December 2, 2024
Chief historian is lost, gotta try to find him. Searched his office, found several lists of places (ID’s) we should look, but the lists don’t align. Need to find out how close they are.
Example Data
Data comes in as space separated lists, so I used the lines_to_matrix function from aochelpers
Pair up the smallest number in the left list with the smallest number in the right list, then the second-smallest left number with the second-smallest right number, and so on.
So, take the separation even further and detach the lists into separate vectors, sorted ascendingly.
Within each pair, figure out how far apart the two numbers are; you’ll need to add up all of those distances.
Sum the absolute differences
Matches the example solution.
The two lists are pretty different. The Historians can’t agree on where the problem is at, but noticed that some of the ID”s are duplicated. So let’s calculate a similarity score.
_ Calculate a total similarity score by adding up each number in the left list after multiplying it by the number of times that number appears in the right list._
So, use grepl to identify the number of times each element in the first vector appears in the second vector. Since it returns a TRUE/FALSE for ever element, add up the trues using sum. Then loop over all rows, add that as a third vector.
Now take the first column, and multiply it by the multiplier, then add the result
Matches example.
─ Session info ───────────────────────────────────────────────────────────────
setting value
version R version 4.4.1 (2024-06-14)
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system aarch64, darwin20
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tz America/Los_Angeles
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─ Packages ───────────────────────────────────────────────────────────────────
package * version date (UTC) lib source
aochelpers * 0.1.0.9000 2024-12-02 [1] Github (EllaKaye/aochelpers@d4ccd91)
dplyr * 1.1.4 2023-11-17 [1] CRAN (R 4.4.0)
[1] /Users/rdonatello/Library/R/arm64/4.4/library
[2] /Library/Frameworks/R.framework/Versions/4.4-arm64/Resources/library
──────────────────────────────────────────────────────────────────────────────---
title: "2024: Day 01 Historian Hysteria"
date: 2024-12-02
categories:
- loop
- grepl
- vectors
draft: false
---
## Setup
[The original challenge](https://adventofcode.com/2024/day/01)
Chief historian is lost, gotta try to find him. Searched his office, found several lists of places (ID's) we should look, but the lists don't align. Need to find out how close they are.
```{r}
library(aochelpers)
library(dplyr)
# other options: aoc_input_data_frame(), aoc_input_matrix()
input <- aoc_input_vector(01, 2024)
head(input)
```
# TLDR; Solutions
## Part 1 ⭐
::: {.callout-danger}
### ❓ What is the total distance between your lists?
:::
```{r}
p1 <- input |> lines_to_matrix(split = " ")
c1 <- sort(as.numeric(p1[,1]))
c2 <- sort(as.numeric(p1[,2]))
sum(abs(c1-c2))
```
## Part 2 ⭐⭐
::: {.callout-danger}
### ❓ Calculate a total similarity score by adding up each number in the left list after multiplying it by the number of times that number appears in the right list.
:::
```{r}
multiplier <- rep(0, NROW(p1))
for(i in 1:NROW(p1)){
multiplier[i] <- grepl(c1[i], c2) |> sum()
}
sum(c1*multiplier)
```
# Walkthrough / Explainer
## Part 1
:::{.callout-exa icon=true}
**Example Data**
Data comes in as space separated lists, so I used the `lines_to_matrix` function from `aochelpers`
```{r}
(exa <- c("3 4", "4 3", "2 5", "1 3", "3 9", "3 3") |> lines_to_matrix())
```
:::
_Pair up the smallest number in the left list with the smallest number in the right list, then the second-smallest left number with the second-smallest right number, and so on._
So, take the separation even further and detach the lists into separate vectors, sorted ascendingly.
```{r}
ex1 <- sort(as.numeric(exa[,1]))
ex2 <- sort(as.numeric(exa[,5]))
```
_Within each pair, figure out how far apart the two numbers are; you'll need to add up all of those distances._
Sum the absolute differences
```{r}
sum(abs(ex1-ex2))
```
Matches the example solution.
## Part 2
The two lists are pretty different. The Historians can't agree on where the problem is at, but noticed that some of the ID"s are duplicated. So let's calculate a similarity score.
_ Calculate a total similarity score by adding up each number in the left list after multiplying it by the number of times that number appears in the right list._
So, use `grepl` to identify the number of times each element in the first vector appears in the second vector. Since it returns a TRUE/FALSE for ever element, add up the trues using `sum`. Then loop over all rows, add that as a third vector.
```{r}
mult <- rep(0, NROW(exa))
for(i in 1:NROW(exa)){
mult[i] <- grepl(ex1[i], ex2) |> sum()
}
```
Now take the first column, and multiply it by the multiplier, then add the result
```{r}
sum(ex1*mult)
```
Matches example.
##### Session info {.appendix}
<details><summary>Toggle</summary>
```{r}
#| echo: false
sessioninfo::session_info(pkgs = "attached")
```
</details>