LISP function to make all possible pairs from a list? - lisp

Question

I'm trying to create a LISP function that creates from a list all possible pairs.
Example of what I'm trying to achieve: (a b c d) --> ((a b) (a c) (a d) (b c) (b d) (c d))
Any advice please? I'm not sure how to approach this problem

Answer 1

Here is a simple solution:
(defun make-couples (x l)
"makes a list of couples whose first element is x and the second is each element of l in turn"
(loop for y in l collect (list x y)))
(defun all-pairs (l)
"makes a list of all the possible pairs of elements of list l"
(loop for (x . y) on l nconc (make-couples x y)))
A recursive solution is:
(defun make-couples (x l)
"makes a list of couples whose first element is x and the second is each element of l in turn"
(if (null l)
nil
(cons (cons x (first l)) (make-couples x (rest l)))))
(defun all-pairs (l)
"makes a list of all the possible pairs of elements of list l"
(if (null l)
nil
(nconc (make-couples (first l) (rest l))
(all-pairs (rest l)))))

Answer 2

Here is a version (this is quite closely related to Gwang-Jin Kim's) which has two nice properties:
it is tail recursive;
it walks no list more than once;
it allocates no storage that it does not use (so there are no calls to append and so on);
it uses no destructive operations.
It does this by noticing that there's a stage in the process where you want to say 'prepend a list of pairs of this element with the elements of this list to this other list' and that this can be done without using append or anything like that.
It does return the results in 'reversed' order, which I believe is inevitable given the above constraints.
(defun all-pairs (l)
(all-pairs-loop l '()))
(defun all-pairs-loop (l results)
(if (null (rest l))
results
(all-pairs-loop (rest l)
(prepend-pairs-to (first l) (rest l) results))))
(defun prepend-pairs-to (e them results)
(if (null them)
results
(prepend-pairs-to e (rest them) (cons (list e (first them))
results))))

Answer 3

the simplest tail recursive variant without explicit loops / mapcar could also look like this:
(defun pairs (data)
(labels ((rec (ls a bs res)
(cond
((null ls) (nreverse res))
((null bs) (rec
(cdr ls)
(car ls)
(cdr ls)
res))
(t (rec
ls
a
(cdr bs)
(cons (cons a (car bs)) res))))))
(rec data nil nil nil)))
CL-USER> (pairs (list 1 2 3 4))
;; ((1 . 2) (1 . 3) (1 . 4) (2 . 3) (2 . 4) (3 . 4))

Answer 4

Tail call recursive solution:
(defun pairs (lst &key (acc '()))
(if (null (cdr lst))
(nreverse acc)
(pairs (cdr lst)
:acc (append (nreverse
(mapcar #'(lambda (el)
(list (car lst) el))
(cdr lst)))
acc))))
Both nreverses are there just for aesthetics (for a nicer looking output). They can be left out.
Try it with:
(pairs '(a b c d))
;; => ((A B) (A C) (A D) (B C) (B D) (C D))
General Combinations
(defun pair (el lst)
"Pair el with each element of lst."
(mapcar (lambda (x) (cons el x)) lst))
(defun dedup (lst &key (test #'eql))
"Deduplicate a list of lists by ignoring order
and comparing the elements by test function."
(remove-duplicates lst :test (lambda (x y) (null (set-difference x y :test test)))))
(defun comb (lst &key (k 3) (acc '()) (test #'eql))
"Return all unique k-mer combinations of the elements in lst."
(labels ((%comb (lst &key (k k) (acc '()) (test #'eql) (total lst))
(let ((total (if total total lst)))
(cond ((or (null (cdr lst)) (zerop k)) (nreverse acc))
((= k 1) (mapcar #'list lst))
(t (let* ((el (car lst))
(rst (remove-if (lambda (x) (funcall test x el)) total)))
(dedup (%comb (cdr lst)
:k k
:total total
:test test
:acc (append (pair el (comb rst :k (1- k) :test test))
acc)))))))))
(%comb lst :k k :acc acc :test test :total lst)))
The number of combinations are calculatable with the combinations formula:
(defun fac (n &key (acc 1) (stop 1))
"n!/stop!"
(if (or (= n stop) (zerop n))
acc
(fac (1- n) :acc (* acc n) :stop stop)))
(defun cnr (n r)
"Number of all r-mer combinations given n elements.
nCr with n and r given"
(/ (fac n :stop r) (fac (- n r))))
We can test and count:
(comb '(a b c d) :k 2)
;; => ((A D) (B D) (B A) (C D) (C B) (C A))
(comb '(a b c d e f) :k 3)
;; => ((B A F) (C B A) (C B F) (C A F) (D C A) (D C B)
;; => (D C F) (D B A) (D B F) (D A F) (E D A) (E D B)
;; => (E D C) (E D F) (E C A) (E C B) (E C F) (E B A)
;; => (E B F) (E A F))
(= (length (comb '(a b c d e f) :k 3)) (cnr 6 3)) ;; => T
(= (length (comb '(a b c d e f g h i) :k 6)) (cnr 9 6)) ;; => T