# 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.

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.

`#include <stdio.h>#include <stdlib.h> typedef unsigned int uint;typedef unsigned long long tree;#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;}`
Output:
```% ./a.out 5
Number of 5-trees: 9
((((()))))
(((()())))
((()(())))
((()()()))
(()((())))
(()(()()))
((())(()))
(()()(()))
(()()()())
```

## D

Translation of: C
`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);}`
Output:
```Number of 5-trees: 9
((((()))))
(((()())))
((()(())))
((()()()))
(()((())))
(()(()()))
((())(()))
(()()(()))
(()()()())```

## Go

Translation of: C
`package main import (    "fmt"    "log"    "os"    "strconv") type tree uint64 var (    list   []tree    offset = [32]uint{1: 1}) func add(t tree) {    list = append(list, 1|t<<1)} func show(t tree, l uint) {    for ; l > 0; t >>= 1 {        l--        var paren byte        if (t & 1) != 0 {            paren = '('        } else {            paren = ')'        }        fmt.Printf("%c", paren)    }} func listTrees(n uint) {    for i := offset[n]; i < offset[n+1]; i++ {        show(list[i], n*2)        fmt.Println()    }} /* assemble tree from subtreesn:   length of tree we want to maket:   assembled parts so farsl:  length of subtree we are looking atpos: offset of subtree we are looking atrem: remaining length to be put together*/ func assemble(n uint, t tree, sl, pos, rem uint) {    if rem == 0 {        add(t)        return    }     if sl > rem { // need smaller sub-trees        sl = rem        pos = offset[sl]    } else if pos >= offset[sl+1] {        // used up sl-trees, try smaller ones        sl--        if sl == 0 {            return        }        pos = offset[sl]    }     assemble(n, t<<(2*sl)|list[pos], sl, pos, rem-sl)    assemble(n, t, sl, pos+1, rem)} func mktrees(n uint) {    if offset[n+1] > 0 {        return    }    if n > 0 {        mktrees(n - 1)    }     assemble(n, 0, n-1, offset[n-1], n-1)    offset[n+1] = uint(len(list))} func main() {    if len(os.Args) != 2 {        log.Fatal("There must be exactly 1 command line argument")    }    n, err := strconv.Atoi(os.Args[1])    if err != nil {        log.Fatal("Argument is not a valid number")    }    if n <= 0 || n > 19 { // stack overflow for n == 20        n = 5    }    // init 1-tree    add(0)     mktrees(uint(n))    fmt.Fprintf(os.Stderr, "Number of %d-trees: %d\n", n, offset[n+1]-offset[n])    listTrees(uint(n))}`
Output:

When passing a command line argument of 5:

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

There probably is a nicer way than the following--

`-- break n down into sum of smaller integersparts 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 thempick _ [] = []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 uptrees 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`
Output:
```((((()))))
(((()())))
((()(())))
((()()()))
((())(()))
(()((())))
(()(()()))
(()()(()))
(()()()())
```

A variant which uses Data.Tree

`import Data.Treeimport Data.List (nub, sortBy, foldl') --' strict variant of foldlimport Data.Ord (comparing) bagPatterns :: Int -> [String]bagPatterns n =  nub \$  (commasFromTree . depthSortedTree . treeFromParentIndices) <\$>  parentIndexPermutations n parentIndexPermutations :: Int -> [[Int]]parentIndexPermutations =  sequenceA . (enumFromTo 0 <\$>) . enumFromTo 0 . subtract 2 treeFromParentIndices :: [Int] -> Tree InttreeFromParentIndices pxs =   foldl' --' strict variant of foldl    go (Node 0 []) (zip [1 .. (length pxs)] pxs)    where      go tree tplIP =        let root = rootLabel tree            nest = subForest tree        in Node             root             (if root == snd tplIP                then nest ++ [Node (fst tplIP) []]                else (`go` tplIP) <\$> nest) depthSortedTree  :: (Num a, Ord a)  => Tree a -> Tree adepthSortedTree = go  where    go tree =      if null (subForest tree)        then Node 0 []        else let xs = go <\$> subForest tree             in Node                  (1 + foldr ((+) . rootLabel) 0 xs)                  (sortBy (flip (comparing rootLabel)) xs) commasFromTree :: Tree a -> StringcommasFromTree = go  where    go tree = "(" ++ concat (go <\$> subForest tree) ++ ")" main :: IO ()main = putStrLn . unlines \$ bagPatterns 5`
Output:
```(()()()())
((())()())
((()())())
((())(()))
(((()))())
((()()()))
(((())()))
(((()())))
((((()))))```

