# Word break problem

Word break problem 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.
Given an input string and a dictionary of words, segment the input string into a space-separated sequence of dictionary words if possible.

## 11l

Translation of: D
```T Node
String val
[String] parsed
F (val, [String] parsed = [String]())
.val = val
.parsed = copy(parsed)

F word_break(s, dictionary)
[[String]] matches
V queue = Deque([Node(s)])
L !queue.empty
V node = queue.pop_left()
I node.val.empty
matches [+]= node.parsed
E
L(word) dictionary
I node.val.starts_with(word)
V val_new = node.val[word.len .< node.val.len]
V parsed_new = copy(node.parsed)
parsed_new [+]= word
queue [+]= Node(val_new, parsed_new)
R matches

F process(d, test_strings)
L(test_string) test_strings
V matches = word_break(test_string, d)
print((‘String = ’test_string‘, Dictionary = ’String(d)‘.  Solutions =’)‘ ’matches.len)
L(match) matches
print(‘  Word Break = ’match)
print()

V d = [‘a’, ‘aa’, ‘b’, ‘ab’, ‘aab’]
process(d, [‘aab’, ‘aa b’])

d = [‘abc’, ‘a’, ‘ac’, ‘b’, ‘c’, ‘cb’, ‘d’]
process(d, [‘abcd’, ‘abbc’, ‘abcbcd’, ‘acdbc’, ‘abcdd’])```
Output:
```String = aab, Dictionary = [a, aa, b, ab, aab].  Solutions = 4
Word Break = [aab]
Word Break = [a, ab]
Word Break = [aa, b]
Word Break = [a, a, b]

String = aa b, Dictionary = [a, aa, b, ab, aab].  Solutions = 0

String = abcd, Dictionary = [abc, a, ac, b, c, cb, d].  Solutions = 2
Word Break = [abc, d]
Word Break = [a, b, c, d]

String = abbc, Dictionary = [abc, a, ac, b, c, cb, d].  Solutions = 1
Word Break = [a, b, b, c]

String = abcbcd, Dictionary = [abc, a, ac, b, c, cb, d].  Solutions = 3
Word Break = [abc, b, c, d]
Word Break = [a, b, cb, c, d]
Word Break = [a, b, c, b, c, d]

String = acdbc, Dictionary = [abc, a, ac, b, c, cb, d].  Solutions = 2
Word Break = [ac, d, b, c]
Word Break = [a, c, d, b, c]

String = abcdd, Dictionary = [abc, a, ac, b, c, cb, d].  Solutions = 2
Word Break = [abc, d, d]
Word Break = [a, b, c, d, d]
```

## Aime

```integer
wordbreak(record dict, text phrase, integer p, list words)
{
integer complete, n;
text s;

complete = 0;
if (rsk_lower(dict, phrase, s)) {
if (s == phrase) {
words.append(s);
complete = 1;
} else {
do {
n = 0;
while (phrase[n] == s[n]) {
n += 1;
}
if (!n) {
break;
}
words.append(cut(s, 0, n));
complete = wordbreak(dict, project(phrase, n), p + 1, words);
if (complete) {
break;
}
words.delete(-1);
} while (rsk_less(dict, s, s));
}
}

if (!p) {
o_(phrase, ":");
if (complete) {
words.ucall(o_, 1, " ");
} else {
o_(" can't break");
}
o_newline();

words.clear;
}

complete;
}

integer
main(void)
{
record dict;

dict.fit("a", 0, "bc", 0, "abc", 0, "cd", 0, "b", 0);
list("abcd", "abbc", "abcbcd", "acdbc", "abcdd").ucall(wordbreak, 1, dict, 0, list());

return 0;
}```
Output:
```abcd: a b cd
abbc: a b bc
abcbcd: abc b cd
acdbc: a cd bc
abcdd: can't break```

## ALGOL 68

Translation of: Nim
```# utility functions: #

PROC includes = ([]STRING row, STRING s) BOOL:
# whether row includes s #
BEGIN
FOR i TO UPB row DO IF s = row[i] THEN true FI OD;
FALSE EXIT
true: TRUE
END;

PRIO & = 6;
OP & = (STRING s, []STRING row) []STRING:
# returns row with s prepended to it #
BEGIN
[UPB row + 1]STRING result;
result[2 : UPB result] := row;
result[1] := s;
result
END;

PROC add = (REF FLEX[]FLEX[]STRING rows, []STRING row) VOID:
# adds row to the end of rows (in place) #
BEGIN
[UPB rows+1]FLEX[0]STRING tmp;
tmp[:UPB rows] := rows;
tmp[UPB tmp] := row;
rows := tmp
END;

# actual solution: #

PROC word breaks = ([]STRING dict, STRING word) [][]STRING:
# build recursively the list of breaks for a word, using the given dictionary #
BEGIN
FLEX[0]FLEX[0]STRING result;
FOR last TO UPB word - 1 DO
STRING part1 := word[1 : last];
IF includes(dict, part1) THEN
STRING part2 := word[last + 1 : ];
IF includes(dict, part2) THEN add(result, (part1, part2)) FI;
[][]STRING sub results = word breaks(dict, part2);
FOR i FROM LWB sub results TO UPB sub results DO
OD
FI
OD;
result
END;

PROC break word = ([]STRING dict, STRING word) VOID:
# find the ways to break a word and display the result #
BEGIN
print((word, ":", newline));
[][]STRING word seqs = word breaks(dict, word);
IF UPB word seqs = 0 THEN
print(("    <no break possible>", newline))
ELSE
FOR i FROM LWB word seqs TO UPB word seqs DO
[]STRING seq = word seqs[i];
print("   ");
FOR j FROM LWB seq TO UPB seq DO
print((" ", seq[j]))
OD;
print(newline)
OD
FI
END;

[]STRING dict = ("a", "bc", "abc", "cd", "b");

[]STRING words = ("abcd", "abbc", "abcbcd", "acdbc", "abcdd");

FOR i TO UPB words DO
break word(dict, words[i])
OD```
Output:
```abcd:
a b cd
abbc:
a b bc
abcbcd:
a bc b cd
abc b cd
acdbc:
a cd bc
abcdd:
<no break possible>
```

## C++

Translation of: Rust
```#include <algorithm>
#include <iostream>
#include <optional>
#include <set>
#include <string>
#include <string_view>
#include <vector>

struct string_comparator {
using is_transparent = void;
bool operator()(const std::string& lhs, const std::string& rhs) const {
return lhs < rhs;
}
bool operator()(const std::string& lhs, const std::string_view& rhs) const {
return lhs < rhs;
}
bool operator()(const std::string_view& lhs, const std::string& rhs) const {
return lhs < rhs;
}
};

using dictionary = std::set<std::string, string_comparator>;

template <typename iterator, typename separator>
std::string join(iterator begin, iterator end, separator sep) {
std::string result;
if (begin != end) {
result += *begin++;
for (; begin != end; ++begin) {
result += sep;
result += *begin;
}
}
return result;
}

auto create_string(const std::string_view& s,
const std::vector<std::optional<size_t>>& v) {
auto idx = s.size();
std::vector<std::string_view> sv;
while (v[idx].has_value()) {
size_t prev = v[idx].value();
sv.push_back(s.substr(prev, idx - prev));
idx = prev;
}
std::reverse(sv.begin(), sv.end());
return join(sv.begin(), sv.end(), ' ');
}

std::optional<std::string> word_break(const std::string_view& str,
const dictionary& dict) {
auto size = str.size() + 1;
std::vector<std::optional<size_t>> possible(size);
auto check_word = [&dict, &str](size_t i, size_t j)
-> std::optional<size_t> {
if (dict.find(str.substr(i, j - i)) != dict.end())
return i;
return std::nullopt;
};
for (size_t i = 1; i < size; ++i) {
if (!possible[i].has_value())
possible[i] = check_word(0, i);
if (possible[i].has_value()) {
for (size_t j = i + 1; j < size; ++j) {
if (!possible[j].has_value())
possible[j] = check_word(i, j);
}
if (possible[str.size()].has_value())
return create_string(str, possible);
}
}
return std::nullopt;
}

int main(int argc, char** argv) {
dictionary dict;
dict.insert("a");
dict.insert("bc");
dict.insert("abc");
dict.insert("cd");
dict.insert("b");
auto result = word_break("abcd", dict);
if (result.has_value())
std::cout << result.value() << '\n';
return 0;
}
```
Output:
```a b cd
```

