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Marco Servetto
Marco Servetto

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Advent of code day 12

Here we are again.
Doing this every day is staring to get tiresome, we do not even get a weekend break! :-P

Anyway, today code is fully relying on immutable data structures.
If there is something I've learnt in those years of programming, is that recursive searches + mutability = bugs.

The key idea for part 2 is to define a 'Forbidden' type,
that is a structured datatype keeping track of the more complex condition to keep visiting.
If I tried to encode the new condition on a simple list of visited, I would still be coding.

The idea is that you can 'push' the new visited node on the forbidden instance, and check if we have got in a 'wrong' state.
Alternatively, I could have thrown an error.

For the fine details, yes, the code after 'next=' should probably go in a separate function to be more elegant.

reuse []
Fs = Load:{reuse[]}
Big ={class method Bool(S that) = that.toStartUp()==that}
Edges =,val=S.List)
Paths = Data.AddList:Data:{S name, This.List next List={}}
Forbidden = Data:{
  S.List visited=\[S"start"]
  Bool bonus=\.false()
  Bool wrong=\.false()
  method This push(S that) = 
    if Big(that) this 
    else \pushSmall(that)
  method This pushSmall(S that) = {
    limit = that==S"start" || that==S"end"
    visited = that in \visited
    if limit && visited return \with(wrong=Bool.true()) 
    if visited && \bonus return \with(wrong=Bool.true())
    if visited return \with(bonus=Bool.true())
    return \with(visited=\visited.withAlso(right=that))
Count = {class method I (Paths that)={
    if"end" && return 1I
    return 0I.acc()(for p in \add(This(p)))
Visit = {
  class method Paths(S that, Edges e, Forbidden forbidden) =   
      next= if that==S"end" \() else Paths.List()(
        for ni in e.val(key=that).val() {
          f = forbidden.push(ni)
          if f.wrong() return void
          return \add(This(ni,e=e,forbidden=f))
  input = Fs.Real.#$of().read(\"input")
  imm edges = Edges()(
    for line in input.split( (
      s = line.split(S"-")
      s1 = s()
      s2 = s()
      on1 =\val(key=s1)
      on2 =\val(key=s2)
      v1 = ( if on1 on1.val() else S.List[] ).withAlso(right=s2)
      v2 = ( if on2 on2.val() else S.List[] ).withAlso(right=s1)
      \put(key=s1 val=v1)
      \put(key=s2 val=v2)
  Debug(Count(res))//3497 and 93686
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Finally, not that the problem can only be well posed if there are never two large caves in direct contact, otherwise we could loop forever.

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