Advent of Code 2021 – Day 3

author: “Barret”title: “Advent of Code 2021 – Day 3, Puzzle 1”image: “img/posts/2021/pexels-any-lane-5727795.jpg”draft: falsedate: 2021-12-03T08:00:00Zdescription: “I’m participating in the Advent of Code 2021. Here’s my solution for Day 3, Puzzle 1 – Binary Diagnostic”tags: [“Advent of Code 2021”]archives: [“2021/12”]

Photo by Any Lane from Pexels

I’m participating in the Advent of Code 2021. Here’s my solution for Day 3, Puzzle 1 – Binary Diagnostic

Problem

Puzzle 1 for Day 3 involves analyzing some binary numbers that are output and using the frequency that a particular bit is 1 vs 0. For each bit position, we get the frequency across all the input data, then use that total to build two final binary numbers.

From the original problem:

```> 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.
> Each bit in the gamma rate can be determined by finding the most common bit in the corresponding position of all numbers in the diagnostic report. For example, given the following diagnostic report:
>
> > 00100
> > 11110
> > 10110
> > 10111
> > 10101
> > 01111
> > 00111
> > 11100
> > 10000
> > 11001
> > 00010
> > 01010
> Considering only the first bit of each number, there are five 0 bits and seven 1 bits. Since the most common bit is 1, the first bit of the gamma rate is 1. The most common second bit of the numbers in the diagnostic report is 0, so the second bit of the gamma rate is 0. The most common value of the third, fourth, and fifth bits are 1, 1, and 0, respectively, and so the final three bits of the gamma rate are 110.
>
> So, the gamma rate is the binary number 10110, or 22 in decimal.
>
> The epsilon rate is calculated in a similar way; rather than use the most common bit, the least common bit from each position is used. So, the epsilon rate is 01001, or 9 in decimal. Multiplying the gamma rate (22) by the epsilon rate (9) produces the power consumption, 198.
>
> 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.)```

Solution

All my solutions are written in C#. You can find all my solutions in my Git repo.

I’m sure there’s an easier way to accomplish this. Working with binary numbers has never been a strong point in my repertoire. So no laughing at my solution. Basically, I’m just splitting each binary input into an array of char, then counting which ones are equal to “1” in a dictionary. Then, if the total count for each bit position is greater than half the total number of rows, it’s the “most common bit”. Finally, I build my gamma rate, then use the opposite bits to determine my epsilon rate, and multiply the two together.

``````using System.Text;
Console.WriteLine("Day 3 - Puzzle 1");
//Solution logic goes here
var counts = new Dictionary<int, int>();
for (int i = 0; i < lines.Length; i++)
{
var split = lines[i].ToArray();
for (var x = 0; x < split.Length; x++)
{
if (split[x] == '1')
{
if (counts.ContainsKey(x))
counts[x] += 1;
else counts[x] = 1;
}
}
}
StringBuilder sbGamma = new StringBuilder();
StringBuilder sbEpsilon = new StringBuilder();
for (var z = 0; z < counts.Count; z++)
{
if (counts.ContainsKey(z) && counts[z] > lines.Length / 2)
{
sbGamma.Append("1");
sbEpsilon.Append("0");
}
else
{
sbGamma.Append("0");
sbEpsilon.Append("1");
}
}
foreach (var item in counts.OrderBy(x=>x.Key))
{
Console.WriteLine(\$"{item.Key}  {item.Value}");
}
Console.WriteLine(sbGamma.ToString());
Console.WriteLine(sbEpsilon.ToString());
var gamma = Convert.ToInt32(sbGamma.ToString(), 2);
var epsilon = Convert.ToInt32(sbEpsilon.ToString(), 2);
Console.WriteLine(gamma);
Console.WriteLine(epsilon);
Console.WriteLine(gamma * epsilon);
//Stop and wait for enter before exiting