## D

Translation of: Java
```import std.algorithm;
import std.range;
import std.stdio;

void main() {
string[] dict = ["a", "aa", "b", "ab", "aab"];
process(dict, ["aab", "aa b"]);

dict = ["abc", "a", "ac", "b", "c", "cb", "d"];
process(dict, ["abcd", "abbc", "abcbcd", "acdbc", "abcdd"]);
}

void process(string[] dict, string[] testStrings) {
foreach (testString; testStrings) {
auto matches = wordBreak(testString, dict);
writeln("String = ", testString, ", Dictionary = ", dict, ".  Solutions = ", matches.length);
foreach (match; matches) {
writeln("  Word Break = ", match);
}
writeln();
}
}

string[][] wordBreak(string s, string[] dictionary) {
string[][] matches;
Node[] queue = [Node(s)];
while (!queue.empty) {
auto node = queue.front;
queue.popFront;
// Check if fully parsed
if (node.val.length == 0) {
matches ~= node.parsed;
} else {
foreach (word; dictionary) {
// Check for match
if (node.val.startsWith(word)) {
auto valNew = node.val[word.length .. node.val.length];
auto parsedNew = node.parsed.dup;
parsedNew ~= word;
queue ~= Node(valNew, parsedNew);
}
}
}
}
return matches;
}

struct Node {
string val;
string[] parsed;

this(string initial) {
val = initial;
}

this(string s, string[] p) {
val = s;
parsed = p;
}
}
```
Output:
```String = aab, Dictionary = ["a", "aa", "b", "ab", "aab"].  Solutions = 4
Word Break = ["aab"]
Word Break = ["a", "ab"]
Word Break = ["aa", "b"]
Word Break = ["a", "a", "b"]

String = aa b, Dictionary = ["a", "aa", "b", "ab", "aab"].  Solutions = 0

String = abcd, Dictionary = ["abc", "a", "ac", "b", "c", "cb", "d"].  Solutions = 2
Word Break = ["abc", "d"]
Word Break = ["a", "b", "c", "d"]

String = abbc, Dictionary = ["abc", "a", "ac", "b", "c", "cb", "d"].  Solutions = 1
Word Break = ["a", "b", "b", "c"]

String = abcbcd, Dictionary = ["abc", "a", "ac", "b", "c", "cb", "d"].  Solutions = 3
Word Break = ["abc", "b", "c", "d"]
Word Break = ["a", "b", "cb", "c", "d"]
Word Break = ["a", "b", "c", "b", "c", "d"]

String = acdbc, Dictionary = ["abc", "a", "ac", "b", "c", "cb", "d"].  Solutions = 2
Word Break = ["ac", "d", "b", "c"]
Word Break = ["a", "c", "d", "b", "c"]

String = abcdd, Dictionary = ["abc", "a", "ac", "b", "c", "cb", "d"].  Solutions = 2
Word Break = ["abc", "d", "d"]
Word Break = ["a", "b", "c", "d", "d"]```

## Delphi

Translation of: Go
```program Word_break_problem;

{\$APPTYPE CONSOLE}

uses
System.SysUtils,
System.Generics.Collections;

type
TDict = TDictionary<string, boolean>;

TPrefix = record
length: integer;
broken: TArray<string>;
constructor Create(len: integer; b: TArray<string>);
end;

const
TESTS: TArray<string> = ['abcd', 'abbc', 'abcbcd', 'acdbc', 'abcdd'];

var
d: TDict;

function newDict(words: TArray<string>): TDict;
var
w: string;
begin
Result := TDict.Create();
for w in words do
end;

function wordBreak(s: string; var broken: TArray<string>): boolean;
var
ed, i: Integer;
w: string;
p: TPrefix;
bp: TArray<TPrefix>;
begin
SetLength(broken, 0);

if s.IsEmpty then
exit(true);

bp := [TPrefix.Create(0, [])];

for ed := 1 to s.Length do
for i := High(bp) downto 0 do
begin

w := s.Substring(bp[i].length, ed - bp[i].length);
if d.ContainsKey(w) then
begin
broken := bp[i].broken + [w];
if ed = s.Length then
exit(true);
p := TPrefix.Create(ed, broken);
bp := bp + [p];
Break;
end;
end;
Result := false;
end;

{ TPrefix }

constructor TPrefix.Create(len: integer; b: TArray<string>);
begin
broken := b;
length := len;
end;

var
s: string;
b: TArray<string>;
ok: boolean;

begin
d := newDict(['a', 'bc', 'abc', 'cd', 'b']);
for s in TESTS do
if wordBreak(s, b) then
Writeln(Format('%s: %s', [s, string.join(' ', b)]))
else
Writeln('can''t break');
d.Free;

end.
```

## Go

```package main

import (
"fmt"
"strings"
)

type dict map[string]bool

func newDict(words ...string) dict {
d := dict{}
for _, w := range words {
d[w] = true
}
return d
}

func (d dict) wordBreak(s string) (broken []string, ok bool) {
if s == "" {
return nil, true
}
type prefix struct {
length int
broken []string
}
bp := []prefix{{0, nil}}
for end := 1; end <= len(s); end++ {
for i := len(bp) - 1; i >= 0; i-- {
w := s[bp[i].length:end]
if d[w] {
b := append(bp[i].broken, w)
if end == len(s) {
return b, true
}
bp = append(bp, prefix{end, b})
break
}
}
}
return nil, false
}

func main() {
d := newDict("a", "bc", "abc", "cd", "b")
for _, s := range []string{"abcd", "abbc", "abcbcd", "acdbc", "abcdd"} {
if b, ok := d.wordBreak(s); ok {
fmt.Printf("%s: %s\n", s, strings.Join(b, " "))
} else {
fmt.Println("can't break")
}
}
}
```
Output:
```abcd: a b cd
abbc: a b bc
abcbcd: a bc b cd
acdbc: a cd bc
can't break
```

Translation of: Javascript
```import Data.List (isPrefixOf, intercalate)
import Data.Tree (Tree(..))

wordBreaks :: [String] -> String -> String
wordBreaks ws = (++) <*> (":\n" ++) . report . fmap go . tokenTrees ws
where
go t
| null (subForest t) = [rootLabel t]
| otherwise = subForest t >>= ((:) (rootLabel t) . go)
report xs
| null xs = "\tNot parseable with these words"
| otherwise = unlines \$ ('\t' :) . intercalate " -> " <\$> xs

tokenTrees :: [String] -> String -> [Tree String]
tokenTrees ws = go
where
go s
| s `elem` ws = [Node s []]
| otherwise = ws >>= next s
next s w
| w `isPrefixOf` s = parse w (go (drop (length w) s))
| otherwise = []
parse w xs
| null xs = []
| otherwise = [Node w xs]

-------------------------- TEST ---------------------------
ws, texts :: [String]
ws = words "a bc abc cd b"

texts = words "abcd abbc abcbcd acdbc abcdd"

main :: IO ()
main = (putStrLn . unlines) \$ wordBreaks ws <\$> texts
```
Output:
```abcd:
a -> b -> cd

abbc:
a -> b -> bc

abcbcd:
a -> bc -> b -> cd
abc -> b -> cd

acdbc:
a -> cd -> bc

abcdd:
Not parseable with these words```

## J

With such short sentences we can find the partition sets, then check that all are words.

```all_partitions=: <@(<;.1)"1 _~  (1,.[:#:[:i.2^<:@:#)  NB. all_partitions 'abcd'
word_break=: ([ #~ 0 = [: #&>@:] -.L:_1 _)~ all_partitions
main=: (] , (;:inv L:_1@:word_break >))"_ 0 boxopen
```
```   NB. demonstrate partitions of four integers
all_partitions i. 4
┌─────────┬─────────┬─────────┬─────────┬─────────┬─────────┬─────────┬─────────┐
│┌───────┐│┌─────┬─┐│┌───┬───┐│┌───┬─┬─┐│┌─┬─────┐│┌─┬───┬─┐│┌─┬─┬───┐│┌─┬─┬─┬─┐│
││0 1 2 3│││0 1 2│3│││0 1│2 3│││0 1│2│3│││0│1 2 3│││0│1 2│3│││0│1│2 3│││0│1│2│3││
│└───────┘│└─────┴─┘│└───┴───┘│└───┴─┴─┘│└─┴─────┘│└─┴───┴─┘│└─┴─┴───┘│└─┴─┴─┴─┘│
└─────────┴─────────┴─────────┴─────────┴─────────┴─────────┴─────────┴─────────┘

NB. demonstrate word_break
NB.   (,L:_1]0;1 2;0 1 2;2 3;1)
NB. ┌─┬───┬─────┬───┬─┐
NB. │0│1 2│0 1 2│2 3│1│
NB. └─┴───┴─────┴───┴─┘
(,L:_1]0;1 2;0 1 2;2 3;1) word_break i. 4
┌─────────┐
│┌─┬─┬───┐│
││0│1│2 3││
│└─┴─┴───┘│
└─────────┘

NB. save and display the dictionary
[dictionary=: ;: 'a bc abc cd b'
┌─┬──┬───┬──┬─┐
│a│bc│abc│cd│b│
└─┴──┴───┴──┴─┘

NB. demonstrate main
dictionary main 'abc'
┌───┬───┬────┐
│abc│abc│a bc│
└───┴───┴────┘

NB. solution
dictionary main ;: 'abcd abbc abcbcd acdbc abcdd'
┌──────┬────────┬─────────┐
│abcd  │a b cd  │         │
├──────┼────────┼─────────┤
│abbc  │a b bc  │         │
├──────┼────────┼─────────┤
│abcbcd│abc b cd│a bc b cd│
├──────┼────────┼─────────┤
│acdbc │a cd bc │         │
├──────┼────────┼─────────┤
│abcdd │        │         │
└──────┴────────┴─────────┘
```

