List rooted trees

List rooted trees is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

You came back from grocery shopping. After putting away all the goods, you are left with a pile of plastic bags, which you want to save for later use, so you take one bag and stuff all the others into it, and throw it under the sink. In doing so, you realize that there are various ways of nesting the bags, with all bags viewed as identical.

If we use a matching pair of parentheses to represent a bag, the ways are:

For 1 bag, there's one way:

 ()	<- a bag

for 2 bags, there's one way:

 (())	<- one bag in another

for 3 bags, there are two:

 ((())) <- 3 bags nested Russian doll style
 (()()) <- 2 bags side by side, inside the third

for 4 bags, four:

 (()()())
 ((())())
 ((()()))
 (((())))

Note that because all bags are identical, the two 4-bag strings ((())()) and (()(())) represent the same configuration.

It's easy to see that each configuration for n bags represents a n-node rooted tree, where a bag is a tree node, and a bag with its content forms a subtree. The outermost bag is the tree root. Number of configurations for given n is given by OEIS A81.

Task: write a program that, when given n, enumerates all ways of nesting n bags. You can use the parentheses notation above, or any tree representation that's unambiguous and preferably intuitive.

This task asks for enumeration of trees only; for counting solutions without enumeration, that OEIS page lists various formulas, but that's not encouraged by this task, especially if implementing it would significantly increase code size.

As an example output, run 5 bags. There should be 9 ways.

C

Trees are represented by integers. When written out in binary with LSB first, 1 is opening bracket and 0 is closing. <lang c>#include <stdio.h>

1. include <stdlib.h>

typedef unsigned int uint; typedef unsigned long long tree;

1. define B(x) (1ULL<<(x))

tree *list = 0; uint cap = 0, len = 0; uint offset[32] = {0, 1, 0};

void append(tree t) { if (len == cap) { cap = cap ? cap*2 : 2; list = realloc(list, cap*sizeof(tree)); } list[len++] = 1 | t<<1; }

void show(tree t, uint len) { for (; len--; t >>= 1) putchar(t&1 ? '(' : ')'); }

void listtrees(uint n) { uint i; for (i = offset[n]; i < offset[n+1]; i++) { show(list[i], n*2); putchar('\n'); } }

/* assemble tree from subtrees n: length of tree we want to make t: assembled parts so far sl: length of subtree we are looking at pos: offset of subtree we are looking at rem: remaining length to be put together

• /

void assemble(uint n, tree t, uint sl, uint pos, uint rem) { if (!rem) { append(t); return; }

if (sl > rem) // need smaller subtrees pos = offset[sl = rem]; else if (pos >= offset[sl + 1]) { // used up sl-trees, try smaller ones if (!--sl) return; pos = offset[sl]; }

assemble(n, t<<(2*sl) | list[pos], sl, pos, rem - sl); assemble(n, t, sl, pos + 1, rem); }

void mktrees(uint n) { if (offset[n + 1]) return; if (n) mktrees(n - 1);

assemble(n, 0, n-1, offset[n-1], n-1); offset[n+1] = len; }

int main(int c, char**v) { int n; if (c < 2 || (n = atoi(v[1])) <= 0 || n > 25) n = 5;

// init 1-tree append(0);

mktrees((uint)n); fprintf(stderr, "Number of %d-trees: %u\n", n, offset[n+1] - offset[n]); listtrees((uint)n);

return 0; }</lang>

Output:

% ./a.out 5
Number of 5-trees: 9
((((()))))
(((()())))
((()(())))
((()()()))
(()((())))
(()(()()))
((())(()))
(()()(()))
(()()()())

D

Translation of: C

<lang d>import std.stdio, std.conv;

alias Tree = ulong,

     TreeList = Tree[],
     Offset = uint[32];

void listTees(in uint n, in ref Offset offset, in TreeList list) nothrow @nogc @safe {

   static void show(in Tree t, in uint len) nothrow @nogc @safe {
       foreach (immutable i; 0 .. len)
           putchar(t & (2 ^^ i) ? '(' : ')');
   }

   foreach (immutable i; offset[n] .. offset[n + 1]) {
       show(list[i], n * 2);
       putchar('\n');
   }

}

void append(in Tree t, ref TreeList list, ref uint len) pure nothrow @safe {

   if (len == list.length)
       list.length = list.length ? list.length * 2 : 2;
   list[len] = 1 | (t << 1);
   len++;

}

/** Assemble tree from subtrees.

