Re: Math operation abstraction

On Mon, 27 May 2024 23:29:56 +0100
Daniel Cerqueira <dan.list@lispclub.com> wrote:

Hi.
 
I have this function
 
```
(defun muo (func a b)
  (cond
    ((eq b 1) a)
    (t (funcall func a (muo func a (1- b))))))
```
 
Which I can use for obtaining the multiplication operation using the
addition. As an example:
 
* (muo #'+ 5 2)
10
 
and "muo #'+" means "use the multiplication".
 
Now if I want the exponentiation I have to do:
 
* (muo #'(lambda (a b) (muo #'+ a b)) 5 2)
25
 
This creates the exponentiation.
 
I would like to obtain the exponentiation using "(muo #'+ 5 2 2)"
instead, being the last argument ("2") the number of times the operation
is applied. In the case of exponentiation, it applies the multiplication
2 times, resulting in the exponentiation math operation.
 
I am hoping I am explaining this concept well enough.
 
Basically, "muo #'+" creates the addition twice, which is the
multiplication, and "muo #'+ a b 2" should create the multiplication
twice, which is the exponentiation. If I use "muo #'+ a b 3" it should
create the exponentiation twice, which is the exponentiation of the
exponentiation.
 
I need help in making this code modification. Seeing code is what I
want. If you can write the code and explain it, better.

You don't give a precise definition but perhaps the following gives you
what you want or at least a starting point :

(defun muo-aux (func a b)
    (cond
        ((eql b 1) a)
        (t (funcall func a (muo-aux func a (1- b))))))

(defun muo (func a b repeats)
    (cond
        ((eql repeats 1)
            (muo-aux func a b))
        ((< 1 repeats)
            (muo-aux (lambda (x y) (muo func x y (1- repeats))) a b))))

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