## J

Support code:

`root=: 1 1 \$ _incr=: ,/@(,"1 0/ [email protected]{:@\$) boxed=: \$:&0 :(<@\:[email protected]([ \$:^:(0 < #@]) [email protected]:=))"1 1 0`

```   ~.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:

`   incr^:3 root_ 0 0 0_ 0 0 1_ 0 0 2_ 0 1 0_ 0 1 1_ 0 1 2`

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:

`   disp=: ('(' , ')' ,~ [: ; [ <@disp"1 0^:(0 < #@]) [email protected]:=) {.   disp incr^:3 root(()()())((())())(()(()))((())())((()()))(((())))`

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
┌────┬─────┬─────┬─────┬─────┬──────┐
│┌┬┬┐│┌──┬┐│┌──┬┐│┌──┬┐│┌───┐│┌────┐│
│││││││┌┐││││┌┐││││┌┐││││┌┬┐│││┌──┐││
│└┴┴┘│││││││││││││││││││││││││││┌┐│││
│    ││└┘││││└┘││││└┘││││└┴┘│││││││││
│    │└──┴┘│└──┴┘│└──┴┘│└───┘│││└┘│││
│    │     │     │     │     ││└──┘││
│    │     │     │     │     │└────┘│
└────┴─────┴─────┴─────┴─────┴──────┘```

## JavaScript

### ES6

Composing a solution from generic functions.

`(() => {    'use strict';     const main = () =>        bagPatterns(5)        .join('\n');     // BAG PATTERNS ---------------------------------------     // bagPatterns :: Int -> [String]    const bagPatterns = n =>        nub(map(            composeList([                commasFromTree,                depthSortedTree,                treeFromParentIndices            ]),            parentIndexPermutations(n)        ));     // parentIndexPermutations :: Int -> [[Int]]    const parentIndexPermutations = n =>        sequenceA(            map(curry(enumFromToInt)(0),                enumFromToInt(0, n - 2)            )        );     // treeFromParentIndices :: [Int] -> Tree Int    const treeFromParentIndices = pxs => {        const go = (tree, tplIP) =>            Node(                tree.root,                tree.root === snd(tplIP) ? (                    tree.nest.concat(Node(fst(tplIP)), [])                ) : map(t => go(t, tplIP), tree.nest)            );        return foldl(            go, Node(0, []),            zip(enumFromToInt(1, pxs.length), pxs)        );    };     // Siblings sorted by descendant count     // depthSortedTree :: Tree a -> Tree Int    const depthSortedTree = t => {        const go = tree =>            isNull(tree.nest) ? (                Node(0, [])            ) : (() => {                const xs = map(go, tree.nest);                return Node(                    1 + foldl((a, x) => a + x.root, 0, xs),                    sortBy(flip(comparing(x => x.root)), xs)                );            })();        return go(t);    };     // Serialisation of the tree structure     // commasFromTree :: Tree a -> String    const commasFromTree = tree => {        const go = t => `(\${concat(map(go, t.nest))})`        return go(tree);    };      // GENERIC FUNCTIONS --------------------------------------     // Node :: a -> [Tree a] -> Tree a    const Node = (v, xs) => ({        type: 'Node',        root: v, // any type of value (but must be consistent across tree)        nest: xs || []    });     // Tuple (,) :: a -> b -> (a, b)    const Tuple = (a, b) => ({        type: 'Tuple',        '0': a,        '1': b,        length: 2    });     // comparing :: (a -> b) -> (a -> a -> Ordering)    const comparing = f =>        (x, y) => {            const                a = f(x),                b = f(y);            return a < b ? -1 : (a > b ? 