Params:

 n   = length of tree we want to make.
 t   = assembled parts so far.
 sl  = length of subtree we are looking at.
 pos = offset of subtree we are looking at.
 rem = remaining length to be put together.

• /

void assemble(in uint n, in Tree t, uint sl, uint pos, in uint rem, in ref Offset offset,

             ref TreeList list, ref uint len) pure nothrow @safe {
   if (!rem) {
       append(t, list, len);
       return;
   }

   if (sl > rem) { // Need smaller subtrees.
       sl = rem;
       pos = offset[sl];
   } else if (pos >= offset[sl + 1]) {
       // Used up sl-trees, try smaller ones.
       sl--;
       if (!sl)
           return;
       pos = offset[sl];
   }

   assemble(n, t << (2 * sl) | list[pos], sl, pos, rem - sl, offset, list, len);
   assemble(n, t, sl, pos + 1, rem, offset, list, len);

}

void makeTrees(in uint n, ref Offset offset,

              ref TreeList list, ref uint len) pure nothrow @safe {
   if (offset[n + 1])
       return;
   if (n)
       makeTrees(n - 1, offset, list, len);

   assemble(n, 0, n - 1, offset[n - 1], n - 1, offset, list, len);
   offset[n + 1] = len;

}

void main(in string[] args) {

   immutable uint n = (args.length == 2) ? args[1].to!uint : 5;
   if (n >= 25)
       return;

   Offset offset;
   offset[1] = 1;

   Tree[] list;
   uint len = 0;

   // Init 1-tree.
   append(0, list, len);

   makeTrees(n, offset, list, len);
   stderr.writefln("Number of %d-trees: %u", n, offset[n + 1] - offset[n]);
   listTees(n, offset, list);

}</lang>

Output:

Number of 5-trees: 9
((((()))))
(((()())))
((()(())))
((()()()))
(()((())))
(()(()()))
((())(()))
(()()(()))
(()()()())

Haskell

There probably is a nicer way than the following-- <lang haskell>-- break n down into sum of smaller integers parts n = f n 1 where f n x | n == 0 = [[]] | x > n = [] | otherwise = f n (x+1) ++ concatMap (\c->map ((c,x):) (f (n-c*x) (x+1))) [1 .. n`div`x]

-- choose n strings out of a list and join them pick _ [] = [] pick 0 _ = [""] pick n aa@(a:as) = map (a++) (pick (n-1) aa) ++ pick n as

-- pick parts to build a series of subtrees that add up to n-1, then wrap them up trees n = map (\x->"("++x++")") $ concatMap (foldr (prod.build) [""]) (parts (n-1)) where build (c,x) = pick c $ trees x prod aa bb = [ a++b | a<-aa, b<-bb ]

main = mapM_ putStrLn $ trees 5</lang>

Output:

((((()))))
(((()())))
((()(())))
((()()()))
((())(()))
(()((())))
(()(()()))
(()()(()))
(()()()())

J

Support code:

<lang J>root=: 1 1 $ _ incr=: ,/@(,"1 0/ i.@{:@$)

boxed=: $:&0 :(<@\:~@([ $:^:(0 < #@]) I.@:=))"1 1 0</lang>

Task:

