I'm having trouble with this regular expression:
Construct a regular expression defining the following language over alphabet
Σ = { a,b }
L6 = {All words that begin and end in different letters}
Here are some examples of regular expressions I was able to solve:
1. L1 = {all words of even length ending in ab}
(aa + ab + ba + bb)*(ab)
2. L2 = {all words that DO NOT have the substring ab}
b*a*
Would this work:
(a.*b)|(b.*a)
Or said in Kleene way:
a(a+b)*b+b(a+b)*a
This should do it:
"^((a.*b)|(b.*a))$"
1- Write a Regular expression for each of the following languages: (a)language of all those strings which end with substrings 'ab' and have odd length. (b)language of all those strings which do not contain the substring 'abb'.
2- Construct a deterministic FSA for each of the following languages: (a)languages of all those strings in which second last symbol is 'b'. (b)language of all those strings whose length is odd,but contain even number if b's.
(aa+ab+ba+bb)∗(a+b)ab
It can choose any number of even length and have any character from a and b, and then end at string ab.
Related
Is there any regex to find a substring with a specific length that contains minimum number of a specific char occurrences?
For example I have a string such as: AABABAAAAA for this string we have a substring with length 5 that contains two B => AABAB so regex should find it.
But for the AAAABAAAAB there is not any substring with length of 5 that contains two B.
Suppose our string just contains A and B and we want to find substring with length of 5 that contains at least two B:
AAAABAAAAB -> Invalid
AAAAAAAABB -> Valid
AAAAAAAAAABAABAAAAAA -> Valid
AAAABAAAAAAABAAAAAAA -> Invalid
Brute force:
.B..B|B...B|..BB.|.B.B.|..B.B|BB...|B.B..|...BB|B..B.|.BB..
Well, I know that such regular expression is not parametrizable. On the other hand it's possible to obtain it programmatically (the example is in Python):
import itertools
def get_regex(char, charnum, strsize):
chars = char * charnum + "." * (strsize - charnum)
return "|".join("".join(x) for x in set(itertools.permutations(chars)))
print get_regex("B", 2, 5)
You can use this regex:
(?=[^B]{0,3}B[^B]{0,3}B).{5}
RegEx Demo
I have to process a comma separated string which contains triplets of values and translate them to runtime types,the input looks like:
"1x2y3z,80r160g255b,48h30m50s,1x3z,255b,1h,..."
So each substring should be transformed this way:
"1x2y3z" should become Vector3 with x = 1, y = 2, z = 3
"80r160g255b" should become Color with r = 80, g = 160, b = 255
"48h30m50s" should become Time with h = 48, m = 30, s = 50
The problem I'm facing is that all the components are optional (but they preserve order) so the following strings are also valid Vector3, Color and Time values:
"1x3z" Vector3 x = 1, y = 0, z = 3
"255b" Color r = 0, g = 0, b = 255
"1h" Time h = 1, m = 0, s = 0
What I have tried so far?
All components optional
((?:\d+A)?(?:\d+B)?(?:\d+C)?)
The A, B and C are replaced with the correct letter for each case, the expression works almost well but it gives twice the expected results (one match for the string and another match for an empty string just after the first match), for example:
"1h1m1s" two matches [1]: "1h1m1s" [2]: ""
"11x50z" two matches [1]: "11x50z" [2]: ""
"11111h" two matches [1]: "11111h" [2]: ""
This isn't unexpected... after all an empty string matches the expression when ALL of the components are empty; so in order to fix this issue I've tried the following:
1 to 3 quantifier
((?:\d+[ABC]){1,3})
But now, the expression matches strings with wrong ordering or even repeated components!:
"1s1m1h" one match, should not match at all! (wrong order)
"11z50z" one match, should not match at all! (repeated components)
"1r1r1b" one match, should not match at all! (repeated components)
As for my last attempt, I've tried this variant of my first expression:
Match from begin ^ to the end $
^((?:\d+A)?(?:\d+B)?(?:\d+C)?)$
And it works better than the first version but it still matches the empty string plus I should first tokenize the input and then pass each token to the expression in order to assure that the test string could match the begin (^) and end ($) operators.
EDIT: Lookahead attempt (thanks to Casimir et Hippolyte)
After reading and (try to) understanding the regex lookahead concept and with the help of Casimir et Hippolyte answer I've tried the suggested expression:
\b(?=[^,])(?=.)((?:\d+A)?(?:\d+B)?(?:\d+C)?)\b
Against the following test string:
"48h30m50s,1h,1h1m1s,11111h,1s1m1h,1h1h1h,1s,1m,1443s,adfank,12322134445688,48h"
And the results were amazing! it is able to detect complete valid matches flawlessly (other expressions gave me 3 matches on "1s1m1h" or "1h1h1h" which weren't intended to be matched at all). Unfortunately it captures emtpy matches everytime a unvalid match is found so a "" is detected just before "1s1m1h", "1h1h1h", "adfank" and "12322134445688", so I modified the Lookahead condition to get the expression below:
\b(?=(?:\d+[ABC]){1,3})(?=.)((?:\d+A)?(?:\d+B)?(?:\d+C)?)\b
It gets rid of the empty matches in any string which doesn't match (?:\d+[ABC]){1,3}) so the empty matches just before "adfank" and "12322134445688" are gone but the ones just before "1s1m1h", "1h1h1h" are stil detected.
So the question is: Is there any regular expression which matches three triplet values in a given order where all component is optional but should be composed of at least one component and doesn't match empty strings?
The regex tool I'm using is the C++11 one.
