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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