Word break problem: Difference between revisions
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abcdd:
(Not parseable with these words)</pre>
=={{header|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".
<lang jq>
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
</lang>
{{out}}
<pre>
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.
</pre>
=={{header|Julia}}==
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