## Java

Accept string to be parsed, and dictionary of words, as method arguments.

```import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Queue;

public class WordBreak {

public static void main(String[] args) {
List<String> dict = Arrays.asList("a", "aa", "b", "ab", "aab");
for ( String testString : Arrays.asList("aab", "aa b") ) {
List<List<String>> matches = wordBreak(testString, dict);
System.out.printf("String = %s, Dictionary = %s.  Solutions = %d:%n", testString, dict, matches.size());
for ( List<String> match : matches ) {
System.out.printf(" Word Break = %s%n", match);
}
System.out.printf("%n");
}
dict = Arrays.asList("abc", "a", "ac", "b", "c", "cb", "d");
for ( String testString : Arrays.asList("abcd", "abbc", "abcbcd", "acdbc", "abcdd") ) {
List<List<String>> matches = wordBreak(testString, dict);
System.out.printf("String = %s, Dictionary = %s.  Solutions = %d:%n", testString, dict, matches.size());
for ( List<String> match : matches ) {
System.out.printf(" Word Break = %s%n", match);
}
System.out.printf("%n");
}
}

private static List<List<String>> wordBreak(String s, List<String> dictionary) {
List<List<String>> matches = new ArrayList<>();
while ( ! queue.isEmpty() ) {
Node node = queue.remove();
//  Check if fully parsed
if ( node.val.length() == 0 ) {
}
else {
for ( String word : dictionary ) {
//  Check for match
if ( node.val.startsWith(word) ) {
String valNew = node.val.substring(word.length(), node.val.length());
List<String> parsedNew = new ArrayList<>();
}
}
}
}
return matches;
}

private static class Node {
private String val;  //  Current unmatched string
private List<String> parsed;  //  Matches in dictionary
public Node(String initial) {
val = initial;
parsed = new ArrayList<>();
}
public Node(String s, List<String> p) {
val = s;
parsed = p;
}
}

}
```
Output:
```String = aab, Dictionary = [a, aa, b, ab, aab].  Solutions = 4:
Word Break = [aab]
Word Break = [a, ab]
Word Break = [aa, b]
Word Break = [a, a, b]

String = aa b, Dictionary = [a, aa, b, ab, aab].  Solutions = 0:

String = abcd, Dictionary = [abc, a, ac, b, c, cb, d].  Solutions = 2:
Word Break = [abc, d]
Word Break = [a, b, c, d]

String = abbc, Dictionary = [abc, a, ac, b, c, cb, d].  Solutions = 1:
Word Break = [a, b, b, c]

String = abcbcd, Dictionary = [abc, a, ac, b, c, cb, d].  Solutions = 3:
Word Break = [abc, b, c, d]
Word Break = [a, b, cb, c, d]
Word Break = [a, b, c, b, c, d]

String = acdbc, Dictionary = [abc, a, ac, b, c, cb, d].  Solutions = 2:
Word Break = [ac, d, b, c]
Word Break = [a, c, d, b, c]

String = abcdd, Dictionary = [abc, a, ac, b, c, cb, d].  Solutions = 2:
Word Break = [abc, d, d]
Word Break = [a, b, c, d, d]
```

## JavaScript

Composing a solution from generic functions.

Translation of: Python
```(() => {
'use strict';

const main = () => {
const
wds = words('a bc abc cd b'),
texts = words('abcd abbc abcbcd acdbc abcdd');

return unlines(
map(wordBreaks(wds),
texts
)
);
};

// WORD BREAKS ----------------------------------------

// tokenTrees :: [String] -> String -> [Tree String]
const tokenTrees = (wds, s) => {
const go = s =>
wds.includes(s) ? (
[Node(s, [])]
) : bindList(wds, next(s));
const next = s => w =>
s.startsWith(w) ? (
parse(w, go(s.slice(w.length)))
) : [];
const parse = (w, xs) =>
0 < xs.length ? [Node(w, xs)] : xs;
return go(s);
};

// wordBreaks :: [String] -> String -> String
const wordBreaks = wds => s => {
const
// go :: Tree a -> [a]
go = t => isNull(t.nest) ? [
t.root
] : bindList(
t.nest,
compose(cons(t.root), go),
),
parses = map(go, tokenTrees(wds, s));
return `\${s}:\n` + (
0 < parses.length ? unlines(
map(x => '\t' + intercalateS(' -> ', x),
parses
)
) : '\t(Not parseable with these words)'
);
};

// GENERIC FUNCTIONS ----------------------------------

// Node :: a -> [Tree a] -> Tree a
const Node = (v, xs) => ({
type: 'Node',
root: v,
nest: xs || []
});

// bindList (>>=) :: [a] -> (a -> [b]) -> [b]
const bindList = (xs, mf) => [].concat.apply([], xs.map(mf));

// compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
const compose = (f, g) => x => f(g(x));

// cons :: a -> [a] -> [a]
const cons = x => xs => [x].concat(xs);

// intercalateS :: String -> [String] -> String
const intercalateS = (s, xs) =>
xs.join(s);

// isNull :: [a] -> Bool
// isNull :: String -> Bool
const isNull = xs =>
Array.isArray(xs) || ('string' === typeof xs) ? (
1 > xs.length
) : undefined;

// isPrefixOf takes two lists or strings and returns
// true iff the first is a prefix of the second.

// isPrefixOf :: [a] -> [a] -> Bool
// isPrefixOf :: String -> String -> Bool
const isPrefixOf = (xs, ys) => {
const pfx = (xs, ys) => {
const intX = xs.length;
return 0 < intX ? (
ys.length >= intX ? xs[0] === ys[0] && pfx(
xs.slice(1), ys.slice(1)
) : false
) : true;
};
return 'string' !== typeof xs ? (
pfx(xs, ys)
) : ys.startsWith(xs);
};

// map :: (a -> b) -> [a] -> [b]
const map = (f, xs) => xs.map(f);

// unlines :: [String] -> String
const unlines = xs => xs.join('\n');

// words :: String -> [String]
const words = s => s.split(/\s+/);

// MAIN ---
return main();
})();
```
Output:
```abcd:
a -> b -> cd
abbc:
a -> b -> bc
abcbcd:
a -> bc -> b -> cd
abc -> b -> cd
acdbc:
a -> cd -> bc
abcdd:
(Not parseable with these words)```

## jq

Works with: jq

Works with gojq, the Go implementation of jq

The solution offered here does not use regular expressions.

The function `string2words` is a generator that can produce all possible parses of a string but which can be stopped after finding the first parse, as in the first demonstration.

In the second demonstration, the generator is used to count the number of possible parses using the same toy dictionary as used in the first demonstration.

In the third demonstration, the well-known dictionary unixdict.txt is used to determine the number of possible parses of the string "totalreturn".

```def words: ["a", "bc", "abc", "cd", "b"];
def strings: ["abcd", "abbc", "abcbcd", "acdbc", "abcdd"];

# input: an array of allowed words
# output: a stream giving all possible parses of the given string into
# the allowed words; each output is an array showing how the string
# has been parsed.
def string2words(string):
. as \$dict
# Input: array of words
# Output: augmented array
| def s2w(s):
if s=="" then .
else \$dict[] as \$word
| (s|startswith(\$word)) as \$ix
| if \$ix
then s[\$word|length:] as \$rest
| (. + [\$word]) | s2w(\$rest)
else empty
end
end;
[] | s2w(string);

def count(s): reduce s as \$x (0; .+1) ;

def demo1:
strings[] as \$s
| words
| (first(string2words(\$s)) // []) as \$parsed
| "\(\$s) => \(\$parsed|join(" "))" ;

def demo2:
strings[] as \$s
| words
| count(string2words(\$s))
| "\(\$s) has \(.) parse\(if . == 1 then "" else "s" end)." ;

# demo3 assumes an invocation along the lines of:
#  jq  -Rrn -f program.jq unixdict.txt
def demo3:
"returntotal" as \$s
| "\(\$s) has \([inputs] |  count(string2words(\$s)) parses."

demo1, " ", demo2, "", demo3```
Output:
```abcd => a b cd
abbc => a b bc
abcbcd => a bc b cd
acdbc => a cd bc
abcdd =>

abcd has 1 parse.
abbc has 1 parse.
abcbcd has 2 parses.
acdbc has 1 parse.
abcdd has 0 parses.

"returntotal" has 99 parses.
```

