OpenEuphoria

1. final message

• Posted by bill May 24, 2012
• 1721 views

I don't believe you guys will ever take any notice of what an outsider has to say, but..

These are thingns of some consequence:

Primary facts:

1 There is no multiplicative inverse of zero.
2 A Greatest Common Divisor has to:
. 1 be a divisor. ie it cannot be equal to zero.
. 2 be greater than any other divisor.

. Defining GCD(0,0) = 0 doesn't satisfy either
. condition 2.1 or 2.2.

No mathematical theory defines GCD(0,0) = 0.

The lattice theory article referenced defines
GCD(0,0) = 1 on the grounds that every integer is
divisible by 1. It doesn't satidfy 2.2.

Practical:

GCD(0,0) is given as 0 by a number of programming
languages because the Knuth algorithm would give
that result. Haskell gives it as undefined.

If used as a divisor it will likely produce an
exception or error. This would not necessarily occur
near where the exceptional result was produced.

There is, of course, the other difficulty that the
language may not have a value to return for
'not an integer'.

0/0:

Not indeterminate. See 1.
Can be defined to be any number with the consequence
of talking nonsense.

intdiv(-8,-3):

Dividing -8 envelopes into sets of maximum -3 is easy
1 set holds -8 envelopes. Not what you wanted?
-1 set holds 8 envelopes? - No you can't do that!
Perhaps I am wrong and you can do this. Show me!

Divide -8 envelopes into -3 sets, not so easy
-2 sets hold 3 envelopes, -1 set holds 2 envelopes or
-4 sets hold 2 envelopes, 1 set holds 0 envelopes.

At some time you guys are going to have to admit that
you are seriously talking shit.

3. Re: final message

• Posted by jimcbrown (admin) May 24, 2012
• 1524 views

Obviously true for basic arithmetic and so on. Simply false for certain mathematical systems (the trivial ring, wheel groups, etc.).

I admit that I'm skeptical with the broad claim you make that no mathematical theory defines GCD(0,0) = 0.

As a practical solution, GCD(0,0) = 0 makes sense to me. If nothing else, it's a simple and elegant way to signal to the caller that 0 was passed into GCD().

I'd imagine that this would be the case most of the time. I can imagine a future version of the language (5.0 maybe) where, with strict_math turned on, an exception is thrown when GCD(0, 0) is called so these errors are caught immediately in programs that need it.

Not sure if this complies with the IEEE spec, but we could always just return NaN.

I think you assume is true what you need to prove. 0/0 is indeterminate, as I have demonstrated elsewhere on this forum.

Or the trivial ring. Or as a simplification of the binomial theorem.

I think I can make a better case against you. Despite my constant and patient replies, you haven't as much as acknowledged my existence. Not to mention your constant attempts to belittle us (you could not think of a more politically correct term than "talking shit" ? Normally, I am sadden to see anyone leave these forums. However, considering your use of language and the misinformation you are trying to put out here, I am forced to admit that I'm quite glad to see you go.

5. Re: final message

• Posted by mattlewis (admin) May 24, 2012
• 1523 views

When I think of integer division, I'm reminded of my elementary school adventures in long division, and the answer should be the quotient, discarding the remainder. Looking at it, it's not immediately obvious to me why (or what for) I'd want to use math:intdiv(). This may simply be conceptual blindness due to my familiarity with integer division as common in other programming languages.

Maybe we could have a version that returns { quotient, remainder }.

I would assume that it would use floor() to go from floating point to integers, and that there should be no need for special sign manipulation.

My expectation (just given the name) would be an implementation of:

public function intdiv(object a, object b)	 
	return floor( floor(a)/ floor(b) ) 
end function 

...and then maybe...

public function intdivr(object a, object b)	 
        object c = floor( floor(a)/ floor(b) ) 
	return { c, a - (b*c) } 
end function 

Matt

7. Re: final message

• Posted by DerekParnell (admin) May 24, 2012
• 1494 views

public function intdiv(object a, object b)	 
	return floor( floor(a)/ floor(b) ) 
end function 

public function intdivr(object a, object b)	 
        object c = floor( floor(a)/ floor(b) ) 
	return { c, a - (b*c) } 
end function 

And mine would more like ...

public function intdiv(object a, object b, integer behave = 1) 
    object t 
    t = abs(a) / abs(b) 
    switch behave do 
       case -1 then 
           return floor(t) 
       case 0 then 
           return round(t) 
       case 1 then 
           return ceil(t) 
       case else 
           fail("Unknown behaviour requested") 
    end switch 
end function 
? intdiv(-8, -3, 1) --> 3 
? intdiv(-8, -3, 0) --> 3 
? intdiv(-8, -3,-1) --> 2 

9. Re: final message

• Posted by DerekParnell (admin) May 25, 2012
• 1446 views

intdiv(-8, 3, 0) --> -2 

Ok, I'm not wedded to abs() in this case. So how about ...

public function intdiv(object a, object b, integer behave = 1) 
    object t 
    t = abs(a) / abs(b) 
    switch behave do 
       case -1 then 
           t = floor(t)  -- Move towards zero. 
       case 0 then 
           t = round(t)  -- Move either way to round 
       case 1 then 
           t = ceil(t)   -- Move away from zero 
       case else 
           fail("Unknown behaviour requested") 
    end switch 
 
    return t * sign(a) * sign(b)  -- Adjust sign for each element in arguments. 
end function 
? intdiv({-8,8}, -3, 1) --> {3,-3} 
? intdiv({-8,8}, -3, 0) --> {3,-3} 
? intdiv({-8,8}, -3,-1) --> {2,-2}