Re: Simplify

*To*: mathgroup at smc.vnet.net*Subject*: [mg98677] Re: [mg98659] Simplify*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Thu, 16 Apr 2009 04:13:23 -0400 (EDT)*References*: <200904150902.FAA08151@smc.vnet.net> <ACCF1343-E098-469B-A599-4CA4F9E1A5E0@mimuw.edu.pl> <49E5CA92.7030005@metrohm.com>

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> >

**References**:**Simplify***From:*dh <dh@metrohm.com>