Re: [Axiom-developer] How to do generic sum using aldor?

[Ralf Hemmecke]

On 01/04/2007 08:34 AM, Gernot Hueber wrote:
> You also can use reduce. E.g. reduce(+, [1,2,3])
> Gernot

Yes, but be careful. The documentation in Collection(S) (aggcat.spad) says:

reduce: ((S,S)->S,%) -> S
  ++ reduce(f,u) reduces the binary operation f across u. For example,
  ++ if u is \axiom{[x,y,...,z]} then \axiom{reduce(f,u)} returns
  ++ \axiom{f(..f(f(x,y),...),z)}.
  ++ Note: if u has one element x, \axiom{reduce(f,u)} returns x.
  ++ Error: if u is empty.

Franz has a bit more information on S. He knows that it is at least a 
Ring, so "sumlist []" will return 0. Otherwise one would have to use

reduce: ((S,S)->S,%,S) -> S
  ++ reduce(f,u,x) reduces the binary operation f across u, where x is
  ++ the identity operation of f.
  ++ Same as \axiom{reduce(f,u)} if u has 2 or more elements.
  ++ Returns \axiom{f(x,y)} if u has one element y,
  ++ x if u is empty.
  ++ For example, \axiom{reduce(+,u,0)} returns the
  ++ sum of the elements of u.

Ralf