All entries for December 2021

December 08, 2021

Advent of Code 2021: First days with R

The Advent of Code is a series of daily programming puzzles running up to Christmas. On 3 December, the Warwick R User Group met jointly with the Manchester R-thritis Statistical Computing Group to informally discuss our solutions to the puzzles from the first two days. Some of the participants shared their solutions in advance as shared in this slide deck.

In this post, Heather Turner (RSE Fellow, Statistics) shares her solutions and how they can be improved based on the ideas put forward at the meetup by David Selby (Manchester R-thritis organizer) and others, while James Tripp (Senior Research Software Engineer, Information and Digital Group) reflects on some issues raised in the meetup discussion.

R Solutions for Days 1 and 2

Heather Turner

Day 1: Sonar Sweep

For a full description of the problem for Day 1, see 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)

First create a vector with the example data:

x <- c(199, 200, 208, 210, 200, 207, 240, 269, 260, 263)

Then the puzzle can be solved with the following R function, that takes the vector x as input, uses diff() to compute differences between consecutive values of x, then sums the differences that are positive:

f01a <- function(x) {
  dx <- diff(x)
  sum(sign(dx) == 1)
}
f01a(x)

## [1] 7

Inspired by David Selby’s solution, this could be made slightly simpler by finding the positive differences with dx > 0, rather than using the sign() function.

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

This can be solved by the following function of x. First, the rolling sums of three consecutive values are computed in a vectorized computation, i.e. creating three vectors containing the first, second and third value in the sum, then adding the vectors together. Then, the function from Part 1 is used to sum the positive differences between these values.

f01b <- function(x) {
  n <- length(x)
  sum3 <- x[1:(n - 2)] + x[2:(n - 1)] + x[3:n]
  f01a(sum3)
}
f01b(x)

## [1] 5

David Schoch put forward a solution that takes advantage of the fact that the difference between consecutive rolling sums of three values is just the difference between values three places apart (the second and third values in the first sum cancel out the first and second values in the second sum). Putting what we’ve learnt together gives this much neater solution for Day 1 Part 2:

f01b_revised <- function(x) {
  dx3 <- diff(x, lag = 3)
  sum(dx3 > 0)
}
f01b_revised(x)

## [1] 5

Day 2: Dive!

For a full description of the problem see 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.

                  horizontal  depth
forward 5   -->            5      -
down 5      -->                   5
forward 8   -->           13
up 3        -->                   2
down 8      -->                  10
forward 2   -->           15

==> horizontal = 15, depth = 10

First create a data frame with the example data

x <- data.frame(direction = c("forward", "down", "forward",
                              "up", "down", "forward"),
                amount = c(5, 5, 8, 3, 8, 2))

Then the puzzle can be solved with the following function, which takes the variables direction and amount as input. The horizontal position is the sum of the amounts where the direction is “forward”. The depth is the sum of the amounts where direction is “down” minus the sum of the amounts where direction is “up”.

f02a <- function(direction, amount) {
  horizontal <- sum(amount[direction == "forward"])
  depth <- sum(amount[direction == "down"]) - sum(amount[direction == "up"])
  c(horizontal = horizontal, depth = depth)
}
f02a(x$direction, x$amount)

## horizontal      depth 
##         15         10

The code above uses logical indexing to select the amounts that contribute to each sum. An alternative approach from David Selby is to coerce the logical indices to numeric (coercing TRUE to 1 and FALSE to 0) and multiply the amount by the resulting vectors as required:

f02a_selby <- function(direction, amount) {
  horizontal_move <- amount * (direction == 'forward')
  depth_move <- amount * ((direction == 'down') - (direction == 'up'))
  c(horizontal = sum(horizontal_move), depth = sum(depth_move))
}

Benchmarking on 1000 datasets of 1000 rows this alternative solution is only marginally faster (an average run-time of 31 μs vs 37 μs), but it has an advantage in Part 2!

