Advent of Code 2017

[stranac]

Just wanted to remind everyone that the third edition of Advent of Code is online.

If you enjoy interesting stories and fun challenges, this should be right up your alley.
I haven't participated in the last year's AoC, but the first one was great fun, and I expect no less from the 2017 challenges.

[snippsat]

AoC door 2:

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from itertools import combinations

with open('2_input.txt') as f:
    lst = [[int(i) for i in line.split()] for line in f]

ans_1 = sum(max(line) - min(line) for line in lst)
ans_2 = sum(x//y for line in lst for y,x in combinations(sorted(line),2) if x%y == 0)
print(ans_1)
print(ans_2)

[snippsat]

AoC door 4:

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# part 1
from collections import Counter

count = 0
with open('4.txt') as f:
    for line in f:
        if Counter(line.split()).most_common(1)[0][-1] == 1:
            count += 1

print(count)

# Part 2
from itertools import combinations

anagrams = 0
with open('4.txt') as f:
    for line_nr,line in enumerate(f, 1):
        for a, b in combinations(line.split(), 2):
            if sorted(a) == sorted(b):
                anagrams += 1
                break

print(line_nr - anagrams)

[nilamo]

#3 is making me feel really dumb lol.  I feel like there's a simple way to get the answer with math, without needing to actually build a circular data structure, but I just can't figure it out.  I tried sleeping on it, but it still looks like magic lol.

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The max distance is the square root of the number.  So for 23, the square root is 4.etc, and you can see that the corners are all 4 moves away from the center or less, while rounding that square root shows that there's 5 rows and columns.  Also, it doesn't really matter where the "end" is, as the end can always be in the top left corner (index 0 in a list), and it'll still be the same distance from the center.  But that doesn't help, as that doesn't let you know if you're in a corner or the center, so you can't just know that it's (number of rows/2) steps.

I feel like I'm thinking way too hard about this, and should just try to brute force it, but that feels ugly...

[buran]

day#3 part1.

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def mid_squares(n):
    min_square = (n-2)**2 + 1
    top_right = min_square + n - 2
    top = top_right + n/2
    left = top + n - 1
    bottom = left + n - 1
    right = top_right - n/2
    return (right, top, left, bottom)
    
puzzle = 289326
h = 0
while (h*2 + 1)**2 <= puzzle:
    h += 1
w = min([abs(puzzle - x) for x in mid_squares(h*2 + 1)])
print('distance is {}'.format(h+w))

still struggling with part2 :-)

[buran]

day#5 part2

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with open('instructions.txt') as f:
    instructions = map(int,f.read().split('\n'))

pos = 0
steps = 0
while pos>=0 and pos < len(instructions):
    if instructions[pos] > 2:
        offset = -1
    else:
        offset = 1
    instructions[pos] += offset
    pos += (instructions[pos] - offset)
    steps += 1
print(steps)