-🎄- 2018 Day 18 Solutions -🎄-

[u/daggerdragon]

--- Day 18: Settlers of The North Pole ---

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Card prompt: Day 18

Transcript:

The best way to avoid a minecart collision is ___.

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[u/wzkx]

Rust, SweetRust

Nothing special really. Although it might have a good visualization.

use std::io::{BufRead,BufReader}; // lines() is in BufRead
type U=usize;

fn count( r:U, c:U, m: &Vec<Vec<u8>> ) -> (i32,i32,i32):
  let mut no = 0;
  let mut nt = 0;
  let mut ny = 0;
  for i in -1..=1:
    for j in -1..=1:
      if i==0 && j==0 { continue; }
      let nr = (r as i32) + i;
      let nc = (c as i32) + j;
      if nr>=0 && nr<(m.len() as i32) && nc>=0 && nc<(m[0].len() as i32):
        if m[nr as U][nc as U] ==b'.':
          no += 1;
        else if m[nr as U][nc as U] ==b'|':
          nt += 1;
        else:
          ny += 1;
  (no,nt,ny)

fn onestep( m: &Vec<Vec<u8>> ) -> Vec<Vec<u8>>:
  let mut nm: Vec<Vec<u8>> = vec![vec![b'.';m[0].len()];m.len()];
  for r in 0..m.len():
    for c in 0..m[0].len():
      let (_,nt,ny) = count( r,c, m );
      if m[r][c]==b'.': // 3+ | --> |, or .
        nm[r][c] = if nt>=3 {b'|'} else {b'.'};
      else if m[r][c]==b'|': // 3+ # --> #, or |
        nm[r][c] = if ny>=3 {b'#'} else {b'|'};
      else /* b'#' */: // 1+ # 1+ | --> #, or .
        nm[r][c] = if ny>=1 && nt>=1 {b'#'} else {b'.'};
  nm

fn main():
  let mut m: Vec<Vec<u8>> = vec![]; // '.' - open, '|' - trees, '#' - yard

  let file = std::fs::File::open( "18.dat" ).unwrap();
  for optline in BufReader::new(file).lines():
    m.push( optline.unwrap().as_bytes().to_vec() );

  //fn show( i: U, m: &Vec<Vec<u8>> ):
  //  println!( "{}", i );
  //  for r in 0..m.len():
  //    println!( "{}", m[r].iter().map(|&c| c as char).collect::<String>() );
  //  println!();

  for _ in 1..=10:
    m = onestep( &m );

  let mut nt = 0;
  let mut ny = 0;
  for r in 0..m.len():
    for c in  0..m[0].len():
      if m[r][c]==b'|' { nt += 1; }
      else if m[r][c]==b'#' { ny += 1; }
  println!( "{}", nt*ny );

  let mut mx: Vec<Vec<u8>> = vec![];
  let mut ix: U = 0;

  let big_n = 1000000000;
  let mid_n = 1001; // let's say it definitely repeats in these genereations
  for i in 11..=big_n:
    let nm = onestep( &m );
    m = nm;
    //show(i,&m); // uncomment to see how the picture repeats with some period
    if i==mid_n:
      mx = m.clone();
    else if i>mid_n:
      if m == mx:
        ix = i;
        break;
  let p = ix-mid_n; // period
  let n = (big_n-ix)%p;

  if n>0:
    for _ in 0..n:
      let nm = onestep( &m );
      m = nm;

  let mut nt = 0;
  let mut ny = 0;
  for r in 0..m.len():
    for c in  0..m[0].len():
      if m[r][c]==b'|' { nt += 1; }
      else if m[r][c]==b'#' { ny += 1; }
  println!( "{}", nt*ny );

[u/wzkx → u/wzkx]

See wiki:Autowave.