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.

                  horizontal  aim  depth
forward 5   -->            5    -      - 
down 5      -->                 5      
forward 8   -->           13          40
up 3        -->                 2
down 8      -->                10
forward 2   -->           15          60

==> horizontal = 15, depth = 60

The following function solves this problem by first computing the sign of the change to aim, which is negative if the direction is “up” and positive otherwise. Then for each change in position, if the direction is “forward” the function adds the amount to the horizontal position and the amount multiplied by aim to the depth, otherwise it adds the sign multiplied by the amount to the aim.

f02b <- function(direction, amount) {
  horizontal <- depth <- aim <- 0
  sign <- ifelse(direction == "up", -1, 1)
  for (i in seq_along(direction)){
    if (direction[i] == "forward"){
      horizontal <- horizontal + amount[i]
      depth <- depth + aim * amount[i]
      next
    }
    aim <- aim + sign[i]*amount[i]
  }
  c(horizontal = horizontal, depth = depth)
}
f02b(x$direction, x$amount)

## horizontal      depth 
##         15         60

As an interpreted language, for loops can be slow in R and vectorized solutions are often preferable if memory is not an issue. David Selby showed that his solution from Part 1 can be extended to solve the problem in Part 2, by using cumulative sums of the value that represented depth in Part 1 to compute the aim value in Part 2.

f02b_revised <- function(direction, amount) {
  horizontal_move <- amount * (direction == "forward")
  aim <- cumsum(amount * (direction == "down") - amount * (direction == "up"))
  depth_move <- aim * horizontal_move
  c(horizontal = sum(horizontal_move), depth = sum(depth_move))
}
f02b_revised(x$direction, x$amount)

## horizontal      depth 
##         15         60

Benchmarking these two solutions on 1000 data sets of 1000 rows, the vectorized solution is ten times faster (on average 58 μs vs 514 μs).

Reflections

James Tripp

How do we solve a problem with code? Writing an answer requires what some educators call computational thinking. We systematically conceptualise the solution to a problem and then work through a series of steps, drawing on coding conventions, to formulate an answer. Each answer is different and, often, a reflection of our priorities, experience, and domains of work. In our meeting, it was wonderful to see people with a wide range of experience and differing interests.

Our discussion considered the criteria of a ‘good solution’.

Speed is one criteria of success - a solution which takes 100 μs (microseconds) is better than a solution taking 150 μs.
Readability for both sharing with others (as done above) and to help future you, lest you forget the intricacies of your own solution.
Good practice such as variable naming and, perhaps, avoiding for loops where possible. Loops are slower and somewhat discouraged in the R community. However, some would argue they are more explicit and helpful for those coming from other languages, such as Python.
Debugging friendly.Some participants, including Heather Turner and David Selby, checked their solutions with tests comparing known inputs and outputs. I drew on my Psychology experience and opted for an explicit DataFrame where I can see each operation. Testing is almost certainly a better solution which I adopt in my packages.
Generalisability. A solution tailored for the Part 1 task on a given day may not be easily generalisable for the Part 2 task. It seemed desirable to refactor one’s code to create a solution which encompasses both tasks. However, the effort and benefits of doing so are certainly debatable.

We also discussed levels of abstraction. The tidyverse family of R packages is powerful, high-level and quite opinionated. Using tidyverse functions returned some intuitive, but slower solutions where we were unsure of the bottlenecks. Solutions built on R base (the functions which come with R) were somewhat faster, though others using libraries such as data.table were also rather quick. These reflections are certainly generalisations and prompted some discussion.

How does one produce fast, readable, debuggable, generalisable code which follows good practice and operates at a suitable level of abstraction? Our discussions did not produce a definitive answer. Instead, our discussions and sharing solutions helped us understand the pros and cons of different approaches and I certainly learned a few useful tricks.

Heather Turner : 08 Dec 2021 10:00 | Tags: Code Programming R | Comments (0) | Close comments | Report a problem