@mmcdara The "directional expansion" works if f is in the class C^n (near [0,0,...,0]) and the result is a polynomial.
Otherwise, the expansion fails or is non-polynomial. mtaylor does basically the same but also checks the polynomiality.

For example,

f:=(x,y)->sin(x^2*y/(x^2+y^2)):
normal(eval(series(f(t*x,t*y),t=0),t=1)); # non-polynomial

because f is not differentiable at [0,0], (although continuous, provided that f(0,0)=0).

@nm Just differentiate (1) wirh respect to t and then take t=1. Using Maple:

f(t*x, t^p*y) = t^(p-1)*f(x,y):
eval(diff(%, t), t=1); p = solve(%, p);

See a nicer solution here.

@acer My answer contains a simplified explanation based on:
-- the information in the help ?simplify/size
-- the output generated by the simplify commands using printlevel:=40;
 For example, we see here:

--> enter \`simplify/size/size/object\`, args = _z1^2*_z2^2+_z1^2*_z3^2

--> enter \`simplify/size/size/object\`, args = _z1^2*(_z2^2+_z3^2)

@mmcdara In newer versions, the size option is by default in simplify.

@lcz Not like this. A compilable procedure is very restrictive. It cannot contain maple structures (except rtable with hardware entries) , so it must be written from scratch, starting e.g. from the adjacency matrix of the graph.

@Carl Love Maybe with a more flexible input:

CyclSum:= proc(f::{procedure,algebraic}, x::{indexed(posint),list(name)})
    local j,k,n,t,g;
    if x::list then t:=x; n:=nops(x)  else t:=op(0,x); n:=op(1,x) fi;
    if f::procedure then g:=f elif x::list then g:=unapply(f,x) else error "2nd param must be a list here" fi; 
    add(g(seq(t[1+irem(j+k-2, n)], j= 1..n)), k= 1..n)
end:

CyclSum((x,y)->x^y, [x,y,z]);

CyclSum(x^y, [x,y,z]);

CyclSum( () -> args[1]^p*args[2]^q, x[7]);

@Carl Love I wonder what would be a functional procedure for constructing a cyclic sum associated to an arbitrary expression, i.e.  Sum( f(x[s(1)], ..., x[s(n)]), s in Cycl(n) ) ,  where Cycl(n) contains the n cyclic permutations of {1, ..., n}.

[ Probably you will enjoy to write it :-) ]

For symmetric polynomials, any such polynomial can be expressed as a polynomial in the symmetric fundamental sums [probably you used the term "basis" for these].
For n=3 variables (let's restrict the question for this case), these are x+y+z, x*y+y*z+z*x and x*y*z.
They are cyclic (for n=3) but obvbiously only symmetric polynomials can be obtained this way.

Probably there is not a finite set of "fundamental" cyclic polynomials. But for an infinite set, it is not difficult to see that your set of polynomials  union {x*y*z}  forms a "fundamental" set. 

Do you mean a basis in the vector space of cyclic polynomials with rational coefficients (say)?
In this case the answer is yes, because your polynomials are linearly independent.

@C_R  Rouben forgot to include:

addcoords(z_cylindrical,[z, r,theta],[r*cos(theta),r*sin(theta),z]);

For a unique solution, you will need to define a well-order (see wiki). 

The attachment contains an image hint. (For the moment mapleprimes refuses to inline it).

pic1.zip

@acer  For testing (without modifying the polygon) it is indeed more convenient to use clicks. In this case, it is enough to replace 

Explore(T(x,y), x=0..80, y=0..100);

with

Explore(T(x,y), x=0..80, y=0..100, markers=[[x,y]]);

@Carl Love I always enjoy your optimized but too cryptic (for my taste) procedures!

You cannot omit the early return when P is in L, otherwise the result will be wrong in most cases.