(*->2022)~made it clear what advent of code I'm doing

4 files changed, 206 insertions, 0 deletions

diff --git a/2022/puzzle-1.lisp b/2022/puzzle-1.lisp
new file mode 100644
index 0000000..3a1510e
--- /dev/null
+++ b/2022/puzzle-1.lisp
@@ -0,0 +1,26 @@
+(defvar input (uiop:read-file-string "2022/1-input"))
+(defvar *sep (format nil "~%~%"))
+
+(defun get-lists (input)
+  (let ((pos (search *sep input)))
+    (with-input-from-string (s (subseq input 0 pos))
+      (let ((converted
+              (loop
+                for line = (read-line s nil nil)
+                while line
+                collect (parse-integer line))))
+        (if (null pos)
+            (list converted)
+            (cons converted
+                  (get-lists (subseq input (+ pos 2)))))))))
+
+(defvar sums (sort (mapcar (lambda (lst) (reduce #'+ lst)) (get-lists input)) #'>))
+
+;; First challenge
+(format t "Top snacks: ~a" (car sums))
+
+;; Second challenge
+(let ((first (car sums))
+      (second (car (cdr sums)))
+      (third (car (cdr (cdr sums)))))
+  (format t "~a,~a,~a:>~a" first second third (+ first second third)))
diff --git a/2022/puzzle-2.lisp b/2022/puzzle-2.lisp
new file mode 100644
index 0000000..b3b0155
--- /dev/null
+++ b/2022/puzzle-2.lisp
@@ -0,0 +1,54 @@
+(defvar input (uiop:read-file-string "2022/2-input"))
+;; Each newline represents a new round, which we should parse on the go
+
+(defun sensible-convert-input (str)
+  (cond
+    ((or (string= str "X") (string= str "A")) 0)
+    ((or (string= str "Y") (string= str "B")) 1)
+    ((or (string= str "Z") (string= str "C")) 2)))
+
+;; Round 1
+(defvar rounds
+  (with-input-from-string (stream input)
+    (loop
+      for strategy = (read-line stream nil)
+      until (null strategy)
+      collect
+      (let ((opponent (subseq strategy 0 1))
+            (yours (subseq strategy 2 3)))
+        (list (sensible-convert-input opponent) (sensible-convert-input yours))))))
+
+(loop
+  for round in rounds
+  until (null round)
+  sum
+  (destructuring-bind (opp you) round
+    (+
+     1 you ;; base score
+     (cond  ; outcome score
+       ((eq you opp) 3)
+       ((eq (mod (+ 1 opp) 3) you) 6)
+       (t 0)))))
+
+;; Round 2.
+
+;; We can still use the same rounds data as previously, just
+;; reinterpret it in when doing the sum.
+
+(defun get-correct-choice (opponent outcome)
+  (case outcome
+    (0 (mod (- opponent 1) 3))
+    (1 opp)
+    (2 (mod (+ 1 opponent) 3))
+    (t 0)))
+
+(loop for round in rounds
+      sum
+      (destructuring-bind (opp you) round
+        (let ((choice (get-correct-choice opp you)))
+          (+ 1 choice
+             (case you ;; outcome -> score
+               (0 0)
+               (1 3)
+               (2 6)
+               (t 0))))))
diff --git a/2022/puzzle-3.lisp b/2022/puzzle-3.lisp
new file mode 100644
index 0000000..c5ee3d3
--- /dev/null
+++ b/2022/puzzle-3.lisp
@@ -0,0 +1,67 @@
+(defvar input (uiop:read-file-string "2022/3-input"))
+
+(defun split-string-in-two (s)
+  (let ((len (length s)))
+    (list (subseq s 0 (/ len 2)) (subseq s (/ len 2)))))
+
+(defvar inputs (with-input-from-string (s input)
+                 (loop
+                   for line = (read-line s nil)
+                   until (null line)
+                   collect (split-string-in-two line))))
+
+(defun string-to-clist (str)
+  (loop for char across str collect char))
+
+(defun common-types (s1 s2)
+  (car (intersection
+        (string-to-clist s1)
+        (string-to-clist s2))))
+
+(defvar shared (mapcar (lambda (x)
+                    (destructuring-bind (s1 s2) x
+                      (common-types s1 s2)))
+                  inputs))
+
+(defun priority-map (c)
+  (if (upper-case-p c)
+      (+ 27 (- (char-code c) (char-code #\A)))
