commit da89a53cbfb2defddee9f1e9eb27d5cc6abbbf58
parent ba46c4a60107dfc1d3e1a741f97f4a43f3521aeb
Author: Luís Ferreira <[email protected]>
Date: Sat, 4 Dec 2021 01:55:39 +0000
posts: Add Zettelkasten 4
Signed-off-by: Luís Ferreira <[email protected]>
Diffstat:
1 file changed, 193 insertions(+), 0 deletions(-)
diff --git a/content/posts/zet-4-aoc-2021-03.md b/content/posts/zet-4-aoc-2021-03.md
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+---
+title: 'Zettelkasten #4: Solution in D for Advent of Code 2021, Day 3'
+date: '2021-12-02T19:02:00+01:00'
+tags: ['zettelkasten', 'zet', 'dlang', 'aoc', 'aoc2021', 'adventofcode']
+description: "This post describes, in detail, my solution in the D programming
+language for the 3rd puzzle of the Advent of Code 2021."
+---
+
+## The challenge
+
+> The submarine has been making some odd creaking noises, so you ask it to
+> produce a diagnostic report just in case.
+>
+> The diagnostic report (your puzzle input) consists of a list of binary
+> numbers which, when decoded properly, can tell you many useful things about
+> the conditions of the submarine. The first parameter to check is the power
+> consumption.
+>
+> You need to use the binary numbers in the diagnostic report to generate two
+> new binary numbers (called the gamma rate and the epsilon rate). The power
+> consumption can then be found by multiplying the gamma rate by the epsilon
+> rate.
+>
+> [...]
+>
+> Use the binary numbers in your diagnostic report to calculate the gamma rate
+> and epsilon rate, then multiply them together. What is the power consumption
+> of the submarine? (Be sure to represent your answer in decimal, not binary.)
+
+You can read the challenge more in depth,
+[here](https://adventofcode.com/2021/day/3).
+
+## Part 1
+
+To make the problem easier to look at, we can think that the input data is a
+matrix of 0s and 1s and we work with the transposed version of the matrix to
+calculate the bit criteria. To transpose a range of ranges in D you can use
+`transposed` template:
+
+```d
+auto tmatrix = input.transposed;
+```
+
+As the problem dictates, the criteria is to select the most common bit for each
+column (in this case, each row, due to the matrix transposition). To do this,
+we shall sort and group the bits, which in practice counts the number of 0s and
+1s in our array. We then convert the group (array of tuples) to an associative
+array, but that is just purely for code aesthetics:
+
+```d
+auto g = tmatrix.sort.group.assocArray;
+```
+
+To extract the common bits you just need to compare the counter and select the
+common bit accordingly. We can solve this with a simple map and a ternary
+operator:
+
+```d
+auto gamma = g.map!"a['0'] > a['1'] ? '0' : '1'";
+```
+
+We now have the encoded version of gamma value. To find epsilon, we just flip
+each bit. We can also use a map and a ternary operator to that job:
+
+```d
+auto epsilon = gamma.map!"a == '0' ? '1' : '0'";
+```
+
+To calculate the result we just decode the values to decimal and multiply them.
+To decode to decimal, you can use `to!int(<base>)` from `std.conv`, where
+`<base>` is the numeric base of the input value. Here I used a `fold` instead
+of repeating two `to` calls:
+
+```d
+auto res = [gamma, epsilon].fold!"b.to!int(2) * a"(a);
+```
+
+### Full solution
+
+```d
+[(cast(char[][])stdin.byLine().map!"a.to!string".array) // input
+ .transposed.map!array // transpose matrix
+ .map!`a.dup.sort.group.assocArray` // sort & group bits
+ .map!"a['0'] > a['1'] ? '0' : '1'".array] // extract common bit
+ .map!(b => [b, b.map!"a == '0' ? '1' : '0'".array]) // flip bits for gamma & epsilon
+ .front.fold!"b.to!int(2) * a"(1).writeln; // decode & multiply gamma & epsilon
+```
+
+## Part 2
+
+> Next, you should verify the life support rating, which can be determined by
+> multiplying the oxygen generator rating by the CO2 scrubber rating.
