Quest 8: The Art of Connection

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  • janAkali@lemmy.sdf.org
    2·
    8 days ago

    Nim

    Part 2 - I just really didn’t want to think that day. So when puzzle asked me to check if lines intersect - I wrote the intersection checking solution with 2D points.

    Part 3 is geometry + bruteforce.

    proc solve_part1*(input: string): Solution =
  let pins = input.split(',').mapIt(parseInt(it))
  for i in 0 ..< pins.high:
    let d = abs(pins[i] - pins[i+1])
    if d == 16: inc result.intVal

proc ccw(A,B,C: Vec2): bool = (C.y-A.y) * (B.x-A.x) > (B.y-A.y) * (C.x-A.x)
proc isIntersection(A,B,C,D: Vec2): bool = ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)

proc solve_part2*(input: string): Solution =
  const two_pi = PI * 2
  const pin_count = 256

  var pins: array[pin_count, Vec2]
  for i in 0 ..< pin_count:
    let angle = two_pi * (i / pin_count)
    let point: Vec2 = (cos(angle), sin(angle))
    pins[i] = point

  let inst = input.split(',').mapIt(parseInt(it))
  var lines: seq[(Vec2, Vec2)]

  for i in 0 ..< inst.high:
    let A = pins[inst[i]-1]
    let B = pins[inst[i+1]-1]

    for (C, D) in lines:
      if isIntersection(A,B,C,D):
        inc result.intVal
    lines.add shortenSegment(A, B, 0.0001)

proc solve_part3*(input: string): Solution =
  const two_pi = PI * 2
  const pin_count = 256

  var pins: array[pin_count, Vec2]
  for i in 0 ..< pin_count:
    let angle = two_pi * (i / pin_count)
    let point: Vec2 = (cos(angle), sin(angle))
    pins[i] = point

  let inst = input.split(',').mapIt(parseInt(it))
  var lines: seq[(Vec2, Vec2)]

  for i in 0 ..< inst.high:
    let A = pins[inst[i]-1]
    let B = pins[inst[i+1]-1]
    lines.add shortenSegment(A, B, 0.0001)

  var bestSum = 0
  for i in 0 ..< pin_count:
    for j in i+1 ..< pin_count:
      let A = pins[i]
      let B = pins[j]
      var sum = 0
      for (C, D) in lines:
        if isIntersection(A,B,C,D): inc sum
      if sum > bestSum: bestSum = sum
  result := bestSum

    Full solution at Codeberg: solution.nim