Advent of Code is a set of small programming puzzles, like Project-Euler and CodeJam.

I used Clojure because it is a practical language: expressive, composable, fun to write, easy to understand, fast, comes with power standard library, intuitive CSP library, usually does not support goto or early return(both I liked very much and considered useful).

Preliminary modeling is assumed, the writeup will focus on my findings: functions to make the code cleaner, proper tools to solve the problem, etc.

Day 1

Warm-up puzzle.

;; lein projects come with `resources` directory, slurp makes REPL clean
(def i1 (slurp "resources/2015/i1.txt"))

;; part 1 does not care about input order
(let [f (frequencies i1)]
  (- (f \() (f \))))

;; part 2 does
(loop [[[i c] & r] (map-indexed vector i1)
       f 0]
  (let [d (if (= c \() 1 -1)
        -f (+ f d)]
    (if (= -1 f)
      (recur r -f))))

map-indexed is equivelant to Python’s for i,v in enumerate(coll).

[[i c] & r] is called destructuring, i(ndex), c(har) and r(est)-as-collection.

Be careful with loop terminating condition, here we are ensured that f(loor) will be -1 before consuming all input, otherwise insert (if (nil? i)) check just before let.

Day 2

Destructure and reduce.

;; 2x3x6 => [2 3 6]
;; [i j k] (map bigdec (clojure.string/split l #"x"))

(let [lines (read-lines "i2.txt")]
  (reduce + (map f lines))) ;; f(line) => value

Day 3

What if we have more than 2 Robo-Santa?

take-nth still works.

(take-nth 2 [1 2 3 4 5]) => (1 3 5)
(take-nth 2 (rest [1 2 3 4 5])) => (2 4)

;; move along Euclidean coordinates
(mapv + [0 1] [99 99]) => [99 100]

Day 4


  • both for and (iterate inc 1) are lazy
  • when when condition is true, comprehension returns first i then stops because of first
(let [k  "laziness"
      is (for [i (iterate inc 1)
               :let [h (md5 (str k i))]
               :when (= (subs h 0 5) "00000")] ;; or 6 "000000"
  (first is)) => 183118

(take 3 is) => (183118 1544928 1571333)

Day 5

partition and some:

(partition 3 1 (map str "abade")) => (("a" "b" "a") ("b" "a" "d") ("a" "d" "e"))
(some #(= (first %) (last %)) (partition 3 1 (map str "abade"))) => true

Day 6

This is a simplified version of 2021-day22, the map is small enough to operate on each cell.

Combine comprehension and anonymous function:

;; function f
  (clojure.string/includes? s "turn on") (fn [x] (inc x))
  (clojure.string/includes? s "turn off") (fn [x] (max 0M (dec x)))
  (clojure.string/includes? s "toggle") (fn [x] (+ x 2)))

;; force realization lazy-for
 (for [x (range x1 (inc x2))
       y (range y1 (inc y2))]
   (update-v! m [y x] f)))

;; m is an atom of the map(nested vector)
;; get-in and assoc-in works similarly
;; (assoc-in [[0 1 2] [3 4 5]] [0 0] 1) => [[1 1 2] [3 4 5]]
(defn update-v!
  [m l f]
  (let [v (get-in @m l)]
    (swap! m assoc-in l (f v))))

Day 7

Circuit building feels like Breadth-First-Search:

  • loop over lines
  • if the current line’s output can be evaluated, record the value; else the line is to-be-evaluated
  • recur with to-be-evaluated lines, until nothing left(or stucked)

Parse line into “registry names” and functions then BFS.

(defn parse-wire
  (let [[op0 op1 op2] splits] ;; only destructure the first 3
      (= (count splits) 3) [op0 nil (fn [x1 _] x1)] ;; {in, 123} -> output
      (= op0 "NOT") [op1 op1 (fn [x1 _] (bit-not x1))]  ;; unary op: NOT d -> a
      (= op1 "OR") [op0 op2 (fn [x1 x2] (bit-or x1 x2))] ;; binary ops
      (= op1 "AND") [op0 op2 (fn [x1 x2] (bit-and x1 x2))]
      (= op1 "LSHIFT") [op0 op2 (fn [x1 x2] (mod (bit-shift-left x1 x2) 16rFFFF))]
      (= op1 "RSHIFT") [op0 op2 (fn [x1 x2] (mod (bit-shift-right x1 x2) 16rFFFF))])))

;; to distinguish registry name and number literal, dangerous but useful
(number? (read-string "a")) => false
(number? (read-string "1")) => true

Day 9

Given that:

  • every point has exactly one route to any other point
  • “cost” of A->B and B->A are equal
  • start from any point
  • search space is still small enough

Just find all costs and point names, calculate costs of all permutation and find min/max.

