List rooted trees
You are encouraged to solve this task according to the task description, using any language you may know.
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.
11l
F bagchain(x, n, bb, start = 0)
I n == 0
R [x]
[(Int, String)] out
L(i) start .< bb.len
V (c, s) = bb[i]
I c <= n
out.extend(bagchain((x[0] + c, x[1]‘’s), n - c, bb, i))
R out
F bags(n)
I n == 0
R [(0, ‘’)]
[(Int, String)] upto
L(x) (n - 1 .< 0).step(-1)
upto.extend(bags(x))
R bagchain((0, ‘’), n - 1, upto).map((c, s) -> (c + 1, ‘(’s‘)’))
F replace_brackets(s)
V depth = 0
[String] out
L(c) s
I c == ‘(’
out.append(‘([{’[depth % 3])
depth++
E
depth--
out.append(‘)]}’[depth % 3])
R out.join(‘’)
L(x) bags(5)
print(replace_brackets(x[1]))- Output:
([{([])}])
([{()()}])
([{()}{}])
([{}{}{}])
([{()}][])
([{}{}][])
([{}][{}])
([{}][][])
([][][][])
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 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
C++
#include <iostream>
#include <vector>
std::vector<long> TREE_LIST;
std::vector<int> OFFSET;
void init() {
for (size_t i = 0; i < 32; i++) {
if (i == 1) {
OFFSET.push_back(1);
} else {
OFFSET.push_back(0);
}
}
}
void append(long t) {
TREE_LIST.push_back(1 | (t << 1));
}
void show(long t, int l) {
while (l-- > 0) {
if (t % 2 == 1) {
std::cout << '(';
} else {
std::cout << ')';
}
t = t >> 1;
}
}
void listTrees(int n) {
for (int i = OFFSET[n]; i < OFFSET[n + 1]; i++) {
show(TREE_LIST[i], 2 * n);
std::cout << '\n';
}
}
void assemble(int n, long t, int sl, int pos, int rem) {
if (rem == 0) {
append(t);
return;
}
auto pp = pos;
auto ss = sl;
if (sl > rem) {
ss = rem;
pp = OFFSET[ss];
} else if (pp >= OFFSET[ss + 1]) {
ss--;
if (ss == 0) {
return;
}
pp = OFFSET[ss];
}
assemble(n, t << (2 * ss) | TREE_LIST[pp], ss, pp, rem - ss);
assemble(n, t, ss, pp + 1, rem);
}
void makeTrees(int n) {
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] = TREE_LIST.size();
}
void test(int n) {
if (n < 1 || n > 12) {
throw std::runtime_error("Argument must be between 1 and 12");
}
append(0);
makeTrees(n);
std::cout << "Number of " << n << "-trees: " << OFFSET[n + 1] - OFFSET[n] << '\n';
listTrees(n);
}
int main() {
init();
test(5);
return 0;
}
- Output:
Number of 5-trees: 9 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
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);
}
- Output:
Number of 5-trees: 9 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
EasyLang
sysconf zero_based
#
trees[] = [ ]
len offs[] 32
offs[1] = 1
proc append t . .
v = bitor 1 bitshift t 1
trees[] &= v
.
proc show t l . .
while l > 0
l -= 1
if t mod 2 = 1
write "("
else
write ")"
.
t = t div 2
.
.
proc list n . .
for i = offs[n] to offs[n + 1] - 1
show trees[i] n * 2
print ""
.
.
proc assemble n t sl pos rem . .
if rem = 0
append t
return
.
if sl > rem
sl = rem
pos = offs[sl]
elif pos >= offs[sl + 1]
sl -= 1
if sl = 0
return
.
pos = offs[sl]
.
h = bitor bitshift t (2 * sl) trees[pos]
assemble n h sl pos rem - sl
assemble n t sl pos + 1 rem
.
proc make n . .
if offs[n + 1] <> 0
return
.
if n > 0
make n - 1
.
assemble n 0 n - 1 offs[n - 1] n - 1
offs[n + 1] = len trees[]
.
proc test n . .
append 0
make n
print "Number of " & n & "-trees: " & offs[n + 1] - offs[n]
list n
.
