Word break problem: Difference between revisions
MaiconSoft (talk | contribs) Added Delphi example |
Alextretyak (talk | contribs) Added 11l |
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{{Template:Strings}} |
{{Template:Strings}} |
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<br><br> |
<br><br> |
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=={{header|11l}}== |
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{{trans|D}} |
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<lang 11l>T Node |
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String val |
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[String] parsed |
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F (val, [String] &parsed = [String]()) |
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.val = val |
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.parsed = parsed |
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F word_break(s, dictionary) |
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[[String]] matches |
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V queue = Deque([Node(s)]) |
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L !queue.empty |
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V node = queue.pop_left() |
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I node.val.empty |
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matches [+]= node.parsed |
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E |
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L(word) dictionary |
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I node.val.starts_with(word) |
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V val_new = node.val[word.len .< node.val.len] |
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V parsed_new = copy(node.parsed) |
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parsed_new [+]= word |
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queue [+]= Node(val_new, &parsed_new) |
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R matches |
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F process(d, test_strings) |
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L(test_string) test_strings |
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V matches = word_break(test_string, d) |
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print((‘String = ’test_string‘, Dictionary = ’String(d)‘. Solutions =’)‘ ’matches.len) |
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L(match) matches |
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print(‘ Word Break = ’match) |
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print() |
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V d = [‘a’, ‘aa’, ‘b’, ‘ab’, ‘aab’] |
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process(d, [‘aab’, ‘aa b’]) |
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d = [‘abc’, ‘a’, ‘ac’, ‘b’, ‘c’, ‘cb’, ‘d’] |
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process(d, [‘abcd’, ‘abbc’, ‘abcbcd’, ‘acdbc’, ‘abcdd’])</lang> |
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{{out}} |
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<pre style="height:45ex;overflow:scroll"> |
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String = aab, Dictionary = [a, aa, b, ab, aab]. Solutions = 4 |
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Word Break = [aab] |
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Word Break = [a, ab] |
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Word Break = [aa, b] |
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Word Break = [a, a, b] |
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String = aa b, Dictionary = [a, aa, b, ab, aab]. Solutions = 0 |
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String = abcd, Dictionary = [abc, a, ac, b, c, cb, d]. Solutions = 2 |
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Word Break = [abc, d] |
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Word Break = [a, b, c, d] |
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String = abbc, Dictionary = [abc, a, ac, b, c, cb, d]. Solutions = 1 |
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Word Break = [a, b, b, c] |
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String = abcbcd, Dictionary = [abc, a, ac, b, c, cb, d]. Solutions = 3 |
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Word Break = [abc, b, c, d] |
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Word Break = [a, b, cb, c, d] |
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Word Break = [a, b, c, b, c, d] |
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String = acdbc, Dictionary = [abc, a, ac, b, c, cb, d]. Solutions = 2 |
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Word Break = [ac, d, b, c] |
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Word Break = [a, c, d, b, c] |
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String = abcdd, Dictionary = [abc, a, ac, b, c, cb, d]. Solutions = 2 |
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Word Break = [abc, d, d] |
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Word Break = [a, b, c, d, d] |
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</pre> |
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=={{header|Aime}}== |
=={{header|Aime}}== |
Revision as of 05:48, 6 May 2021
- Task
- Given an input string and a dictionary of words, segment the input string into a space-separated sequence of dictionary words if possible.
- Metrics
- Counting
- Word frequency
- Letter frequency
- Jewels and stones
- I before E except after C
- Bioinformatics/base count
- Count occurrences of a substring
- Count how many vowels and consonants occur in a string
- Remove/replace
- XXXX redacted
- Conjugate a Latin verb
- Remove vowels from a string
- String interpolation (included)
- Strip block comments
- Strip comments from a string
- Strip a set of characters from a string
- Strip whitespace from a string -- top and tail
- Strip control codes and extended characters from a string
- Anagrams/Derangements/shuffling
- Word wheel
- ABC problem
