Determine if a number is perfect, abundant, or deficient based on Nicomachus' (60 - 120 CE) classification scheme for natural numbers.
The Greek mathematician Nicomachus devised a classification scheme for natural numbers, identifying each as belonging uniquely to the categories of perfect, abundant, or deficient based on their aliquot sum. The aliquot sum is defined as the sum of the factors of a number not including the number itself. For example, the aliquot sum of 15 is (1 + 3 + 5) = 9
Implement a way to determine whether a given number is perfect. Depending on your language track, you may also need to implement a way to determine whether a given number is abundant or deficient.
See this guide for instructions on how to setup your local R environment.
In each problem folder, there is a file named <exercise_name>.R
containing a function that returns a NULL
value. Place your implementation inside the body of the function.
Inside of RStudio, simply execute the test_<exercise_name>.R
script. This can be conveniently done with testthat's auto_test
function. Because exercism code and tests are in the same folder, use this same path for both code_path
and test_path
parameters. On the command-line, you can also run Rscript test_<exercise_name>.R
.
Taken from Chapter 2 of Functional Thinking by Neal Ford. http://shop.oreilly.com/product/0636920029687.do
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
source("./perfect-numbers.R")
library(testthat)
context("perfect numbers")
test_that("Smallest perfect number is classified correctly", {
n <- 6
expect_equal(number_type(n), "perfect")
})
test_that("Medium perfect number is classified correctly", {
n <- 28
expect_equal(number_type(n), "perfect")
})
test_that("Large perfect number is classified correctly", {
n <- 33550336
expect_equal(number_type(n), "perfect")
})
test_that("Smallest abundant number is classified correctly", {
n <- 12
expect_equal(number_type(n), "abundant")
})
test_that("Medium abundant number is classified correctly", {
n <- 30
expect_equal(number_type(n), "abundant")
})
test_that("Large abundant number is classified correctly", {
n <- 33550335
expect_equal(number_type(n), "abundant")
})
test_that("Smallest prime deficient number is classified correctly", {
n <- 2
expect_equal(number_type(n), "deficient")
})
test_that("Smallest non-prime deficient number is classified correctly", {
n <- 4
expect_equal(number_type(n), "deficient")
})
test_that("Medium deficient number is classified correctly", {
n <- 32
expect_equal(number_type(n), "deficient")
})
test_that("Large deficient number is classified correctly", {
n <- 33550337
expect_equal(number_type(n), "deficient")
})
test_that("Edge case (no factors other than itself) is classified correctly", {
n <- 1
expect_equal(number_type(n), "deficient")
})
test_that("Zero is rejected (not a natural number)", {
n <- 0
expect_error(number_type(n))
})
test_that("Negative integer is rejected (not a natural number)", {
n <- -1
expect_error(number_type(n))
})
message("All tests passed for exercise: perfect-numbers")
library(magrittr)
is_perfect <- function(n){
# catch edge cases
if (n <= 0)
stop("Only natural number can be classified here!")
if (n <= 2)
return("deficient")
# find n's factors, incl. 1 but not n (aliquots)
factor <- function(i)
if (n %% i == 0)
i
# calculate sum and classify n
lapply(1:(n/2), factor) %>%
unlist %>%
sum ->
sum
dplyr::case_when(
sum == n ~ "perfect",
sum < n ~ "deficient",
sum > n ~ "abundant"
)
}
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