2024, dag 22

Author: Oddsor

input/2024/input21.nippy

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input/2024/input22.nippy

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+(ns y2024.d22
+  (:require
+   [advent-of-code-clj.input :as input]))
+
+; # 2024, dag 22
+
+; ## Del 1
+
+; Vi skal kalkulere det 2000'nde hemmelige nummeret i en sekvens
+; basert på et par regler. Reglene er som følger:
+
+(defn mix
+  "Mixing a number means bitwise-xor'ing it with another number"
+  [other-number secret-number]
+  (bit-xor other-number secret-number))
+
+(= 37 (mix 15 42))
+
+(defn prune
+  "Pruning a number means applying modulo 16777216 to it"
+  [secret-number] (mod secret-number 16777216))
+
+(= 16113920 (prune 100000000))
+
+; Nå kan vi lage algoritmen for å finne neste hemmelige siffer:
+
+(defn next-secret-number [secret-number]
+  (let [multiplied-by-64 (* secret-number 64)
+        new-secret-number (prune (mix multiplied-by-64 secret-number))
+        divided-by-32 (quot new-secret-number 32)
+        new-secret-number (prune (mix divided-by-32 new-secret-number))
+        multiplied-by-2048 (* new-secret-number 2048)]
+    (prune (mix multiplied-by-2048 new-secret-number))))
+
+; `iterate`-funksjonen lar oss hente ut sekvensen av hemmelige tall, slik
+; at vi enten kan hente x antall siffer, eller det n'te sifferet:
+
+(= [15887950 16495136 527345 704524 1553684 12683156 11100544 12249484 7753432 5908254]
+   (take 10 (drop 1 (iterate next-secret-number 123))))
+
+(= [8685429 4700978 15273692 8667524]
+   (mapv #(nth (iterate next-secret-number %) 2000) [1 10 100 2024]))
+
+; Dermed har vi løsningen på del 1 for test-input:
+
+(defn part-1 [input]
+  (transduce (comp
+              (map parse-long)
+              (map #(nth (iterate next-secret-number %) 2000)))
+             +
+             (re-seq #"\d+" input)))
+
+(part-1 "1 10 100 2024")
+
+; Og på reell input:
+
+(part-1 (input/get-input 2024 22))
+
+; ## Del 2
+
+; TODO