- Sattolo cycle
- Knuth shuffle
- Ordered words
- Superpermutation minimisation
- Textonyms (using a phone text pad)
- Anagrams
- Anagrams/Deranged anagrams
- Permutations/Derangements
- Find/Search/Determine
- ABC words
- Odd words
- Word ladder
- Semordnilap
- Word search
- Wordiff (game)
- String matching
- Tea cup rim text
- Alternade words
- Changeable words
- State name puzzle
- String comparison
- Unique characters
- Unique characters in each string
- Extract file extension
- Levenshtein distance
- Palindrome detection
- Common list elements
- Longest common suffix
- Longest common prefix
- Compare a list of strings
- Longest common substring
- Find common directory path
- Words from neighbour ones
- Change e letters to i in words
- Non-continuous subsequences
- Longest common subsequence
- Longest palindromic substrings
- Longest increasing subsequence
- Words containing "the" substring
- Sum of the digits of n is substring of n
- Determine if a string is numeric
- Determine if a string is collapsible
- Determine if a string is squeezable
- Determine if a string has all unique characters
- Determine if a string has all the same characters
- Longest substrings without repeating characters
- Find words which contains all the vowels
- Find words which contain the most consonants
- Find words which contains more than 3 vowels
- Find words whose first and last three letters are equal
- Find words with alternating vowels and consonants
- Formatting
- Substring
- Rep-string
- Word wrap
- String case
- Align columns
- Literals/String
- Repeat a string
- Brace expansion
- Brace expansion using ranges
- Reverse a string
- Phrase reversals
- Comma quibbling
- Special characters
- String concatenation
- Substring/Top and tail
- Commatizing numbers
- Reverse words in a string
- Suffixation of decimal numbers
- Long literals, with continuations
- Numerical and alphabetical suffixes
- Abbreviations, easy
- Abbreviations, simple
- Abbreviations, automatic
- Song lyrics/poems/Mad Libs/phrases
- Mad Libs
- Magic 8-ball
- 99 bottles of beer
- The Name Game (a song)
- The Old lady swallowed a fly
- The Twelve Days of Christmas
- Tokenize
- Text between
- Tokenize a string
- Word break problem
- Tokenize a string with escaping
- Split a character string based on change of character
- Sequences
11l
<lang 11l>T Node
String val [String] parsed F (val, [String] &parsed = [String]()) .val = val .parsed = 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’])</lang>
- 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
<lang 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;
}</lang>
- Output:
abcd: a b cd abbc: a b bc abcbcd: abc b cd acdbc: a cd bc abcdd: can't break
C++
<lang cpp>#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;
} </lang>
- Output:
a b cd
D
<lang d>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; }
}</lang>
- 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
<lang Delphi> 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('cant break'); d.Free;
Readln;
end.</lang>
Go
<lang 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 := []prefixTemplate: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") } }
}</lang>
- Output:
abcd: a b cd abbc: a b bc abcbcd: a bc b cd acdbc: a cd bc can't break
Haskell
<lang haskell>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</lang>
- 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.
<lang J> 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 </lang>
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. <lang Java> 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; } }
} </lang>
- 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.
<lang javascript>(() => {
'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();
})();</lang>
- 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)
Julia
Some extra loops to record and print all solutions.<lang Julia> 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 $s : No fitting matches found.") end end
end
wordbreak()</lang>
- Output:
1 Solution(s) for abcd:
Solution: SubString{String}["cd", "b", "a"]1 Solution(s) for abbc:
Solution: SubString{String}["b", "a", "bc"]2 Solution(s) for abcbcd:
Solution: SubString{String}["cd", "b", "a", "bc"] Solution: SubString{String}["cd", "abc", "b"]1 Solution(s) for acdbc:
Solution: SubString{String}["cd", "a", "bc"]No solutions for abcdd : No fitting matches found.
Kotlin
I've downloaded the free dictionary at http://www.puzzlers.org/pub/wordlists/unixdict.txt for this task. All single letters from 'a' to 'z' are considered to be words by this dictionary but 'bc' and 'cd' which I'd have expected to be present are not. <lang scala>// version 1.1.3
import java.io.File
val partitions = mutableListOf<List<String>>()
fun partitionString(s: String, ml: MutableList<String>, level: Int) {
for (i in s.length - 1 downTo 1) { val part1 = s.substring(0, i) val part2 = s.substring(i) ml.add(part1) ml.add(part2) partitions.add(ml.toList()) if (part2.length > 1) { ml.removeAt(ml.lastIndex) partitionString(part2, ml, level + 1) } while (ml.size > level) ml.removeAt(ml.lastIndex) }
}
fun main(args: Array<String>) {
val words = File("unixdict.txt").readLines() val strings = listOf("abcd", "abbc", "abcbcd", "acdbc", "abcdd") for (s in strings) { partitions.clear() partitions.add(listOf(s)) val ml = mutableListOf<String>() partitionString(s, ml, 0) val solutions = mutableListOf<List<String>>() for (partition in partitions) { var allInDict = true for (item in partition) { if (words.indexOf(item) == -1) { allInDict = false break } } if (allInDict) solutions.add(partition) } val plural = if (solutions.size == 1) "" else "s" println("$s: ${solutions.size} solution$plural") for (solution in solutions) { println(" ${solution.joinToString(" ")}") } println() }
}</lang>
- Output:
abcd: 2 solutions abc d a b c d abbc: 1 solution a b b c abcbcd: 3 solutions abc b c d a b cb c d a b c b c d acdbc: 2 solutions ac d b c a c d b c abcdd: 2 solutions abc d d a b c d d
Lua
<lang lua>-- a specialized dict format is used to minimize the -- possible candidates for this particalur problem function genDict(ws)
