1
 for i in range(len(now)):
 now[i] = now[i].go(d)
 if now[i].endswith("Z"):
 ends.append(nb)
 now[i] = None
 for i in range(now.count(None)):
 now.remove(None)
 if len(now) == 0:
 break
 return lcm(*ends)
def solve1(puzzle,testing=False):
 s=0
 d, nodes = parse_node(puzzle)
 s = run(d, tuple(beforerun(nodes)))
 if testing:
 print(s)
 return s
def solve2(puzzle,testing=False):
 s = 0
 d, nodes = parse_node(puzzle)
 s = runboth(d, tuple(beforerun(nodes)))
 if testing:
 print(s)
 return s
test1 = """RL
AAA = (BBB, CCC)
BBB = (DDD, EEE)
CCC = (ZZZ, GGG)
DDD = (DDD, DDD)
EEE = (EEE, EEE)
GGG = (GGG, GGG)
ZZZ = (ZZZ, ZZZ)
"""
result1 = 2
test2 = """LR
11A = (11B, XXX)
11B = (XXX, 11Z)
11Z = (11B, XXX)
22A = (22B, XXX)
22B = (22C, 22C)
22C = (22Z, 22Z)
22Z = (22B, 22B)
XXX = (XXX, XXX)
"""
result2 = 6
# for cli invocation
def solve(short=False,one=True,two=True):
 s1, s2 = False , False
 if one:
 print("----Part 1----")
 if short == False:
 if solve1(test1.split("\n"),testing=True) != result1:
 print("Not working.")
 return False
 else :
 print("Maybe working?")
 with open("input.txt",'r') as file:
 lines = file.read().split("\n")
 s1 = solve1(lines)
 print(s1)
 if two:
 print("----Part 2----")
 if short == False:
 if solve2(test2.split("\n"),testing=True) != result2:
 print("Not working.")
 return False
 else :
 print("Maybe working?")
 with open("input.txt",'r') as file:
 lines = file.read().split("\n")
 s2 = solve2(lines)
 print(s2)
 return s1, s2
if __name__ == "__main__":
 from sys import argv
 s = False
 one = True
 two = True
 if len(argv) >= 2:
 if "-s" in argv or "--summary" in argv or "--short" in argv:
 s = True
 if "1" in argv:
 two = False
 one = True
 elif "2" in argv:
 one = False
 two = True
 if "2" in argv:
 two = True
 solve(short=s,one=one,two=two)

L'informatique n'est pas une science exacte, on n'est jamais à l'abri d'un succès