## Julia

Some extra loops to record and print all solutions.

```words = ["a", "bc", "abc", "cd", "b"]
strings = ["abcd", "abbc", "abcbcd", "acdbc", "abcdd"]

subregex = join(words, ")|(")
regexes = ["\^\(\(\$subregex\)\)\{\$i}\\$" for i in 6:-1:1]

function wordbreak()
for s in strings
solutions = []
for regex in regexes
rmat = match(Regex(regex), s)
if rmat != nothing
push!(solutions, ["\$w" for w in Set(rmat.captures) if w != nothing])
end
end
if length(solutions) > 0
println("\$(length(solutions)) Solution(s) for \$s:")
for sol in solutions
println("   Solution: \$(sol)")
end
else
println("No solutions for \$s : No fitting matches found.")
end
end
end

wordbreak()
```
Output:
```
1 Solution(s) for abcd:

Solution: SubString{String}["cd", "b", "a"]

1 Solution(s) for abbc:

Solution: SubString{String}["b", "a", "bc"]

2 Solution(s) for abcbcd:

Solution: SubString{String}["cd", "b", "a", "bc"]
Solution: SubString{String}["cd", "abc", "b"]

1 Solution(s) for acdbc:

Solution: SubString{String}["cd", "a", "bc"]

No solutions for abcdd : No fitting matches found.```

## Kotlin

I've downloaded the free dictionary at http://www.puzzlers.org/pub/wordlists/unixdict.txt for this task. All single letters from 'a' to 'z' are considered to be words by this dictionary but 'bc' and 'cd' which I'd have expected to be present are not.

```// version 1.1.3

import java.io.File

val partitions = mutableListOf<List<String>>()

fun partitionString(s: String, ml: MutableList<String>, level: Int) {
for (i in s.length - 1 downTo 1) {
val part1 = s.substring(0, i)
val part2 = s.substring(i)
if (part2.length > 1) {
ml.removeAt(ml.lastIndex)
partitionString(part2, ml, level + 1)
}
while (ml.size > level) ml.removeAt(ml.lastIndex)
}
}

fun main(args: Array<String>) {
val strings = listOf("abcd", "abbc", "abcbcd", "acdbc", "abcdd")
for (s in strings) {
partitions.clear()
val ml = mutableListOf<String>()
partitionString(s, ml, 0)
val solutions = mutableListOf<List<String>>()
for (partition in partitions) {
var allInDict = true
for (item in partition) {
if (words.indexOf(item) == -1) {
allInDict = false
break
}
}
}
val plural = if (solutions.size == 1) "" else "s"
println("\$s: \${solutions.size} solution\$plural")
for (solution in solutions) {
println("    \${solution.joinToString(" ")}")
}
println()
}
}
```
Output:
```abcd: 2 solutions
abc d
a b c d

abbc: 1 solution
a b b c

abcbcd: 3 solutions
abc b c d
a b cb c d
a b c b c d

acdbc: 2 solutions
ac d b c
a c d b c

abcdd: 2 solutions
abc d d
a b c d d
```

## Lua

```-- a specialized dict format is used to minimize the
-- possible candidates for this particalur problem
function genDict(ws)
for w in ws:gmatch"%w+" do
local lw = w:lower()
if not dup[lw] then
local len = #rest
end
end
end
return d
end

-- sample default dict
local defWords = "a;bc;abc;cd;b"
local defDict = genDict(defWords)

function wordbreak(w, dict)
if type(w)~='string' or w:len()==0 then
return nil,'emprty or not a string'
end

dict = type(dict)=='string' and genDict(dict) or dict or defDict

local r, len = {}, #w

-- backtracking
local function try(i)
if i>len then return true end
if not d then return end
for j=math.min(d.n, len-i),0,-1 do -- prefer longer first
if d[j] then
local rest = w:sub(i+1,i+j):lower()
if d[j][rest] then
r[1+#r] = w:sub(i,i+j)
if try(i+j+1) then
return true
else
r[#r]=nil
end
end
end
end
end

if try(1) then
return table.unpack(r)
else
return nil,'-no solution-'
end
end

-- test
local test = {'abcd','abbc','abcbcd','acdbc','abcdd'  }
for i=1,#test do
print(test[i],wordbreak(test[i]))
end
```
Output:
```abcd	a	b	cd
abbc	a	b	bc
abcbcd	abc	b	cd
acdbc	a	cd	bc
abcdd	nil	-no solution-```

## Nim

We provide two ways to initialize the dictionary: either explicitly by providing the list of words or by providing a file name from which extract the words. In the second case, it is possible to specify the minimal length of the words to keep. This is useful mainly to eliminate all the one letter words we get with “unixdict.txt”.

```import sequtils, sets, strutils

type
Dict = HashSet[string]
WordSeq = seq[string]

proc initDict(words: openArray[string]): Dict =
## Initialize a dictionary from a list of words.
words.toHashSet

proc initDict(fileName: string; minlength = 0): Dict =
## Initialize a dictionary with words from a file.
## Only words with minimal length are retained.
for word in filename.lines:
if word.len >= minLength:
result.incl word

func wordBreaks(dict: Dict; word: string): seq[WordSeq] =
## Build recursively the list of breaks for a word, using the given dictionary.
for last in 0..<word.high:
let part1 = word[0..last]
if part1 in dict:
let part2 = word[last+1..^1]
if part2 in dict: result.add(@[part1, part2])

proc breakWord(dict: Dict; word: string) =
## Find the ways to break a word and display the result.
echo word, ": "
let wordSeqs = dict.wordBreaks(word)
if wordSeqs.len == 0:
echo "    <no break possible>"
else:
for wordSeq in wordSeqs:
echo "    ", wordSeq.join(" ")

when isMainModule:

const EDict = ["a", "bc", "abc", "cd", "b"]
echo "Using explicit dictionary: ", EDict
var dict = initDict(EDict)
for s in ["abcd", "abbc", "abcbcd", "acdbc", "abcdd"]:
dict.breakWord(s)

echo("\nUsing “unixdict.txt” dictionary without single letter words.")
dict = initDict("unixdict.txt", 2)
dict.breakWord("because")
dict.breakWord("software")
```
Output:

We used the same dictionary and words than in the other languages solutions and also added two examples using “unixdict.txt”.

```Using explicit dictionary: ["a", "bc", "abc", "cd", "b"]
abcd:
a b cd
abbc:
a b bc
abcbcd:
a bc b cd
abc b cd
acdbc:
a cd bc
abcdd:
<no break possible>

Using “unixdict.txt” dictionary without single letter words.
because:
be cause
be ca use
software:
so ft ware
so ft wa re
soft ware
soft wa re```

## Perl

```use strict;
use warnings;

my @words = <a o is pi ion par per sip miss able>;
print "\$_: " . word_break(\$_,@words) . "\n" for <a aa amiss parable opera operable inoperable permission mississippi>;

sub word_break {
my(\$word,@dictionary) = @_;
my @matches;
my \$one_of = join '|', @dictionary;
@matches = \$word =~ /^ (\$one_of) (\$one_of)? (\$one_of)? (\$one_of)? \$/x; # sub-optimal: limited number of matches
return join(' ', grep {\$_} @matches) || "(not possible)";
}
```
Output:
```a: a
aa: a a
amiss: a miss
parable: par able
opera: o per a
operable: o per able
inoperable: in o per able
permission: per miss ion
permissible: Not possible
mississippi: miss is sip pi```

