Discussion on: Challenge: find 'Kaprekar numbers'

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Replies for: Solution in Haskell: I didn't want to take the route to String representation and back, I considered it as a computational problem and I wanted to...

I tried to compute the Kaprekar numbers up to 10^8. A simple improvement in taking the interesting part and dropping the rest: If quotient or remainder are > n the sum of both must be > n. This reduced execution time by 50 %.

f base n =
  map ((+) <$> fst <*> snd) $
  takeWhile ((<n).snd) $ dropWhile ((>=n).fst) $
  map ((quotRem (n^2)).(base^)) [1..]

kaprekar n base = (1:) $ filter (elem <*> f base) [1..n]

0nomat0 • Jul 15 '19 • Edited

~~Could I trouble someone to comment this code? Oops, I guess that's in the parent comment, thanks!~~ I don't know Haskell but I'm interested in it. Does this solution work for the 'strange' ones like 4789, 5292?

Trying to wrap my head around how to interpret the definition of a Kaprekar number, in a way that includes these oddball numbers.

I'm currently studying Ruby -- I'll necro-post my Ruby solution if I manage to get it working with the oddballs.