Solution to problem 1 in Python

src/Year_2018/P1.py

@@ -0,0 +1,113 @@
+# --- Day 1: Chronal Calibration ---
+
+# "We've detected some temporal anomalies," one of Santa's Elves at the
+# Temporal Anomaly Research and Detection Instrument Station tells you. She
+# sounded pretty worried when she called you down here. "At 500-year intervals
+# into the past, someone has been changing Santa's history!"
+
+# "The good news is that the changes won't propagate to our time stream for
+# another 25 days, and we have a device" - she attaches something to your wrist
+# - "that will let you fix the changes with no such propagation delay. It's
+# configured to send you 500 years further into the past every few days; that
+# was the best we could do on such short notice."
+
+# "The bad news is that we are detecting roughly fifty anomalies throughout
+# time; the device will indicate fixed anomalies with stars. The other bad news
+# is that we only have one device and you're the best person for the job! Good
+# lu--" She taps a button on the device and you suddenly feel like you're
+# falling. To save Christmas, you need to get all fifty stars by December 25th.
+
+# Collect stars by solving puzzles. Two puzzles will be made available on each
+# day in the Advent calendar; the second puzzle is unlocked when you complete
+# the first. Each puzzle grants one star. Good luck!
+
+# After feeling like you've been falling for a few minutes, you look at the
+# device's tiny screen. "Error: Device must be calibrated before first use.
+# Frequency drift detected. Cannot maintain destination lock." Below the
+# message, the device shows a sequence of changes in frequency (your puzzle
+# input). A value like +6 means the current frequency increases by 6; a value
+# like -3 means the current frequency decreases by 3.
+
+# For example, if the device displays frequency changes of +1, -2, +3, +1, then
+# starting from a frequency of zero, the following changes would occur:
+
+#     Current frequency  0, change of +1; resulting frequency  1.
+#     Current frequency  1, change of -2; resulting frequency -1.
+#     Current frequency -1, change of +3; resulting frequency  2.
+#     Current frequency  2, change of +1; resulting frequency  3.
+
+# In this example, the resulting frequency is 3.
+
+# Here are other example situations:
+
+#     +1, +1, +1 results in  3
+#     +1, +1, -2 results in  0
+#     -1, -2, -3 results in -6
+
+# Starting with a frequency of zero, what is the resulting frequency after all
+# of the changes in frequency have been applied?
+
+with open("files/P1.txt") as f:
+    frequencies = [int(number) for number in f.read().strip().split()]
+
+
+def part_1() -> None:
+    init = 0
+    for change in frequencies:
+        init += change
+
+    print(f"The resulting frequency is {init}")
+
+
+# --- Part Two ---
+
+# You notice that the device repeats the same frequency change list over and
+# over. To calibrate the device, you need to find the first frequency it
+# reaches twice.
+
+# For example, using the same list of changes above, the device would loop as
+# follows:
+
+#     Current frequency  0, change of +1; resulting frequency  1.
+#     Current frequency  1, change of -2; resulting frequency -1.
+#     Current frequency -1, change of +3; resulting frequency  2.
+#     Current frequency  2, change of +1; resulting frequency  3.
+#     (At this point, the device continues from the start of the list.)
+#     Current frequency  3, change of +1; resulting frequency  4.
+#     Current frequency  4, change of -2; resulting frequency  2, which has
+# already been seen.
+
+# In this example, the first frequency reached twice is 2. Note that your
+# device might need to repeat its list of frequency changes many times before a
+# duplicate frequency is found, and that duplicates might be found while in the
+# middle of processing the list.
+
+# Here are other examples:
+
+#     +1, -1 first reaches 0 twice.
+#     +3, +3, +4, -2, -4 first reaches 10 twice.
+#     -6, +3, +8, +5, -6 first reaches 5 twice.
+#     +7, +7, -2, -7, -4 first reaches 14 twice.
+
+# What is the first frequency your device reaches twice?
+
+
+def part_2() -> None:
+    visited = set()
+    init = 0
+    visited.add(init)
+    repeated = False
+    while not repeated:
+        for change in frequencies:
+            init += change
+            if init not in visited:
+                visited.add(init)
+            else:
+                repeated = True
+                break
+    print(f"The frequency reached twice is {init}")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()