Re: [External] : Re: How to make M-x TAB not work on (interactive) declaration?

[Jean Louis]

* Drew Adams <drew.adams@oracle.com> [2023-01-16 20:07]:
> I think the replies you've gotten already might well help you
> understand.  Please consider rereading them.

I do not dispute that group theory exists, or identity, and it is part
of mathematic.

I do not see relation between identity and lisp function which is
supposed to give product of multiplication or sum of addition, and I
can't imagine that Lisp authors did that to those functions for reasons
to avoid `apply' function to fail.

There are X number of mathematical stuff that is not represented in
Lisp because it should not be there, a function should do one thing well.

What is missing in my understanding is:

- purpose of (*) ➜ 1, (-) ➜ 0 and (+) ➜ 0 as I do not understand how I
  would apply it else but making a funny Christmass tree in obscured
  programming and relation to Lisp. I have tried searching for
  references but can't find.

(defun m (n)
  (let ((m))
  (dotimes (b n)
    (setq m (cons "(*)" m)))
  (concat "(+" (string-join m) ")")))

(let ((first 1))
  (insert "\n")
  (while (<= first 10)
    (let ((second 1))
      (while (<= second 10)
        (insert "(*" (m first)(m second) ")\n")
        (setq second (1+ second)))
      (setq first (1+ first)))))

One reference to it I find in "Common Lisp - A Gentle Introduction to
Symbolic Computing":

> Suppose x and y are lists. (REDUCE #’+ (APPEND x y)) should produce
> the same value as the sum of (REDUCE #’+ x) and (REDUCE #’+ y). If y
> is NIL, then (APPEND x y) equals x, so (REDUCE #’+ y) has to return
> zero. Zero is the identity value for addition. That’s why calling +
> with no arguments returns zero. Similarly, calling * with no arguments
> returns one because one is the multiplicative identity.

Let's compare:

(let ((x '(3 3 3))
      (y '(4 4 4)))
  (reduce #'* (append x y))) ➜ 1728
Same in Common Lisp

(let ((x '(3 3 3))
      (y '(4 4 4)))
  (+ (reduce #'+ x) (reduce #'+ y))) ➜ 21
Same in Common Lisp

Am I wrong here making sum how it said I should make?

(let ((x '(3 3 3))
      (y nil))
  (append x y)) ➜ (3 3 3)
I cannot see that (append x y) equals x -- I cannot follow the example

(let ((x '(3 3 3))
      (y nil))
  (reduce #'+ y)) ➜ 0
This is correct as in example

Again I read "That's why calling + with no arguments returns zero" and do they 
mean that "why" is with reason to make function `reduce' work similarly like to 
make `apply' work?

-- 
Jean

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