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Re: Pattern matching problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg88539] Re: [mg88516] Pattern matching problem
*From*: "W_Craig Carter" <ccarter at mit.edu>
*Date*: Thu, 8 May 2008 04:11:28 -0400 (EDT)
*References*: <200805071107.HAA15035@smc.vnet.net>
Hello Charlie,
I have an inelegant solution. I'll put it in steps so that you can see
what it is doing:
(*representative data*)
poly =1 + u[x] +
Sum[Product[
D[u[x], {x, RandomInteger[{1, 6}]}], {i, 1,
RandomInteger[{1, 4}]}], {4}]
InputForm[poly]
(*Take each termp in poly and turn it into a list, @@@ applies list a
first level*)
temp = Flatten[Apply[List, Flatten[List @@@ poly]]]
(*Replace each derivative it its order, so that powers are counted*)
temp2 = temp /. Derivative[n_][u][x] :> n
(*take the product of all the integers*)
Times @@ Cases[temp2, _Integer]
You could put it all together and "make it look" more elegant and
(even) more unreadable...
Craig
On Wed, May 7, 2008 at 7:07 AM, Charlie Brummitt <cbrummitt at wisc.edu> wrote:
> Hi all,
> Here is my problem: Given a polynomial in the variables u[x,t] and its
> spatial derivatives (for example, the polynomial could be 1 + u +
> u_xx*u_xxx^2), count the number of spatial derivatives with multiplicity.
> That is, after writing the above polynomial as
>
>
> 1 + u + u_xx * u_xxx * u_xxx
>
> the output should be 2 + 3 + 3 (basically, you count the number of x's).
>
> I have tried implementing this using a pattern matching approach. Here is
> what I tried:
>
> f[equation_ ] := Sum[ k * j * Count[ equation , D[ u[x, t], {x, j} ] ^ k ,
> {0, \infinity} ], {j, 1, 50}, {k, 1, 50}]
>
> This fails to work on, for example, the input u_xx^2, because it outputs 6
> when it should output 4. This is because the u_xx is counted (+2 to the
> sum), and the u_xx^2 is counted (+4 to the sum). This is because the u_xx is
> nested inside the Power[ , 2] in its representation in Mathematica and so
> it gets counted too many times in my formula. I can't seem to figure out a
> way to use the "provided that" operator /; to make this formula work.
>
> I've also tried doing some replacement methods, but to no success.
>
> Thanks for any help you may be able to provide.
>
> -Charlie
>
>
>
--
W. Craig Carter
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