Day 5: Print Queue

Megathread guidelines

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FAQ

  • mykl@lemmy.world
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    10 days ago

    Uiua

    Well it’s still today here, and this is how I spent my evening. It’s not pretty or maybe even good, but it works on the test data…

    spoiler

    Uses Kahn’s algorithm with simplifying assumptions based on the helpful nature of the data.

    Try it here

    Data  ()⊸≠@\n "47|53\n97|13\n97|61\n97|47\n75|29\n61|13\n75|53\n29|13\n97|29\n53|29\n61|53\n97|53\n61|29\n47|13\n75|47\n97|75\n47|61\n75|61\n47|29\n75|13\n53|13\n\n75,47,61,53,29\n97,61,53,29,13\n75,29,13\n75,97,47,61,53\n61,13,29\n97,13,75,29,47"
    Rs    ≡◇(⊜⋕⊸≠@|)▽⊸≡◇(⧻⊚⌕@|)Data
    Ps    ≡⍚(⊜⋕⊸≠@,)▽⊸≡◇(¬⧻⊚⌕@|)Data
    
    NoPred   ⊢▽:((=0/+⌕)⊙¤)◴♭⟜≡⊣                # Find entry without predecessors.
    GetLead  (:((¬/+=))⊙¤)NoPred             # Remove that leading entry.
    Rules    ⇌⊂⊃(⇌⊢°□⊢|≡°□↘1)[□⍢(GetLead|≠1)] Rs # Repeatedly find rule without predecessors (Kaaaaaahn!).
    
    Sorted    ⊏⍏⊗,Rules
    IsSorted  /×>0≡/-◫2⊗°□: Rules
    MidVal    :(⌊÷ 2)
    
    ⇌⊕□⊸≡IsSorted Ps        # Group by whether the pages are in sort order.
    ≡◇(/+≡◇(MidVal Sorted)) # Find midpoints and sum.
    
    
      • mykl@lemmy.world
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        9 days ago

        Ah, but the terseness of the code allows the beauty of the underlying algorithm to shine through :-)

    • mykl@lemmy.world
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      9 days ago

      Oh my. I just watched yernab’s video, and this becomes so much easier:

      # Order is totally specified, so sort by number of predecessors,
      # check to see which were already sorted, then group and sum each group.
      Data  (□⊜□⊸≠@\n)(¬⦷"\n\n")"47|53\n97|13\n97|61\n97|47\n75|29\n61|13\n75|53\n29|13\n97|29\n53|29\n61|53\n97|53\n61|29\n47|13\n75|47\n97|75\n47|61\n75|61\n47|29\n75|13\n53|13\n\n75,47,61,53,29\n97,61,53,29,13\n75,29,13\n75,97,47,61,53\n61,13,29\n97,13,75,29,47"
      Rs    ≡◇(⊜⋕⊸≠@|)°□⊢Data
      Ps    ≡⍚(⊜⋕⊸≠@,)°□⊣Data
      (/+≡◇(⊡⌊÷2⧻.))¬≡≍⟜:≡⍚(⊏⍏/+⊞(Rs)..).Ps