How to score Rock Paper Scissors
This article is originally published at https://tjmahr.github.io/
Ho ho ho, it is the most wonderful time of the year: Advent of code!
AOC is a yearly collection of programming puzzles throughout the first 25 days of December. I like it… so much so that I wrote an R package for completing my puzzles using the structure of an R package. The puzzles start out easy and get progressively more elaborate or devious in their requirements. But I am going to talk about an easy puzzle in this post, and specifically, one little trick I used in my solution.
Day 2 of 2022 requires us to score games of Rock Paper Scissors. The
moves are encoded using letters, where our opponent’s moves are coded as
A, B, C and ours are coded as X, Y, Z. So, an input
describing three moves will look like the following:
example_input <- c(
"A Y",
"B X",
"C Z"
)
Where the letters mean the following:
move_codes <- c(
"A" = "rock",
"B" = "paper",
"C" = "scissors",
"X" = "rock",
"Y" = "paper",
"Z" = "scissors"
)
This encoding seems like a weird bit of indirection thrown on, and it is, because the puzzle changes the meanings of the letters in Part 2. Still, it is straightforward to parse the input into a list of roshambo moves.
input <- example_input |>
strsplit(" ") |>
# Use character subsetting to convert letters to moves
lapply(function(x) unname(move_codes[x]))
# Our character's move is the second element in each vector
str(input)
#> List of 3
#> $ : chr [1:2] "rock" "paper"
#> $ : chr [1:2] "paper" "rock"
#> $ : chr [1:2] "scissors" "scissors"
Now, for the point of this post, how do we score each game?
The naive approach is to start typing away furiously
before eventually noping the hell out of there.
What we have is a decision tree: we need to follow a branch for player one and another branch for player two. And here’s the main point of this post: nested lists are trees. (Yes, I love lists—see this post where I use them in my knitr reporting.) The top (outer) level of the list will be all of the player one options, and then the bottom (inner) level will be all the player two options. The nodes of the tree (bottom level values) are the outcomes of the games.
run_game <- function(pair) {
# nested lists are trees
rules <- list(
rock = list(
rock = "draw",
scissors = "lose",
paper = "win"
),
scissors = list(
scissors = "draw",
rock = "win",
paper = "lose"
),
paper = list(
paper = "draw",
scissors = "win",
rock = "lose"
)
)
# Because `rules[[pair[1]]][[pair[2]]]` is unsightly:
rules |>
getElement(pair[1]) |>
getElement(pair[2])
}
At this point, we could take a second to ponder how the structure of
several nested if-elses—the actual shape of the code, indenting in and
out in and in again—resembles the structure and the shape of the
nested list, and ponder further about how the regular, orderly shape of
code could be the whispers of hidden data, saying “list() me, list()
me”. Or, we could run the code and see it in action.
input |>
lapply(run_game)
#> [[1]]
#> [1] "win"
#>
#> [[2]]
#> [1] "lose"
#>
#> [[3]]
#> [1] "draw"
# Or to repeat the input
input |>
stats::setNames(input) |>
lapply(run_game)
#> $`c("rock", "paper")`
#> [1] "win"
#>
#> $`c("paper", "rock")`
#> [1] "lose"
#>
#> $`c("scissors", "scissors")`
#> [1] "draw"
Earlier in the post, I used character subsetting to convert letters into moves. This process turned a matching/replacement problem into a data lookup problem. The Rock Paper Scissors are the same trick again: converting a decision tree into a data lookup problem.
Last knitted on 2022-12-06. Source code on GitHub.1
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