test 5
- Output:
Number of 5-trees: 9 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
FreeBASIC
Dim Shared As String*2 list = "()"
Dim Shared As String addstr()
Dim Shared As Boolean flag = False
Dim Shared As String cad, newcad
Sub result(list As String, pn As Integer)
flag = False
newcad = list
While Instr(newcad, "()") <> 0
If list = "()" Or list = "(())" Then
flag = True
Exit Sub
End If
Dim As Integer num = Instr(newcad, "()")
newcad = Left(newcad, num - 1) & Mid(newcad, num + 2)
If Left(list, 2) = "()" Or Right(list, 2) = "()" Or Left(list, 4) = "(())" Or Right(list, 4) = "(())" Then
flag = False
Exit Sub
Else
If Len(list) <> 2 And Len(list) <> 4 And newcad = "" Then
flag = True
Exit Sub
End If
End If
Wend
End Sub
Sub permutation(list As String, pn As Integer)
Redim addstr(0)
Do
cad = ""
For n As Integer = 1 To pn
Dim As Integer rand = Int(Rnd * 2) + 1
cad &= Mid(list, rand, 1)
Next
Dim As Integer found = False
For m As Integer = 0 To Ubound(addstr)
If addstr(m) = cad Then
found = True
Exit For
End If
Next
If Not found Then
Redim Preserve addstr(Ubound(addstr) + 1)
addstr(Ubound(addstr)) = cad
End If
If Ubound(addstr) + 1 = 2 ^ pn Then Exit Do
Loop
End Sub
Sub listroot(pn As Integer)
For n As Integer = 0 To Ubound(addstr)
result(addstr(n), pn)
If flag Then Print "" & addstr(n)
Next
End Sub
' test cases
Dim As Integer np(5) = {0, 1, 2, 3, 4, 9}
For nr As Integer = 0 To Ubound(np)
Dim As String bg1 = Iif(nr = 1, "bag", "bags")
Print !"\nfor " & np(nr) & " " & bg1 & " :"
permutation(list, nr * 2)
listroot(nr * 2)
Next
Sleep
- Output:
for 0 bags : for 1 bag : () for 2 bags : (()) for 3 bags : (()()) ((())) for 4 bags : (()()()) ((())()) (((()))) ((()())) (()(())) for 9 bags : (((()))()) ((()(()))) ((())()()) (((()()))) ((()())()) (()()()()) (()(())()) ((())(())) (()((()))) ((((())))) (((())())) (()()(())) ((()()())) (()(()()))
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))
}
- Output:
When passing a command line argument of 5:
Number of 5-trees: 9 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
Haskell
There probably is a nicer way than the following--
-- break n down into sum of smaller integers
parts :: Int -> [[(Int, Int)]]
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 :: Int -> [String] -> [String]
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 :: Int -> [String]
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 :: IO ()
main = mapM_ putStrLn $ trees 5
- Output:
((((())))) (((()()))) ((()(()))) ((()()())) ((())(())) (()((()))) (()(()())) (()()(())) (()()()())
A variant expressed in terms of Data.Tree
import Data.List (foldl', nub, sortOn) --' strict variant of foldl
import Data.Ord (comparing)
import Data.Tree (Tree (..), foldTree)
-------------------- LIST ROOTED TREES -------------------
bagPatterns :: Int -> [String]
bagPatterns n =
nub $
foldTree asBrackets
. foldTree depthSorted
. treeFromParentIndices
<$> parentIndexPermutations n
--------------------------- TEST -------------------------
main :: IO ()
main = putStrLn . unlines $ bagPatterns 5
----------------------- DEFINITIONS ----------------------
asBrackets :: a -> [String] -> String
asBrackets = const (('(' :) . (<> ")") . concat)
depthSorted :: a -> [Tree Int] -> Tree Int
depthSorted = const (Node <$> length <*> sortOn rootLabel)
parentIndexPermutations :: Int -> [[Int]]
parentIndexPermutations =
traverse
(enumFromTo 0)
. enumFromTo 0
. subtract 2
treeFromParentIndices :: [Int] -> Tree Int
treeFromParentIndices ixs =
foldl' --' strict variant of foldl
go
(Node 0 [])
(zip [1 .. length ixs] ixs)
where
go tree (i, x) = Node root forest
where
root = rootLabel tree
nest = subForest tree
forest
| root == x = nest <> [Node i []]
| otherwise = (`go` (i, x)) <$> nest
- Output:
(()()()()) (()()(())) (()(()())) ((())(())) (()((()))) ((()()())) ((()(()))) (((()()))) ((((()))))
J
Support code:
root=: 1 1 $ _
incr=: ,/@(,"1 0/ i.@{:@$)
boxed=: $:&0 :(<@\:~@([ $:^:(0 < #@]) I.@:=))"1 1 0
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:
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 < #@]) I.@:=) {.
