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comp / comp.lang.scheme / Cprod (Cartesian Product) in Lisp (or Scheme)

SubjectAuthor
* Cprod (Cartesian Product) in Lisp (or Scheme)HenHanna
`- Re: Cprod (Cartesian Product) in Lisp (or Scheme)Kaz Kylheku

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Subject: Cprod (Cartesian Product) in Lisp (or Scheme)
From: HenHanna
Newsgroups: comp.lang.lisp, comp.lang.scheme
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Date: Tue, 21 May 2024 19:18 UTC
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From: HenHanna@devnull.tb (HenHanna)
Newsgroups: comp.lang.lisp,comp.lang.scheme
Subject: Cprod (Cartesian Product) in Lisp (or Scheme)
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How would you write this in Lisp (or Scheme) ?

in Python... (writing this out: itertools.product([0, 1], repeat=N )

The value can be a list or a Tuple.

cprod([0, 1], 1) => ((0) (1))

cprod([0, 1], 2) => ((0,0) (0,1) (1,0) (1,1))

This works:

def cprod(x, c):
if c==1: return [[i] for i in x]
Sub= cprod(x, c-1)
return [i for F in x for i in [[F]+R for R in Sub]]

---------- Is there another (better) way to write [F]+R ???

it seems odd, compared to CONS in Lisp

Other ways to improve it?

Subject: Re: Cprod (Cartesian Product) in Lisp (or Scheme)
From: Kaz Kylheku
Newsgroups: comp.lang.lisp, comp.lang.scheme
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From: 643-408-1753@kylheku.com (Kaz Kylheku)
Newsgroups: comp.lang.lisp,comp.lang.scheme
Subject: Re: Cprod (Cartesian Product) in Lisp (or Scheme)
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On 2024-05-21, HenHanna <HenHanna@devnull.tb> wrote:
>
>
> How would you write this in Lisp (or Scheme) ?
>
>
>
> in Python... (writing this out: itertools.product([0, 1], repeat=N )
>
> The value can be a list or a Tuple.
>
> cprod([0, 1], 1) => ((0) (1))
>
> cprod([0, 1], 2) => ((0,0) (0,1) (1,0) (1,1))

Another name for this is "repeating permutations": permutations
of the (0 1) elements, such that repetitions are allowed.

How I would write this is by having it built into the language.

This is the TXR Lisp interactive listener of TXR 294.
Quit with :quit or Ctrl-D on an empty line. Ctrl-X ? for cheatsheet.
Everything you type here can and will be used against you in
comp.lang.lisp.
1> (rperm '(0 1) 2)
((0 0) (0 1) (1 0) (1 1))

I would have it as a lazy list, so we can ask for the first 5
items of an incredibly long instance of such a sequence.

2> (take 5 (rperm #\A..#\Z 15))
((#\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A)
(#\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\B)
(#\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\C)
(#\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\D)
(#\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\A #\E))
3> (take 5 (rperm (join #\A..#\Z) 15))
("AAAAAAAAAAAAAAA" "AAAAAAAAAAAAAAB" "AAAAAAAAAAAAAAC" "AAAAAAAAAAAAAAD"
"AAAAAAAAAAAAAAE")

That reminds me; I should probably implement iterators which
step over these sequences, to complement the lazy list implementation.

The implementation of rperm starts here:

https://www.kylheku.com/cgit/txr/tree/combi.c?h=txr-294#n264

The heart of it is the rperm_gen_fun function, which updates
a permutation vector to the next permutation.

The state consists of a vector of lists, and a reset list.

For instance, if we are generating triplets of (A B C D), the
vector gets initialized to a copy of the list in every position:

#((A B C D)
(A B C D)
(A B C D))

We take the first repeating permutation by taking the car
of every list: (A A A A). Then to generate the next permutation,
we pop the last list:

#((A B C D)
(A B C D)
(B C D)) ;; pop!

When we pop the last list empty, we restore it back to (A B C D),
and pop the next one:

#((A B C D)
(A B C D)
(B))

#((A B C D)
(A B C D)
()) ;; pop! oops!

#((A B C D)
(B C D) ;; pop!
(A B C D)) ;; whump! (restored)

When we pop the first list down to nil, then we are done.
The rperm_while_fun tests for this condition.

It's a very simple algorithm compared to the nonrepeating
permutations, and repeating or nonrepeating combinations.

--
TXR Programming Language: http://nongnu.org/txr
Cygnal: Cygwin Native Application Library: http://kylheku.com/cygnal
Mastodon: @Kazinator@mstdn.ca

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