You may have seen the memes going around about fun ways to program the
straightforward function isEven() which returns TRUE if the input is even,
and FALSE otherwise. I had a play with this and it turned into enough for a
blog post, and a nice walk through some features of R.
The ‘traditional’ way to check if an integer is even is to check if it is
divisible by 2. This can be achieved with the modulo operator %% which
gives the remainder after dividing by another number. For example, 5 modulo 2
or 5 %% 2 gives 1 because 2 goes into 5 twice with 1 leftover. If a number x is
even, it is an exact multiple of 2, and so x %% 2 == 0.
5 %% 2## [1] 16 %% 2## [1] 0A function which tests values of x for this property could be written as
## 1
isEven <- function(x) {
## traditional modulo check
x %% 2 == 0
}The == operation checks that the left side is equal to the right side (but not
necessarily identical, e.g. the classes can be different) and returns either TRUE
or FALSE (or NA, but that’s not an issue for the cases we’re looking at here).
I’ve also relied on the fact that the result of the last statement in a function
body is used as the return value if no explicit return() is used.
Confirming that this works is as easy as trying some values. It’s always good to check that your function produces results you expect. It’s also a good idea to try some odd values to ensure you don’t hit edge-cases.
isEven(0)## [1] TRUEisEven(3)## [1] FALSEisEven(4)## [1] TRUEisEven(-1)## [1] FALSEisEven(-6)## [1] TRUEIn the process of playing with this function I refined how I tested my code. I started with a set of confirming evaluations like above. Then I wanted to confirm that they actually gave the results I expect, so I refined it to
test_isEven <- function() {
all(
isEven(0) == TRUE,
isEven(3) == FALSE,
isEven(4) == TRUE,
isEven(-1) == FALSE,
isEven(-6) == TRUE
)
}Now I just had one function to call which confirmed that all these tests gave
the expected results. Once I had explained this layout to myself with the word
‘expected’, I realised what I actually wanted was a test suite, and testthat
is a great candidate for that. Refactoring the above into a series of expectations
might look like
library(testthat)
test_isEven <- function() {
test_that("isEven peforms as expected", {
expect_true(isEven(0))
expect_false(isEven(3))
expect_true(isEven(4))
expect_false(isEven(-1))
expect_true(isEven(-6))
})
}Now I can test any implementation of isEven() with just one function call, and if
one of the expectations fails I’ll know which it is. Running this with the above
isEven() produces no output, so the tests succeeded
test_isEven()The ‘no output’ might be concerning, so we can also run a negative control to make
sure it breaks when things are broken. Let’s break the isEven() by reversing the
test
## 1a
isEven <- function(x) {
## (broken) traditional modulo check
x %% 2 == 1
}
test_isEven()## Error: Test failed: 'isEven peforms as expected'
## * <text>:4: isEven(0) isn't true.
## * <text>:5: isEven(3) isn't false.
## * <text>:6: isEven(4) isn't true.
## * <text>:7: isEven(-1) isn't false.
## * <text>:8: isEven(-6) isn't true.So, we can trust that if we make a mistake or don’t implement this properly, we’ll know. Typically a function you write would have a lot more safety checking, such as ensuring that we actually passed a value, and that it’s an integer, but for the sake of this post I’m going to assume that these are both guaranteed to be true.
This version of isEven() is simple and it works, but that’s not what the internet
wants - a common challenge is to make a version of isEven() which doesn’t use
modulo. Now we need to think a little more, but we can at least check any
implementation with our tests.
