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katrinleinweber's solution

to Perfect Numbers in the R Track

Published at Jul 13 2018 · 0 comments
Instructions
Test suite
Solution

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

  • Perfect: aliquot sum = number
    • 6 is a perfect number because (1 + 2 + 3) = 6
    • 28 is a perfect number because (1 + 2 + 4 + 7 + 14) = 28
  • Abundant: aliquot sum > number
    • 12 is an abundant number because (1 + 2 + 3 + 4 + 6) = 16
    • 24 is an abundant number because (1 + 2 + 3 + 4 + 6 + 8 + 12) = 36
  • Deficient: aliquot sum < number
    • 8 is a deficient number because (1 + 2 + 4) = 7
    • Prime numbers are deficient

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.

Installation

See this guide for instructions on how to setup your local R environment.

How to implement your solution

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.

How to run tests

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.

Source

Taken from Chapter 2 of Functional Thinking by Neal Ford. http://shop.oreilly.com/product/0636920029687.do

Submitting Incomplete Solutions

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

test_perfect-numbers.R

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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