Solution to problem 8 in Python

src/Year_2021/P8.py

@@ -106,6 +106,8 @@
 
 # In the output values, how many times do digits 1, 4, 7, or 8 appear?
 
+from ast import Num
+
 with open("files/P8.txt") as f:
     output = [line for line in f.read().strip().split("\n")]
 
@@ -124,5 +126,118 @@ def part_1() -> None:
     print(f"There are {easy_digits} ocurrences in the output values")
 
 
+# --- Part Two ---
+
+# Through a little deduction, you should now be able to determine the remaining
+# digits. Consider again the first example above:
+
+# acedgfb cdfbe gcdfa fbcad dab cefabd cdfgeb eafb cagedb ab |
+# cdfeb fcadb cdfeb cdbaf
+
+# After some careful analysis, the mapping between signal wires and segments
+# only make sense in the following configuration:
+
+#  dddd
+# e    a
+# e    a
+#  ffff
+# g    b
+# g    b
+#  cccc
+
+# So, the unique signal patterns would correspond to the following digits:
+
+#     acedgfb: 8
+#     cdfbe: 5
+#     gcdfa: 2
+#     fbcad: 3
+#     dab: 7
+#     cefabd: 9
+#     cdfgeb: 6
+#     eafb: 4
+#     cagedb: 0
+#     ab: 1
+
+# Then, the four digits of the output value can be decoded:
+
+#     cdfeb: 5
+#     fcadb: 3
+#     cdfeb: 5
+#     cdbaf: 3
+
+# Therefore, the output value for this entry is 5353.
+
+# Following this same process for each entry in the second, larger example
+# above, the output value of each entry can be determined:
+
+#     fdgacbe cefdb cefbgd gcbe: 8394
+#     fcgedb cgb dgebacf gc: 9781
+#     cg cg fdcagb cbg: 1197
+#     efabcd cedba gadfec cb: 9361
+#     gecf egdcabf bgf bfgea: 4873
+#     gebdcfa ecba ca fadegcb: 8418
+#     cefg dcbef fcge gbcadfe: 4548
+#     ed bcgafe cdgba cbgef: 1625
+#     gbdfcae bgc cg cgb: 8717
+#     fgae cfgab fg bagce: 4315
+
+# Adding all of the output values in this larger example produces 61229.
+
+# For each entry, determine all of the wire/segment connections and decode the
+# four-digit output values. What do you get if you add up all of the output
+# values?
+
+
+def solve_line(numbers: str) -> dict[str, list[str]]:
+    numbers_dic = {}
+    for num in numbers.split(" "):
+        # first populate with easy numbers
+        if len(num) == 2:
+            numbers_dic["1"] = sorted(num)
+        elif len(num) == 3:
+            numbers_dic["7"] = sorted(num)
+        elif len(num) == 4:
+            numbers_dic["4"] = sorted(num)
+        elif len(num) == 7:
+            numbers_dic["8"] = sorted(num)
+    # populate with subsets of easy numbers
+    for num in numbers.split(" "):
+        if len(num) == 6:
+            if set(numbers_dic["4"]).issubset(set(num)):
+                numbers_dic["9"] = sorted(num)
+            elif set(numbers_dic["1"]).issubset(set(num)):
+                numbers_dic["0"] = sorted(num)
+            else:
+                numbers_dic["6"] = sorted(num)
+    # missing numbers of length 5
+    for num in numbers.split(" "):
+        if len(num) == 5:
+            if set(num).issubset(set(numbers_dic["6"])):
+                numbers_dic["5"] = sorted(num)
+            elif set(numbers_dic["1"]).issubset(set(num)):
+                numbers_dic["3"] = sorted(num)
+            else:
+                numbers_dic["2"] = sorted(num)
+
+    return numbers_dic
+
+
+def part_2() -> None:
+    decoded_output = 0
+    for line in output:
+        patterns, out_val = line.split(" | ")
+        _numbers_dic = solve_line(patterns)
+        entry_value = []
+        for out in out_val.split(" "):
+            for k, v in _numbers_dic.items():
+                if sorted(out) == v:
+                    entry_value.append(k)
+
+        decoded_output += int("".join(entry_value))
+
+    print(f"All the output values sum {decoded_output}")
+
+
 if __name__ == "__main__":
     part_1()
+    part_2()