(defn calc-dist
  [dm coll] ;; distance-map, city names
  (reduce +
  (for [r (partition 2 1 coll)]
    (dm r))))

;; thread-last macro ->> is very expressive
(let [[places dm] (build-dist)
      perms (permutations places)]
  (->> perms
       (map (partial calc-dist dm))
       (apply max)))

Day 10


(partition-by identity [1 1 1 2 2 1]) => ((1 1 1) (2 2) (1))

Day 12

Recursion, parse input json into map using cheshire (parse-string i12 true):

;; part 1
(->> (re-seq #"-?\d+" i12)
     (map bigdec)
     (reduce +))

(defn calc-sum
    (or (nil? m) (string? m))

    (or (decimal? m) (integer? m))
    (bigdec m)

    (map? m)
    (if ((into #{} (vals m)) "red")
      (reduce + (map calc-sum (vals m))))

    :whatevercoll ;; "else" in your favourite form
    (reduce + (map calc-sum m))))

Day 13

After “you” joined the table, the optimal happiness decreased, which makes sense.

Just like Day 9, but the cost is a sum of A->B and B->A, it can be negative as well.

Day 14

Compose functions and data:

  • distance(speed, battery, rest, time), for part 1
  • distance over time(M-athletes * N-seconds) matrix
  • for each second, find lead distance(maximum number), increment the corresponding counters(index of leaders -> lead count)
  • find max value in counter

Day 15

Think of 3/8 calories as 2 groups,

  • a+b=100, 3a+8b=500
  • 60*3+40*8

The rest is still comprehension.

;; transpose to mapv * and sum
(apply mapv vector [[1 2 3] [4 5 6]]) => [[1 4] [2 5] [3 6]]
(reduce + (mapv * [a b c d] x))

Day 16

Laziness and compare function,

(let [su {"children" "3" "cats" "7" "samoyeds" "2"
          "pomeranians" "3" "akitas" "0" "vizslas" "0"
          "goldfish" "5" "trees" "3" "cars" "2" "perfumes" "1"}]
  (defn kv-comp1 [x] (= (second x) (su (first x))))
  (defn kv-comp2
    (let [[k v] x]
        (= k "cats") (> (bigdec v) 7)
        (= k "trees") (> (bigdec v) 3)
        (= k "pomeranians") (< (bigdec v) 3)
        (= k "goldfish") (< (bigdec v) 5)

        (kv-comp1 x)))))

;; mismatch? accepts compare function, returns true on mismatch, aunt index on all pass
;; (if (some false? (map kv-comp2 kvs)) true aunt-idx)
(->> (for [l ls] (mismatch? l kv-comp2))
     (drop-while true?)

Day 17

Is this a 0-1 knapsack problem?

(def bss (atom [])) ;; ways to take bottles

(defn-memo calc-water
  [k ws bs] ;; k target value, ws water bottles left, bs bottles taken
  (let [[f & r] ws]
    (when (= k 0)
      (swap! bss conj bs))
    (if (= k 0) ;; exactly k
      (if (nil? f) ;; failed to add to k, does not count
        (if (> f k) ;; take next one or not?
          (calc-water k r bs) ;; this is a "tail position" to recur
          (+ (calc-water (- k f) r (conj bs f)) (calc-water k r bs)))))))

;; part 2: least bottles to 150
(let [mm (apply min (map count @bss))]
  (->> @bss
       (filter #(= mm (count %)))

Some caveats about recursion and memoize:

  • recur from tail position to “constant-space recursive looping by rebinding and jumping to the nearest enclosing loop or function frame”(to avoid stackoverflow)
  • calc-water is deterministic, its calculation can be cached
  • core.memoize for concurrent use, “unlike the built-in memoize function, ensures that in the case of concurrent calls with the same arguments, the memoized function is only invoked once”

Day 19

Analytical solution of part 2 worths reading.