+      (+ 1 (- (char-code c) (char-code #\a)))))
+
+(defvar round-1-answer (reduce #'+ (mapcar #'priority-map shared)))
+
+;; Round 2
+
+;; Simple recursive algorithm which produces consecutive groups of 3 elements
+(defun group-by-3 (lst)
+  (if (null lst)
+      nil
+      (cons
+       (list (car lst) (car (cdr lst)) (car (cdr (cdr lst))))
+       (group-by-3 (cdr (cdr (cdr lst)))))))
+
+;; Note the use of group-by-3 here
+(defvar inputs (group-by-3
+                (with-input-from-string (s input)
+                  (loop
+                    for line = (read-line s nil)
+                    until (null line)
+                    collect line))))
+
+;; Extend intersection to three
+(defun common-types-3 (s1 s2 s3)
+  (car
+   (intersection
+    (string-to-clist s1)
+    (intersection
+     (string-to-clist s2)
+     (string-to-clist s3)))))
+
+;; Extend the destructuring bind and use of common-types-3
+(defvar shared (mapcar (lambda (x)
+                    (destructuring-bind (s1 s2 s3) x
+                      (common-types-3 s1 s2 s3)))
+                  inputs))
+
+;; Same as before
+(defvar round-2-answer (reduce #'+ (mapcar #'priority-map shared)))
diff --git a/2022/puzzle-4.lisp b/2022/puzzle-4.lisp
new file mode 100644
index 0000000..04c7bfb
--- /dev/null
+++ b/2022/puzzle-4.lisp
@@ -0,0 +1,59 @@
+;; Example input: a-b,c-d which denotes [a,b] and [c,d]
+
+;; We want to find if [c,d] < [a,b] or vice versa (complete inclusion)
+;; and since we're working with integers, it's simply checking if the
+;; bounds are included i.e. c in [a,b] and d in [a,b]
+
+(defvar input (uiop:read-file-string "2022/4-input"))
+
+(defun parse-bound (str)
+  "Given STR=\"a-b\" return (a b)"
+  (let* ((sep (search "-" str))
+         (first (subseq str 0 sep))
+         (second (subseq str (+ sep 1))))
+    (list (parse-integer first) (parse-integer second))))
+
+(defvar completed-parse
+  (with-input-from-string (s input)
+    (loop for line = (read-line s nil)
+          until (null line)
+          collect
+          ;; given a-b,c-d we want ((a b) (c d))
+          (let* ((sep (search "," line))
+                 (first-bound (subseq line 0 sep))
+                 (second-bound (subseq line (+ sep 1))))
+            (list (parse-bound first-bound) (parse-bound second-bound))))))
+
+(defun complete-inclusion (first-bound second-bound)
+  (destructuring-bind (a b) first-bound
+    (destructuring-bind (c d) second-bound
+      (or
+       (and
+        (>= a c) (<= a d)
+        (>= b c) (<= b d))
+       (and
+        (>= c a) (<= c b)
+        (>= d a) (<= d b))))))
+
+(defvar round-1-answer (length (remove-if #'null
+                                          (mapcar (lambda (pair)
+                                               (destructuring-bind (first second) pair
+                                                 (complete-inclusion first second)))
+                                             completed-parse))))
+
+;; Round 2: any overlap at all.  Basically just overhaul the inclusion
+;; function and then do the same answer checking.
+(defun any-inclusion (first second)
+  (destructuring-bind (a b) first
+    (destructuring-bind (c d) second
+      ;; How about doing this through negation?  [a,b] does not overlap with [c,d] at all if either b < c or a > d.
+      (not
+       (or
+        (< b c)
+        (> a d))))))
+
+(defvar round-2-answer (length (remove-if #'null
+                                          (mapcar (lambda (pair)
+                                               (destructuring-bind (first second) pair
+                                                 (any-inclusion first second)))
+                                             completed-parse))))