   ~.boxed incr^:4 root
┌─────┬──────┬──────┬───────┬───────┬──────┬───────┬───────┬────────┐
│┌┬┬┬┐│┌──┬┬┐│┌───┬┐│┌──┬──┐│┌────┬┐│┌────┐│┌─────┐│┌─────┐│┌──────┐│
││││││││┌┐│││││┌┬┐││││┌┐│┌┐│││┌──┐││││┌┬┬┐│││┌──┬┐│││┌───┐│││┌────┐││
│└┴┴┴┘│││││││││││││││││││││││││┌┐│││││││││││││┌┐││││││┌┬┐│││││┌──┐│││
│     ││└┘│││││└┴┘││││└┘│└┘│││││││││││└┴┴┘│││││││││││││││││││││┌┐││││
│     │└──┴┴┘│└───┴┘│└──┴──┘│││└┘││││└────┘│││└┘││││││└┴┘││││││││││││
│     │      │      │       ││└──┘│││      ││└──┴┘│││└───┘│││││└┘││││
│     │      │      │       │└────┴┘│      │└─────┘│└─────┘│││└──┘│││
│     │      │      │       │       │      │       │       ││└────┘││
│     │      │      │       │       │      │       │       │└──────┘│
└─────┴──────┴──────┴───────┴───────┴──────┴───────┴───────┴────────┘

Explanation: while building the trees, we are using the parent index representation of a tree. The tree is represented as a sequence of indices of the parent nodes. We use _ to represent the root node (so our root node has no parent).

In the boxed representation we use here, each square box represents a bag.

boxed represents a single tree structure in a nested boxed form, with each box representing a bag. Here, we sort each sequence of boxes (which we are thinking of as bags), so we can recognize mechanically different tree structures which happen to represent the same bag structures.

And for the task example, we want four bags into the outside containing bag, and also we want to eliminate redundant representations...

So, for example, here is what some intermediate results would look like for the four bag case:

<lang J> incr^:3 root _ 0 0 0 _ 0 0 1 _ 0 0 2 _ 0 1 0 _ 0 1 1 _ 0 1 2</lang>

Each row represents a bag with another three bags stuffed into it. Each column represents a bag, and each index is the column of the bag that it is stuffed into. (The first bag isn't stuffed into another bag.)

But some of these are equivalent, we can see that if we use our parenthesis notation and think about how they could be rearranged:

<lang J> disp=: ('(' , ')' ,~ [: ; [ <@disp"1 0^:(0 < #@]) I.@:=) {.

  disp incr^:3 root

(()()()) ((())()) (()(())) ((())()) ((()())) (((())))</lang>

But that's not a convenient way of finding the all of the duplicates. So we use a boxed representation - with all boxes at each level in a canonical order (fullest first) - and that makes the duplicates obvious:

   boxed incr^:3 root
┌────┬─────┬─────┬─────┬─────┬──────┐
│┌┬┬┐│┌──┬┐│┌──┬┐│┌──┬┐│┌───┐│┌────┐│
│││││││┌┐││││┌┐││││┌┐││││┌┬┐│││┌──┐││
│└┴┴┘│││││││││││││││││││││││││││┌┐│││
│    ││└┘││││└┘││││└┘││││└┴┘│││││││││
│    │└──┴┘│└──┴┘│└──┴┘│└───┘│││└┘│││
│    │     │     │     │     ││└──┘││
│    │     │     │     │     │└────┘│
└────┴─────┴─────┴─────┴─────┴──────┘

Perl 6

Bags are represented by Perl 6 type Bag.

<lang perl6>use v6;

multi expand-tree ( Bag $tree ) {

   bag(bag(bag()) (+) $tree) (+)
   [(+)] (
       $tree.keys ==> map {
           $^a.&expand-tree.map: * (+) ( $tree (-) bag($^a) )
       }
   );

}

multi expand-trees ( Bag $trees ) {

   [(+)] $trees.keys.map:  { $_.&expand-tree } ;

}

my $n = 5; for ( bag(), bag(bag()), *.&expand-trees ... * )[$n] {

   print ++$,".\t";
   .say

}; </lang>

Output:

