From: sto.mar@...
Date: 2013-05-31T18:34:20+09:00
Subject: Re: Eloquent Ruby Kaprekar's Number

Am 31.05.2013 01:59, schrieb Alphonse 23:
> Which way is the eloquent ruby way to write this algorithm?
>
> Kaprekar Number
>
> 9 is a Kaprekar number since
> 9 ^ 2 = 81 and 8 + 1 = 9
>
> 297 is also Kaprekar number since
> 297 ^ 2 = 88209 and 88 + 209 = 297.
>
> In short, for a Kaprekar number k with n-digits, if you square it and
> add the right n digits to the left n or n-1 digits, the resultant sum is
> k.
>
> !Spoiler Below!
>
> # way one
> def kaprekar?(k)
>    ks = k**2
>    lenf = (ks.to_s.length)-1
>    lenh = (ks.to_s.length / 2) - 1
>    a = ks.to_s[0..lenh].to_i
>    b = ks.to_s[lenh+1..lenf].to_i
>    k == (a+b)
> end
>
> # way two
> def kaprekar?(k)
>    a = (k**2).to_s[0..((((k**2).to_s.length) / 2) - 1)].to_i
>    b = (k**2).to_s[(((((k**2).to_s.length) / 2) -
> 1)+1)..(((k**2).to_s.length)-1)].to_i
>    k == (a+b)
> end
>
> # way three
> def kaprekar?(k)
>    k == ((k**2).to_s[0..((((k**2).to_s.length) / 2) -
> 1)].to_i)+((k**2).to_s[(((((k**2).to_s.length) / 2) -
> 1)+1)..(((k**2).to_s.length)-1)].to_i)
> end
>
> def  simple_test(i)
>    if kaprekar?(i) == true then p "success on #{i}" else "fail on #{i}"
> end
> end
>
> simple_test(9)
> simple_test(297)
>
> All three pass my tests, but which of the three would be considered the
> most eloquent written ruby way? As in, which would be consider best
> style-wise? The first takes up the most lines, but uses less nesting.
> The last uses only one line but uses lots of nesting.

To me, as long as there are no performance considerations involved,
generally the best version is the one I understand in the shortest
amount of time. Cryptic one-liners might be nice as brain-teasers
but not for real applications.

Also, I highly prefer telling variable names over ks, a, b, etc.
Why not use square, digits, left_number, ... or similar?
IMO that make a big difference in readability.


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