Re: Re: Re: how to find n in expression x^n using a pattern?

*To*: mathgroup at smc.vnet.net*Subject*: [mg58536] Re: [mg58512] Re: [mg58487] Re: [mg58426] how to find n in expression x^n using a pattern?*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Wed, 6 Jul 2005 03:11:21 -0400 (EDT)*References*: <200507020806.EAA01589@smc.vnet.net> <200507040624.CAA05758@smc.vnet.net> <26A87A3E-2A81-483F-B880-FD4E30020592@gmail.com> <42C956F0.6090509@umbc.edu> <200507050557.BAA29462@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Andrzej Kozlowski wrote: > On 5 Jul 2005, at 00:34, Pratik Desai wrote: > > >>Andrzej Kozlowski wrote: >> >> >> >>>On 4 Jul 2005, at 15:24, Pratik Desai wrote: >>> >>> >>> >>>>steve wrote: >>>> >>>> >>>> >>>> >>>>>I am learning patterns in Mathemtica, and I thought this will >>>>>be easy. >>>>> >>>>>suppose you are given a factored expression, such as x^n(x^2+2) >>>>>(x^3 +4) >>>>>where 'n' above can be any number, including zero (i.e. there is no >>>>>(x) as a separate factor). I need to find 'n', which can be >>>>>0,1,2,3,..etc.. >>>>> >>>>>i.e I need to find the power of the factor x by itself if it >>>>>exist in >>>>>the expression. not (x^n+anything), just (x^n) byitself. >>>>> >>>>>For example >>>>> >>>>>p= x (x^2+2) ---> here n=1 >>>>> >>>>>I tried this >>>>> >>>>>p /. x^n_(any_) -> n >>>>> >>>>>This did not work since x^1 does not match pattern x^n_. I >>>>>would have >>>>>worked if I had x^2 (x^3+2) for example. >>>>> >>>>>I know that I need to use something like x^_:1 to make it >>>>>match x, >>>>>which >>>>>is really x^1, but I do not know how to use it in a replacement >>>>>rule, >>>>>becuase >>>>>when I write >>>>> >>>>>p /. (x^n_:1)(any_) -> n >>>>> >>>>>it is an error. >>>>> >>>>>If I write >>>>> >>>>>p /. (x^_:1)(any_) -> _ >>>>> >>>>>I do not get back anything. >>>>> >>>>>Next I tried the Cases command, and again had no luck. >>>>> >>>>>The main problem is that Mathematica makes a difference between >>>>>x, x^1, and x^2, etc... when it comes to matching with pattern x^n_ >>>>> >>>>>any idea how to do this? >>>>>thanks, >>>>>steve >>>>> >>>>> >>>>> >>>>> >>>>> >>>> >>>>Why not do something like this , it sounds quite simplistic but >>>>it works :) >>>>Clear[p] >>>>p[x_] = x ^5(x^3 + 2x^2 + 3x - 13) >>>>sol2 = Solve[p[x] == 0, x] >>>>y = x /. sol2 >>>>n = Count[y, 0] >>>> >>>>Best regards >>>> >>>>-- >>>>Pratik Desai >>>>Graduate Student >>>>UMBC >>>>Department of Mechanical Engineering >>>>Phone: 410 455 8134 >>>> >>>> >>>> >>>> >>> >>>In fact after seeing this posting I understood what the original >>>poster wanted (probably). (He wrote that he was learning patterns >>>so I thought this was meant just to be an exercise in pattern >>>matching). However, using algebraic method, like the one above, >>>could be unnecessarily complicated if the second factor of the >>>equation is a high degree polynomial or something even more >>>complicated. It is in fact easier to use patterns. Compare: >>> >>>p[x_] = x^5*(x^32 + 2x^21 + 3x^17 - 13); >>> >>> >>>Timing[Cases[p[x], x^(n_.) -> n, {1}]] >>> >>> >>>{0.0005019999999973379*Second, {5}} >>> >>>(as I wrote in my first response, giving the correct level >>>specification is important). >>> >>> >>>Timing[Count[x /. Solve[p[x] == 0, x], 0]] >>> >>> >>>{0.08699899999999872*Second, 5} >>> >>> >>> >>>Andrzej Kozlowski >>> >>>Chiba, Japan >>> >>> >>> >> >>Hi Andrzej >> >>I have had occasion to use patterns before and by far I have found >>it the most difficult feature of mathematica to work with and the >>help on it leaves a lot to be desired. I agree the poster seems to >>be trying to learn patterns, I was just trying to point out, as you >>have on your earlier reply, that there are easier ways to do the >>same job (i.e finding the highest common exponent in x). Again not >>to put a finer point on it, it seems to me that the use of >>patterning as you have shown seems only to work when the expression >>is in its factored form (which I admit was the posters original >>query) whereas using Solve and count works irrespective of the >>form of the expression. Or am I wrong? > > > You are of course right. Mathematica's pattern matching is (with some > rudimentary exceptions) only syntactic, which means that expressions > are matched "literally" and not according to their "mathematical > meaning" (that would be "semantic" pattern matching). A factored > expression and an unfactored oen are quite different as far as > pattern matching is concerned. Algebraic methods can do much more > than pattern matching but they tend to be very slow and suitable > mostly for "small" problems. So both approaches have their strong > and weak points and one or the other should be preferred depending on > what one is trying to do. Obviously there is no panacea for all types > of situations, which is why posters should try to state exactly what > they are really trying to do. > > Andrzej Kozlowski > > > > > > >>p[x_] = x^5*(x^32 + 2x^21 + 3x^17 - 13x^5); >>Timing[Cases[p[x], x^(n_.) -> n, {1}]] >> >>>>{0. Second, {5}} >> >>Timing[Count[x /. Solve[p[x] == 0, x], 0]] >> >>>>{0.01 Second, 10} >> >>Now if you use simplify >>p[x_] = x^5*(x^32 + 2x^21 + 3x^17 - 13x^5) // Simplify; >>Timing[Cases[p[x], x^(n_.) -> n, {1}]] >>{0. Second, {10}} >> >>Best regards >>Pratik >> >>-- >>Pratik Desai >>Graduate Student >>UMBC >>Department of Mechanical Engineering >>Phone: 410 455 8134 A method that is reasonably fast but (for most purposes) more robust than syntactic approaches is to use Exponent with a third argument. In[3]:= Exponent[x^5*(x^32 + 2x^21 + 3x^17 - 13x^5), x, Min] Out[3]= 10 Since the topic was broached, there was some Mathematica work on semantic pattern matching a few years back. You might want to look at Jason Harris. Semantica: semantic pattern matching in Mathematica. The Mathematica Journal 7(3):329-360. Further information about the article may be found at http://library.wolfram.com/infocenter/Articles/3902/ Daniel Lichtblau Wolfram Research

**References**:**how to find n in expression x^n using a pattern?***From:*"steve" <nma124@hotmail.com>

**Re: how to find n in expression x^n using a pattern?***From:*Pratik Desai <pdesai1@umbc.edu>

**Re: Re: how to find n in expression x^n using a pattern?***From:*Andrzej Kozlowski <akozlowski@gmail.com>