Word break problem

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

Other tasks related to string operations:

Metrics

Counting

Remove/replace

Anagrams/Derangements/shuffling

Find/Search/Determine

Formatting

Song lyrics/poems/Mad Libs/phrases

Tokenize

Sequences

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
          add(result, part1 & sub results[i])
        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

Library: System.SysUtils

Library: System.Generics.Collections

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
    Result.AddOrSetValue(w, true);
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;

  Readln;
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

Haskell

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.LinkedList;
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<>();
        Queue<Node> queue = new LinkedList<>();
        queue.add(new Node(s));
        while ( ! queue.isEmpty() ) {
            Node node = queue.remove();
            //  Check if fully parsed
            if ( node.val.length() == 0 ) {
                matches.add(node.parsed);
            }
            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<>();
                        parsedNew.addAll(node.parsed);
                        parsedNew.add(word);
                        queue.add(new Node(valNew, parsedNew));
                    }
                }
            }
        }
        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.

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)
        ml.add(part1)
        ml.add(part2)
        partitions.add(ml.toList())
        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 words = File("unixdict.txt").readLines()
    val strings = listOf("abcd", "abbc", "abcbcd", "acdbc", "abcdd")
    for (s in strings) {
        partitions.clear()
        partitions.add(listOf(s))
        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
                }
            }
            if (allInDict) solutions.add(partition)
        }
        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)
  local d,dup,head,rest = {},{}
  for w in ws:gmatch"%w+" do
    local lw = w:lower()
    if not dup[lw] then
      dup[lw], head,rest = true, lw:match"^(%w)(.-)$"
      d[head] = d[head] or {n=-1}
      local len = #rest
      d[head][len] = d[head][len] or {}
      d[head][len][rest] = true
      if len>d[head].n then 
        d[head].n = len 
      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
    local head = w:sub(i,i):lower()
    local d = dict[head]
    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])
      result.add dict.wordBreaks(part2).mapIt(part1 & it)

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
ado: ad o
amiss: a miss
admission: ad miss ion
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

load "stdlib.ring"
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)
        ready = []
        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
                       pos = find(ready,strsave)               
                       if pos = 0              
                          add(ready,strsave)
                          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))
      copy(children = children + (s.head -> child.add(s.tail)))
    }
  }
}

def buildTrie(xs: String*): TrieNode = {
  xs.foldLeft(TrieNode(false, Map.empty))(_.add(_))
}

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 {
    ch <- s.headOption
    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 = {
      def child = children.withDefaultValue(empty)(s.head)

      if (s.isEmpty) copy(isWord = true)
      else copy(children = children.updated(s.head, child.add(s.tail)))
    }
  }

  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 =
         (for {ch <- s.headOption
              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_val As String
        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
                matches.Add(node.Parsed)
            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)
                        parsedNew.AddRange(node.Parsed)
                        parsedNew.Add(word)
                        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
                b.add(w)
                if (end == s.count) return [b, true]
                bp.add(Prefix.new(end, b))
                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