Problem 2

Puzzle 2 for Day 3 involves analyzing the same set of binary numbers from Part 1, but looking for a different analysis result. This time we’ll analyze the bits positions one at a time. For each one, we’ll reduce the pool of input data to only those who have a value in that position that matches either the most popular (Oxygen) or least popular (C02) value in that position.

```> 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.
> For example, to determine the oxygen generator rating value using the same example diagnostic report from above:
>
> Start with all 12 numbers and consider only the first bit of each number. There are more 1 bits (7) than 0 bits (5), so keep only the 7 numbers with a 1 in the first position: 11110, 10110, 10111, 10101, 11100, 10000, and 11001.
> Then, consider the second bit of the 7 remaining numbers: there are more 0 bits (4) than 1 bits (3), so keep only the 4 numbers with a 0 in the second position: 10110, 10111, 10101, and 10000.
> In the third position, three of the four numbers have a 1, so keep those three: 10110, 10111, and 10101.
> In the fourth position, two of the three numbers have a 1, so keep those two: 10110 and 10111.
> In the fifth position, there are an equal number of 0 bits and 1 bits (one each). So, to find the oxygen generator rating, keep the number with a 1 in that position: 10111.
> As there is only one number left, stop; the oxygen generator rating is 10111, or 23 in decimal.
>
> Then, to determine the CO2 scrubber rating value from the same example above:
> Start again with all 12 numbers and consider only the first bit of each number. There are fewer 0 bits (5) than 1 bits (7), so keep only the 5 numbers with a 0 in the first position: 00100, 01111, 00111, 00010, and 01010.
> Then, consider the second bit of the 5 remaining numbers: there are fewer 1 bits (2) than 0 bits (3), so keep only the 2 numbers with a 1 in the second position: 01111 and 01010.
> In the third position, there are an equal number of 0 bits and 1 bits (one each). So, to find the CO2 scrubber rating, keep the number with a 0 in that position: 01010.
> As there is only one number left, stop; the CO2 scrubber rating is 01010, or 10 in decimal.
> Finally, to find the life support rating, multiply the oxygen generator rating (23) by the CO2 scrubber rating (10) to get 230.
>
> 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.)  ```

Solution

All my solutions are written in C#. You can find all my solutions in my Git repo.

As with Puzzle 1, there’s probably an easier way to do this, but this is where I ended up. For each stat we’re looking at, Oxygen and CO2, I created a copy of the source data array as a List object. I then iterate through each bit position and filter that list smaller and smaller based on matches to the value in that matching position.

``````Console.WriteLine("Advent of Code 2021");
Console.WriteLine("Day 3 - Puzzle 1");
//Solution logic goes here
var oxyLines = lines.ToList();
for (var a = 0; a < 12; a++)
{
if (oxyLines.Count > 1)
{
var count = oxyLines.Count(x => x.Substring(a, 1) == "1");
if (count >= oxyLines.Count/2)
{
oxyLines = oxyLines.Where(x => x.Substring(a, 1) == "1").ToList();
} else
{
oxyLines = oxyLines.Where(x => x.Substring(a, 1) == "0").ToList();
}
}
}
var oxy = 0;
if(oxyLines.Count == 1)
{
oxy = Convert.ToInt32(oxyLines[0], 2);
} else
{
throw new Exception("Didn't end with just one row");
}
var co2Lines = lines.ToList();
for (var b = 0; b < 12; b++)
{
if (co2Lines.Count > 1)
{
var count = co2Lines.Count(x => x.Substring(b, 1) == "0");
if (count <= co2Lines.Count / 2)
{
co2Lines = co2Lines.Where(x => x.Substring(b, 1) == "0").ToList();
}
else
{
co2Lines = co2Lines.Where(x => x.Substring(b, 1) == "1").ToList();
}
}
}
var co2 = 0;
if (co2Lines.Count == 1)
{
co2 = Convert.ToInt32(co2Lines[0], 2);
}
else
{
throw new Exception("Didn't end with just one row");
}
Console.WriteLine(\$"{oxy} * {co2} = {oxy * co2}");
//Stop and wait for enter before exiting