1 : 0);        };     // composeList :: [(a -> a)] -> (a -> a)    const composeList = fs =>        x => fs.reduceRight((a, f) => f(a), x, fs);     // concat :: [[a]] -> [a]    // concat :: [String] -> String    const concat = xs =>        xs.length > 0 ? (() => {            const unit = typeof xs[0] === 'string' ? '' : [];            return unit.concat.apply(unit, xs);        })() : [];         // concatMap :: (a -> [b]) -> [a] -> [b]    const concatMap = (f, xs) => []        .concat.apply(            [],            (Array.isArray(xs) ? (                xs            ) : xs.split('')).map(f)        );         // cons :: a -> [a] -> [a]    const cons = (x, xs) =>  [x].concat(xs);     // Flexibly handles two or more arguments, applying    // the function directly if the argument array is complete,    // or recursing with a concatenation of any existing and    // newly supplied arguments, if gaps remain.    // curry :: ((a, b) -> c) -> a -> b -> c    const curry = (f, ...args) => {        const go = xs => xs.length >= f.length ? (            f.apply(null, xs)        ) : function() {            return go(xs.concat(Array.from(arguments)));        };        return go(args);    };     // enumFromToInt :: Int -> Int -> [Int]    const enumFromToInt = (m, n) =>        n >= m ? (            iterateUntil(x => x >= n, x => 1 + x, m)        ) : [];     // flip :: (a -> b -> c) -> b -> a -> c    const flip = f => (a, b) => f.apply(null, [b, a]);     // foldl :: (a -> b -> a) -> a -> [b] -> a    const foldl = (f, a, xs) => xs.reduce(f, a);     // fst :: (a, b) -> a    const fst = tpl => tpl[0];     // isNull :: [a] -> Bool    // isNull :: String -> Bool    const isNull = xs =>        Array.isArray(xs) || typeof xs === 'string' ? (            xs.length < 1        ) : undefined;     // iterateUntil :: (a -> Bool) -> (a -> a) -> a -> [a]    const iterateUntil = (p, f, x) => {        let vs = [x],            h = x;        while (!p(h))(h = f(h), vs.push(h));        return vs;    };     // liftA2List :: (a -> b -> c) -> [a] -> [b] -> [c]    const liftA2List = (f, xs, ys) =>        concatMap(x => concatMap(y => [f(x, y)], ys), xs);     // map :: (a -> b) -> [a] -> [b]    const map = (f, xs) => xs.map(f);     // nub :: [a] -> [a]    const nub = xs => nubBy((a, b) => a === b, xs);     // nubBy :: (a -> a -> Bool) -> [a] -> [a]    const nubBy = (p, xs) => {        const go = xs => xs.length > 0 ? (() => {            const x = xs[0];            return [x].concat(                go(xs.slice(1)                    .filter(y => !p(x, y))                )            )        })() : [];        return go(xs);    };     // sequenceA :: (Applicative f, Traversable t) => t (f a) -> f (t a)    const sequenceA = tfa =>        traverseList(x => x, tfa);     // traverseList :: (Applicative f) => (a -> f b) -> [a] -> f [b]    const traverseList = (f, xs) => {        const lng = xs.length;        return 0 < lng ? (() => {            const                vLast = f(xs[lng - 1]),                t = vLast.type || 'List';            return xs.slice(0, -1).reduceRight(                (ys, x) => liftA2List(cons, f(x), ys),                liftA2List(cons, vLast, [[]])            );        })() : [            []        ];    };     // snd :: (a, b) -> b    const snd = tpl => tpl[1];     // sortBy :: (a -> a -> Ordering) -> [a] -> [a]    const sortBy = (f, xs) =>        xs.slice()        .sort(f);     // zip :: [a] -> [b] -> [(a, b)]    const zip = (xs, ys) =>        xs.slice(0, Math.min(xs.length, ys.length))        .map((x, i) => Tuple(x, ys[i]));     // MAIN ---    return main()})();`
Output:
```(()()()())
((())()())
((()())())
((())(()))
(((()))())
((()()()))
(((())()))
(((()())))
((((()))))```

