Say there is any given list in Scheme. This list is ‘(2 3 4)
I want to find all possible partitions of this list. This means a partition where a list is separated into two subsets such that every element of the list must be in one or the other subsets but not both, and no element can be left out of a split.
So, given the list ‘(2 3 4), I want to find all such possible partitions. These partitions would be the following: {2, 3} and {4}, {2, 4} and {3}, and the final possible partition being {3, 4} and {2}.
I want to be able to recursively find all partitions given a list in Scheme, but I have no ideas on how to do so. Code or psuedocode will help me if anyone can provide it for me!
I do believe I will have to use lambda for my recursive function.
I discuss several different types of partitions at my blog, though not this specific one. As an example, consider that an integer partition is the set of all sets of positive integers that sum to the given integer. For instance, the partitions of 4 is the set of sets ((1 1 1 1) (1 1 2) (1 3) (2 2) (4)).
The process is building the partitions is recursive. There is a single partition of 0, the empty set (). There is a single partition of 1, the set (1). There are two partitions of 2, the sets (1 1) and (2). There are three partitions of 3, the sets (1 1 1), (1 2) and (3). There are five partitions of 4, the sets (1 1 1 1), (1 1 2), (1 3), (2 2), and (4). There are seven partitions of 5, the sets (1 1 1 1 1), (1 1 1 2), (1 2 2), (1 1 3), (1 4), (2 3) and (5). And so on. In each case, the next-larger set of partitions is determined by adding each integer x less than or equal to the desired integer n to all the sets formed by the partition of n − x, eliminating any duplicates. Here's how I implement that:
Petite Chez Scheme Version 8.4
Copyright (c) 1985-2011 Cadence Research Systems
> (define (set-cons x xs)
(if (member x xs) xs
(cons x xs)))
> (define (parts n)
(if (zero? n) (list (list))
(let x-loop ((x 1) (xs (list)))
(if (= x n) (cons (list n) xs)
(let y-loop ((yss (parts (- n x))) (xs xs))
(if (null? yss) (x-loop (+ x 1) xs)
(y-loop (cdr yss)
(set-cons (sort < (cons x (car yss)))
xs))))))))
> (parts 6)
((6) (3 3) (2 2 2) (2 4) (1 1 4) (1 1 2 2) (1 1 1 1 2)
(1 1 1 1 1 1) (1 1 1 3) (1 2 3) (1 5))
I'm not going to solve your homework for you, but your solution will be similar to the one given above. You need to state your algorithm in recursive fashion, then write code to implement that algorithm. Your recursion is going to be something like this: For each item in the set, add the item to each partition of the remaining items of the set, eliminating duplicates.
That will get you started. If you have specific questions, come back here for additional help.
EDIT: Here is my solution. I'll let you figure out how it works.
(define range (case-lambda ; start, start+step, ..., start+step<stop
((stop) (range 0 stop (if (negative? stop) -1 1)))
((start stop) (range start stop (if (< start stop) 1 -1)))
((start stop step) (let ((le? (if (negative? step) >= <=)))
(let loop ((x start) (xs (list)))
(if (le? stop x) (reverse xs) (loop (+ x step) (cons x xs))))))
(else (error 'range "unrecognized arguments"))))
(define (sum xs) (apply + xs)) ; sum of elements of xs
(define digits (case-lambda ; list of base-b digits of n
((n) (digits n 10))
((n b) (do ((n n (quotient n b))
(ds (list) (cons (modulo n b) ds)))
((zero? n) ds)))))
(define (part k xs) ; k'th lexicographical left-partition of xs
(let loop ((ds (reverse (digits k 2))) (xs xs) (ys (list)))
(if (null? ds) (reverse ys)
(if (zero? (car ds))
(loop (cdr ds) (cdr xs) ys)
(loop (cdr ds) (cdr xs) (cons (car xs) ys))))))
(define (max-lcm xs) ; max lcm of part-sums of 2-partitions of xs
(let ((len (length xs)) (tot (sum xs)))
(apply max (map (lambda (s) (lcm s (- tot s)))
(map sum (map (lambda (k) (part k xs))
(range (expt 2 (- len 1)))))))))
(display (max-lcm '(2 3 4))) (newline) ; 20
(display (max-lcm '(2 3 4 6))) (newline) ; 56
You can find all 2-partitions of a list using the built-in combinations procedure. The idea is, for every element of a (len-k)-combination, there will be an element in the k-combination that complements it, producing a pair of lists whose union is the original list and intersection is the empty list.