+
+> Both the oxygen generator rating and the CO2 scrubber rating are values that
+> can be found in your diagnostic report - finding them is the tricky part.
+> Both values are located using a similar process that involves filtering out
+> values until only one remains. Before searching for either rating value,
+> start with the full list of binary numbers from your diagnostic report and
+> consider just the first bit of those numbers. Then:
+>
+> - Keep only numbers selected by the bit criteria for the type of rating value
+> for which you are searching. Discard numbers which do not match the bit
+> criteria.
+> - If you only have one number left, stop; this is the rating value for which
+> you are searching.
+> - Otherwise, repeat the process, considering the next bit to the right.
+>
+> The bit criteria depends on which type of rating value you want to find:
+>
+> - To find oxygen generator rating, determine the most common value (0 or 1)
+> in the current bit position, and keep only numbers with that bit in that
+> position. If 0 and 1 are equally common, keep values with a 1 in the
+> position being considered.
+> - To find CO2 scrubber rating, determine the least common value (0 or 1) in
+> the current bit position, and keep only numbers with that bit in that
+> position. If 0 and 1 are equally common, keep values with a 0 in the
+> position being considered.
+>
+> [...]
+>
+> Use the binary numbers in your diagnostic report to calculate the oxygen
+> generator rating and CO2 scrubber rating, then multiply them together. What
+> is the life support rating of the submarine? (Be sure to represent your
+> answer in decimal, not binary.)
+
+I got impressed with the second part, as it is quite big, although not too
+complicated, just more restrictive bit criteria and more rules for each
+criteria.
+
+To facilitate comprehension and avoid code duplication, I decided to create a
+rate function to calculate the rate for oxygen and CO2. The rate calculation
+needs similar approaches to the first part, although, each iteration is
+dependent of the previous one, so you need to recalculate the bit counter for
+each state.
+
+To the basic iteration logic you will need to recalculate the counter which
+includes transpose the state matrix, sort and group the bits of each column
+(row, for transposed matrix). You will also need to add a stop condition when
+only one value is present in the state matrix. The basic logic will end up
+being something like this:
+
+```d
+auto rate(char[][] input) {
+ auto ret = input; // state matrix
+ foreach(n, _; input[0]) {
+ auto b = ret.dup.transposed.map!array // transpose matrix
+ .map!`a.dup.sort.group.assocArray`.array; // sort & group bits
+ if(ret.length == 1) break; // stop on one bitarray
+ }
+ return ret.front.to!int(2); // decode bits
+}
+```
+
+To add the bit criteria and filter the values we can use the `filter` template
+along with a comparison with a ternary operator:
+
+```d
+auto newState = oldState.filter!(f => // filter result
+ (b[n]['1'] >= b[n]['0'] ? c : (c == '0' ? '1' : '0')) == f[n] // bit criteria
+ ).array;
+```
+
+More in detail, `b` will include the transposed and grouped bits, `n` is the
+column position and `c` is the bit criteria corresponding to the desired rate,
+which will be a template parameter, in our case. Now we just need to add this
+filter in the loop and we have a working rate function. To calculate the final
+result, we just instantiate each rate template and multiply the results:
+
+```d
+auto result = rate!'1'(input) * rate!'0'(input);
+```
+
+### Full solution
+
+```d
+int rate(char c)(char[][] input) {
+ auto ret = input;
+ foreach(n, _; input[0]) {
+ auto b = ret.dup.transposed.map!array // transpose matrix
+ .map!`a.dup.sort.group.assocArray`.array; // sort & group bits
+ ret = ret.filter!(f => // filter result
+ (b[n]['1'] >= b[n]['0'] ? c : (c == '0' ? '1' : '0')) == f[n] // bit criteria
+ ).array;
+ if(ret.length == 1) break; // stop on one bitarray
+ }
+ return ret.front.to!int(2); // decode bits
+}
+
+void main() {
+ auto i = cast(char[][])stdin.byLine().map!(to!string).array; // input
+ writeln(rate!'1'(i) * rate!'0'(i)); // calculate result
+}
+```