1.	bag(bag(), bag(bag()(2))) => 2
2.	bag(bag(bag()(3))) => 1
3.	bag(bag(bag(bag()), bag())) => 2
4.	bag(bag(bag(bag(bag())))) => 1
5.	bag(bag(bag())(2)) => 1
6.	bag(bag(bag(bag()(2)))) => 1
7.	bag(bag(), bag(bag(bag()))) => 2
8.	bag(bag(bag()), bag()(2)) => 2
9.	bag(bag()(4)) => 1

The bag bag(bag(bag()), bag()(2)) coresponds with ((())()()). There are two independent ways how we can get it by nesting 4 bags.

Python

<lang python>def bags(n,cache={}): if not n: return [(0, "")]

upto = sum([bags(x) for x in range(n-1, 0, -1)], []) return [(c+1, '('+s+')') for c,s in bagchain((0, ""), n-1, upto)]

def bagchain(x, n, bb, start=0): if not n: return [x]

out = [] for i in range(start, len(bb)): c,s = bb[i] if c <= n: out += bagchain((x[0] + c, x[1] + s), n-c, bb, i) return out

1. Maybe this lessens eye strain. Maybe not.

def replace_brackets(s): depth,out = 0,[] for c in s: if c == '(': out.append("([{"[depth%3]) depth += 1 else: depth -= 1 out.append(")]}"[depth%3]) return "".join(out)

for x in bags(5): print(replace_brackets(x[1]))</lang>

Output:

([{([])}])
([{()()}])
([{()}{}])
([{}{}{}])
([{()}][])
([{}{}][])
([{}][{}])
([{}][][])
([][][][])

Another method by incrementing subtrees: <lang python>treeid = {(): 0}

Successor of a tree. The predecessor p of a tree t is:

 1. if the smallest subtree of t is a single node, then p is t minus that node
 2. otherwise, p is t with its smalles subtree "m" replaced by m's predecessor

Here "smaller" means the tree is generated earlier, as recorded by treeid. Obviously, predecessor to a tree is unique. Since every degree n tree has a unique degree (n-1) predecessor, inverting the process leads to the successors to tree t:

 1. append a single node tree to t's root, or
 2. replace t's smallest subtree by its successors

We need to keep the trees so generated canonical, so when replacing a subtree, the replacement must not be larger than the next smallest subtree.

Note that trees can be compared by other means, as long as trees with fewer nodes are considered smaller, and trees with the same number of nodes have a fixed order. def succ(x):

   yield(((),) + x)
   if not x: return

   if len(x) == 1:
       for i in succ(x[0]): yield((i,))
       return

   head,rest = x[0],tuple(x[1:])
   top = treeid[rest[0]]

   for i in [i for i in succ(head) if treeid[i] <= top]:
       yield((i,) + rest)

def trees(n):

   if n == 1:
       yield()
       return

   global treeid
   for x in trees(n-1):
       for a in succ(x):
           if not a in treeid: treeid[a] = len(treeid)
           yield(a)

def tostr(x): return "(" + "".join(map(tostr, x)) + ")"

for x in trees(5): print(tostr(x))</lang>

Racket

<lang racket>#lang racket (require racket/splicing data/order)

(define (filtered-cartesian-product #:f (fltr (λ (cand left) #t)) l . more-ls)

 (let inr ((lls (cons l more-ls)) (left null))
   (match lls
     [(list) '(())]
     [(cons lla lld)
      (for*/list ((a (in-list (filter (curryr fltr left) lla)))
                  (d (in-list (inr lld (cons a left)))))
        (cons a d))])))

The "order" of an LRT

(define LRT-order (match-lambda [(list (app LRT-order o) ...) (apply + 1 o)]))

Some order for List Rooted Trees

(define LRT<=

 (match-lambda**
  [(_ (list)) #t]
  [((and bar (app LRT-order baro)) (cons (and badr (app LRT-order badro)) bddr))
   (and (or (< baro badro) (not (eq? '> (datum-order bar badr)))) (LRT<= badr bddr))]))

