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MathGroup Archive 2006

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Re: Simplify UnitStep expressions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69316] Re: [mg69195] Simplify UnitStep expressions
  • From: Adam Strzebonski <adams at wolfram.com>
  • Date: Wed, 6 Sep 2006 04:29:02 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Andrzej Kozlowski wrote:
> 
> On 5 Sep 2006, at 16:20, Adam Strzebonski wrote:
> 
>> Andrzej Kozlowski wrote:
>>
>>> On 1 Sep 2006, at 11:41, L. Dwynn Lafleur wrote:
>>>
>>>> The following is transcribed from a Mathematica 5.2 notebook in   
>>>> Windows XP:
>>>>
>>>> In[1]:= Simplify[UnitStep[a-x/b], a-x/b > 0]
>>>> Out[1]= 1
>>>>
>>>> In[2]:= Simplify[UnitStep[a-Pi/b], a-Pi/b > 0]
>>>> Out[2]= UnitStep[a-Pi/b]
>>>>
>>>> Why does the second output different from the first?  I know it has
>>>> something to do with the fact that Pi is internally defined in   
>>>> Mathematica
>>>> because a similar result occurs Pi is replaced with E, but what   
>>>> logic is
>>>> being followed?
>>>>
>>>> -- 
>>>> ======================================
>>>>  L. Dwynn Lafleur
>>>>  Professor of Physics
>>>>  University of Louisiana at Lafayette
>>>>  lafleur at louisiana.edu
>>>> ======================================
>>>>
>>> Curiously, if you use FullSimplify rather then Simplify you will get:
>>> FullSimplify[UnitStep[a-Pi/b], a-Pi/b > 0]
>>> 1
>>> The same holds if Pi is replaces by E, or indeed by explicit   
>>> functions of E or Pi such as Pi^2, E^Pi etc. In all such cases   
>>> FullSimplify works but Simplify does not work. Strange.
>>> Andrzej Kozlowski
>>
>>
>> The cylindrical algebraic decomposition (CAD) algorithm used by  Simplify
>> to prove inference requires polynomial inequalities with rational  number
>> coefficients. a-x/b > 0 is equivalent to a polynomial inequality
>> -(a*b^2) + b*x < 0 which has rational number coefficients.
>> a-Pi/b > 0 is equivalent to a polynomial inequality -(a*b^2) + b*Pi  < 0
>> which has a numeric coefficient Pi which is not a rational number.
>>
>> Mathematica has two ways of dealing with nonrational numeric
>> coefficients in CAD. One is to replace each nonrational coefficient
>> with a new variable. This method always allows to decide inference
>> (modulo the ability to zero-test the exact numeric constants), but
>> it is potentially very expensive - CAD has a doubly exponential
>> complexity in the number of variables and we add a new variable for
>> each nonrational coefficient. The second method replaces nonrational
>> numeric coefficients with their approximations. This is much less
>> expensive, but in some cases it fails to decide inference.
>> Simplify uses the second method which in this case is insufficient.
>>
>> FullSimplify uses more transformations, and one of the additional
>> transformations succeeds.
>>
>> Best Regards,
>>
>> Adam Strzebonski
>> Wolfram Research
>>
>>
>>
> 
> Thanks for the explanation. I feel I should have guessed it, but  there 
> is one thing that still puzzles me. What exactly makes the  rational 
> approximation fail for Pi or E, since it seems to work fine  for 
> algebraic numbers such as Sqrt[2] or 3^(1/3)? It certainly can't  be 
> anything to do with Pi or E not being algebraic, so presumably it  is 
> something to do with the way the rational approximation is chosen?  This 
> sounds very interesting; could you explain the exact reason why  the 
> rational approximation in this case doe snot work and in what  other 
> cases will it not work in general? It sounds like the reason  might be 
> mathematically interesting (?).
> 
> Andrzej Kozlowski
> 

For Pi or E the assumption mechanism uses inexact (bignum)
approximations and constructs CAD with inexact sample point
coordinates. If we have an inequality f(X)<0 and we do not
find a cell with a sample point P for which f(P)<0, but we
do find a cell with a sample point P for which f(P) is
a bignum zero (for instance 0``20) then we cannot tell
whether f(X)<0 has any solutions or not.

For algebraic numbers the assumption mechanism does not use
approximations. It replaces the algebraic numbers with new
variables, because in this case it does not contribute that
much to the complexity. We do not need to compute projections
wrt. the new variables. Instead we make the variables last in
the projection ordering, and in the lifting phase we only lift
the one-point cell which corresponds to the new variables being
equal to the corresponding algebraic numbers.

Best Regards,

Adam Strzebonski
Wolfram Research





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