## Kotlin

Translation of: C
`// version 1.1.3 typealias Tree = Long val treeList = mutableListOf<Tree>()val offset = IntArray(32) { if (it == 1) 1 else 0 }  fun append(t: Tree) {    treeList.add(1L or (t shl 1))} fun show(t: Tree, l: Int) {    var tt = t    var ll = l    while (ll-- > 0) {        print(if (tt % 2L == 1L) "(" else ")")        tt = tt ushr 1    }} fun listTrees(n: Int) {    for (i in offset[n] until offset[n + 1]) {        show(treeList[i], n * 2)        println()    }} /* 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*/ fun assemble(n: Int, t: Tree, sl: Int, pos: Int, rem: Int) {    if (rem == 0) {        append(t)        return    }     var pp = pos    var ss = sl     if (sl > rem) { // need smaller subtrees        ss = rem        pp = offset[ss]    }    else if (pp >= offset[ss + 1]) {        // used up sl-trees, try smaller ones        ss--        if(ss == 0) return        pp = offset[ss]    }     assemble(n, (t shl (2 * ss)) or treeList[pp], ss, pp, rem - ss)    assemble(n, t, ss, pp + 1, rem)} fun makeTrees(n: Int) {    if (offset[n + 1] != 0) return    if (n > 0) makeTrees(n - 1)    assemble(n, 0, n - 1, offset[n - 1], n - 1)    offset[n + 1] = treeList.size} fun main(args: Array<String>) {    if (args.size != 1) {        throw IllegalArgumentException("There must be exactly 1 command line argument")    }    val n = args[0].toIntOrNull()    if (n == null) throw IllegalArgumentException("Argument is not a valid number")    // n limited to 12 to avoid overflowing default stack     if (n !in 1..12) throw IllegalArgumentException("Argument must be between 1 and 12")     // init 1-tree    append(0)     makeTrees(n)    println("Number of \$n-trees: \${offset[n + 1] - offset[n]}")     listTrees(n)}`
Output:
```Number of 5-trees: 9
((((()))))
(((()())))
((()(())))
((()()()))
(()((())))
(()(()()))
((())(()))
(()()(()))
(()()()())
```

## Perl 6

Bags are represented by Perl 6 type `Bag`.

`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}; `
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

`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 # 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]))`
Output:
```([{([])}])
([{()()}])
([{()}{}])
([{}{}{}])
([{()}][])
([{}{}][])
([{}][{}])
([{}][][])
([][][][])
```

Another method by incrementing subtrees:

`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 aunique degree (n-1) predecessor, inverting the process leads to the successorsto 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 nodesare 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))`

