Nifty free tool for brushing up on Ruby skills

[google...@jamesgagne.com]

Dear Rubyists (Rubyistas? Ruby fans? Rubyphiles?),
Teach Me to Code has a neat, quick screencast about rubykoans.com, a
nifty site for deepening one's knowledge of Ruby. You download a group
of files that run tests. The tests fail until you insert the correct
Ruby code.

Here's the screencast describing rubykoans.com:
http://teachmetocode.com/screencasts/ruby-koans/

[nate]

On the triangle problem, is there a better way to write

if a == b || b == c || c == a

I came up with

[a, b, c].all? { |side| side == a }

It's doesn't save any typing until you get past three values that need
to be compared

Nate

On Aug 13, 9:58 pm, "google_nos...@jamesgagne.com"

[nate]

Forgot to ask if anyone knows an alternative to this one:

if a == b || b == c || c == a

[Scott Mueller]

Hi Nate,

I'm not following the book club and don't know the triangle problem, but your answer below isn't correct.  You're only testing the values against a (including testing a against a).  If you're looking for a solution to test if any values are equal to each other in an arbitrary size set of values, you'll want to sort the values first and then compare them sequentially like this:

def anything_equal?(*values)
  values.sort.inject do |previous, current|
    if previous == current
      return true
    else
      current
    end
  end
  false
end

[Bruce]

On Aug 15, 11:22 pm, Scott Mueller <sc...@gig.io> wrote:
>
> def anything_equal?(*values)
>   values.sort.inject do |previous, current|
>     if previous == current
>       return true
>     else
>       current
>     end
>   end
>   false
> end
>

Since a set is a collection of unique values:

require 'set'
def anything_equal?(*values)
values.size != values.to_set.size
end

[Dan Loewenherz]

I prefer this solution. It uses the built-in combination method on the Array class.

values.combination(2).any? {|i,j| i==j}

Then you can monkey patch Array itself if you're really into this keystroke-saving thing.

class Array

  def anything_equal?

    self.combination(2).any? {|i,j| i==j}

  end

end

The result:

> [1,2,3].anything_equal?

=> false

> [1,1,3].anything_equal?

=> true

-Dan

[Scott Mueller]

I like the monkey patching.  Though that solution seems less efficient as you might have to compare all possible pairs of values.

[Scott Mueller]

That's pretty convenient!  I just looked at the source for Set's enum.to_set method.  Basically it adds the array's elements individually to a hash.  Certainly satisfies the original concern of saving typing.  But not so elegant given the different needs of a set and therefore inefficient underlying implementation.

On Mon, Aug 15, 2011 at 11:50 PM, Bruce <bruce...@gmail.com> wrote:

[Meng]

I placed all sides into an array

case [a, b, c].uniq
when 3
when 2
when 1
end

[Meng]

case [a, b, c].uniq.size

[Dan Loewenherz]

Actually, combination only shoots out distinct pairs.

> [1,2,3,4].combination(2).to_a

=> [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]

[Bruce Hobbs]

On Tue, Aug 16, 2011 at 6:31 AM, Meng <meng...@gmail.com> wrote:

case [a, b, c].uniq.size

That certainly simplifies it further:

def anything_equal?(*values)  values.size != values.uniq.size end

[Meng Fung]

if you want to keep track of sides and duplicates.

sides = [a, b, c].inject(Hash.new(0)) { |h, side| h[side] += 1; h}

case sides.size

when 3

when 2

when 1

end

[Scott Mueller]

Yes, I meant all possible unique pairs.  If there are no equal pairs, it will have to go through n*(n-1)/2 pairs.  Exponentially worse performance as the set gets bigger.

[nate]

The triangle problem is in the ruby koans code. Given three values,
the code is supposed to determine if the the triangle sides make a
scalene, isosceles or equilateral triangle, or if it's not a legal
triangle.

I meant to indicate that I wanted all values equal, not just any two
values.

a == b && b == c && c == a

I was hoping for parallel comparison, like a == b == c

I also needed the other answer, so thanks for all the replies.

On Aug 15, 11:22 pm, Scott Mueller <sc...@gig.io> wrote:

[Bruce Hobbs]

On Tue, Aug 16, 2011 at 10:44 AM, nate <nate.b...@gmail.com> wrote:

The triangle problem is in the ruby koans code. Given three values,
the code is supposed to determine if the the triangle sides make a
scalene, isosceles or equilateral triangle, or if it's not a legal
triangle.

I meant to indicate that I wanted all values equal, not just any two
values.

a == b && b == c && c == a

This is probably a stupid question, but isn't it sufficient to just check that the first two comparisons are true? If that's the case then there's another, admittedly longer, alternative:

a == b ? b == c : false 

I was hoping for parallel comparison, like a == b == c

Unfortunately the standard == operator/method returns a boolean and overriding that probably wouldn't be an optimal solution. If the sides are all Fixnums you could override a much less-used operator/method, say =~, for example:

class Fixnum

  def =~(b,c)

    [self,b,c].uniq.size == 1

  end

end

I can't figure out how to get the usual syntactic sugar, but this seems to work:

a.=~(b,c)

[nate]

It wasn't really about saving keystrokes, it was about making multiple
comparisons scalable.

Nate

[nate]

So my two favorite answers are:

"a == b ? b == c: false" and "case sides_array.uniq.size"

I wasn't thinking about the redundancy of checking a against c if the
first two were true. The case is great because it captures all three
triangle option with one comparison.

Nate

[nate]

Here's the class solution

class Triangle
def initialize(a, b, c)
@sides = [a, b, c]
end

def is_valid?
shortest, middle, longest = @sides.sort
shortest + middle > longest
end

def is_scalene?
@sides.uniq.count == 3
end

def is_isoceles?
@sides.uniq.count == 2
end

def is_equilateral?
@sides.uniq.count == 1
end

def triangle_type
return :invalid unless is_valid?
return :scalene if is_scalene?
return :isoceles if is_isoceles?
:equilateral
end
end