local d,dup,head,rest = {},{} for w in ws:gmatch"%w+" do local lw = w:lower() if not dup[lw] then dup[lw], head,rest = true, lw:match"^(%w)(.-)$" 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</lang>
- Output:
abcd a b cd abbc a b bc abcbcd abc b cd acdbc a cd bc abcdd nil -no solution-
Perl
<lang 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)";
}</lang>
- 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
See talk page <lang Phix>procedure populate_dict(sequence s) ?s
for i=1 to length(s) do setd(s[i],0) end for
end procedure
--/* function valid_word(string word)
if length(word)=1 then return find(word,{"a","i"})!=0 end if if find(word,{"sis","sst","se"}) then return false end if -- hack for i=1 to length(word) do integer ch = word[i] if ch<'a' or ch>'z' then return false end if end for return true
end function --*/
populate_dict(split("a bc abc cd b")) --/* integer fn = open("demo\\unixdict.txt","r") sequence words = get_text(fn,GT_LF_STRIPPED) close(fn) for i=length(words) to 1 by -1 do
if not valid_word(words[i]) then words[i] = words[$] words = words[1..$-1] end if
end for populate_dict(words) --*/
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(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) sequence wordstarts = repeat({},l), wordends = repeat(0,l) integer wordend = 1 -- (pretend a word just ended at start)
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 wordends[i+wl-1] -= 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 wordends[i+wl-1] -= 1 worthwhile = true end if end for wordstarts[i] = {} end if wordend = wordends[i] end for end while
--?{wordstarts,wordends}
if sum(wordends)=0 then printf(1,"%s: not possible",{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
constant tests = {"abcd","abbc","abcbcd","acdbc","abcdd"} --constant tests = {"wordsisstringofspaceseparatedwords"} for i=1 to length(tests) do test(tests[i]) end for</lang>
- 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
PicoLisp
<lang 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))</lang>
- 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).
<lang python>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()</lang>
- 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>#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)))</lang>
- 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)
This implementation does not necessarily find every combination, it returns the one with the longest matching tokens.
<lang perl6>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" }</lang>
- 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. <lang rexx>/*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</lang>
- 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
<lang 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
</lang> Output:
abcd = a b cd abbc = a b bc abcbcd = abc b cd acdbc = a cd bc
Rust
Dynamic programming
<lang rust>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());
}</lang>
- Output:
"a b cd"
Scala
First solution
Finds all possible solutions recursively, using a trie representation of the dictionary: <lang scala>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, "")
}</lang> Calling it with some example strings: <lang scala>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 ${solutions.size} solution(s):") for (words <- solutions) { println("\t" + words.mkString(" ")) }
}</lang>
- Output:
abcd has 1 solution(s): a b cd abbc has 1 solution(s): a b bc abcbcd has 2 solution(s): a bc b cd abc b cd acdbc has 1 solution(s): a cd bc abcdd has 0 solution(s):
Combined solution
- Output:
Best seen running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).
<lang Scala>object WordBreak extends App {
val dict = buildTrie("a", "bc", "abc", "cd", "b") lazy val empty = TrieNode(isWord = false, Map.empty) // lazy or in a companion object
case class TrieNode(isWord: Boolean, children: Map[Char, TrieNode]) {
def add(s: String): TrieNode = { def child = children.withDefaultValue(empty)(s.head)
if (s.isEmpty) copy(isWord = true) else copy(children = children.updated(s.head, child.add(s.tail))) } }
def buildTrie(xs: String*): TrieNode = xs.foldLeft(empty)(_.add(_))
def wordBreak(s: String, dict: TrieNode): List[List[String]] = {
def wordBreakRec(s: String, root: TrieNode, currentPos: TrieNode, soFar: String): List[List[String]] = {
def usingCurrentWord = if (currentPos.isWord) if (s.isEmpty) List(List(soFar)) else wordBreakRec(s, root, root, "").map(soFar :: _) else Nil
def usingCurrentPrefix = (for {ch <- s.headOption child <- currentPos.children.get(ch) } yield wordBreakRec(s.tail, root, child, soFar + ch)) .getOrElse(Nil)
usingCurrentWord ++ usingCurrentPrefix }
wordBreakRec(s, dict, dict, "") }
// Calling it with some example strings: List("abcd", "abbc", "abcbcd", "acdbc", "abcdd").foreach(s => { val solutions = wordBreak(s, dict)
println(s"$s has ${solutions.size} solution(s):") solutions.foreach(words => println(words.mkString("\t", " ", ""))) })
}</lang>
Seed7
<lang seed7>$ 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;</lang>
- 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
<lang ruby>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)')
}</lang>
- 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
<lang swift>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")")
}</lang>
- 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 (we could easily stop looking after one is found, just add a condition to do nothing if a done-flag is set) <lang tailspin> 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 </lang>
- 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
<lang vbnet>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</lang>
- 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
<lang ecmascript>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") }
}</lang>
- Output:
abcd: a b cd abbc: a b bc abcbcd: a bc b cd acdbc: a cd bc can't break
zkl
<lang 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(" ")
}</lang> <lang zkl>foreach text in (T("abcd","abbc","abcbcd","acdbc","abcdd")){
println(text,": ",wordBreak(text,"a bc abc cd b"))
}</lang>
- Output:
abcd: a b cd abbc: a b bc abcbcd: a bc b cd acdbc: a cd bc abcdd: not possible