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 ┌────┬─────┬─────┬─────┬─────┬──────┐ │┌┬┬┐│┌──┬┐│┌──┬┐│┌──┬┐│┌───┐│┌────┐│ │││││││┌┐││││┌┐││││┌┐││││┌┬┐│││┌──┐││ │└┴┴┘│││││││││││││││││││││││││││┌┐│││ │ ││└┘││││└┘││││└┘││││└┴┘│││││││││ │ │└──┴┘│└──┴┘│└──┴┘│└───┘│││└┘│││ │ │ │ │ │ ││└──┘││ │ │ │ │ │ │└────┘│ └────┴─────┴─────┴─────┴─────┴──────┘
Java
import java.util.ArrayList;
import java.util.List;
public class ListRootedTrees {
private static final List<Long> TREE_LIST = new ArrayList<>();
private static final List<Integer> OFFSET = new ArrayList<>();
static {
for (int i = 0; i < 32; i++) {
if (i == 1) {
OFFSET.add(1);
} else {
OFFSET.add(0);
}
}
}
private static void append(long t) {
TREE_LIST.add(1 | (t << 1));
}
private static void show(long t, int l) {
while (l-- > 0) {
if (t % 2 == 1) {
System.out.print('(');
} else {
System.out.print(')');
}
t = t >> 1;
}
}
private static void listTrees(int n) {
for (int i = OFFSET.get(n); i < OFFSET.get(n + 1); i++) {
show(TREE_LIST.get(i), n * 2);
System.out.println();
}
}
private static void assemble(int n, long t, int sl, int pos, int rem) {
if (rem == 0) {
append(t);
return;
}
var pp = pos;
var ss = sl;
if (sl > rem) {
ss = rem;
pp = OFFSET.get(ss);
} else if (pp >= OFFSET.get(ss + 1)) {
ss--;
if (ss == 0) {
return;
}
pp = OFFSET.get(ss);
}
assemble(n, t << (2 * ss) | TREE_LIST.get(pp), ss, pp, rem - ss);
assemble(n, t, ss, pp + 1, rem);
}
private static void makeTrees(int n) {
if (OFFSET.get(n + 1) != 0) {
return;
}
if (n > 0) {
makeTrees(n - 1);
}
assemble(n, 0, n - 1, OFFSET.get(n - 1), n - 1);
OFFSET.set(n + 1, TREE_LIST.size());
}
private static void test(int n) {
if (n < 1 || n > 12) {
throw new IllegalArgumentException("Argument must be between 1 and 12");
}
append(0);
makeTrees(n);
System.out.printf("Number of %d-trees: %d\n", n, OFFSET.get(n + 1) - OFFSET.get(n));
listTrees(n);
}
public static void main(String[] args) {
test(5);
}
}
- Output:
Number of 5-trees: 9 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
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:
(()()()()) ((())()()) ((()())()) ((())(())) (((()))()) ((()()())) (((())())) (((()()))) ((((()))))
jq
Works with gojq, the Go implementation of jq
In this entry, a straightforward approach is used: bags are added one at a time, employing the one-liner `sortmap` to select canonical representations.
`rootedtrees($n)` displays the rooted trees of order $n as a stream of JSON arrays. `count_rootedtrees($n) displays a table of [$n, $count] values, where $count is the number of $n-trees.
For trees of order up to about 12, the algorithm is quite snappy (*). Beyond 17, the results are very slow in coming, and hence the limit in the displayed table.
(*) Subsecond times in the case of the C implementation of jq.
