List rooted trees

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.

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.

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

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

There ought to be a more elegant way of doing this:

<lang J>incr=:3 :0

r=.
for_bag.y do.
  in=.>bag
  r=. r,<a:,in
  for_j.in i. ~.in do.
    for_more. incr j { in do.
      r=. r,< /:~ more j} in
    end.
  end.
end.
~.r

)

disp=: [: ; ('(',')',~disp)each^:(0<L.)</lang>

Here, we use boxes to represent bags, and a: is an empty box... er... bag. Each use of incr adds a new bag in each of the distinct possible ways.

We actually overdo that slightly - finding all the distinct "one more bag" possibilities for each of our argument cases, and then eliminate duplicates because sometimes adding a bag in one place in one configuration does the same thing as adding a bag in a different place for a different configuration. This deduping requires a certain amount of normalization (at each level we define a canonical order).

(The python example looks like it takes advantage of this canonical structure from the start, probably worth reading...)

Finally, we use disp to convert our tree structures to the desired "parenthesis" notation.

Required example:

<lang J> disp"0 incr^:4 a: (()()()()) (()()(())) ((())(())) (()(()())) (()((()))) ((()()())) ((()(()))) (((()()))) ((((()))))</lang>

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>