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.
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>
- 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; }</lang>
- Output:
% ./a.out 5 Number of 5-trees: 9 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
D
<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 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
Go
<lang go>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 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
- /
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))
}</lang>
- Output:
When passing a command line argument of 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:
((((())))) (((()()))) ((()(()))) ((()()())) ((())(())) (()((()))) (()(()())) (()()(())) (()()()())
A variant which uses Data.Tree
<lang haskell>import Data.List (nub, sortBy, foldl') --' strict variant of foldl import Data.Ord (comparing) import Data.Tree
bagPatterns :: Int -> [String] bagPatterns n =
nub $ bracketsFromTree . depthSortedTree . treeFromParentIndices <$> parentIndexPermutations n
parentIndexPermutations :: Int -> Int parentIndexPermutations = traverse (enumFromTo 0) . enumFromTo 0 . subtract 2
treeFromParentIndices :: [Int] -> Tree Int treeFromParentIndices pxs =
foldl' --' strict variant of foldl
go
(Node 0 [])
(zip [1 .. (length pxs)] pxs)
where
go tree tplIP =
let root = rootLabel tree
nest = subForest tree
forest
| root == snd tplIP = nest ++ [Node (fst tplIP) []]
| otherwise = (`go` tplIP) <$> nest
in Node root forest
depthSortedTree
:: (Num a, Ord a) => Tree a -> Tree a
depthSortedTree = go
where
go tree
| null (subForest tree) = Node 0 []
| otherwise =
let xs = go <$> subForest tree
in Node
(1 + foldr ((+) . rootLabel) 0 xs)
(sortBy (flip (comparing rootLabel)) xs)
bracketsFromTree :: Tree a -> String bracketsFromTree = foldNest (\xs -> '(' : (concat xs ++ ")"))
foldNest :: ([b] -> b) -> Tree a -> b foldNest f =
let go (Node _ ts) = f (map go ts) in go
main :: IO () main = putStrLn . unlines $ bagPatterns 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 ┌────┬─────┬─────┬─────┬─────┬──────┐ │┌┬┬┐│┌──┬┐│┌──┬┐│┌──┬┐│┌───┐│┌────┐│ │││││││┌┐││││┌┐││││┌┐││││┌┬┐│││┌──┐││ │└┴┴┘│││││││││││││││││││││││││││┌┐│││ │ ││└┘││││└┘││││└┘││││└┴┘│││││││││ │ │└──┴┘│└──┴┘│└──┴┘│└───┘│││└┘│││ │ │ │ │ │ ││└──┘││ │ │ │ │ │ │└────┘│ └────┴─────┴─────┴─────┴─────┴──────┘
JavaScript
ES6
Composing a solution from generic functions. <lang javascript>(() => {
'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()
})();</lang>
- Output:
(()()()()) ((())()()) ((()())()) ((())(())) (((()))()) ((()()())) (((())())) (((()()))) ((((()))))
Julia
<lang julia>bags(n,cache="") = n < 1 ? [(0, "")] :
[(c + 1, "(" * s * ")") for (c, s) in bagchain((0, ""), n - 1,
n < 2 ? [] : reduce(append!, [bags(x) for x in n-1:-1:1]))]
bagchain(x, n, bb, start=1) = n < 1 ? [x] :
reduce(append!, [bagchain((x[1] + bb[i][1], x[2] * bb[i][2]),
n - bb[i][1], bb, i) for i in start:length(bb) if bb[i][1] <= n])
for bag in bags(5)
println(bag[2])
end
</lang>
- Output:
((((())))) (((()()))) (((())())) ((()()())) (((()))()) ((()())()) ((())(())) ((())()()) (()()()())
Kotlin
<lang scala>// 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: ${offset[n + 1] - offset[n]}")
listTrees(n)
}</lang>
- Output:
Number of 5-trees: 9 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
Perl
<lang perl>use strict; use warnings; use feature 'say';
sub bagchain {
my($x, $n, $bb, $start) = @_; return [@$x] unless $n;
my @sets;
$start //= 0;
for my $i ($start .. @$bb-1) {
my($c, $s) = @{$$bb[$i]};
push @sets, bagchain([$$x[0] + $c, $$x[1] . $s], $n-$c, $bb, $i) if $c <= $n
}
@sets
}
sub bags {
my($n) = @_; return [0, ] unless $n;
my(@upto,@sets);
push @upto, bags($_) for reverse 1 .. $n-1;
for ( bagchain([0, ], $n-1, \@upto) ) {
my($c,$s) = @$_;
push @sets, [$c+1, '(' . $s . ')']
}
@sets;
}
sub replace_brackets {
my $bags;
my $depth = 0;
for my $b (split //, $_[0]) {
if ($b eq '(') { $bags .= (qw<( [ {>)[$depth++ % 3] }
else { $bags .= (qw<) ] }>)[--$depth % 3] }
}
$bags
}
say replace_brackets $$_[1] for bags(5);</lang>
- Output:
([{([])}])
([{()()}])
([{()}{}])
([{}{}{}])
([{()}][])
([{}{}][])
([{}][{}])
([{}][][])
([][][][])
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.