# add one bag somewhere
def addbag:
def sortmap: map(sortmap) | sort;
if . == null then [] # one bag
else ([[]] + .),
(paths as $p
| getpath($p) as $x
| setpath($p; [[]] + $x)
| sortmap # indistinguishability of bags
)
end ;
# emit a stream of the distinct rooted trees of order $n > 0
def rootedtrees($n):
if $n==1 then []
else foreach range(0; $n-1) as $i ([[]];
[.[] | addbag] | unique;
select($i == $n-2))
| .[]
end;
# emit $n arrays of the form [$i, $count] for 0 < $i <= $n
def count_rootedtrees($n):
[1, 1],
foreach range(0; $n - 1) as $i ([[]];
[.[] | addbag] | unique;
[$i + 2, length]) ;
rootedtrees(5),
"",
count_rootedtrees(17)- Output:
[[],[],[],[]] [[],[],[[]]] [[],[[],[]]] [[],[[[]]]] [[[]],[[]]] [[[],[],[]]] [[[],[[]]]] [[[[],[]]]] [[[[[]]]]] [1,1] [2,1] [3,2] [4,4] [5,9] [6,20] [7,48] [8,115] [9,286] [10,719] [11,1842] [12,4766] [13,12486] [14,32973] [15,87811] [16,235381] [17,634847]
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
- Output:
((((())))) (((()()))) (((())())) ((()()())) (((()))()) ((()())()) ((())(())) ((())()()) (()()()())
Kotlin
// 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 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
Lua
tree_list = {}
offset = {}
function init()
for i=1,32 do
if i == 2 then
table.insert(offset, 1)
else
table.insert(offset, 0)
end
end
end
function append(t)
local v = 1 | (t << 1)
table.insert(tree_list, v)
end
function show(t, l)
while l > 0 do
l = l - 1
if (t % 2) == 1 then
io.write('(')
else
io.write(')')
end
t = t >> 1
end
end
function listTrees(n)
local i = offset[n]
while i < offset[n + 1] do
show(tree_list[i + 1], n * 2)
print()
i = i + 1
end
end
function assemble(m, t, sl, pos, rem)
if rem == 0 then
append(t)
return
end
local pp = pos
local ss = sl
if sl > rem then
ss = rem
pp = offset[ss]
elseif pp >= offset[ss + 1] then
ss = ss - 1
if ss == 0 then
return
end
pp = offset[ss]
end
assemble(n, t << (2 * ss) | tree_list[pp + 1], ss, pp, rem - ss)
assemble(n, t, ss, pp + 1, rem)
end
function makeTrees(n)
if offset[n + 1] ~= 0 then
return
end
if n > 0 then
makeTrees(n - 1)
end
assemble(n, 0, n - 1, offset[n - 1], n - 1)
offset[n + 1] = #tree_list
end
function test(n)
append(0)
makeTrees(n)
print(string.format("Number of %d-trees: %d", n, offset[n+1] - offset[n]))
listTrees(n)
end
init()
test(5)
- Output:
Number of 5-trees: 9 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
Mathematica / Wolfram Language
The following defines functions which create a nest of functions, bags wrapped inside a main bag which are stored inside a "cabinet".
Addbags[configs_List] :=
DeleteDuplicates[
Map[LexicographicSort,
Catenate[AddbagAll /@ configs], \[Infinity]]]
AddbagAll[config_] :=
Addbag[config, #] & /@ Position[config, bag[___], \[Infinity]]
Addbag[config_, pos_] :=
ReplacePart[config, pos -> Append[Extract[config, pos], bag[]]]
With[{n = 5}, Nest[Addbags, {cabinet[bag[]]}, n - 1] // Column]
The output can be viewed as a tree graph by replacing the last line with
With[{n = 5}, TreeForm /@ Nest[Addbags, {cabinet[bag[]]}, n - 1]]
- Output:
cabinet[bag[bag[bag[bag[bag[]]]]]] cabinet[bag[bag[bag[bag[],bag[]]]]] cabinet[bag[bag[bag[],bag[bag[]]]]] cabinet[bag[bag[],bag[bag[bag[]]]]] cabinet[bag[bag[bag[],bag[],bag[]]]] cabinet[bag[bag[],bag[bag[],bag[]]]] cabinet[bag[bag[bag[]],bag[bag[]]]] cabinet[bag[bag[],bag[],bag[bag[]]]] cabinet[bag[bag[],bag[],bag[],bag[]]]
Nim
import os, strformat, strutils
type
Tree = int
Trees = object
list: seq[Tree]
offsets: array[32, int]
func isOdd(n: int): bool = (n and 1) != 0
func append(trees: var Trees; tree: Tree) =
trees.list.add(1 or tree shl 1)
proc show(tree: Tree; n: int) =
var tree = tree
var n = n
while n > 0:
dec n
stdout.write if tree.isOdd: '(' else: ')'
tree = tree shr 1
stdout.write '\n'
proc print(trees: Trees; n: int) =
for i in trees.offsets[n]..<trees.offsets[n + 1]:
trees.list[i].show(n * 2)
#[ 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(trees: var Trees; n: int; t: Tree; sl, pos, rem: int) =
if rem == 0:
trees.append(t)
return
var p = pos
var s = sl
if s > rem:
s = rem
p = trees.offsets[s]
elif p >= trees.offsets[s + 1]:
