Re: Cprod (Cartesian Product) in Lisp (or Scheme)

On 2024-05-21, HenHanna <HenHanna@devnull.tb> wrote:

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How would you write this in Lisp (or Scheme) ?
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in Python...    (writing this out: itertools.product([0, 1], repeat=N )
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The value can be a list or a Tuple.
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                 cprod([0, 1], 1) => ((0) (1))
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                 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.

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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.

--
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