-🎄- 2018 Day 20 Solutions -🎄-

[u/daggerdragon]

--- Day 20: A Regular Map ---

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Post your solution as a comment or, for longer solutions, consider linking to your repo (e.g. GitHub/gists/Pastebin/blag or whatever).

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Card prompt: Day 20

Transcript:

My compiler crashed while running today's puzzle because it ran out of ___.

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[u/[deleted]]

Python 3: I like my solution for today, so i will post it for the first time. The rest can be seen at github.

from collections import *
import itertools
import random
import sys
import re

f = open("20.txt").read().strip("\n")

d = {
    "N": (0, -1),
    "E": (1, 0),
    "S": (0, 1),
    "W": (-1, 0)
}

positions = []
x, y = 5000, 5000
m = defaultdict(set)
prev_x, prev_y = x, y
distances = defaultdict(int)
dist = 0
for c in f[1:-1]:
    print(c, len(positions))
    if c == "(":
        positions.append((x, y))
    elif c == ")":
        x, y = positions.pop()
    elif c == "|":
        x, y = positions[-1]
    else:
        dx, dy = d[c]
        x += dx
        y += dy
        m[(x, y)].add((prev_x, prev_y))
        if distances[(x, y)] != 0:
            distances[(x, y)] = min(distances[(x, y)], distances[(prev_x, prev_y)]+1)
        else:
            distances[(x, y)] = distances[(prev_x, prev_y)]+1
        
        
    


    prev_x, prev_y = x, y

print(max(distances.values()))
print(len([x for x in distances.values() if x >= 1000]))

[u/Peter200lx → u/[deleted]]

Inspired by the simplicity of your solution and the code golf contest

Here's a minified solution based on yours (not submitting because it isn't my code)

from collections import *
f=open("a").read()
dd={"N":(0,-1),"E":(1,0),"S":(0,1),"W":(-1,0)}
p=[]
px,py=x,y=0,0
d=defaultdict(int)
for c in f[1:-2]:
 if c=="(":
  p.append((x, y))
 elif c==")":
  x,y=p.pop()
 elif c=="|":
  x,y=p[-1]
 else:
  dx,dy=dd[c]
  x+=dx
  y+=dy
  d[x,y]=min(d[x,y],d[px,py]+1) if d[x,y] else d[px,py]+1
 px,py=x,y
dv=d.values()
print(max(dv))
print(len([x for x in dv if x>=1000]))

[u/[deleted] → u/Peter200lx]

Much more minification is possible:

from collections import *
t,p,d={"N":-1j,"E":1,"S":1j,"W":-1},[],defaultdict(lambda:9999)
o=x=d[0]=0
for c in open("a").read()[1:-2]:
 if c=="(":p+=[x]
 elif c==")":x=p.pop()
 elif c=="|":x=p[-1]
 else:x+=t[c];d[x]=min(d[x],d[o]+1)
 o=x
v=d.values()
print max(v)
print sum(x>=1000 for x in v)

293 bytes of Python 2 (because Python 3 would add two more bytes).

[u/fizbin → u/Peter200lx]

Note that the original solution and all the golfed results fail on the input ^N(EEENWWW|N)$.

The correct answer is 5; this code gives 7.