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comp / comp.lang.lisp / Math operation abstraction

SubjectAuthor
* Math operation abstractionDaniel Cerqueira
+- Re: Math operation abstractionB. Pym
+- Re: Math operation abstractionSpiros Bousbouras
`- Re: Math operation abstractionSpiros Bousbouras

1
Subject: Math operation abstraction
From: Daniel Cerqueira
Newsgroups: comp.lang.lisp
Organization: A noiseless patient Spider
Date: Mon, 27 May 2024 22:29 UTC
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From: dan.list@lispclub.com (Daniel Cerqueira)
Newsgroups: comp.lang.lisp
Subject: Math operation abstraction
Date: Mon, 27 May 2024 23:29:56 +0100
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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.

Subject: Re: Math operation abstraction
From: B. Pym
Newsgroups: comp.lang.lisp
Organization: A noiseless patient Spider
Date: Mon, 27 May 2024 23:54 UTC
References: 1
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From: No_spamming@noWhere_7073.org (B. Pym)
Newsgroups: comp.lang.lisp
Subject: Re: Math operation abstraction
Date: Mon, 27 May 2024 23:54:43 -0000 (UTC)
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On 5/27/2024, Daniel Cerqueira wrote:

> * (muo #'(lambda (a b) (muo #'+ a b)) 5 2)

Pascal Bourguignon wrote:

> > #'(lambda (letter)
> > (add-counter letters letter))
> > word))
> > (read-words infile)))
>
> (function (lambda ...)) is a pleonasm. lambda is a macro that
> already expands to (function (lambda ...)).

Subject: Re: Math operation abstraction
From: Spiros Bousbouras
Newsgroups: comp.lang.lisp
Organization: A noiseless patient Spider
Date: Tue, 28 May 2024 21:10 UTC
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From: spibou@gmail.com (Spiros Bousbouras)
Newsgroups: comp.lang.lisp
Subject: Re: Math operation abstraction
Date: Tue, 28 May 2024 21:10:11 -0000 (UTC)
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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))))

--
vlaho.ninja/menu

Subject: Re: Math operation abstraction
From: Spiros Bousbouras
Newsgroups: comp.lang.lisp
Organization: A noiseless patient Spider
Date: Tue, 28 May 2024 21:17 UTC
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From: spibou@gmail.com (Spiros Bousbouras)
Newsgroups: comp.lang.lisp
Subject: Re: Math operation abstraction
Date: Tue, 28 May 2024 21:17:55 -0000 (UTC)
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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))))))

By the way , I don't think that in Common Lisp it is guaranteed that
(eq 1 1) returns T so it's best to use EQL to compare numbers.

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