# Used up sl-trees, try smaller ones.
dec s
if s == 0: return
p = trees.offsets[s]
trees.assemble(n, t shl ( 2 * s) or trees.list[p], s, p, rem - s)
trees.assemble(n, t, s, p + 1, rem)
func make(trees: var Trees; n: int) =
if trees.offsets[n + 1] != 0: return
if n > 0: trees.make(n - 1)
trees.assemble(n, 0, n - 1, trees.offsets[n - 1], n - 1)
trees.offsets[n + 1] = trees.list.len
when isMainModule:
if paramCount() != 1:
raise newException(ValueError, "there must be exactly one command line argument")
let n = try:
paramStr(1).parseInt()
except ValueError:
raise newException(ValueError, "argument is not a valid number")
# Insure "n" is limited to 12 to avoid overflowing default stack.
if n notin 1..12:
raise newException(ValueError, "argument must be between 1 and 12")
# Init 1-tree.
var trees: Trees
trees.offsets[1] = 1
trees.append(0)
trees.make(n)
echo &"Number of {n}-trees: {trees.offsets[n + 1] - trees.offsets[n]}"
trees.print(n)
- Output:
For instance, for n = 5:
Number of 5-trees: 9 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
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);
- Output:
([{([])}])
([{()()}])
([{()}{}])
([{}{}{}])
([{()}][])
([{}{}][])
([{}][{}])
([{}][][])
([][][][])
Phix
with javascript_semantics
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 iff(platform()=JS?12:20) do
main(i)
end for
- 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. Under pwa/p2js 14-trees exceeded the JavaScript call stack limit, so I knocked a couple more off for safety.
fast analytical version
Matches https://oeis.org/A000081/b000081.txt, a(1000) takes about half a second.
include mpfr.e
sequence acache = {mpz_init(1)},
ajs = {}
function a(integer n)
assert(n>=0)
if n=0 then return mpz_init(0) end if
for l=length(acache)+1 to n do
mpz {aj,ak} = mpz_inits(2)
for d in factors(l-1,1) do
mpz_addmul_ui(aj,acache[d],d)
end for
ajs &= aj
for j=1 to l-1 do
mpz_addmul(ak,ajs[j],acache[l-j])
end for
mpz_divexact_ui(ak,ak,l-1)
acache &= ak
end for
return acache[n]
end function
for n in tagset(20,0)&1000 do
printf(1,"Number of %d-trees: %s\n",{n,mpz_get_short_str(a(n),comma_fill:=true)})
end for
- 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 Number of 17-trees: 634,847 Number of 18-trees: 1,721,159 Number of 19-trees: 4,688,676 Number of 20-trees: 12,826,228 Number of 1000-trees: 6,506,773,573,532,28...,820,228,861,274,503 (466 digits)
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 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))
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
Raku
(formerly Perl 6)
Bags are represented by Raku 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.
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 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)"► "; o = 'o' /*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 /*this a duplicate bag stuffing? */
say @ changestr('()', translate(t, "()", 10), o) /*show a compact display with oh.*/
#= # + 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 = 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
endOutput:
for 1 bag: () for 2 bags: (()) for 3 bags: ((())) (()()) for 4 bags: (()()()) ((())()) ((()())) (((())))
Ruby
TREE_LIST = []
OFFSET = []
for i in 0..31
if i == 1 then
OFFSET << 1
else
OFFSET << 0
end
end
def append(t)
TREE_LIST << (1 | (t << 1))
end
def show(t, l)
while l > 0
l = l - 1
if t % 2 == 1 then
print '('
else
print ')'
end
t = t >> 1
end
end
def listTrees(n)
for i in OFFSET[n] .. OFFSET[n + 1] - 1
show(TREE_LIST[i], n * 2)
print "\n"
end
end
def assemble(n, t, sl, pos, rem)
if rem == 0 then
append(t)
return
end
if sl > rem then
sl = rem
pos = OFFSET[sl]
elsif pos >= OFFSET[sl + 1] then
sl = sl - 1
if sl == 0 then
return
end
pos = OFFSET[sl]
end
assemble(n, t << (2 * sl) | TREE_LIST[pos], sl, pos, rem - sl)