## 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`
Output:
```(() () () ())
((()) () ())
((()) (()))
((() ()) ())
(((())) ())
((() () ()))
(((()) ()))
(((() ())))
((((()))))
#t```

## REXX

This REXX version uses (internally) a binary string to represent nodes on a tree   (0   is a left parenthesis,   1   is a right parenthesis).   A   ()   is translated to a   O.

`/*REXX program lists  n─node  rooted trees  (by enumerating all ways of nesting N bato).*/parse arg N .                                    /*obtain optional argument from the CL.*/if N=='' | N==","  then N=5                      /*Not specified?  Then use the default.*/if N>5  then do;  say N  "isn't supported for this program at this time.";   exit 13;  endnn= N + N - 1                                    /*power of 2 that is used for dec start*/numeric digits 200                               /*ensure enough digs for the next calc.*/numeric digits max(9, 1 + length( x2b( d2x(2**(nn+1) - 1) ) ) )  /*limit decimal digits.*/start=2**nn    +    (2**nn) % 2                  /*calculate the starting decimal number*/if N==1  then start= 2**1                        /*treat the start for unity as special.*/_= copies('─', 20)"► "                           /*demonstrative literal for solutions. */#=0                                              /*count of ways to nest bags (so far). */\$=                                               /*string holds possible duplicious strs*/     do j=start + start//2  to 2**(nn+1)-1  by 2 /*limit the search, smart start and end*/     t= x2b( d2x(j) )   +   0                    /*convert dec number to a binary string*/     z= length( space( translate(t, , 0), 0) )   /*count the number of zeros in bit str.*/     if z\==n  then iterate                      /*Not enough zeroes?  Then skip this T.*/     if N>1  then if left(t,N)==right(t,N)  then iterate       /*left side ≡ right side?*/     if N>2  then if right(t,2)==    10  then iterate  /*has a right-most isolated bag ?*/     if N>3  then if right(t,4)==  1100  then iterate  /* "  "      "         "     "  ?*/     if N>4  then if right(t,6)==111000  then iterate  /* "  "      "         "     "  ?*/     if N>4  then if right(t,6)==110100  then iterate  /* "  "      "         "     "  ?*/     if N>4  then if right(t,6)==100100  then iterate  /* "  "      "         "     "  ?*/     if wordpos(t, \$)\==0  then iterate                        /*duplicate bag stuffing?*/     say _  changestr('()', translate(t, "()", 10),  'O')      /*show a compact display.*/     #= # + 1                                    /*bump count of ways of nesting bags.  */     \$=\$  translate( reverse(t), 01, 10)         /*save a (possible) duplicious string. */     end   /*j*/say                                              /*separate number─of─ways with a blank.*/say # ' is the number of ways to nest' n "bags." /*stick a fork in it,  we're all done. */`
output   when using the default input:
```────────────────────►  (OOOO)
────────────────────►  (OO(O))
────────────────────►  (O(OO))
────────────────────►  (O((O)))
────────────────────►  ((O)(O))
────────────────────►  ((OOO))
────────────────────►  ((O(O)))
────────────────────►  (((OO)))
────────────────────►  ((((O))))

9  is the number of ways to nest 5 bags.
```

## Ring

` # Project : List rooted trees list = "()"addstr = []flag = 0newstr = []str = []np = [1,2,3,4]for nr = 1 to len(np)      if nr = 1         bg1 = "bag"       else          bg1 = "bags"      ok      see "for " + nr + " " + bg1 + " :" + nl     permutation(list,nr*2)     listroot(nr*2)nextsee "ok" + nl func listroot(pn)        for n = 1 to len(addstr)             result(addstr[n],pn)             if flag = 1                see "" + addstr[n] + nl                addstr[n]             ok        next func result(list,pn)        flag = 0        newstr = list        while substr(newstr, "()") != 0                 if list = "()" or list = "(())"                    flag = 1                    exit                 ok                 num = substr(newstr, "()")                 newstr = substr(newstr, "()", "")                 if left(list,2) = "()" or right(list,2) = "()" or left(list,4) = "(())" or right(list,4) = "(())"                    flag = 0                    exit                 else                     if len(list) != 2 and len(list) != 4 and newstr = ""                       flag = 1                       exit                    ok                 ok        end func permutation(list,pn)       addstr = []       while true               str = ""               for n = 1 to pn                    rnd = random(1) + 1                    str = str + list[rnd]               next               add(addstr,str)               for m = 1 to len(addstr)                    for p = m + 1 to len(addstr) - 1                         if addstr[m] = addstr[p]                            del(addstr,p)                         ok                    next               next               if len(addstr) = pow(2,pn)                  exit               ok       end `

Output:

```for 1 bag:
()
for 2 bags:
(())
for 3 bags:
((()))
(()())
for 4 bags:
(()()())
((())())
((()()))
(((())))
```

## Sidef

Translation of: Python
`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])}`
Output:
```([{([])}])
([{()()}])
([{()}{}])
([{}{}{}])
([{()}][])
([{}{}][])
([{}][{}])
([{}][][])
([][][][])
```

## zkl

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

Translation of: Python
`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}# 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());`
Output:
```([{([])}])
([{()()}])
([{()}{}])
([{}{}{}])
([{()}][])
([{}{}][])
([{}][{}])
([{}][][])
([][][][])
or
L("((((()))))","(((()())))","(((())()))","((()()()))","(((()))())","((()())())","((())(()))","((())()())","(()()()())")9
```