## Phix

The distributed version also contains a few experiments with applying a real dictionary, largely unsuccessful.

```--
-- demo\rosetta\Word_break_problem.exw
-- ===================================
--
with javascript_semantics
procedure populate_dict(sequence s)
for i=1 to length(s) do setd(s[i],0) end for
end procedure

populate_dict(split("a bc abc cd b"))

function prrec(sequence wordstarts, integer idx, sequence sofar, bool show)
if idx>length(wordstarts) then
if show then
?sofar
end if
return 1
end if
integer res = 0
for i=1 to length(wordstarts[idx]) do
string w = wordstarts[idx][i]
res += prrec(wordstarts,idx+length(w),append(deep_copy(sofar),w),show)
end for
return res
end function

function flattens(sequence s)
-- remove all nesting and empty sequences from a nested sequence of strings
sequence res = {}, si
for i=1 to length(s) do
si = s[i]
if string(si) then
res = append(res,si)
else
res &= flattens(si)
end if
end for
return res
end function

procedure test(string s)
integer l = length(s),
wordend = 1 -- (pretend a word just ended at start)
sequence wordstarts = repeat({},l),
wordends = repeat(0,l)
for i=1 to l do
if wordend then
for j=i to l do
object pkey = getd_partial_key(s[i..j])
if string(pkey) and length(pkey)>j-i and s[i..j]=pkey[1..j-i+1] then
if length(pkey)=j-i+1 then
-- exact match
wordstarts[i] = append(wordstarts[i],pkey)
wordends[j] += 1
end if
else
exit
end if
end for
end if
wordend = wordends[i]
end for
bool worthwhile = true
while worthwhile do
worthwhile = false
wordend = 1 -- (pretend a word just ended at start)
for i=1 to l do
if wordend then
-- eliminate any words that end before a wordstarts of {}.
for j=length(wordstarts[i]) to 1 by -1 do
integer wl = length(wordstarts[i][j])
if i+wl<=l and wordstarts[i+wl]={} then
integer wedx = i+wl-1
wordends[wedx] -= 1
wordstarts[i][j..j] = {}
worthwhile = true
end if
end for
else
-- elimitate all words that start here.
for j=1 to length(wordstarts[i]) do
integer wl = length(wordstarts[i][j])
if i+wl<=l then
integer wedx = i+wl-1
wordends[wedx] -= 1
worthwhile = true
end if
end for
wordstarts[i] = {}
end if
wordend = wordends[i]
end for
end while
if sum(wordends)=0 then
printf(1,"%s: not possible\n",{s})
else
integer count = prrec(wordstarts,1,{},false)
if count=1 then
printf(1,"%s: 1 solution: %s\n",{s,join(flattens(wordstarts))})
elsif count>20 then
printf(1,"%s: %d solution(s): (too many to show)\n",{s,count})
pp({wordstarts,wordends})
else
printf(1,"%s: %d solution(s):\n",{s,count})
count = prrec(wordstarts,1,{},true)
end if
end if
end procedure

papply({"abcd","abbc","abcbcd","acdbc","abcdd"},test)

{} = wait_key()
```
Output:
```{"a","bc","abc","cd","b"}
abcd: 1 solution: a b cd
abbc: 1 solution: a b bc
abcbcd: 2 solution(s):
{"a","bc","b","cd"}
{"abc","b","cd"}
acdbc: 1 solution: a cd bc
abcdd: not possible
```

## Picat

### Non-determinism using member/2

`s/2` checks all possible combinations. It uses `member/2` to select some element from Dict and backtracks if no solution is found.

```main =>
Tests = [["aab", ["a", "aa", "b", "ab", "aab"]],
["abba", ["a", "aa", "b", "ab", "aab"]],
["aa b", ["a", "aa", "b", "ab", "aab"]],
["abcd", ["abc", "a", "ac", "b", "c", "cb", "d"]],
["abbc", ["abc", "a", "ac", "b", "c", "cb", "d"]],
["abcbcd", ["abc", "a", "ac", "b", "c", "cb", "d"]],
["acdbc", ["abc", "a", "ac", "b", "c", "cb", "d"]],
["abcdd", ["abc", "a", "ac", "b", "c", "cb", "d"]]
],
foreach([String,Dict] in Tests)
println([string=String,dict=Dict]),
All = findall(S, S = s(Dict,String)),
println(All),
nl
end,
nl.

s(Dict,String) = S =>
S = [],
while(S.flatten != String, S.flatten.len < String.len)
member(D,Dict), % pick some element in Dict
S := S ++ [D]
end,
S.flatten==String.```
Output:
```[string = aab,dict = [a,aa,b,ab,aab]]
[[a,a,b],[a,ab],[aa,b],[aab]]

[string = abba,dict = [a,aa,b,ab,aab]]
[[a,b,b,a],[ab,b,a]]

[string = aa b,dict = [a,aa,b,ab,aab]]
[]

[string = abcd,dict = [abc,a,ac,b,c,cb,d]]
[[abc,d],[a,b,c,d]]

[string = abbc,dict = [abc,a,ac,b,c,cb,d]]
[[a,b,b,c]]

[string = abcbcd,dict = [abc,a,ac,b,c,cb,d]]
[[abc,b,c,d],[a,b,c,b,c,d],[a,b,cb,c,d]]

[string = acdbc,dict = [abc,a,ac,b,c,cb,d]]
[[a,c,d,b,c],[ac,d,b,c]]

[string = abcdd,dict = [abc,a,ac,b,c,cb,d]]
[[abc,d,d],[a,b,c,d,d]]```

### Non-determinism using member/2 and append/3

`s2/2` is more efficient. It uses `append/3` to extract the found prefix from the string.

```s2(Dict,String) = S =>
String2 = copy_term(String),
S = [],
while(S.flatten != String, S.flatten.len < String.len)
% Pick an element from Dict
member(D,Dict),

% Check that it fits and remove it from the string.
% If not: backtrack
append(D,String3,String2),

% ok!
String2 := String3,
S := S ++ [D]
end,
S.flatten==String.```

### Using recursion

`s3/2` use the same idea as `s2/2` but use recursion. Neater and more efficient.

```s3(Dict,String) = S =>
s3(Dict,String,S).

s3(_Dict,[],[]).
s3(Dict,String,[E|S]) :-
member(E,Dict),
append(E,String2,String),
s3(Dict,String2,S).```

### Some tests

Here is the test engine.

```main =>
garbage_collect(300_000_000),
_ = random2(),
Chars = "ab",
Dict = ["a", "aa", "b", "ab", "aab"],
println(dict=Dict),
String = [Chars[random(1,Chars.len)] : _ in 1..11],
% The tested strings
% String := "abbaabba",
% String := "babbbbaabbababb",
% String := "babbbbaabbababbaaaaababbaaabbb",
% String := "babbbbaabbababbaaaaababbaaabbbbaaabbbaba",
% String := "aabababbbaaabaababaaabababaaabbabbaabbba",
println(string=String),
All = findall(S, S = s3(Dict,String)),
Len = All.len,
if Len < 100 then
println(All)
end,
println(len=All.len),
nl.```

As can be seen by this summary, `s3/2` is the most efficient of these versions. `s/2` is too slow but for the simplest examples.

Output:
```dict = [a,aa,b,ab,aab]
string = abbaabba
[[a,b,b,a,a,b,b,a],[a,b,b,a,ab,b,a],[a,b,b,aa,b,b,a],[a,b,b,aab,b,a],[ab,b,a,a,b,b,a],[ab,b,a,ab,b,a],[ab,b,aa,b,b,a],[ab,b,aab,b,a]]
len = 8

s: 0.012s
s2: 0.0s
s3: 0.0

Dict = ["a", "aa", "b", "ab", "aab"],
string = babbbbaabbababb
[[b,a,b,b,b,b,a,a,b,b,a,b,a,b,b],[b,a,b,b,b,b,a,a,b,b,a,b,ab,b],[b,a,b,b,b,b,a,a,b,b,ab,a,b,b],[b,a,b,b,b,b,a,a,b,b,ab,ab,b],[b,a,b,b,b,b,a,ab,b,a,b,a,b,b],[b,a,b,b,b,b,a,ab,b,a,b,ab,b],[b,a,b,b,b,b,a,ab,b,ab,a,b,b],[b,a,b,b,b,b,a,ab,b,ab,ab,b],[b,a,b,b,b,b,aa,b,b,a,b,a,b,b],[b,a,b,b,b,b,aa,b,b,a,b,ab,b],[b,a,b,b,b,b,aa,b,b,ab,a,b,b],[b,a,b,b,b,b,aa,b,b,ab,ab,b],[b,a,b,b,b,b,aab,b,a,b,a,b,b],[b,a,b,b,b,b,aab,b,a,b,ab,b],[b,a,b,b,b,b,aab,b,ab,a,b,b],[b,a,b,b,b,b,aab,b,ab,ab,b],[b,ab,b,b,b,a,a,b,b,a,b,a,b,b],[b,ab,b,b,b,a,a,b,b,a,b,ab,b],[b,ab,b,b,b,a,a,b,b,ab,a,b,b],[b,ab,b,b,b,a,a,b,b,ab,ab,b],[b,ab,b,b,b,a,ab,b,a,b,a,b,b],[b,ab,b,b,b,a,ab,b,a,b,ab,b],[b,ab,b,b,b,a,ab,b,ab,a,b,b],[b,ab,b,b,b,a,ab,b,ab,ab,b],[b,ab,b,b,b,aa,b,b,a,b,a,b,b],[b,ab,b,b,b,aa,b,b,a,b,ab,b],[b,ab,b,b,b,aa,b,b,ab,a,b,b],[b,ab,b,b,b,aa,b,b,ab,ab,b],[b,ab,b,b,b,aab,b,a,b,a,b,b],[b,ab,b,b,b,aab,b,a,b,ab,b],[b,ab,b,b,b,aab,b,ab,a,b,b],[b,ab,b,b,b,aab,b,ab,ab,b]]
len = 32
s: 22.513s
s2: 0s
s3: 0s

Dict = ["a", "aa", "b", "ab", "aab"],
string = babbbbaabbababbaaaaababbaaabbb
len = 6144
s2: 0.088s
s3: 0.012s

