From: tamouse mailing lists <tamouse.lists@...>
Date: 2013-05-31T11:20:23+09:00
Subject: Re: Eloquent Ruby Kaprekar's Number

On Thu, May 30, 2013 at 9:08 PM, tamouse mailing lists
<tamouse.lists@gmail.com> wrote:
> On Thu, May 30, 2013 at 6:59 PM, Alphonse 23 <lists@ruby-forum.com> wrote:
>> 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.
>>
>> --
>> Posted via http://www.ruby-forum.com/.
>>
>
> This is the way I'd do it:
>
> def kaprekar?(k)
>
>   n = k.to_s.length
>   ks = (k*k).to_s
>   l = ((n-1)-(n%2))
>   l = 0 if l < 0
>
>   nl = ks[0..l].to_i
>   nr = ks[-n..-1].to_i
>
>   k == nl + nr
>
> end
>
>
> I could not say if that's more rubyist or not, though. I imagine there
> is probably a way to use map/reduce and pass a block or something ;)
>

Well, that will teach me not to test more thoroughly.

def kaprekar?(k)

  n = k.to_s.length
  ks = (k*k).to_s
  l = ((n-1)-(ks.length%2))
  l = 0 if l < 0

  nl = ks[0..l].to_i
  nr = ks[-n..-1].to_i

  if k == nl + nr
    puts "k=#{k}, n=#{n}, ks=#{ks}, l=#{l}, nl=#{nl}, nr=#{nr}, nl+nr=#{nl+nr}"
    true
  else
    false
  end


end

and some tests:

irb(main):056:0> 0.upto(1000).each {|i| puts "#{i} is kaprekar" if kaprekar?(i)}
k=0, n=1, ks=0, l=0, nl=0, nr=0, nl+nr=0
0 is kaprekar
k=9, n=1, ks=81, l=0, nl=8, nr=1, nl+nr=9
9 is kaprekar
k=45, n=2, ks=2025, l=1, nl=20, nr=25, nl+nr=45
45 is kaprekar
k=55, n=2, ks=3025, l=1, nl=30, nr=25, nl+nr=55
55 is kaprekar
k=99, n=2, ks=9801, l=1, nl=98, nr=1, nl+nr=99
99 is kaprekar
k=297, n=3, ks=88209, l=1, nl=88, nr=209, nl+nr=297
297 is kaprekar
k=703, n=3, ks=494209, l=2, nl=494, nr=209, nl+nr=703
703 is kaprekar
k=999, n=3, ks=998001, l=2, nl=998, nr=1, nl+nr=999
999 is kaprekar