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Re: Simplify

  • To: mathgroup at smc.vnet.net
  • Subject: [mg98678] Re: [mg98659] Simplify
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 16 Apr 2009 04:13:34 -0400 (EDT)
  • References: <200904150902.FAA08151@smc.vnet.net> <ACCF1343-E098-469B-A599-4CA4F9E1A5E0@mimuw.edu.pl> <49E5CA92.7030005@metrohm.com> <7379A405-70A4-404B-8ACB-B4367F9A5A2A@mimuw.edu.pl>

On 15 Apr 2009, at 22:13, Andrzej Kozlowski wrote:

> The reason for the problem is the following "canonical form":   
> expressions like  (a^k1*b^k2*c^k3...)^z *(a*l1*b^l2*...) where a, b,  
> c are positive are re-written as a^(k1+l1)*b^(k2+l2)*....
>
> For example
>
> (2^u*5^v)*(2*3^w*5^z)
> 2^(u + 1) 3^w 5^(v + z)


I did not state this "rule" correctly. The rule is that in products of  
powers of integers are factored, with the factors being powers of  
primes.

Andrzej





> Yes, you are right but I still think it is beside the point.  
> Consider this rather more striking (in my opinion) example.
>
> let
>
> expr = 2^(z + 1)*3^z
>
> You can easily see that this can be re-written as 2*6^z and that:
>
> LeafCount[Hold[2*6^z]]
> 6
>
> while
>
> LeafCount[expr]
> 9
>
> so  2*6^z has a much smaller LeafCount than expr, but you can never  
> Simplify expr to 2*6^z because the Evaluator will always rewrite it  
> again as expr. This has a rather unpleasant consequence:
>
> FullSimplify[2*6^z - 6^z]
> 2^(z + 1)*3^z - 6^z
>
> Simplify cannot see that the simplest answer if 6^z (no matter what  
> ComplexityFunction you use) because immediately on evaluation 2*6^z  
> is converted to expr above and then it is too late; Simplify can't  
> see that the first term is simply twice the second term. Yet  
> Simplify has no problem noticing that:
>
> Simplify[2*6^z - 6^z == 6^z]
> True
>
> The reason for the problem is the following "canonical form":   
> expressions like  (a^k1*b^k2*c^k3...)^z *(a*l1*b^l2*...) where a, b,  
> c are positive are re-written as a^(k1+l1)*b^(k2+l2)*....
>
> For example
>
> (2^u*5^v)*(2*3^w*5^z)
> 2^(u + 1) 3^w 5^(v + z)
>
> (actually the rule used is note general, but I just want to give an  
> example of a "canonical form" that causes trouble).
>
> Since these reductions are made by the Evaluator, Simplify has no  
> effect on them. But because of them we get the following  
> inconsistent behaviour:
>
> 2*5^z - 5^z
> 5^z
>
> but
>
> 2*6^z - 6^z
> 2^(z + 1)*3^z - 6^z
>
>
> This, of course, cannot be changed by applying Simplify with any  
> ComplexityFunction because canonical forms can't be changed by  
> Simplify (unless you apply Hold).
>
> Andrzej
>
>
>
>
>
>
>
>
> On 15 Apr 2009, at 20:52, dh wrote:
>
>> Hi Andrzej,
>> thanks for your helpfull explanations.
>> Let me just mention that there is an additional reason, internally  
>> the expression x^2/y^2 is rewritten and has the the same leaf count  
>> as (x/y)^2. Consider:
>> FullForm[Hold[x^2/y^2]]
>> FullForm[x^2/y^2]
>> Daniel
>>
>>
>> Andrzej Kozlowski wrote:
>>> Actually, this point has been explained many times (by me ;-))
>>> (I like to think of Mathematica's evaluation process in terms of  
>>> something called "The Evaluator", which I think I first found in  
>>> David Wagner's book "Power Programming with Mathematica". I think  
>>> it is only an abstraction, along with "the Parser", "the  
>>> Typesetter" etc, but a convenient one when one is thinking about  
>>> the evaluation process. )
>>> The "Evaluator" always evaluates
>>> (x/y)^2
>>> to
>>> x^2/y^2
>>> This happens before Simplify takes any effect. Even if Simplify  
>>> converted x^2/y^2 to (x/y)^2  the Evaluator would kick in and  
>>> again convert  it back to x^2/y^2. Since the Evaluator always  
>>> overrides Simplify there is no way to get (x/y)^2 as the output  
>>> without using Hold.
>>> Perhaps you are asking why Mathematica (or the "Evaluator")  
>>> automatically converts (x/y)^2 to  x^2/y^2. It's because of  
>>> something called "canonical forms" or "standard forms". Basically,  
>>> in order to optimize performance in computer algebra systems one  
>>> want to reduce to 0 as quickly as possible as many expressions  
>>> that are actually equal. The earlier you do this the better the  
>>> performance. If you allow expressions to contain a large number of  
>>> subexpressions that are actually 0 until you apply Simplify, it  
>>> may very seriously impair performance. "Canonical forms" (or  
>>> "normal forms", there is a slight difference between them but I  
>>> shall ignore it) are certain unique forms to which various  
>>> expressions are reduced automatically by the Evaluator (before  
>>> applying Simplify). This has the effect causing cancellations to  
>>> occur early. Also, these "canonical forms" have to be independent  
>>> of any particular ComplexityFunction used, hence the reduction has  
>>> to be performed outside Simplify. The advantage of using canonical  
>>> forms independent of ComplexityFunction is that they often enable  
>>> Mathematica to identify two expressions as equal even if Simplify  
>>> can't fine a sequence of Complexity reducing transformations that  
>>> will convert one expression into the other.
>>> Not surprisingly, using "canonical forms" can sometimes produce  
>>> undesirable side-effects and this is one of them (a rather minor  
>>> one, worse ones do occur).
>>> Andrzej Kozlowski
>>> Andrzej Kozlowski
>>> On 15 Apr 2009, at 18:02, dh wrote:
>>>>
>>>>
>>>> Hi,
>>>>
>>>> can somebody explain, why
>>>>
>>>> Simplify[x^2/y^2,ComplexityFunction->LeafCount]
>>>>
>>>> does not simplify to (x/y)^2, although the LeafCount is:
>>>>
>>>> LeafCount[Hold[x^2/y^2]] gives 10
>>>>
>>>> and
>>>>
>>>> LeafCount[Hold[(x/y)^2]] gives 8
>>>>
>>>>
>>>>
>>>> Daniel
>>>>
>>>>
>>>>
>>
>>
>> -- 
>>
>> Daniel Huber
>> Metrohm Ltd.
>> Oberdorfstr. 68
>> CH-9100 Herisau
>> Tel. +41 71 353 8585, Fax +41 71 353 8907
>> E-Mail:<mailto:dh at metrohm.com>
>> Internet:<http://www.metrohm.com>
>>
>



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