Finish day 8

2021/common.nim

@@ -50,6 +50,10 @@ proc doDay*[T](
     doAssert testInput.part1() == expectedPart1
     doAssert testInput.part2() == expectedPart2
 
-proc reduce*[T](s: openArray[T], op: (T, T) -> T, init: T): T =
+proc reduce*[T, X](s: openArray[T], op: (X, T) -> X, init: X): X =
   result = init
   for n in s: result = op(result, n)
+
+proc findFirst*[T](s: openArray[T], op: (T) -> bool): T =
+  for n in s:
+    if op(n): return n

2021/eight.nim

@@ -1,12 +1,67 @@
-import ./common, std/[strutils, sequtils, strformat, sugar]
+import ./common, std/[strutils, sequtils, strformat, sugar, sets, math]
 
-proc p1(c: string): int =
+proc parseLine(s: string): seq[seq[string]] = s.split(" | ").mapIt(it.split(" "))
-  return 26
 
-proc p2(c: string): int =
+proc p1(c: seq[string]): int =
-  result = 0
+  c.reduce((a: int, l: string) =>
+    a + l.parseLine()[1].reduce((b: int, h) => b + int(h.len() in [2, 3, 4, 7]), 0), 0)
 
-doDay(8, (n) => n.loadInputText(),
+proc numShared(s1: string, s2: string): int = (s1.toHashSet() * s2.toHashSet()).len()
+proc p2(c: seq[string]): int =
+  for oLine in c.map(parseLine):
+    let four = oLine[0].findFirst((s) => s.len() == 4)
+    let seven = oLine[0].findFirst((s) => s.len() == 3)
+    for i, output in oLine[1].pairs():
+      let outputVal = case output.len():
+        of 2: 1
+        of 3: 7
+        of 4: 4
+        of 7: 8
+        of 5:
+          case output.numShared(four) + output.numShared(seven):
+            of 4: 2
+            of 6: 3
+            else: 5
+        else:
+          case output.numShared(four) + output.numShared(seven):
+            of 6: 0
+            of 5: 6
+            else: 9
+      let x = (abs(oLine[1].len() - i) - 1)
+      result += outputVal * int(10 ^ x)
+
+# Explanation:
+# 1, 7, 8, and 4 are known and may be identified just by length, so we only
+# need to figure out the 5- and 6- inputs, 0, 2, 3, 5, 6, and 9
+# we should be able to do this by seeing how many segments are shared with the
+# known digits
+# 5-inputs: 235
+#   2: shares 2 with 4, 1 with 1, 2 with 7
+#   3: shares 3 with 4, 2 with 1, 3 with 7
+#   5: shares 3 with 4, 1 with 1, 2 with 7
+# 6-inputs: 069
+#   0: shares 3 with 4, 2 with 1, 3 with 7
+#   6: shares 3 with 4, 1 with 1, 2 with 7
+#   9: shares 4 with 4, 2 with 1, 3 with 7
+# from this information, we can identify each digit simply by knowing the
+# number of shared segments with 4 and 7 alone:
+# 2: 22
+# 3: 33
+# 5: 32
+# 0: 33
+# 6: 32
+# 9: 43
+
+doDay(8, (n) => n.loadInput(),
   p1,
   p2,
-  "acedgfb cdfbe gcdfa fbcad dab cefabd cdfgeb eafb cagedb ab | cdfeb fcadb cdfeb cdbaf", 26, -1)
+  """be cfbegad cbdgef fgaecd cgeb fdcge agebfd fecdb fabcd edb | fdgacbe cefdb cefbgd gcbe
+edbfga begcd cbg gc gcadebf fbgde acbgfd abcde gfcbed gfec | fcgedb cgb dgebacf gc
+fgaebd cg bdaec gdafb agbcfd gdcbef bgcad gfac gcb cdgabef | cg cg fdcagb cbg
+fbegcd cbd adcefb dageb afcb bc aefdc ecdab fgdeca fcdbega | efabcd cedba gadfec cb
+aecbfdg fbg gf bafeg dbefa fcge gcbea fcaegb dgceab fcbdga | gecf egdcabf bgf bfgea
+fgeab ca afcebg bdacfeg cfaedg gcfdb baec bfadeg bafgc acf | gebdcfa ecba ca fadegcb
+dbcfg fgd bdegcaf fgec aegbdf ecdfab fbedc dacgb gdcebf gf | cefg dcbef fcge gbcadfe
+bdfegc cbegaf gecbf dfcage bdacg ed bedf ced adcbefg gebcd | ed bcgafe cdgba cbgef
+egadfb cdbfeg cegd fecab cgb gbdefca cg fgcdab egfdb bfceg | gbdfcae bgc cg cgb
+gcafb gcf dcaebfg ecagb gf abcdeg gaef cafbge fdbac fegbdc | fgae cfgab fg bagce""".split('\n'), 26, 61229)

2021/readme.md

@@ -58,7 +58,7 @@ Or for Zig programs:
 - [x] Day 5: [Nim](./five.nim)
 - [x] Day 6: [Nim](./six.nim)
 - [x] Day 7: [Nim](./seven.nim)
-- [ ] Day 8
+- [x] Day 8: [Nim](./eight.nim)
 - [ ] Day 9
 - [ ] Day 10
 - [ ] Day 11