assemble(n, t, sl, pos + 1, rem)
end
def makeTrees(n)
if OFFSET[n + 1] != 0 then
return
end
if n > 0 then
makeTrees(n - 1)
end
assemble(n, 0, n - 1, OFFSET[n - 1], n - 1)
OFFSET[n + 1] = TREE_LIST.length()
end
def test(n)
if n < 1 || n > 12 then
raise ArgumentError.new("Argument must be between 1 and 12")
end
append(0)
makeTrees(n)
print "Number of %d-trees: %d\n" % [n, OFFSET[n + 1] - OFFSET[n]]
listTrees(n)
end
test(5)
- Output:
Number of 5-trees: 9 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
Rust
use std::env;
use std::str::FromStr;
fn add(list: &mut Vec<usize>, t: usize) {
list.push(1 | t << 1);
}
fn show(mut t: usize, mut len: usize) {
while len > 0 {
len -= 1;
print!("{}", if (t & 1) != 0 { '(' } else { ')' });
t >>= 1;
}
}
fn list_trees(list: &Vec<usize>, offset: &Vec<usize>, n: usize) {
for i in offset[n]..offset[n+1] {
show(list[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
*/
fn assemble(list: &mut Vec<usize>, offset: &mut Vec<usize>, n: usize, t: usize, mut sl: usize, mut pos: usize, rem: usize) {
if rem == 0 {
add(list, 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 -= 1;
if sl == 0 {
return;
}
pos = offset[sl];
}
assemble(list, offset, n, (t << (2*sl)) | list[pos], sl, pos, rem-sl);
assemble(list, offset, n, t, sl, pos+1, rem);
}
fn mktrees(list: &mut Vec<usize>, offset: &mut Vec<usize>, n: usize) {
if offset[n+1] > 0 {
return;
}
if n > 0 {
mktrees(list, offset, n - 1);
}
assemble(list, offset, n, 0, n-1, offset[n-1], n-1);
offset[n+1] = list.len();
}
fn main() {
let list: &mut Vec<usize> = &mut vec![];
let offset: &mut Vec<usize> = &mut vec![0_usize; 32];
offset[1] = 1_usize;
let mut n = 5;
let args: Vec<String> = env::args().collect();
if args.len() == 2 {
n = usize::from_str(args[1].as_str()).unwrap_or(5);
n = if n <= 0 || n > 19 { 5 } else { n };
}
add(list, 0);
mktrees(list, offset, n);
println!("Number of {}-trees: {}", n, offset[n+1]-offset[n]);
list_trees(list, offset, n);
}
- Output:
Number of 5-trees: 9 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
Sidef
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:
([{([])}])
([{()()}])
([{()}{}])
([{}{}{}])
([{()}][])
([{}{}][])
([{}][{}])
([{}][][])
([][][][])
Wren
import "./long" for ULong
import "os" for Process
var treeList = []
var offset = List.filled(32, 0)
offset[1] = 1
var append = Fn.new { |t|
treeList.add(ULong.one | (t << 1))
}
var show = Fn.new { |t, len|
while (len > 0) {
len = len - 1
System.write(t.isOdd ? "(" : ")")
t = t >> 1
}
}
var listTrees = Fn.new { |n|
for (i in offset[n]...offset[n+1]) {
show.call(treeList[i], n * 2)
System.print()
}
}
/* 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
*/
var assemble // recursive
assemble = Fn.new { |n, t, sl, pos, rem|
if (rem == 0) {
append.call(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
sl = sl - 1
if (sl == 0) return
pos = offset[sl]
}
assemble.call(n, (t << (2 * sl)) | treeList[pos], sl, pos, rem - sl)
assemble.call(n, t, sl, pos + 1, rem)
}
var makeTrees // recursive
makeTrees = Fn.new { |n|
if (offset[n + 1] != 0) return
if (n > 0) makeTrees.call(n - 1)
assemble.call(n, ULong.zero, n - 1, offset[n - 1], n - 1)
offset[n + 1] = treeList.count
}
var args = Process.arguments
if (args.count != 1) Fiber.abort("There must be exactly 1 command line argument.")
var n = Num.fromString(args[0])
if (!n) Fiber.abort("Argument is not a valid number.")
if (n < 1 || n > 12) Fiber.abort("Argument must be between 1 and 12.")
// init 1-tree
append.call(ULong.zero)
makeTrees.call(n)
System.print("Number of %(n)-trees: %(offset[n + 1] - offset[n])")
listTrees.call(n)
- Output:
Number of 5-trees: 9 ((((())))) (((()()))) ((()(()))) ((()()())) (()((()))) (()(()())) ((())(())) (()()(())) (()()()())
zkl
Note that "( () (()) () )" the same as "( (()) () () )"
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