From: Josh Cheek <josh.cheek@...>
Date: 2013-05-31T12:14:58+09:00
Subject: Re: Eloquent Ruby Kaprekar's Number

--001a11c2af3ad612dc04ddfb0458
Content-Type: text/plain; charset=ISO-8859-1

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/.
>
>

I'd write it like this:

def kaprekar?(num)
  return false if num.zero? || num.to_s =~ /^10+$/ # IDK why 0 and 10.. are
disallowed when 1 is allowed, but w/e
  square   = (num * num).to_s
  midpoint = square.size / 2
  (0..midpoint).any? do |separation_index|
    first  = square[0...separation_index].to_i
    second = square[separation_index..-1].to_i
    num == first + second
  end
end

RSpec.configure { |config| config.fail_fast = true }

# numbers taken from http://en.wikipedia.org/wiki/Kaprekar_number
describe 'kaprekar number' do
  def check_kaprekars(kaprekars)
    (0..kaprekars.last).select { |n| kaprekar? n }.should == kaprekars
  end

  example 'quick match' do
    check_kaprekars [1, 9, 45, 55, 99, 297, 703, 999]
  end

  example 'medium match' do
    check_kaprekars [1, 9, 45, 55, 99, 297, 703, 999 , 2223, 2728, 4879,
4950, 5050, 5292, 7272, 7777, 9999]
  end

  example 'these should match' do
    check_kaprekars [1, 9, 45, 55, 99, 297, 703, 999 , 2223, 2728, 4879,
4950, 5050, 5292, 7272, 7777, 9999 , 17344, 22222, 38962, 77778, 82656,
95121, 99999, 142857, 148149, 181819, 187110, 208495, 318682, 329967,
351352, 356643, 390313, 461539, 466830, 499500, 500500, 533170]
  end
end

--001a11c2af3ad612dc04ddfb0458
Content-Type: text/html; charset=ISO-8859-1
Content-Transfer-Encoding: quoted-printable

On Thu, May 30, 2013 at 6:59 PM, Alphonse 23 <span dir=3D"ltr">&lt;<a href=
=3D"mailto:lists@ruby-forum.com" target=3D"_blank">lists@ruby-forum.com</a>=
&gt;</span> wrote:<br><div><div class=3D"gmail_quote"><blockquote class=3D"=
gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-=
left:1ex">
Which way is the eloquent ruby way to write this algorithm?<br>
<br>
Kaprekar Number<br>
<br>
9 is a Kaprekar number since<br>
9 ^ 2 =3D 81 and 8 + 1 =3D 9<br>
<br>
297 is also Kaprekar number since<br>
297 ^ 2 =3D 88209 and 88 + 209 =3D 297.<br>
<br>
In short, for a Kaprekar number k with n-digits, if you square it and<br>
add the right n digits to the left n or n-1 digits, the resultant sum is<br=
>
k.<br>
<br>
!Spoiler Below!<br>
<br>
# way one<br>
def kaprekar?(k)<br>
=A0 ks =3D k**2<br>
=A0 lenf =3D (ks.to_s.length)-1<br>
=A0 lenh =3D (ks.to_s.length / 2) - 1<br>
=A0 a =3D ks.to_s[0..lenh].to_i<br>
=A0 b =3D ks.to_s[lenh+1..lenf].to_i<br>
=A0 k =3D=3D (a+b)<br>
end<br>
<br>
# way two<br>
def kaprekar?(k)<br>
=A0 a =3D (k**2).to_s[0..((((k**2).to_s.length) / 2) - 1)].to_i<br>
=A0 b =3D (k**2).to_s[(((((k**2).to_s.length) / 2) -<br>
1)+1)..(((k**2).to_s.length)-1)].to_i<br>
=A0 k =3D=3D (a+b)<br>
end<br>
<br>
# way three<br>
def kaprekar?(k)<br>
=A0 k =3D=3D ((k**2).to_s[0..((((k**2).to_s.length) / 2) -<br>
1)].to_i)+((k**2).to_s[(((((k**2).to_s.length) / 2) -<br>
1)+1)..(((k**2).to_s.length)-1)].to_i)<br>
end<br>
<br>
def =A0simple_test(i)<br>
=A0 if kaprekar?(i) =3D=3D true then p &quot;success on #{i}&quot; else &qu=
ot;fail on #{i}&quot;<br>
end<br>
end<br>
<br>
simple_test(9)<br>
simple_test(297)<br>
<br>
All three pass my tests, but which of the three would be considered the<br>
most eloquent written ruby way? As in, which would be consider best<br>
style-wise? The first takes up the most lines, but uses less nesting.<br>
The last uses only one line but uses lots of nesting.<br>
<span class=3D"HOEnZb"><font color=3D"#888888"><br>
--<br>
Posted via <a href=3D"http://www.ruby-forum.com/" target=3D"_blank">http://=
www.ruby-forum.com/</a>.<br>
<br>
</font></span></blockquote></div><br></div><div><br></div><div>I&#39;d writ=
e it like this:<div><br></div><div><div>def kaprekar?(num)</div><div>=A0 re=
turn false if num.zero? || num.to_s =3D~ /^10+$/ # IDK why 0 and 10.. are d=
isallowed when 1 is allowed, but w/e</div>
<div>=A0 square =A0 =3D (num * num).to_s</div><div>=A0 midpoint =3D square.=
size / 2</div><div>=A0 (0..midpoint).any? do |separation_index|</div><div>=
=A0 =A0 first =A0=3D square[0...separation_index].to_i</div><div>=A0 =A0 se=
cond =3D square[separation_index..-1].to_i</div>
<div>=A0 =A0 num =3D=3D first + second</div><div>=A0 end</div><div>end</div=
><div><br></div><div>RSpec.configure { |config| config.fail_fast =3D true }=
</div><div><br></div><div># numbers taken from <a href=3D"http://en.wikiped=
ia.org/wiki/Kaprekar_number">http://en.wikipedia.org/wiki/Kaprekar_number</=
a></div>
<div>describe &#39;kaprekar number&#39; do</div><div>=A0 def check_kaprekar=
s(kaprekars)</div><div>=A0 =A0 (0..kaprekars.last).select { |n| kaprekar? n=
 }.should =3D=3D kaprekars</div><div>=A0 end</div><div>=A0=A0</div><div>=A0=
 example &#39;quick match&#39; do</div>
<div>=A0 =A0 check_kaprekars [1, 9, 45, 55, 99, 297, 703, 999]</div><div>=
=A0 end</div><div>=A0=A0</div><div>=A0 example &#39;medium match&#39; do</d=
iv><div>=A0 =A0 check_kaprekars [1, 9, 45, 55, 99, 297, 703, 999 , 2223, 27=
28, 4879, 4950, 5050, 5292, 7272, 7777, 9999]</div>
<div>=A0 end</div><div>=A0=A0</div><div>=A0 example &#39;these should match=
&#39; do</div><div>=A0 =A0 check_kaprekars [1, 9, 45, 55, 99, 297, 703, 999=
 , 2223, 2728, 4879, 4950, 5050, 5292, 7272, 7777, 9999 , 17344, 22222, 389=
62, 77778, 82656, 95121, 99999, 142857, 148149, 181819, 187110, 208495, 318=
682, 329967, 351352, 356643, 390313, 461539, 466830, 499500, 500500, 533170=
] =A0 =A0</div>
<div>=A0 end</div><div>end</div></div></div>

--001a11c2af3ad612dc04ddfb0458--