class: center, middle, inverse, title-slide # Advent of Code 2021 ## Joint discussion ### David Selby, Heather Turner & James Tripp ### R-thritis & Warwick RUG ### 3 December 2021 --- class: center, middle, inverse <div> <style type="text/css">.xaringan-extra-logo { width: 67px; height: 50px; z-index: 0; background-image: url(https://www.cfe.manchester.ac.uk/media/mhs/bmh-faculty/images/CfE-logo-150px.jpg); background-size: contain; background-repeat: no-repeat; position: absolute; bottom:.3em;left:0.3em; } </style> <script>(function () { let tries = 0 function addLogo () { if (typeof slideshow === 'undefined') { tries += 1 if (tries < 10) { setTimeout(addLogo, 100) } } else { document.querySelectorAll('.remark-slide-content:not(.title-slide):not(.inverse):not(.hide_logo)') .forEach(function (slide) { const logo = document.createElement('div') logo.classList = 'xaringan-extra-logo' logo.href = null slide.appendChild(logo) }) } } document.addEventListener('DOMContentLoaded', addLogo) })()</script> </div> # Day 1: Sonar Sweep <https://adventofcode.com/2021/day/1> --- ## Day 1 - Part 1 How many measurements are larger than the previous measurement? ``` 199 (N/A - no previous measurement) 200 (increased) 208 (increased) 210 (increased) 200 (decreased) 207 (increased) 240 (increased) 269 (increased) 260 (decreased) 263 (increased) ``` --- ## Day 1 - Part 2 How many **sliding three-measurement sums** are larger than the previous sum? ``` 199 A 607 (N/A - no previous sum) 200 A B 618 (increased) 208 A B C 618 (no change) 210 B C D 617 (decreased) 200 E C D 647 (increased) 207 E F D 716 (increased) 240 E F G 769 (increased) 269 F G H 792 (increased) 260 G H 263 H ``` --- class: center, middle ## Day 1 - Solutions --- ### Day 1: How to make life harder for yourself By Claire Little .pull-right[.red[10,000 μs]] ```r library(tidyverse) # Part 1: input %>% mutate(change = V1 - lag(V1), sign = change / abs(change)) %>% group_by(sign) %>% summarise(n = n()) # Part 2: input %>% mutate(window = V1 + lag(V1) + lag(V1, 2), change = window - lag(window), sign = change / abs(change)) %>% group_by(sign) %>% summarise(n = n()) ``` ??? Day 1 was well easy but I clearly need to get better at "I wonder if there's a function for that" rather than trying to bosh something together i.e. `change/abs(change)` Would have tried to do it without tidyverse if I had more time but it was just calling out for some tidyverseness --- ### Day 1: base R using `diff()` By David Selby .pull-right[.red[~100 μs]] ```r increases <- function(depths) { sum(diff(depths) > 0) } rolling_sum <- function(depths) { head(depths, -2) + tail(depths, -2) + head(tail(depths, -1), -1) } input1 <- scan('input01.txt') increases(input1) increases(rolling_sum(input1)) ``` .footnote[Or you could use `zoo::rollsum(depths, 3)`] ??? 100 μs is around 100 times faster than 10 ms (= 10,000 μs) --- ### Day 1: one-liner using `diff(lag = n)` By David Schoch .pull-right[.red[~100 μs]] ```r sapply(c(1, 3), function(n) sum(diff(input1, lag = n) > 0)) ``` -- Or with R's new anonymous function `\(x)` syntax: ```r sapply(c(1, 3), \(n) sum(diff(input1, n) > 0)) ``` ??? 100 μs is around 100 times faster than 10 ms (= 10,000 μs) Worth pointing out that you can't attempt part 2 until you've completed and submitted part 1, so nobody is going to have a one-liner as their first solution! --- class: center, middle, inverse # Day 2: Dive! <https://adventofcode.com/2021/day/2> --- ## Day 2 - Part 1 - `forward X` increases the horizontal position by `X` units. - `down X` increases the depth by `X` units. - `up X` decreases the depth by `X` units. ``` x depth forward 5 --> 5 - down 5 --> 5 forward 8 --> 13 up 3 --> 2 down 8 --> 10 forward 2 --> 15 ==> x = 15, depth = 10 ``` --- ## Day 2 - Part 2 - `down X` increases your aim by `X` units. - `up X` decreases your aim by `X` units. - `forward X` does two things: - It increases your horizontal position by `X` units. - It increases your depth by your aim **multiplied by** `X`. ``` x aim depth forward 5 --> 5 - - down 5 --> 5 forward 8 --> 13 40 up 3 --> 2 down 8 --> 10 forward 2 --> 15 60 ==> x = 15, depth = 60 ``` --- class: center, middle ## Day 2 - Solutions --- ### Day 2: I don't care if loops are inefficient By Claire Little .pull-right[.red[60,000 μs]] ```r inst = read.table("path/to/Day2.csv") ninst = dim(inst)[1] # Day 2 - Part 1 position = c(0, 0) for (i in 1:ninst) { if (inst[i, 1] == 'forward') { position = position + c(inst[i, 2], 0) } else if (inst[i, 1] == 'down') { position = position + c(0, inst[i, 2]) } else { position = position - c(0, inst[i, 2]) } } answer1 = position[1] * position[2] ``` ??? Day 2 - I forgot how many loops & if statements I used last year and I liked using them but also appreciate it wouldn't have worked very well with bigger datasets. But it was efficient enough for what we needed --- ### Day 2: I don't care if loops are inefficient By Claire Little .pull-right[.red[60,000 μs]] ```r # Day 2 - Part 2 position = c(0, 0, 0) for (i in 1:ninst) { if (inst[i, 1] == 'forward') { position = position + c(inst[i, 2], 0, 0) position = position + c(0, (inst[i, 2] * position[3]), 0) } else if (inst[i, 1] == 'down') { position = position + c(0, 0, inst[i, 2]) } else { position = position - c(0, 0, inst[i, 2]) } } answer2 = position[1] * position[2] ``` ??? Day 2 - I forgot how many loops & if statements I used last year and I liked using them but also appreciate it wouldn't have worked very well with bigger datasets. But it was efficient enough for what we needed --- ### Day 2: no ifs, no loops By David Selby .pull-right[.red[100 μs]] ```r course <- read.table('input.txt', col.names = c('dir', 'value')) with(course, { x <- (dir == 'forward') * value y <- ((dir == 'down') - (dir == 'up')) * value sum(x) * sum(y) }) ``` ```r with(course, { x <- (dir == 'forward') * value y <- ((dir == 'down') - (dir == 'up')) * value depth <- cumsum(y) * x sum(x) * sum(depth) }) ``` ??? 100 μs is around **600 times faster** than 60 ms (= 60,000 μs) --- ### Day 2: tidyverse-style By David Selby .pull-right[.red[3000 μs]] ```r course %>% mutate(x = if_else(dir == 'forward', value, 0L), y = if_else(dir == 'down', value, -value) * !x, depth = cumsum(y) * x) %>% summarise(part1 = sum(x) * sum(y), part2 = sum(x) * sum(depth)) ``` ??? At 3 ms (= 3000 μs), this tidyverse solution is around 30 times slower than the base R (100 μs), but 20 times faster than loops (60,000 μs / 60 ms) --- layout: false class: center, middle # Thanks! `david.selby@manchester.ac.uk` .small[For these slides: search `"rthritis" manchester`] --- class: center, middle ## Next meeting Friday 14 January 2022 .big[Writing your own R package] _To be confirmed_