(splicing-letrec ((t# (make-hash '((1 . (())))))

                 (p# (make-hash '((0 . (()))))))
 ;; positive-integer -> (listof (listof positive-integer))
 (define (partitions N)
   (hash-ref! p# N
              (λ () (for*/list ((m (in-range 1 (add1 N)))
                                (p (partitions (- N m)))
                                #:when (or (null? p) (>= m (car p))))
                      (cons m p)))))
 
 ;; positive-integer -> (listof trees)
 (define (LRTs N)
   (hash-ref! t# N
              (λ ()
                ;; sub1 because we will use the N'th bag to wrap the lot!
                (define ps (partitions (sub1 N)))
                (append*
                 (for/list ((p ps))
                   (apply filtered-cartesian-product (map LRTs p) #:f LRT<=)))))))

(module+ main

 (for-each displayln (LRTs 5))
 (equal? (map (compose length LRTs) (range 1 (add1 13)))
         '(1 1 2 4 9 20 48 115 286 719 1842 4766 12486))) ;; https://oeis.org/A000081</lang>

Output:

(() () () ())
((()) () ())
((()) (()))
((() ()) ())
(((())) ())
((() () ()))
(((()) ()))
(((() ())))
((((()))))
#t

Sidef

Translation of: Python

<lang ruby>func bagchain(x, n, bb, start=0) {

   n || return [x]

   var out = []
   for i in (start .. bb.end) {
       var (c, s) = bb[i]...
       if (c <= n) {
           out += bagchain([x[0] + c, x[1] + s], n-c, bb, i)
       }
   }

   return out

}

func bags(n) {

   n || return 0, ""
   var upto = []
   for i in (n ^.. 1) { upto += bags(i) }
   bagchain([0, ""], n-1, upto).map{|p| [p[0]+1, '('+p[1]+')'] }

}

func replace_brackets(s) {

   var (depth, out) = (0, [])
   for c in s {
       if (c == '(') {
           out.append(<( [ {>[depth%3])
           ++depth
       }
       else {
           --depth
           out.append(<) ] }>[depth%3])
       }
   }
   return out.join

}

for x in (bags(5)) {

   say replace_brackets(x[1])

}</lang>

Output:

([{([])}])
([{()()}])
([{()}{}])
([{}{}{}])
([{()}][])
([{}{}][])
([{}][{}])
([{}][][])
([][][][])

zkl

Note that "( () (()) () )" the same as "( (()) () () )"

Translation of: Python

<lang zkl>fcn bags(n){

  if(not n) return(T(T(0,"")));

  [n-1 .. 1, -1].pump(List,bags).flatten() :
  bagchain(T(0,""), n-1, _).apply(fcn([(c,s)]){ T(c+1,String("(",s,")")) })

} fcn bagchain(x,n,bb,start=0){

  if(not n) return(T(x));

  out := List();
  foreach i in ([start..bb.len()-1]){
     c,s := bb[i];
     if(c<=n) out.extend(bagchain(L(x[0]+c, x[1]+s), n-c, bb, i));
  }
  out

}

1. Maybe this lessens eye strain. Maybe not.

fcn replace_brackets(s){

  depth,out := 0,Sink(String);
  foreach c in (s){
     if(c=="("){

out.write("([{"[depth%3]); depth += 1;

     }else{

depth -= 1; out.write(")]}"[depth%3]);

     }
  }
  out.close()

} foreach x in (bags(5)){ println(replace_brackets(x[1])) } println("or"); b:=bags(5); b.apply("get",1).println(b.len());</lang>

Output:

([{([])}])
([{()()}])
([{()}{}])
([{}{}{}])
([{()}][])
([{}{}][])
([{}][{}])
([{}][][])
([][][][])
or
L("((((()))))","(((()())))","(((())()))","((()()()))","(((()))())","((()())())","((())(()))","((())()())","(()()()())")9