def part2(lines):
t = 754018
# print(t % 67 == 0)
# print(t % 7 == 6)
# print(t % 59 == 57)
# print(t % 61 == 58)
# print('Part 2:')
n = []
a = []
for i, bus in enumerate(lines[1].split(',')):
try:
n1 = int(bus)
rem = (n1 - i) % n1
except ValueError:
pass
else:
if i == 0 or True: #i != n[0]:
n.append(n1)
a.append(rem)
else:
print('skipping', n[0], n1, rem, i)
for n1, n2 in combinations(n, 2):
if gcd(n1, n2) != 1: print(' ', n1, n2)
print(n)
print(a)
result = int(chinese_remainder(n, a))
for x, y in zip(n, a):
print(x, result % x, y)
print(result)
def useCRT(self):
"Solve the puzzle using the Chineese Remainder Theorem"
# 1. Adjust the positions
adjusted = []
for offset, position in enumerate(self.positions):
adjusted.append(-position - offset)
# 2. Use CRT
result = crt.chinese_remainder(self.sizes, adjusted)
# 3. Drop time is one less
if result == 0:
return None
return result - 1
def solve(self):
chosen_holders = random.sample(self.__holders, self.__min_holder)
if self.__verbose:
print("Chosen holders :")
for holder in chosen_holders:
print(holder)
## Solver CRT
modulo_list = [holder.modulo for holder in chosen_holders]
remainder_list = [holder.secret for holder in chosen_holders]
solution = chinese_remainder(modulo_list, remainder_list)
if self.__verbose:
print("CRT Solution : " + str(solution))
return (solution) % self.__sequence[0]
def main():
""" solutions """
timestamp = 1005162
puzzle_input = "19,x,x,x,x,x,x,x,x,41,x,x,x,x,x,x,x,x,x,823,x,x,x,x,x,x,x,23,x,x,x,x,x,x,x,x,"\
"17,x,x,x,x,x,x,x,x,x,x,x,29,x,443,x,x,x,x,x,37,x,x,x,x,x,x,13"
# convert input to tuple list of positions in input and bus_ids
bus_ids = [(x[0], int(x[1]))
for x in enumerate(puzzle_input.replace('x', '1').split(','))]
bus_ids = [*filter(lambda x: x[1] != 1, bus_ids)]
# calculate departure time relative to timestamp
times = [(x[0], (math.ceil(timestamp / x[1]) * x[1]) - timestamp)
for x in bus_ids]
# calculate the earliest departure time
min_time = sorted(times, key=lambda x: x[1])[0]
print("part 1:", min_time[1] * bus_ids[times.index(min_time)][1])
# find the earliest timestamp such that the first bus ID departs at that time and each
# subsequent listed bus ID departs at that subsequent minute.
# this boils down to a chinese remainder theorem problem find the congruent modulo
seq_n = [x[1] for x in bus_ids]
coprime_a = [x[1] - x[0] for x in bus_ids]
print("part 2:", chinese_remainder(seq_n, coprime_a))
def test_rosetta_code_examples(self):
"Test examples from rosettacode"
self.assertEqual(crt.chinese_remainder([3, 5, 7], [2, 3, 2]), 23)
self.assertEqual(crt.chinese_remainder([5, 13], [2, 3]), 42)
self.assertEqual(crt.chinese_remainder([100, 23], [19, 0]), 1219)
self.assertEqual(crt.chinese_remainder([11, 12, 13], [10, 4, 12]),
1000)
self.assertEqual(crt.chinese_remainder([5, 7, 9, 11], [1, 2, 3, 4]),
1731)
self.assertEqual(
crt.chinese_remainder([
17353461355013928499, 3882485124428619605195281,
13563122655762143587
], [
7631415079307304117, 1248561880341424820456626,
2756437267211517231
]), 937307771161836294247413550632295202816)
import crt
input = open('input.txt').read().split()
ids = [int(e) for e in input[1].replace('x', '0').split(',')]
# print(ids)
frequency = [e for e in ids if e]
# print(frequency)
# print([ ids.index(e)for e in frequency ])
rem = [e - ids.index(e) for e in frequency]
# print(rem)
print(crt.chinese_remainder(frequency, rem))
def test_part_one_example(self):
"Test example from part one description [disc sizes], [initial values]"
self.assertEqual(crt.chinese_remainder([5, 2], [-4, -1 - 1]), 5 + 1)