       Re: Pattern matching problem

• To: mathgroup at smc.vnet.net
• Subject: [mg88565] Re: Pattern matching problem
• From: Szabolcs Horvát <szhorvat at gmail.com>
• Date: Thu, 8 May 2008 04:16:15 -0400 (EDT)
• Organization: University of Bergen
• References: <fvs2kf\$eo4\$1@smc.vnet.net>

```Charlie Brummitt 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.
>
>

One possibility is replacing the Power[] with an explicit product:

In:= ee=1+u[x,t]+D[u[x,t],{x,2}] D[u[x,t],{x,3}]^2
Out= 1+u[x,t]+(u^(2,0))[x,t] (u^(3,0))[x,t]^2

In:=
ee1 = ee /. expr_^n_Integer /; n>1 :>
(HoldForm@Times[##]&) @@ ConstantArray[expr,n]

Out= 1+u[x,t]+((u^(3,0))[x,t] (u^(3,0))[x,t]) (u^(2,0))[x,t]

In:= Cases[ee1, HoldPattern[Derivative[n_,_][u][__]] :> n, Infinity]
Out= {3,3,2}

In:= Plus@@%
Out= 8

Just a consistency check:

In:= ReleaseHold[ee1] == ee

Out= True

```

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