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Marco Servetto
Marco Servetto

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Advent of code Day 21

The two parts for the problem of today are well divided, so I will present them one at a time.
Or, if you prefer, you can just look to my comments on you tube:

Part 1 is really trivial.
Note the trick with the modulo arithmetic, where we store a position that is 1 less than the intended position of the player.

D100 = Data:{
  var I rolled = 0I
  var I seed = 0I
  mut method I roll()=(
    \seed((if \seed>=100I 0I else \seed)+1I)
Player = Data:{
  var I pos
  var I points = 0I
  mut method Void turn(mut D100 that) =(
    tot = that.roll()+that.roll()+that.roll()
    \pos((\pos+tot).mod(10I)) //pos from 0--10, real pos is +1
  read method Bool won() = \points>=1000I
  p1 = Player(pos=0I) //1-1
  p2 = Player(pos=2I) //3-1
  dice = D100()
  while !p1.won() && !p2.won() (
    if !p1.won() (p2.turn(dice))
  if p1.won() (
    Debug(S"p2 loses for %p2.points(), %dice.rolled(), %(p2.points()*dice.rolled())")
  if p2.won() (
    Debug(S"p1 loses for %p1.points(), %dice.rolled(), %(p1.points()*dice.rolled())")
  //p1 loses for 671, 1338, 897798
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Part 2 is much more interesting, and it was feasible thanks to @Cache.Lazy; in this way our 'recursive' call may simply return a pre-computed value.

Report = Data:{
  Num p1Wins
  Num p2Wins
  method Bool open() = \p1Wins+\p2Wins==0Num
  method This +(This that) = This(
State = Data:{
  I p1Pos
  I p2Pos
  I p1Score = 0I
  I p2Score = 0I
  method This turn(I p1Roll) = (
    (p1Pos0,p2Pos0,p1Score0,p2Score0) = this
    p1Pos = (p1Pos0+p1Roll).mod(10I) //from 0--10, real pos is +1
    p1Score = p1Score0+p1Pos+1I
  method This turn(I p2Roll) = (
    (p1Pos0,p2Pos0,p1Score0,p2Score0) = this
    p2Pos = (p2Pos0+p2Roll).mod(10I) //from 0--10, real pos is +1
    p2Score = p2Score0+p2Pos+1I
  method Report directWin() = {
    (p1Pos,p2Pos,p1Score,p2Score) = this
    X[p1Score<21I || p2Score<21I]
    if p1Score>=21I return \(p1Wins=1\ p2Wins=0\)
    if p2Score>=21I return \(p1Wins=0\ p2Wins=1\)
    return \(p1Wins=0\ p2Wins=0\)
  @Cache.Lazy method Report wins() = {
    var res = this.directWin()
    if ! return res
    r = Range(1I to=4I)      
    for a1 in r,for a2 in r,for a3 in r {
      stateA = this.turn(p1Roll=a1+a2+a3)
      resA = stateA.directWin()
      if ! return res+=resA
      return for b1 in r,for b2 in r,for b3 in r {
        stateB = stateA.turn(p2Roll=b1+b2+b3)
        resB = stateB.directWin()
        if ! return res+=resB
        return res+=stateB.wins()
    return res
Main = (
  s = State(p1Pos=0I,p2Pos=2I)
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What do you think? with minor modifications I could combine automatic parallelism and automatic caching for an even faster version.

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