Day 5: Print Queue
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View all comments 4 points
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.
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.
2 points
*
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
3 points
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