Dict = [a,aa,b,ab,aab]
string = babbbbaabbababbaaaaababbaaabbbbaaabbbaba
len = 73728
s2: 1.466s
s3: 0.341

sdict = [a,aa,b,ab,aab]
string = aabababbbaaabaababaaabababaaabbabbaabbba
len = 884736
s2: 19.431 seconds
s3: 3.81 seconds```

## PicoLisp

```(setq *Dict (quote "a" "bc" "abc" "cd" "b"))
(setq *Dict2
(quote
"mobile" "samsung" "sam" "sung" "man" "mango"
"icecream" "and" "go" "i" "like" "ice" "cream" ) )

(de word (Str D)
(let
(Str (chop Str)
Len (length Str)
DP (need (inc Len))
Res (need (inc Len))
B 1 )
(set DP 0)
(map
'((L)
(and
(get DP B)
(for N (length L)
(let Str (pack (head N L))
(when (member Str D)
(set (nth Res (+ B N))
(copy (get Res B)) )
(queue (nth Res (+ B N)) Str)
(set (nth DP (+ B N))
(inc (get DP B)) ) ) ) ) )
(inc 'B) )
Str )
(last Res) ) )

(println (word "abcd" *Dict))
(println (word "abbc" *Dict))
(println (word "abcbcd" *Dict))
(println (word "acdbc" *Dict))
(println (word "abcdd" *Dict))
(println (word "ilikesamsung" *Dict2))
(println (word "iii" *Dict2))
(println (word "ilikelikeimangoiii" *Dict2))
(println (word "samsungandmango" *Dict2))
(println (word "samsungandmangok" *Dict2))
(println (word "ksamsungandmango" *Dict2))```
Output:
```("a" "b" "cd")
("a" "b" "bc")
("a" "bc" "b" "cd")
("a" "cd" "bc")
NIL
("i" "like" "sam" "sung")
("i" "i" "i")
("i" "like" "like" "i" "man" "go" "i" "i" "i")
("sam" "sung" "and" "man" "go")
NIL
NIL
```

## Python

### Functional

The tokenTrees function recursively builds a tree of possible token sequences, using a list monad (concatMap with a function which returns its result wrapped in a list – an empty list where a parse has failed) to discard all branches which lead to dead ends. This allows us to return more than one possible word-break parse for a given lexicon and input string. (Searches for 'monadic parsing in Python' will yield references to more sophisticated uses of this general approach).

Works with: Python version 3.7
```'''Parsing a string for word breaks'''

from itertools import (chain)

# stringParse :: [String] -> String -> Tree String
def stringParse(lexicon):
'''A tree of strings representing a parse of s
in terms of the tokens in lexicon.
'''
return lambda s: Node(s)(
tokenTrees(lexicon)(s)
)

# tokenTrees :: [String] -> String -> [Tree String]
def tokenTrees(wds):
'''A list of possible parse trees for s,
based on the lexicon supplied in wds.
'''
def go(s):
return [Node(s)([])] if s in wds else (
concatMap(nxt(s))(wds)
)

def nxt(s):
return lambda w: parse(
w, go(s[len(w):])
) if s.startswith(w) else []

def parse(w, xs):
return [Node(w)(xs)] if xs else xs

return lambda s: go(s)

# showParse :: Tree String -> String
def showParse(tree):
'''Multi line display of a string followed by any
possible parses of it, or an explanatory
message, if no parse was possible.
'''
def showTokens(x):
xs = x['nest']
return ' ' + x['root'] + (showTokens(xs[0]) if xs else '')
parses = tree['nest']
return tree['root'] + ':\n' + (
'\n'.join(
map(showTokens, parses)
) if parses else ' ( Not parseable in terms of these words )'
)

# TEST -------------------------------------------------
# main :: IO ()
def main():
'''Parse test and display of results.'''

lexicon = 'a bc abc cd b'.split()
testSamples = 'abcd abbc abcbcd acdbc abcdd'.split()

print(unlines(
map(
showParse,
map(
stringParse(lexicon),
testSamples
)
)
))

# GENERIC FUNCTIONS ---------------------------------------

# Node :: a -> [Tree a] -> Tree a
def Node(v):
'''Contructor for a Tree node which connects a
value of some kind to a list of zero or
more child trees.'''
return lambda xs: {'type': 'Node', 'root': v, 'nest': xs}

# concatMap :: (a -> [b]) -> [a] -> [b]
def concatMap(f):
'''A concatenated list over which a function has been mapped.
The list monad can be derived by using a function f which
wraps its output in a list,
(using an empty list to represent computational failure).'''
return lambda xs: list(
chain.from_iterable(map(f, xs))
)

# unlines :: [String] -> String
def unlines(xs):
'''A single string derived by the intercalation
of a list of strings with the newline character.'''
return '\n'.join(xs)

# MAIN ---
if __name__ == '__main__':
main()
```
Output:
```abcd:
a b cd
abbc:
a b bc
abcbcd:
a bc b cd
abc b cd
acdbc:
a cd bc
abcdd:
( Not parseable in terms of these words )```

## Racket

This returns all the possible splits (and null list if none is possible). Who's to say which is the best?

```#lang racket

(define render-phrases pretty-print)

(define dict-1 (list "a" "bc" "abc" "cd" "b"))
(define dict-2 (list "mobile" "samsung" "sam" "sung" "man" "mango"
"icecream" "and" "go" "i" "like" "ice" "cream"))

(define (word-splits str d)
(let ((memo (make-hash)))
(let inr ((s str))
(hash-ref! memo s
(λ () (append* (filter-map (λ (w)
(and (string-prefix? s w)
(if (string=? w s)
(list s)
(map (λ (tl) (string-append w " " tl))
(inr (substring s (string-length w)))))))
d)))))))

(module+ main
(render-phrases (word-splits "abcd" dict-1))
(render-phrases (word-splits "abbc" dict-1))
(render-phrases (word-splits "abcbcd" dict-1))
(render-phrases (word-splits "acdbc" dict-1))
(render-phrases (word-splits "abcdd" dict-1))
(render-phrases (word-splits "ilikesamsung" dict-2))
(render-phrases (word-splits "iii" dict-2))
(render-phrases (word-splits "ilikelikeimangoiii" dict-2))
(render-phrases (word-splits "samsungandmango" dict-2))
(render-phrases (word-splits "samsungandmangok" dict-2))
(render-phrases (word-splits "ksamsungandmango" dict-2)))
```
Output:
```'("a b cd")
'("a b bc")
'("a bc b cd" "abc b cd")
'("a cd bc")
'()
'("i like samsung" "i like sam sung")
'("i i i")
'("i like like i man go i i i" "i like like i mango i i i")
'("samsung and man go"
"samsung and mango"
"sam sung and man go"
"sam sung and mango")
'()
'()```

## Raku

(formerly Perl 6)

Works with: Rakudo version 2017.04

This implementation does not necessarily find every combination, it returns the one with the longest matching tokens.

```my @words = <a bc abc cd b>;
my \$regex = @words.join('|');

put "\$_: ", word-break(\$_) for <abcd abbc abcbcd acdbc abcdd>;

sub word-break (Str \$word) { (\$word ~~ / ^ (<\$regex>)+ \$ /)[0] // "Not possible" }
```
Output:
```abcd: a b cd
abbc: a b bc
abcbcd: abc b cd
acdbc: a cd bc
abcdd: Not possible```

## REXX

This REXX version allows the words to be tested (and the dictionary words) to be specified on the command line.

```/*REXX program breaks up a  word  (or string)  into a  list of words  from a dictionary.*/
parse arg a '/' x;   a=space(a);     x=space(x)  /*get optional args; elide extra blanks*/
if a=='' | a==","  then a= 'abcd abbc abcbcd acdbc abcdd' /*Not specififed?  Use default*/
if x=='' | x==","  then x= 'a bc abc cd b'                /* "      "        "      "   */
na= words(a)                                     /*the number of words to be tested.    */
nx= words(x)                                     /* "     "    "   "    " the dictionary*/
say nx  ' dictionary words: '   x                /*display the words in the dictionary. */
aw= 0                                            /*maximum word width obtained (so far).*/
say                                              /*display a blank line to the terminal.*/
do i=1  for na;           _= word(a, i)    /*obtain a word that will be tested.   */
aw= max(aw, length(_) )                    /*find widest width word being tested. */
end   /*i*/                                /* [↑]  AW  is used to align the output*/
@.= 0                                            /*initialize the dictionary to "null". */
xw= 0
do i=1  for nx;           _= word(x, i)    /*obtain a word from the dictionary.   */
xw= max(xw, length(_) );  @._= 1           /*find widest width dictionary word.   */
end   /*i*/                                /* [↑]  define a dictionary word.      */
p= 0                                             /* [↓]  process a word in the  A  list.*/
do j=1  for na;           yy= word(a, j)   /*YY:   test a word  from the  A  list.*/
do t=(nx+1)**(xw+1)  by -1  to 1  until y=='';  y= yy    /*try word possibility.*/
\$=                                       /*nullify the (possible) result list.  */
do try=t  while y\=''                 /*keep testing until  Y  is exhausted. */
p= (try + p)  // xw    + 1            /*use a possible width for this attempt*/
p= fw(y, p);  if p==0  then iterate t /*is this part of the word not found ? */
\$= \$ ?                                /*It was found. Add partial to the list*/
y= substr(y,  p + 1)                  /*now, use and test the rest of word.  */
end   /*try*/
end      /*t*/

if t==0  then \$= '(not possible)'          /*indicate that the word isn't possible*/
say right(yy, aw)    '───►'    strip(\$)    /*display the result to the terminal.  */
end   /*j*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
fw: parse arg z,L;  do k=L  by -1  for L; ?=left(z,k); if @.?  then leave; end;   return k
```
output   when using the default inputs:
```5  dictionary words:  a bc abc cd b

abcd ───► a b cd
abbc ───► a b bc
abcbcd ───► abc b cd
acdbc ───► a cd bc
abcdd ───► (not possible)
```