I came up with a few, both from borrowing from other solutions and on my own. Let’s see…
If the last digit is any of 0, 2, 4, 6, or 8, then it’s an even number
## 2
isEven <- function(x) {
## ends with an even digit
grepl("[02468]$", x)
}
test_isEven()With that same idea, if the least significant bit (binary) is unset then it’s even
## 3
isEven <- function(x) {
## least significant bit is unset
x == 0 || !bitwAnd(x, 1)
}
test_isEven()Continuing down the bitwise path, if we can shift left and right and get back to the original number, then it’s even
## 4
isEven <- function(x) {
## bitwise shift right then left
!(x-(bitwShiftL(bitwShiftR(x, 1), 1)))
}
test_isEven()If we alternate FALSE and TRUE counting from 0 to x then we get our answer
## 5
isEven <- function(x) {
## alternate TRUE/FALSE
y <- FALSE
for (i in 0:x) {
y <- !y
}
return(y)
}
test_isEven()We could do the same thing with recursion
## 6
isEven <- function(x) {
## recursion, n-1 is odd if n is even
if (x == 0) return(TRUE)
!isEven(abs(x) - 1)
}
test_isEven()Not quite using modulo, integer division by 2, doubled, should return the original value
## 7
isEven <- function(x) {
## integer division, doubled
2*(x %/% 2) == x
}
test_isEven()Similarly, the result of regular division cast to integer, doubled, should return the original value
## 7a
isEven <- function(x) {
## normal division, doubled
2*as.integer(x/2) == x
}
test_isEven()If we start from a number and count towards 0 by twos then we will hit 0 if the number is even
## 8
isEven <- function(x) {
## moving by 2s towards 0 ends at 0
y <- x
repeat({
if (y == 0) return(TRUE)
if (sign(x) != sign(y)) return(FALSE)
y <- y - sign(x)*2
})
}
test_isEven()We can write that a bit simpler if we only use the absolute value of x
## 8a (abs version)
isEven <- function(x) {
## moving by 2s towards 0 ends at 0
y <- abs(x)
repeat({
if (y == 0) return(TRUE)
if (y < 0) return(FALSE)
y <- y - 2
})
}
test_isEven()Exploiting mathematical properties, we know that -1 to any even power returns 1
## 9
isEven <- function(x) {
## -1 to an even power is 1
(-1)**x == 1
}
test_isEven()The relationship \[\cos(2x) = -\cos(x)\] can also be exploited
## 10
isEven <- function(x) {
## cos(2x) == -cos(x)
cos(x*pi) == -cos(pi)
}
test_isEven()Now for some more R-specific solutions… R rounds towards even integers, and we can exploit that
## 11
isEven <- function(x) {
## R rounds even real numbers down
round(x + 0.5) == x
}
test_isEven()We can create a vector of ‘every other integer’ and check whether a value is in there
## 12
isEven <- function(x) {
## is x in set of 'every other integer'?
abs(x) %in% (0:abs(x))[c(TRUE, FALSE)]
}
test_isEven()Creating a vector of TRUE and FALSE we can extract the element corresponding to
a value
## 13
isEven <- function(x) {
## even/odd sequence
if (x == 0) return(TRUE)
rep(c(FALSE, TRUE), (abs(x)/2) + 1)[abs(x)]
}
test_isEven()Then, starting to get really absurd, we could solve the equation \[2n = x\] which
will have an integer n if x is even
## 14
isEven <- function(x) {
## integer solution to 2n = x?
n <- solve(2, x)
as.integer(n) == n
}
test_isEven()And, lastly, for the truly absurd, we can use the fact that “zero” and “eight” are the only single digits written as English words with an “e”. This requires a couple of extra packages, but can be done.
## 15
isEven <- function(x) {
## zero and eight are the only odd
## last digit as words with an e
last <- english::words(as.integer(stringr::str_sub(x, -1, -1)))
last == "zero" || last == "eight" || !grepl("e", last)
}
test_isEven()This isn’t an exhaustive list, but it seemed like a good place to stop looking. If you can think of more then add them to this thread on Twitter.
I hope this demonstrates the usefulness of writing functions and testing them with
testthat. Plus, if the %% operator ever breaks, you have plenty of alternatives.
devtools::session_info()
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## language en_AU:en
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## ctype en_AU.UTF-8
## tz Australia/Adelaide
## date 2020-03-09
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