For example:
(define (2-partitions lst)
(define (combine left right)
(map (lambda (x y) (list x y)) left right))
(let loop ((m (sub1 (length lst)))
(n 1))
(cond
((< m n) '())
((= m n)
(let* ((l (combinations lst m))
(half (/ (length l) 2)))
(combine (take l half)
(reverse (drop l half)))))
(else
(append
(combine (combinations lst m)
(reverse (combinations lst n)))
(loop (sub1 m) (add1 n)))))))
then you can build the partitions as:
(2-partitions '(2 3 4))
=> '(((2 3) (4))
((2 4) (3))
((3 4) (2)))
(2-partitions '(4 6 7 9))
=> '(((4 6 7) (9))
((4 6 9) (7))
((4 7 9) (6))
((6 7 9) (4))
((4 6) (7 9))
((4 7) (6 9))
((6 7) (4 9)))
Furthermore, you can find the max lcm of the partitions:
(define (max-lcm lst)
(define (local-lcm arg)
(lcm (apply + (car arg))
(apply + (cadr arg))))
(apply max (map local-lcm (2-partitions lst))))
For example:
(max-lcm '(2 3 4))
=> 20
(max-lcm '(4 6 7 9))
=> 165
To partition a list is straightforward recursive non-deterministic programming.
Given an element, we put it either into one bag, or the other.
The very first element will go into the first bag, without loss of generality.
The very last element must go into an empty bag only, if such is present at that time. Since we start by putting the first element into the first bag, it can only be the second:
(define (two-parts xs)
(if (or (null? xs) (null? (cdr xs)))
(list xs '())
(let go ((acc (list (list (car xs)) '())) ; the two bags
(xs (cdr xs)) ; the rest of list
(i (- (length xs) 1)) ; and its length
(z '()))
(if (= i 1) ; the last element in the list is reached:
(if (null? (cadr acc)) ; the 2nd bag is empty:
(cons (list (car acc) (list (car xs))) ; add only to the empty 2nd
z) ; otherwise,
(cons (list (cons (car xs) (car acc)) (cadr acc)) ; two solutions,
(cons (list (car acc) (cons (car xs) (cadr acc))) ; adding to
z))) ; either of the two bags;
(go (list (cons (car xs) (car acc)) (cadr acc)) ; all solutions after
(cdr xs) ; adding to the 1st bag
(- i 1) ; and then,
(go (list (car acc) (cons (car xs) (cadr acc))) ; all solutions
(cdr xs) ; after adding
(- i 1) ; to the 2nd instead
z))))))
And that's that!
In writing this I was helped by following this earlier related answer of mine.
Testing:
(two-parts (list 1 2 3))
; => '(((2 1) (3)) ((3 1) (2)) ((1) (3 2)))
(two-parts (list 1 2 3 4))
; => '(((3 2 1) (4))
; ((4 2 1) (3))
; ((2 1) (4 3))
; ((4 3 1) (2))
; ((3 1) (4 2))
; ((4 1) (3 2))
; ((1) (4 3 2)))
It is possible to reverse the parts before returning, or course; I wanted to keep the code short and clean, without the extraneous details.
edit: The code makes use of a technique by Richard Bird, of replacing (append (g A) (g B)) with (g' A (g' B z)) where (append (g A) y) = (g' A y) and the initial value for z is an empty list.
Another possibility is for the nested call to go to be put behind lambda (as the OP indeed suspected) and activated when the outer call to go finishes its job, making the whole function tail recursive, essentially in CPS style.
(define (all-sublists buffer n)
(cond ((= n 0) n)
((all-sublists (append buffer (list (list n)) (map (lambda (x) (append (list n) x)) buffer)) (- n 1)))))
the result looks like this:
(all-sublists '((3) (2) (2 3) (1) (1 3) (1 2) (1 2 3)) 0)
when there is only one list around n:
(define (all-sublists buffer n)
(cond ((= n 0) n)
((all-sublists (append buffer (list n) (map (lambda (x) (append (list n) x)) buffer)) (- n 1)))))
the results get a dotted pair:
(all-sublists '(3 2 (2 . 3) 1 (1 . 3) (1 . 2) (1 2 . 3)) 0)
Is not that you have "to surround n with list twice to get the proper result", the truth is that there are several problems with your code, for starters: the last condition of a cond should start with an else, and you're using append incorrectly. If I understood correctly, you just want the powerset of a list:
(define (powerset aL)
(if (empty? aL)
'(())
(let ((rst (powerset (rest aL))))
(append (map (lambda (x) (cons (first aL) x))
rst)
rst))))
Like this:
(powerset '(1 2 3))
=> '((1 2 3) (1 2) (1 3) (1) (2 3) (2) (3) ())
The problem is when lists have a different length, any idea of how to do it?
I have to use functions like map or something like that
This is the code I wrote so far, it works with lists of the same length but it also needs to work with lists of different lengths. Thank you.
(define (interleave list1 list2)
(flatten [map (lambda (x y) (cons x (cons y null))) list1 list2]))
if lists have different length this is what I get:
map: all lists must have same size; arguments were: # '(1 2 3 4 5) '(a b c)
I'm trying to get (1 a 2 b 3 c 4 5)
#lang racket
(define (weave xs ys)
(match (list xs ys)
[(list (cons x xs) (cons y ys)) (cons x (cons y (weave xs ys)))]
[(list '() ys) ys]
[(list xs '()) xs]))
I'm assuming your desired behavior is that the lists are interleaved for as long as this is possible, and then whatever is left over from the nonempty list is appended to the end. In that case one possible implementation is
(define (interleave a b)
(if (null? a)
b
(cons (car a)
(interleave b (cdr a)))))
I think this is probably the simplest possible way to write what you're looking for.