## Ring

```# Project : Word break problem

list = ["a", "bc", "abc", "cd", "b"]
inslist = list
for n = 1 to len(inslist) - 1
for m = len(inslist) to 1 step -1
insert(list,0,inslist[m])
next
next
strings = ["abcd", "abbc", "abcbcd", "acdbc", "abcdd"]
ind = len(list)
items = newlist(pow(2,len(list))-1,ind)
powerset(list,ind)

for p = 1 to len(strings)
showarray(items,strings[p])
next

func powerset(list,ind)
num = 0
num2 = 0
items = newlist(pow(2,len(list))-1,2*ind)
for i = 2 to (2 << len(list)) - 1 step 2
num2 = 0
num = num + 1
for j = 1 to len(list)
if i & (1 << j)
num2 = num2 + 1
if list[j] != 0
items[num][num2] = list[j]
ok
ok
next
next
return items

func showarray(items,par)
for n = 1 to len(items)
for m = n + 1 to len(items) - 1
flag = 0
str = ""
for x = 1 to len(items[n])
if items[n][x] != 0
str = str + items[n][x] + " "
ok
next
str = left(str, len(str) - 1)
strsave = str
str = substr(str, " ", "")
if str = par
if pos = 0
flag = 1
see par + " = " + strsave + nl
ok
if flag != 1
del(items,m)
ok
ok
next
next```

Output:

```abcd = a b cd
abbc = a b bc
abcbcd = abc b cd
acdbc = a cd bc
```

## Ruby

```def split_text_with_dict(text, dict, splited=[])
solutions = []
dict.each do |word|
if text.start_with? word
new_text = text.delete_prefix word
new_splited = splited.dup<< word
if new_text.empty?
solutions << new_splited
else
sols = split_text_with_dict(new_text, dict, new_splited)
sols.each { |s| solutions << s }
end
end
end
return solutions
end
```
Output:
```dict = ["a", "abc", "b", "bc", "cd"]

abcd => a b cd
abbc => a b bc
abcbcd
=> a bc b cd
=> abc b cd
acdbc => a cd bc
abcdd => No solution

```

## Rust

### Dynamic programming

```use std::collections::HashSet;
fn create_string(s: &str, v: Vec<Option<usize>>) -> String {
let mut idx = s.len();
let mut slice_vec = vec![];
while let Some(prev) = v[idx] {
slice_vec.push(&s[prev..idx]);
idx = prev;
}
slice_vec.reverse();
slice_vec.join(" ")

}

fn word_break(s: &str, dict: HashSet<&str>) -> Option<String> {
let size = s.len() + 1;
let mut possible = vec![None; size];

let check_word = |i,j| dict.get(&s[i..j]).map(|_| i);

for i in 1..size {
possible[i] = possible[i].or_else(|| check_word(0,i));

if possible[i].is_some() {
for j in i+1..size {
possible[j] = possible[j].or_else(|| check_word(i,j));
}

if possible[s.len()].is_some() {
return Some(create_string(s, possible));
}

};
}
None
}

fn main() {
let mut set = HashSet::new();
set.insert("a");
set.insert("bc");
set.insert("abc");
set.insert("cd");
set.insert("b");
println!("{:?}", word_break("abcd", set).unwrap());
}
```
Output:
`"a b cd"`

## Scala

### First solution

Finds all possible solutions recursively, using a trie representation of the dictionary:

```case class TrieNode(isWord: Boolean, children: Map[Char, TrieNode]) {
def add(s: String): TrieNode = s match {
case "" => copy(isWord = true)
case _ => {
val child = children.getOrElse(s.head, TrieNode(false, Map.empty))
}
}
}

def buildTrie(xs: String*): TrieNode = {
}

def wordBreakRec(s: String, root: TrieNode, currentPos: TrieNode, soFar: String): List[List[String]] = {
val usingCurrentWord = if (currentPos.isWord) {
if (s.isEmpty) {
List(List(soFar))
} else {
wordBreakRec(s, root, root, "").map(soFar :: _)
}
} else {
List.empty[List[String]]
}
val usingCurrentPrefix = (for {
child <- currentPos.children.get(ch)
} yield wordBreakRec(s.tail, root, child, soFar + ch)).getOrElse(List.empty)
usingCurrentWord ++ usingCurrentPrefix
}

def wordBreak(s: String, dict: TrieNode): List[List[String]] = {
wordBreakRec(s, dict, dict, "")
}
```

Calling it with some example strings:

```val dict = buildTrie("a", "bc", "abc", "cd", "b")
val testCases = List("abcd", "abbc", "abcbcd", "acdbc", "abcdd")
for (s <- testCases) {
val solutions = wordBreak(s, dict)
println(s"\$s has \${solutions.size} solution(s):")
for (words <- solutions) {
println("\t" + words.mkString(" "))
}
}
```
Output:
```abcd has 1 solution(s):
a b cd
abbc has 1 solution(s):
a b bc
abcbcd has 2 solution(s):
a bc b cd
abc b cd
acdbc has 1 solution(s):
a cd bc
abcdd has 0 solution(s):```

### Combined solution

Output:

Best seen running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).

```object WordBreak extends App {
val dict = buildTrie("a", "bc", "abc", "cd", "b")
lazy val empty = TrieNode(isWord = false, Map.empty) // lazy or in a companion object

case class TrieNode(isWord: Boolean, children: Map[Char, TrieNode]) {

def add(s: String): TrieNode = {

if (s.isEmpty) copy(isWord = true)
}
}

def buildTrie(xs: String*): TrieNode = xs.foldLeft(empty)(_.add(_))

def wordBreak(s: String, dict: TrieNode): List[List[String]] = {

def wordBreakRec(s: String,
root: TrieNode,
currentPos: TrieNode,
soFar: String): List[List[String]] = {

def usingCurrentWord =
if (currentPos.isWord)
if (s.isEmpty) List(List(soFar))
else wordBreakRec(s, root, root, "").map(soFar :: _)
else Nil

def usingCurrentPrefix =
child <- currentPos.children.get(ch)
} yield wordBreakRec(s.tail, root, child, soFar + ch))
.getOrElse(Nil)

usingCurrentWord ++ usingCurrentPrefix
}

wordBreakRec(s, dict, dict, "")
}

// Calling it with some example strings:
List("abcd", "abbc", "abcbcd", "acdbc", "abcdd").foreach(s => {
val solutions = wordBreak(s, dict)

println(s"\$s has \${solutions.size} solution(s):")
solutions.foreach(words => println(words.mkString("\t", " ", "")))
})

}
```

## Seed7

```\$ include "seed7_05.s7i";

const func boolean: wordBreak (in string: stri, in array string: words, in string: resultList) is func
result
var boolean: found is FALSE;
local
var string: word is "";
begin
if stri = "" then
writeln(resultList);
found := TRUE;
else
for word range words do
if startsWith(stri, word) and
wordBreak(stri[succ(length(word)) ..], words, resultList & " " & word) then
found := TRUE;
end if;
end for;
end if;
end func;

const proc: main is func
local
const array string: words is [] ("a", "bc", "abc", "cd", "b");
var string: stri is "";
var string: resultList is "";
begin
for stri range [] ("abcd", "abbc", "abcbcd", "acdbc", "abcdd") do
write(stri <& ": ");
if not wordBreak(stri, words, resultList) then
writeln("can't break");
end if;
end for;
end func;```
Output:
```abcd:  a b cd
abbc:  a b bc
abcbcd:  a bc b cd
abc b cd
acdbc:  a cd bc
abcdd: can't break
```

## Sidef

Translation of: zkl
```func word_break (str, words) {

var r = ->(str, arr=[]) {
return true if str.is_empty
for word in (words) {
str.begins_with(word) || next
if (__FUNC__(str.substr(word.len), arr)) {
arr << word
return arr
}
}
return false
}(str)

r.kind_of(Array) ? r.reverse : nil
}

var words = %w(a o is pi ion par per sip miss able)
var strs = %w(a amiss parable opera operable inoperable permission mississippi)

for str in (strs) {
printf("%11s: %s\n", str, word_break(str, words) \\ '(not possible)')
}
```
Output:
```          a: ["a"]
amiss: ["a", "miss"]
parable: ["par", "able"]
opera: ["o", "per", "a"]
operable: ["o", "per", "able"]
inoperable: (not possible)
permission: ["per", "miss", "ion"]
mississippi: ["miss", "is", "sip", "pi"]
```

## Swift

Translation of: Rust
```infix operator ??= : AssignmentPrecedence

