2
$\begingroup$

I want to define a function evaluate[f_[x__]] which acts on an arbitrary function f[a,b,c,d]. The function f takes any kind of argument and the function evaluate[f_[x__]] has to deal with the following properties

  • The function f is defined for specific values for a,b,c,d, which may be numbers, list, functions etc.. For example,consider the definitions f[1,2,3,4]=1 and f[1,2,p[3],p[4]]=2 where p[a_] are undefined functions/tensors. Notice the order matters.

I want the function evaluate returning the function f evaluated on the specific permutation for which I have given a definition of f itself. For example,

  • evaluate[f[3,4,1,2]] must return f[1,2,3,4]=1 since among the permutations of {3,4,1,2}, f is only defined on {1,2,3,4}.
  • evaluate[f[p[3],p[4],1,2]] must return f[1,2,p[3],p[4]]=2 since among the permutations of {p[3],p[4],1,2}, f is only defined on {1,2,p[3],p[4]}.
  • evaluate[f[x__]] must return False is multiple permutations are defined and if no permutations are matched.

Is there any intelligent and clever way to write a function with these properties?

Improve this question
$\endgroup$
7
  • $\begingroup$ Please clarify: do you want evaluate to permute arguments of f(or any function?) to a form for which term rewriting finds a rule for arguments of f? What if there are multiple permutations which are permissible? $\endgroup$ Commented May 31, 2019 at 17:35
  • $\begingroup$ @kirma I didn't understand your first question. I want evaluate to permute arguments of only the function f. It is supposed there are no multiple permutations which are permissible. I can add the requirement that if multiple permutations are permissible, return False. $\endgroup$ Commented May 31, 2019 at 17:39
  • $\begingroup$ OK. What if no permutation matches? $\endgroup$ Commented May 31, 2019 at 17:48
  • $\begingroup$ @kirma It may return False. Let me edit the question $\endgroup$ Commented May 31, 2019 at 17:55
  • $\begingroup$ In addition to my accepted answer, I would suggest taking a look at Orderless, that is SetAttribute[f, Orderless] for f. This doesn't handle no-matches or multiple-matches cases, but can auto-permute arguments when evaluating the function. $\endgroup$ Commented Jun 1, 2019 at 7:37

1 Answer 1

Reset to default
1
$\begingroup$

Maybe this is what you want:

ClearAll@evaluate; evaluate[f_[args___]] := With[{perms = Select[Unevaluated@*f @@@ Permutations[{args}], ValueQ]}, If[Length@perms == 1, perms[[1, 1]], False]];

(args are evaluated normally when evaluate is called.)

Now, let's define a function:

ClearAll@f; f[2, 4, 3, 1] = 3; f[a, b, c, d] = 1; f[b, c, d, a] = 1; f[5, x_?Positive] := x + 10;

This results:

evaluate[f[1, 2, 3, 4]]

3

evaluate[f[b, a, c, d]]

False

evaluate[f[foo, bar]]

False

evaluate[f[-1, 5]]

False

evaluate[f[1, 5]]

11

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.