Phix
<lang Phix>atom t0 = time() sequence list = {1},
offset = repeat(0,32)
offset[1..2] = 1
function show(integer t, l)
string res = repeat('?',l)
integer level = 0
for i=l to 1 by -1 do
integer r2 = remainder(t,2)
res[i] = "[}(]{)"[mod(level-r2,6)+1]
level += r2*4-2
t = floor(t/2)
end for
if level!=0 then ?9/0 end if
return res
end function
procedure listTrees(integer n)
for i:=offset[n+1]+1 to offset[n+2] do
printf(1,"%s\n",{show(list[i], n*2)})
end for
end procedure
procedure assemble(atom t, integer n, sl, pos, rem) -- -- assemble tree from subtrees -- t: assembled parts so far -- n: length of tree we want to make -- sl: length of subtree we are looking at -- pos: offset of subtree we are looking at -- rem: remaining length to be put together --
if rem == 0 then
list = append(list, t*2+1)
else
if sl>rem then -- need smaller sub-trees
sl = rem
pos = offset[sl+1]
elsif pos>=offset[sl+2] then
-- used up sl-trees, try smaller ones
if sl == 1 then return end if
pos = offset[sl]
sl -= 1
end if
atom u = or_bits(t*power(2,2*sl),list[pos+1])
assemble(u, n, sl, pos, rem-sl)
assemble(t, n, sl, pos+1, rem)
end if
end procedure
procedure mktrees(integer n)
if offset[n+2]=0 then
if n>0 then
mktrees(n - 1)
end if
assemble(0, n, n-1, offset[n], n-1)
offset[n+2] = length(list)
end if
end procedure
procedure main(integer n)
mktrees(n)
atom nt = offset[n+2]-offset[n+1],
td = time()-t0
string e = iff(td>0.1?" ("&elapsed(td)&")":"")
printf(1,"Number of %d-trees: %,d%s\n", {n, nt, e})
if n<=5 then listTrees(n) end if
end procedure for i=0 to 20 do
main(i)
end for</lang>
- Output:
Number of 0-trees: 0
Number of 1-trees: 1
()
Number of 2-trees: 1
({})
Number of 3-trees: 2
({[]})
({}{})
Number of 4-trees: 4
({[()]})
({[][]})
({[]}{})
({}{}{})
Number of 5-trees: 9
({[({})]})
({[()()]})
({[()][]})
({[][][]})
({[()]}{})
({[][]}{})
({[]}{[]})
({[]}{}{})
({}{}{}{})
Number of 6-trees: 20
Number of 7-trees: 48
Number of 8-trees: 115
Number of 9-trees: 286
Number of 10-trees: 719
Number of 11-trees: 1,842
Number of 12-trees: 4,766
Number of 13-trees: 12,486
Number of 14-trees: 32,973
Number of 15-trees: 87,811
Number of 16-trees: 235,381 (0.2s)
Number of 17-trees: 634,847 (0.5s)
Number of 18-trees: 1,721,159 (1.3s)
Number of 19-trees: 4,688,676 (4.0s)
Number of 20-trees: 12,826,228 (13.6s)
Beyond that it gets extremely slow.
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
- 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
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. <lang rexx>/*REXX program lists n─node rooted trees (by enumerating all ways of nesting N bags).*/ 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; end nn= N + N - 1 /*power of 2 that is used for dec start*/ numeric digits 200 /*use enough digs for next calculation.*/ 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. */</lang>
- 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
<lang ring>
- Project : List rooted trees
list = "()" addstr = [] flag = 0 newstr = [] 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)
next see "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
</lang> Output:
for 1 bag: () for 2 bags: (()) for 3 bags: ((())) (()()) for 4 bags: (()()()) ((())()) ((()())) (((())))
Sidef
<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 "( (()) () () )"
<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
}
- 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