Then I wrote a greedy one:

  • find the longest replace-able string, replace once, increment counter
  • if not “e” then recur
  • return counter

Day 20

Part 1 is to find the least integer, sum of whose factors(including 1 and itself) * 10 >= itself.

Part 2 is about pruning:

  • the last house to reach 29000000 is (quot 29000000 11) => 2636363
  • the irrelevant houses to ignore (-> 29000000 (/ 50) (/ 11) int) => 52727
(let [v (atom {})]
  (loop [i 1]
    (doall ;; send presents to houses
     (for [h (map #(* i %) (take 50 (iterate inc 1)))
           :when (<= 52000 h 2640000)]
       (swap! v assoc h (+ (* 11 i) (or (@v h) 0)))))

    (if (> (or (@v i) 0) 29000000)
      (recur (inc i)))))

Day 21

“What is the most amount of gold you can spend and still lose the fight?”

Permutation of choices, where weapons are required, 0-1 armor, 0-2 rings:

  • append [0 0 0](cost atk def) to armor and rings
  • additional index for the optional second ring, prevent choosing the same valid ring
(defn win-21?
  [a1 d1 a2 d2]
  (let [-d1 (max 1 (- a1 d2)) ;; minimal damage 1
        -d2 (max 1 (- a2 d1))]
    ;; because user attack first
    (>= -d1 -d2)))

;; for part 1, apply min, change to if win, and 0M->10000M
(apply max
(for [i1 (range 0 5)
      i2 (range 0 6)
      i3 (range 0 7)
      i4 (range 0 7)]
  (let [[cost atk def] (mapv + (nth weapons i1)
                             (nth armors i2)
                             (nth rings i3)
                             (if (and (= i3 i4) (not= i3 6))
                               [0 0 0] (nth rings i4)))]
    (if-not (win-21? atk def 8 2)
      cost 0M))))

Day 22

Invite your friend to solve this puzzle, ask them about Cyberpunk 2077 afterward.

There are several states to keep track of:

  • ours: HP, def, mana
  • boss’s: HP
  • 3 buffs: turns left
  • stop signal for path pruning: if this sequence of spells will result in {out of mana,re-cast buff,no spell provided before game end}, then there’s no need to explore further

Intuitively one might want to apply DoT, cast mana regenerating, increase defense, before casting actual damage spells. But the boss’s attack is high, intuition does not lead to a conclusion directly.

(defn explore-spells
  (let [spells (map str s)
        [win? stop?] (game-win? spells)]
    (if win?
      [true s] ;; when win just return the current spells
      (if stop? ;; if early terminated, then its "univisited paths" is empty
        [false []]
        [false (map #(str s %) "RSPDM")]))))

(defn path->cost
  (let [costs {"M" 53 "D" 73 "S" 113 "P" 173 "R" 229}
        spells (map str p)]
    (reduce + (map costs spells))))

(loop [init ["R" "S" "P" "D" "M"]]
  (let [results (map explore-spells init)
        win? (some true? (map first results))
        explore (mapcat second results)]
    (if-not win?
      (recur explore)
      (let [paths (map second (filter #(true? (first %)) results))
            costs (map path->cost paths)]
        (apply min costs)))))

game-win? can be implemented functionally, but I used several atoms and a large loop instead:

  • pre-fight buff effects(add player’s dot here for part 2)
  • check hp
  • boss attacks / player tries to cast a spell, path pruning checks here
  • check hp
  • if the game ends or spells are invalid, return [win? stop?], else recur into the opponent’s turn

The good thing about BFS is that, when modeled properly, the search results are also intuitively minimal costs.

Day 23

An example of optimized recursion, naively calc-23 will possibly cause stackoverflow.

(defn calc-23
  [is i a b] ;; instructions, pointer index, a b value
  (if (>= i (count is))
    [a b]
    (recur is new-i new-a new-b)))

Use condp or core.match to simplify calculation of new i,a,b values.

Day 25

The first row is Triangular Number:

  • imagine a virtual diagonal line from input [a b] to point p [x 0] in first row, where x=a+b
  • k=x*(x+1)/2
  • n=k-b, loop n times for final module result

Be careful with offset-by-one errors.