From 2b40cdd77a247af463d7c4bd210b282aba9dec13 Mon Sep 17 00:00:00 2001
From: daviddoji <daviddoji@pm.me>
Date: Sat, 5 Mar 2022 09:22:49 +0100
Subject: [PATCH] Solution to problem 6 part 1 in Python

---
 src/Year_2016/P6.py | 56 +++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 56 insertions(+)
 create mode 100644 src/Year_2016/P6.py

diff --git a/src/Year_2016/P6.py b/src/Year_2016/P6.py
new file mode 100644
index 0000000..d4c7ea0
--- /dev/null
+++ b/src/Year_2016/P6.py
@@ -0,0 +1,56 @@
+# --- Day 6: Signals and Noise ---
+
+# Something is jamming your communications with Santa. Fortunately, your signal
+# is only partially jammed, and protocol in situations like this is to switch
+# to a simple repetition code to get the message through.
+
+# In this model, the same message is sent repeatedly. You've recorded the
+# repeating message signal (your puzzle input), but the data seems quite
+# corrupted - almost too badly to recover. Almost.
+
+# All you need to do is figure out which character is most frequent for each
+# position. For example, suppose you had recorded the following messages:
+
+# eedadn
+# drvtee
+# eandsr
+# raavrd
+# atevrs
+# tsrnev
+# sdttsa
+# rasrtv
+# nssdts
+# ntnada
+# svetve
+# tesnvt
+# vntsnd
+# vrdear
+# dvrsen
+# enarar
+
+# The most common character in the first column is e; in the second, a; in the
+# third, s, and so on. Combining these characters returns the error-corrected
+# message, easter.
+
+# Given the recording in your puzzle input, what is the error-corrected version
+# of the message being sent?
+
+from collections import Counter
+
+with open("files/P6.txt") as f:
+    signals = [line for line in f.read().strip().split()]
+
+signals_transp = list(zip(*signals))
+
+
+def part_1() -> None:
+    _message = []
+    for column in signals_transp:
+        _message.append(Counter(column).most_common(1))
+
+    message = "".join((char[0][0]) for char in _message)
+    print(f"The message is {message}")
+
+
+if __name__ == "__main__":
+    part_1()