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.
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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.
{-# LANGUAGE RecordWildCards #-}
import Data.Foldable (for_)
import Test.Hspec (Spec, describe, it, shouldBe)
import Test.Hspec.Runner (configFastFail, defaultConfig, hspecWith)
import PerfectNumbers
main :: IO ()
main = hspecWith defaultConfig {configFastFail = True} specs
specs :: Spec
specs = describe "classify" $ for_ cases test
where
test Case{..} = it description assertion
where
assertion = classify number `shouldBe` expected
data Case = Case { description :: String
, number :: Int
, expected :: Maybe Classification
}
cases :: [Case]
cases = [ Case { description = "Smallest perfect number is classified correctly"
, number = 6
, expected = Just Perfect
}
, Case { description = "Medium perfect number is classified correctly"
, number = 28
, expected = Just Perfect
}
, Case { description = "Large perfect number is classified correctly"
, number = 33550336
, expected = Just Perfect
}
, Case { description = "Smallest abundant number is classified correctly"
, number = 12
, expected = Just Abundant
}
, Case { description = "Medium abundant number is classified correctly"
, number = 30
, expected = Just Abundant
}
, Case { description = "Large abundant number is classified correctly"
, number = 33550335
, expected = Just Abundant
}
, Case { description = "Smallest prime deficient number is classified correctly"
, number = 2
, expected = Just Deficient
}
, Case { description = "Smallest non-prime deficient number is classified correctly"
, number = 4
, expected = Just Deficient
}
, Case { description = "Medium deficient number is classified correctly"
, number = 32
, expected = Just Deficient
}
, Case { description = "Large deficient number is classified correctly"
, number = 33550337
, expected = Just Deficient
}
, Case { description = "Edge case (no factors other than itself) is classified correctly"
, number = 1
, expected = Just Deficient
}
, Case { description = "Zero is rejected (not a natural number)"
, number = 0
, expected = Nothing
}
, Case { description = "Negative integer is rejected (not a natural number)"
, number = -1
, expected = Nothing
}
]
module PerfectNumbers (classify, Classification(..)) where
import Math.NumberTheory.ArithmeticFunctions
data Classification = Deficient | Perfect | Abundant deriving (Eq, Show)
classify :: Int -> Maybe Classification
classify n
| n < 1 = Nothing
| otherwise = case compare . sum . divisors <*> (2*) $ n of
LT -> Just Deficient
EQ -> Just Perfect
GT -> Just Abundant
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Community comments
This is way beyond anyone else's solution. Someday I hope to understand how divisors gives what is wanted.
If you're going to factor n out of compare and 2*, then surely you can factor Just out of the case statement.