Neither map nor fold-right would work because they either signal an error when one list is smaller than the other or they tend to stop at the shortest list. eg. SRFI-1's map (interleave '(1 2 3 4) (circular-list 9 8)) ; ==> (1 9 2 8 3 9 4 8). For a different behavior you need to roll your own.
A solution using simple list manipulation functions might be:
(define (interleave list1 list2)
(cond ((empty? list1) list2)
((empty? list2) list1)
(else
(append
(list (car list1) (car list2))
(interleave (cdr list1) (cdr list2))))))
Testing...
> (interleave '(1 2 3 4 5) '(a b c))
(1 a 2 b 3 c 4 5)
> (interleave '(1 2 3 4 5) '())
(1 2 3 4 5)
> (interleave '() '(a b c))
(a b c)
>
I think it is fairly self-documenting.
"There ain't nothin' you can't not do with fold-right and some of them con-tin-uations thingies", said a cowboy to another, spittin' into the campfire and puffin' on his cigar in the evening, sippin' his black coffee from his rugged banged up tin mug. "Yessa, nothin' in the whole darn world."
(define (interleave xs ys)
;; interleave xs ys = foldr g n xs ys
;; where
;; g x r (y:ys) = x : y : r ys
;; g x r [] = x : r []
;; n ys = ys
((foldr
(lambda (x r)
(lambda (ys)
(cond ((null? ys) (cons x (r '())))
(else (apply (lambda (y . ys)
(cons x (cons y (r ys))))
ys)))))
(lambda (ys) ys)
xs)
ys))
I'm trying to write a procedure that takes 2 streams and that puts together their pairs and then interleaves them. Right now it's not producing the correct output. Here's the code I have:
(define (interleave-pairs s t)
(cons-stream (cons (stream-car s) (stream-car t))
(cons-stream
(stream-map (lambda (x) (cons (stream-car s) x))
(stream-cdr t))
(interleave-pairs t (stream-cdr s)))))
To obtain the values I wrote a procedure:
(define (take n s)
(if (= n 0)
'()
(cons (stream-car s) (take (- n 1) (stream-cdr s)))))
And this is for the stream of integers:
(define integers (cons-stream 1 (add-streams ones integers)))
So when I call
(take 6 (pairs integers integers))
I'm getting:
((1 . 1)
((1 . 2) . #<promise>)
(1 . 2)
((1 . 3) . #<promise>)
(2 . 2)
((2 . 3) . #<promise>))
Whereas I want (1 1) (1 1) (1 2) (1 3) (2 2) (2 3)
The base case in your procedure is missing (what happens if one of the streams ends before the other?). The procedure is rather simple, just take one pair of elements (one from each stream) and then interleave them, no need to use stream-map here:
(define (pairs s1 s2)
(if (stream-null? s1)
s2
(stream-cons (list (stream-car s1) (stream-car s2))
(pairs s2 (stream-cdr s1)))))
I have two procedures, one for counting an element in the list and the other one for removing the same element from the same list. What should I do for counting and removing at the same time? I am trying it for long time but nothing is working. I work with this list: (list 1 2 3 2 1 2 3), finally it should be like: ((1 . 2) (2 . 3) (3 . 2)). The first number of pair is an element and second number of pair is sum of first pair's number from all list.
My try:
1) it works only with counting and result is: ((1 . 2) (2 . 3) (3 . 2) (2 . 2) (1 . 1) (2 . 1) (3 . 1))
2) it works only with removing and result is: ((1 . 2) 2 3 2 2 3)
Where is the problem?
This is for counting:
(define count-occurrences
(lambda (x ls)
(cond
[(memq x ls) =>
(lambda (ls)
(+ (count-occurrences x (cdr ls)) 1))]
[else 0])))
(count-occurrences '2 (list 1 2 3 2 1 2 3)) -> 3
This is for removing:
(define (remove-el p s)
(cond ((null? s) '())
((equal? p (car s)) (remove-el p (cdr s)))
(else (cons (car s) (remove-el p (cdr s))))))
(remove-el '2 (list 1 2 3 2 1 2 3)) -> (1 3 1 3)
Just return the count and the removed list at once. I call this routine
count-remove. (Pardon to all schemers for not idiomatic or efficient style)
(define (count-remove ls x)
(letrec ([loop (lambda (count l removed)
(cond
[(eq? l '()) (list count removed)]
[(eq? (car l) x) (loop (+ 1 count) (cdr l) removed)]
[else (loop count (cdr l) (cons (car l) removed))]))])
(loop 0 ls '())))
(define (count-map ls)
(cond
[(eq? ls '()) '()]
[else
(letrec ([elem (car ls)]
[cr (count-remove ls elem)])
(cons (cons elem (car cr)) (count-map (cadr cr))))]))
Here is some usage:
(count-map '(1 1 2 3 2))
((1 . 2) (2 . 2) (3 . 1))