@inlinable
public func ??= <T>(lhs: inout T?, rhs: T?) {
lhs = lhs ?? rhs
}

private func createString(_ from: String, _ v: [Int?]) -> String {
var idx = from.count
var sliceVec = [Substring]()

while let prev = v[idx] {
let s = from.index(from.startIndex, offsetBy: prev)
let e = from.index(from.startIndex, offsetBy: idx)

sliceVec.append(from[s..<e])
idx = prev
}

return sliceVec.reversed().joined(separator: " ")
}

public func wordBreak(str: String, dict: Set<String>) -> String? {
let size = str.count + 1
var possible = [Int?](repeating: nil, count: size)

func checkWord(i: Int, j: Int) -> Int? {
let s = str.index(str.startIndex, offsetBy: i)
let e = str.index(str.startIndex, offsetBy: j)

return dict.contains(String(str[s..<e])) ? i : nil
}

for i in 1..<size {
possible[i] ??= checkWord(i: 0, j: i)

guard possible[i] != nil else {
continue
}

for j in i+1..<size {
possible[j] ??= checkWord(i: i, j: j)
}

if possible[str.count] != nil {
return createString(str, possible)
}
}

return nil
}

let words = [
"a",
"bc",
"abc",
"cd",
"b"
] as Set

let testCases = [
"abcd",
"abbc",
"abcbcd",
"acdbc",
"abcdd"
]

for test in testCases {
print("\(test):")
print("  \(wordBreak(str: test, dict: words) ?? "did not parse with given words")")
}
```
Output:
```abcd:
a b cd
abbc:
a b bc
abcbcd:
a bc b cd
acdbc:
a cd bc
abcdd:
did not parse with given words```

## Tailspin

Does a depth-first search (after generating all possibilities on the current level). We could stop looking further after one is found, just add a condition to do nothing if a done-flag is set.

```templates wordBreaks&{dict:}
composer starts&{with:}
<does|not>
rule does: (<=\$with>) <'.*'>
rule not: <'.*'> -> \(!VOID\)
end starts

'Breaking "\$;" by \$dict;:\$#10;' -> !OUT::write
{ words:[], remaining: \$} -> #
'--done--\$#10;' !

when <{remaining: <=''>}> do
\$.words -> \spaceSeparate[i](when <?(\$i <=1>)> do \$! otherwise ' ' ! \$ ! \spaceSeparate) -> '\$...;\$#10;' !

otherwise
def base: \$;
\$dict... -> \(
def word: \$;
\$base.remaining -> starts&{with: \$word} -> {words: [\$base.words..., \$word], remaining: \$} !
\) -> #
end wordBreaks

'ababab' -> wordBreaks&{dict: ['a', 'ab', 'bab']} -> !OUT::write
'abcbab' -> wordBreaks&{dict: ['a', 'ab', 'bab']} -> !OUT::write```
Output:
```Breaking "ababab" by [a, ab, bab]:
a bab ab
ab a bab
ab ab ab
--done--
Breaking "abcbab" by [a, ab, bab]:
--done--
```

## Visual Basic .NET

Translation of: Java
```Module Module1

Structure Node
Private ReadOnly m_parsed As List(Of String)

Sub New(initial As String)
m_val = initial
m_parsed = New List(Of String)
End Sub

Sub New(s As String, p As List(Of String))
m_val = s
m_parsed = p
End Sub

Public Function Value() As String
Return m_val
End Function

Public Function Parsed() As List(Of String)
Return m_parsed
End Function
End Structure

Function WordBreak(s As String, dictionary As List(Of String)) As List(Of List(Of String))
Dim matches As New List(Of List(Of String))
Dim q As New Queue(Of Node)
q.Enqueue(New Node(s))
While q.Count > 0
Dim node = q.Dequeue()
REM check if fully parsed
If node.Value.Length = 0 Then
Else
For Each word In dictionary
REM check for match
If node.Value.StartsWith(word) Then
Dim valNew = node.Value.Substring(word.Length, node.Value.Length - word.Length)
Dim parsedNew As New List(Of String)
q.Enqueue(New Node(valNew, parsedNew))
End If
Next
End If
End While
Return matches
End Function

Sub Main()
Dim dict As New List(Of String) From {"a", "aa", "b", "ab", "aab"}
For Each testString In {"aab", "aa b"}
Dim matches = WordBreak(testString, dict)
Console.WriteLine("String = {0}, Dictionary = {1}. Solutions = {2}", testString, dict, matches.Count)
For Each match In matches
Console.WriteLine(" Word Break = [{0}]", String.Join(", ", match))
Next
Console.WriteLine()
Next

dict = New List(Of String) From {"abc", "a", "ac", "b", "c", "cb", "d"}
For Each testString In {"abcd", "abbc", "abcbcd", "acdbc", "abcdd"}
Dim matches = WordBreak(testString, dict)
Console.WriteLine("String = {0}, Dictionary = {1}. Solutions = {2}", testString, dict, matches.Count)
For Each match In matches
Console.WriteLine(" Word Break = [{0}]", String.Join(", ", match))
Next
Console.WriteLine()
Next
End Sub

End Module
```
Output:
```String = aab, Dictionary = System.Collections.Generic.List`1[System.String]. Solutions = 4
Word Break = [aab]
Word Break = [a, ab]
Word Break = [aa, b]
Word Break = [a, a, b]

String = aa b, Dictionary = System.Collections.Generic.List`1[System.String]. Solutions = 0

String = abcd, Dictionary = System.Collections.Generic.List`1[System.String]. Solutions = 2
Word Break = [abc, d]
Word Break = [a, b, c, d]

String = abbc, Dictionary = System.Collections.Generic.List`1[System.String]. Solutions = 1
Word Break = [a, b, b, c]

String = abcbcd, Dictionary = System.Collections.Generic.List`1[System.String]. Solutions = 3
Word Break = [abc, b, c, d]
Word Break = [a, b, cb, c, d]
Word Break = [a, b, c, b, c, d]

String = acdbc, Dictionary = System.Collections.Generic.List`1[System.String]. Solutions = 2
Word Break = [ac, d, b, c]
Word Break = [a, c, d, b, c]

String = abcdd, Dictionary = System.Collections.Generic.List`1[System.String]. Solutions = 2
Word Break = [abc, d, d]
Word Break = [a, b, c, d, d]```

## Wren

Translation of: Go
Library: Wren-fmt
```import "./fmt" for Fmt

class Prefix {
construct new(length, broken) {
_length = length
_broken = broken
}
length { _length }
broken { _broken }
}

var wordBreak = Fn.new { |d, s|
if (s == "") return [[], true]
var bp = [Prefix.new(0, [])]
for (end in 1..s.count) {
for (i in bp.count-1..0) {
var w = s[bp[i].length...end]
if (d[w]) {
var b = bp[i].broken.toList
if (end == s.count) return [b, true]
break
}
}
}
return [[], false]
}

var words = ["a", "bc", "abc", "cd", "b"]
var d = {}
words.each { |w| d[w] = true }
for (s in ["abcd", "abbc", "abcbcd", "acdbc", "abcdd"]) {
var res = wordBreak.call(d, s)
if (res[1]) {
Fmt.print("\$s: \$s", s, res[0].join(" "))
} else {
System.print("can't break")
}
}
```
Output:
```abcd: a b cd
abbc: a b bc
abcbcd: a bc b cd
acdbc: a cd bc
can't break
```

## zkl

```fcn wordBreak(str,words){	// words is string of space seperated words
words=words.split(" ");	// to list of words
r:=fcn(str,words,sink){	// recursive search, easy to collect answer
foreach word in (words){
if(not str) return(True);  // consumed string ie matched everything
if(str.find(word)==0){     // word starts str, 0 so answer is ordered
z:=word.len();
if(self.fcn(str.del(0,z),words,sink)) return(sink.write(word));
}
}
False		// can't make forward progress, back out & retry
}(str,words,List());		// run the lambda
if(False==r) return("not possible");
r.reverse().concat(" ")
}```
```foreach text in (T("abcd","abbc","abcbcd","acdbc","abcdd")){
println(text,": ",wordBreak(text,"a bc abc cd b"))
}```
Output:
```abcd: a b cd
abbc: a b bc
abcbcd: a bc b cd
acdbc: a cd bc
abcdd: not possible
```