From: Robert Klemme <shortcutter@...>
Date: 2007-05-25T23:50:07+09:00
Subject: Re: Array troubles

On 25.05.2007 16:26, Hakusa@gmail.com wrote:
> I'm converting a Java calculator program I wrote into Ruby--which,
> just so you know, has been unbelievable with how quickly the language
> can be learned and implemented and how easy it made programming what
> had taken months of Java (but that was during school-time so
> development was, admittedly, very slow)--but I've run into a problem.
> 
> In a term, if you have q/q^2/4/8, it turns into q^-1/.5, but I'm
> getting q^-10! I've narrowed it down to one part of one function:
> division in Term#simplify.
> 
> There's a lot of code involved, so I'll only post relevant data. I
> REALLY do think you'll need any more than this. Also: when I say
> algebraic calculator, I mean that half the time you're working with
> letters (variables), not numbers.
> 
> Term#simplify
> 
>>> def simplify
>>>        count=0 # count will keep track of iterations.
>>>        temp = [] # This array will be our temp for @items.
>>>       @items.each do |item|
>>>
>>>            # Why check for this?
>>>            # The only time strings are use here is as an operator.
>>>            if item.class == String
>>>                # Defining these really just make the code easier for further on.
>>>                # They're the items before and after the operator.
>>>                num1 = @items[count-1]
>>>                num2 = @items[count+1]
>>>
>>>                # So if the operator is division...
>>>                if item.eql?('/')
>>>                    # You can't do much if the operands aren't of the same class.
>>>                    if num1.class.eql?( num2.class )
>>>                        puts 'temp pre ' + (temp).to_s
>>>                        # And use the / method of whatever class they are.
>>>                        puts num1.to_s + item + num2.to_s + " = " +
>>>                            (num1/num2).to_s

You don't need all the #to_s, instead you can do

print num1, item, num2, " = ", (num1/num2), "\n"

or use printf or puts with string interpolation.

>>>                        temp << num1/num2
>>>                        puts 'temp post ' + (temp).to_s
>>>                    end
>>>                end
>>>            end
>>>            count += 1
>>>        end
>>>        puts temp[1]
>>>        temp
>>>    end
> 
> With q/q^2/4/8 it prints out
> 
>>> temp pre
>>> q/q^2 = q^-1
>>> temp post q^-1
>>> temp pre q^-1
>>> 4/8 = 0
>>> temp post q^-10
>>> 0
> 
> Interestingly enough, 4/8=0. I don't know how the hell that works.

That's standard behavior for integer math in programming languages.  You 
might want to look into mathn / Rational:

irb(main):001:0> 4/8
=> 0
irb(main):002:0> require 'mathn'
=> true
irb(main):003:0> 4/8
=> 1/2
irb(main):004:0> (4/8).class
=> Rational
irb(main):005:0> (4/8).to_f
=> 0.5
irb(main):006:0> "%10f" % (4/8)
=> "  0.500000"

> Also, when it goes around for the second time to get 4/8, it puts it
> in the first place in the array (array[0]) and not the next space,
> which is where I thought this method is supposed to put it. ri must
> have this wrong too.
> 
> I don't think you need to know how I divide variables, but just in
> case you do:
> 
>>>    # Return the result of this instance and var1.
>>>    # Assumes that the two are like.
>>>    def /(var)
>>>        Variable.new(self.base, (self.exponent-var.exponent))
>>>    end
> 
> A variable is just a string base and integer exponent.
> 
> PS: I would accept some criticism as long as it's productive, but I do
> know that my code is lacking in certain areas (like how it doesn't
> check whether or not the variables have like bases), but it is a WIP.

Hope the math hint gets you started.

Kind regards

	robert