# Advent of Code 2021 -- Day 3: Binary Diagnostic

3 December 2021

Looks like the submarine is making some concerning noises and we have to write some code to figure out what’s wrong with it. We are provided with the input as a list of bitstrings that represents the submarine’s diagnostic report. For the first part we are building up a bitstring, call it gamma, using the most common bit at each index. That is, for each index from 0 up to the length of the input bitstrings, we must find the most common bit (0 or 1) and then write it at that index in gamma.

Nothing too hairy in this loop and there are no hidden tricks: it was useful to use a lambda in order to collect all of the bits at a given index and then pass them to Lisp’s count function. I also considered at first using gamma as a list, then at each iteration pushing the most common bit to this list, though this would have entailed an extra reversal (since push would mean elements are stored in the reverse order) as well as a conversion from list to number. Instead, gamma is considered a number from the get-go and at each iteration, depending on whether we found 0 or 1 to be most common at the current index, we either left-shift it once (to “write a zero” to the bitstring) or left-shift it once and add 1 (to “write a one” to the bitstring).

The answer to the puzzle was then the product of gamma and epsilon, where epsilon was obtained in exactly the same way except by finding the least common bit at each index. Instead of executing the same loop again and flipping the inequality, this could be found from gamma by just flipping its bits. Unfortunately I couldn’t just call (lognot gamma) since this was resulting in negative numbers1, so instead it was found with an XOR between gamma and a bitstring of all 1s, or $$2^n-1$$ where $$n$$ is the number of bits in gamma (simply the length of bitstrings given in the input).

(loop with input = (get-file filename)
with n = (length (first input))
with m = (length input)
for i from 0 below n
with gamma = 0
for zeroes = (count #\0 (mapcar (lambda (bitstring) (char bitstring i)) input))
do (setf gamma (if (>= zeroes (/ m 2)) (ash gamma 1) (1+ (ash gamma 1))))
finally (return (* gamma (logxor (1- (expt 2 n)) gamma))))


Part two was a bit more complex. We have to find two bitstrings, generator and scrubber, by successively filtering out bitstrings in the input that do not meet certain criteria. For generator, for each index we must find the most common bit among all remaining candidates (breaking ties in favour of 1) then filter out those without that bit at that index. For scrubber, we do the same but for the least common bit and break ties in favour of 0. Although simple to sate, I found it took a fair bit of code to implement the first time around as I was performing the filtering for generator and scrubber all within the same loop, which meant I had to keep track of separate variables for each. While this worked, I figured I could do better. Since the rules for filtering out candidates for the two bitstrings differed only slightly, I decided to write a function (filter-candidates candidates filter-rule) that took a list of candidate bitstrings and repeatedly applied the function filter-rule to remove invalid elements. Note that in the code snippet below I also use a function nth-char=, which simply determines whether the nth char of the given string is equal to the given character.

(labels ((nth-char= (string n character) ...)
(filter-candidates (candidates filter-rule)
(loop for i from 0 below (length (first candidates))
with remaining = candidates
for zeroes = (count #\0 (mapcar (lambda (bitstring) (char bitstring i)) remaining))
for ones = (- (length remaining) zeroes)
for char = (funcall filter-rule zeroes ones)
until (= 1 (length remaining))
do (setf remaining (remove-if-not (lambda (bitstring) (nth-char= bitstring i char)) remaining))
finally (return (first remaining)))))


filter-rule is a function that takes two numbers and decides whether to return the character 0 or 1. This is then run at each iteration to determine which character we would filter the remaining candidates on (i.e., whether to look for a 0 or a 1 at the current index under consideration). For generator this was the rule specifying that, if the number of zeroes at the given index exceeds the number of ones then we remove all candidates without a 0 at the given index; for scrubber we remove all candidates without a 1 at the given index if the number of ones exceeds the number of zeroes. Actually filtering out all the invalid candidates is then simply a case of calling filter-candidates on the puzzle input, once for generator and once for scrubber, and a lambda function corresponding to the appropriate rule:

;; filter out invalid bitstrings for generator
(filter-candidates input (lambda (zeroes ones) (if (> zeroes ones) #\0 #\1)))

;; filter out invalid bitstrings for scrubber
(filter-candidates input (lambda (zeroes ones) (if (<= zeroes ones) #\0 #\1)))


As I mentioned, I ended up taking two attempts at this. The first one completed the puzzle in a single loop but required extra variables to keep track of generator and scrubber candidates separately, and ended up reusing a lot of the code. I’m much happier with my second attempt since it makes the solution more generic and about ten lines more concise. I was also pleased to make use of higher-order functions since it isn’t something I use often and it always feels powerful when it works. I already spent quite a bit more time on today’s puzzles than the previous two days, so I’m preparing for the steady increase in difficulty and inevitable loss of more time as the month progresses.

## Footnotes

1. I think this is probably because of the way numbers are represented, e.g. two’s complement.