From: Robert Klemme <shortcutter@...>
Date: 2011-09-05T17:30:18+09:00
Subject: Re: Beginner needing help - Writing right-angle triangle program

On Sun, Sep 4, 2011 at 6:43 AM, Josh Cheek <josh.cheek@gmail.com> wrote:
> On Sat, Sep 3, 2011 at 11:23 PM, Kane Williams <theburrick@hotmail.com>wrote:
>
>> Thanks for the reply!
>>
>> I wanted to get something out of the way first, some other guy made a
>> thread exactly the same as my one and it has generated 6 replys somehow.
>> It is located here http://www.ruby-forum.com/topic/2497160#new although
>> you can probably just see it for yourself on the frontpage.
>>
>> We are in no way connected, in fact I don't really know any other
>> programmers :O. So I would assume this is a glitch in the system. I hope
>> there is some way to merge these two threads together.
>>
>> Now to reply to the questions:
>>
>> > Is this no good?
>> > [7, 7.058823529, 9.941176471]
>> >
>> > or this?
>> > [6.5, 7.542857143, 9.957142857]
>> >
>> >
>> > Are you requiring only integers?
>> > Can you show some code?
>>
>> Yea for simplicity's sake I just want full integers, although I don't
>> really mind either way.
>>
>> Here is the Haskell code:
>> ghci> let rightTriangles' = [ (a,b,c) | c <- [1..10], b <- [1..c], a <-
>> [1..b], a^2 + b^2 == c^2, a+b+c == 24]
>> ghci> rightTriangles'
>> [(6,8,10)]
>>
>> Which was found at the bottom of this tutorial:
>> http://learnyouahaskell.com/starting-out#tuples
>>
>> And I don't have any ruby code, that's why I'm posting here! :3 Just
>> wondering how I would start about to make a program which uses
>> Pythagoras Theorem.
>>
>> --
>> Posted via http://www.ruby-forum.com/.
>>
>>
> A direct translation of that would look like this:
>
> def right_triangles
>  results = []
>  (1..10).each do |c|
>    (1..c).each do |b|
>      (1..b).each do |a|
>        results << [a, b, c] if a+b+c == 24 && a**2 + b**2 == c**2
>      end
>    end
>  end
>  results
> end
>
> right_triangles # => [[6, 8, 10]]

I'd rather stick with a more traditional approach

for c in 1 .. perimeter / 2
  cq = c * c

  for b in 1 ... c
    a = perimeter - c - b
    p [a, b, c] if a * a + b * b == cq
  end
end

Note: the third side can be calculated from the first two sides so
iterating is not necessary for that.

Kind regards

robert


-- 
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/