Re: generate all possible math expr of one term

On 15 Mai, 01:49, "xah...@xxxxxxxxx" <xah...@xxxxxxxxx> wrote:
Here's a example of Expressiveness of a Language.

The following is Mathematica code that generates all possible
equations of one term involving trig function. (tweak the funList and
nesting level to define what “all possible” means. if nesting level is
2, it takes about 20 minutes and returns a list of 2876 terms on a
2004 personal computer.

<< DiscreteMath`Combinatorica`
funList = {Sin, Tan, Power[#, -1] &};
Nest[Module[{li = #},
 (Union[#, SameTest -> (Module[{arg1 = #1, arg2 = #2},
   If[(*both expression contains both x and y*)
     And @@ (((((StringMatchQ[#, "*x*"] &&
     StringMatchQ[#, "*y*"]) &)@
     ToString@#) &) /@ {arg1, arg2})
     , SameQ[arg1, arg2 /. {x -> y, y -> x}],
     SameQ[arg1, arg2]]
   ] &)] &)@
   Union@Flatten@(Table[(Times @@ # &) /@ KSubsets[#, i], {i, 1,
   2}] &)@Flatten@{li, Outer[#1@#2 &, funList, li]}
 ] &, {x, y}, 1];
Select[%, (StringMatchQ[ToString@#, "*x*"] &&
 StringMatchQ[ToString@#, "*y*"]) &]

The problem is this: generate a list of all possible math expressions
using the following combination and rules:

• the math expression involves both x and y. (must have both present)
• you can use any of the 6 trig functions (you must, since the goal is
to create all possibilities)
• The binary operations you can use are multiply, divide. (no addition
or substraction, since that can make the result 2 terms)
• a sub-expression (such as x or y) can be replaced by a more
complicated one. (i.e. you can nest)

For example, these are first few items from the above code:

{1/(x^2*y^2), 1/(x*y^2), x/y^2, 1/(x*y), x/y, x^2/y, x*y, x^2*y,
x^2*y^2, Cos[x]/y^2, Cos[x]/y, Cos[x]/(x*y), (x*Cos[x])/y, y*Cos[x],
(y*Cos[x])/x, x*y*Cos[x], y^2*Cos[x], Cos[x]*Cos[y], Cot[x]/y^2,

For a gallery of selection plots of these equations, see

The above i wrote in 2002. If there are requests, i'll translate the
above code into emacs lisp. The result lisp expression should match
Mathematica's, token for token. (the code make a lot use of nested
lambda and or apply and or map) If you are interested, you could
translate the above into lisp too, it's not difficult (though the
number of lines will increase maybe 10 fold. And if Common Lisp
doesn't have combinatorics library providing KSubsets, and also since
CL doesn't have Outer, so the above in CL might be few hundred lines).
(see here for a example of how to:

PS as a after-thought, i decided to post this to perl, python, and
java too. This will take about the same number of lines in perl as in
Common Lisp. Probably llightly more in Python due to syntax. In Java,
it will be one million lines.

Gratuitous poem of the day:

in the climb to geekdom,
you have few rungs to catch,
before you see my ass.

 —Xah Lee, 2005


The thing is stuff like that is relatively in Lisp. Let's see your
Emacs Lisp version!

In Common Lisp you get the benefit of nicer code.

To give a hint:

(defun example ()
(cross-product '(* +)
(cons 'x (cross-product '(sin cos tan /) '(x)))
(cons 'y (cross-product '(sin cos tan /) '(y))))))

CL-USER 33 > (time (example))
Timing the evaluation of (EXAMPLE)

((+ X Y)
(+ X (/ Y))
(+ X (TAN Y))
(+ X (COS Y))
(+ X (SIN Y))
(+ (/ X) Y)
(+ (/ X) (/ Y))
(+ (/ X) (TAN Y))
(+ (/ X) (COS Y))
(+ (/ X) (SIN Y))
(+ (TAN X) Y)
(+ (TAN X) (/ Y))
(+ (TAN X) (TAN Y))
(+ (TAN X) (COS Y))
(+ (TAN X) (SIN Y))
(+ (COS X) Y)
(+ (COS X) (/ Y))
(+ (COS X) (TAN Y))
(+ (COS X) (COS Y))
(+ (COS X) (SIN Y))
(+ (SIN X) Y)
(+ (SIN X) (/ Y))
(+ (SIN X) (TAN Y))
(+ (SIN X) (COS Y))
(+ (SIN X) (SIN Y))
(* X Y)
(* X (/ Y))
(* X (TAN Y))
(* X (COS Y))
(* X (SIN Y))
(* (/ X) Y)
(* (/ X) (/ Y))
(* (/ X) (TAN Y))
(* (/ X) (COS Y))
(* (/ X) (SIN Y))
(* (TAN X) Y)
(* (TAN X) (/ Y))
(* (TAN X) (TAN Y))
(* (TAN X) (COS Y))
(* (TAN X) (SIN Y))
(* (COS X) Y)
(* (COS X) (/ Y))
(* (COS X) (TAN Y))
(* (COS X) (COS Y))
(* (COS X) (SIN Y))
(* (SIN X) Y)
(* (SIN X) (/ Y))
(* (SIN X) (TAN Y))
(* (SIN X) (COS Y))
(* (SIN X) (SIN Y)))

User time = 0.002

That should be easy to extend to get a solution for your problem.

Then you map a simplifier over the result. (mapcar 'simplify result)
and you get a nice list of simplified forms.
With some added recursion you can change the depth of the term