Yes, you can add a lookahead at the begining to ensure there is at least one character:
^(?=.)((?:\d+A)?(?:\d+B)?(?:\d+C)?)$
If you need to find this kind of substring in a larger string (so without to tokenize before), you can remove the anchors and use a more explicit subpattern in a lookahead:
(?=\d+[ABC])((?:\d+A)?(?:\d+B)?(?:\d+C)?)
In this case, to avoid false positive (since you are looking for very small strings that can be a part of something else), you can add word-boundaries to the pattern:
\b(?=\d+[ABC])((?:\d+A)?(?:\d+B)?(?:\d+C)?)\b
Note: in a comma delimited string: (?=\d+[ABC]) can be replaced by (?=[^,])
I think this might do the trick.
I am keying on either the beginning of the string to match ^ or the comma separator , for fix the start of each match: (?:^|,).
Example:
#include <regex>
#include <iostream>
const std::regex r(R"~((?:^|,)((?:\d+[xrh])?(?:\d+[ygm])?(?:\d+[zbs])?))~");
int main()
{
std::string test = "1x2y3z,80r160g255b,48h30m50s,1x3z,255b";
std::sregex_iterator iter(test.begin(), test.end(), r);
std::sregex_iterator end_iter;
for(; iter != end_iter; ++iter)
std::cout << iter->str(1) << '\n';
}
Output:
1x2y3z
80r160g255b
48h30m50s
1x3z
255b
Is that what you are after?
EDIT:
If you really want to go to town and make empty expressions unmatched then as far as I can tell you have to put in every permutation like this:
const std::string A = "(?:\\d+[xrh])";
const std::string B = "(?:\\d+[ygm])";
const std::string C = "(?:\\d+[zbs])";
const std::regex r("(?:^|,)(" + A + B + C + "|" + A + B + "|" + A + C + "|" + B + C + "|" + A + "|" + B + "|" + C + ")");
Is it possible to do a find/replace using regular expressions on a string of dna such that it only considers every 3 characters (a codon of dna) at a time.
for example I would like the regular expression to see this:
dna="AAACCCTTTGGG"
as this:
AAA CCC TTT GGG
If I use the regular expressions right now and the expression was
Regex.Replace(dna,"ACC","AAA") it would find a match, but in this case of looking at 3 characters at a time there would be no match.
Is this possible?
Why use a regex? Try this instead, which is probably more efficient to boot:
public string DnaReplaceCodon(string input, string match, string replace) {
if (match.Length != 3 || replace.Length != 3)
throw new ArgumentOutOfRangeException();
var output = new StringBuilder(input.Length);
int i = 0;
while (i + 2 < input.Length) {
if (input[i] == match[0] && input[i+1] == match[1] && input[i+2] == match[2]) {
output.Append(replace);
} else {
output.Append(input[i]);
output.Append(input[i]+1);
output.Append(input[i]+2);
}
i += 3;
}
// pick up trailing letters.
while (i < input.Length) output.Append(input[i]);
return output.ToString();
}
Solution
It is possible to do this with regex. Assuming the input is valid (contains only A, T, G, C):
Regex.Replace(input, #"\G((?:.{3})*?)" + codon, "$1" + replacement);
DEMO
If the input is not guaranteed to be valid, you can just do a check with the regex ^[ATCG]*$ (allow non-multiple of 3) or ^([ATCG]{3})*$ (sequence must be multiple of 3). It doesn't make sense to operate on invalid input anyway.
Explanation
The construction above works for any codon. For the sake of explanation, let the codon be AAA. The regex will be \G((?:.{3})*?)AAA.
The whole regex actually matches the shortest substring that ends with the codon to be replaced.
\G # Must be at beginning of the string, or where last match left off
((?:.{3})*?) # Match any number of codon, lazily. The text is also captured.
AAA # The codon we want to replace
We make sure the matches only starts from positions whose index is multiple of 3 with:
\G which asserts that the match starts from where the previous match left off (or the beginning of the string)
And the fact that the pattern ((?:.{3})*?)AAA can only match a sequence whose length is multiple of 3.
Due to the lazy quantifier, we can be sure that in each match, the part before the codon to be replaced (matched by ((?:.{3})*?) part) does not contain the codon.
In the replacement, we put back the part before the codon (which is captured in capturing group 1 and can be referred to with $1), follows by the replacement codon.
NOTE
As explained in the comment, the following is not a good solution! I leave it in so that others will not fall for the same mistake
You can usually find out where a match starts and ends via m.start() and m.end(). If m.start() % 3 == 0 you found a relevant match.
Given an input string such as " word1 word2 word3 word4 ", what would be the best approach to split this as an array of strings in Go? Note that there can be any number of spaces or unicode-spacing characters between each word.
In Java I would just use someString.trim().split("\\s+").
(Note: possible duplicate Split string using regular expression in Go doesn't give any good quality answer. Please provide an actual example, not just a link to the regexp or strings packages reference.)
The strings package has a Fields method.
someString := "one two three four "
words := strings.Fields(someString)
fmt.Println(words, len(words)) // [one two three four] 4
DEMO: http://play.golang.org/p/et97S90cIH
From the docs:
Fields splits the string s around each instance of one or more consecutive white space characters, as defined by unicode.IsSpace, returning a slice of substrings of s or an empty slice if s contains only white space.
If you're using tip: regexp.Split
func (re *Regexp) Split(s string, n int) []string
Split slices s into substrings separated by the expression and returns
a slice of the substrings between those expression matches.
The slice returned by this method consists of all the substrings
of s not contained in the slice returned by FindAllString. When called
on an expression that contains no metacharacters, it is equivalent to strings.SplitN.
Example:
s := regexp.MustCompile("a*").Split("abaabaccadaaae", 5)
// s: ["", "b", "b", "c", "cadaaae"]
The count determines the number of substrings to return:
n > 0: at most n substrings; the last substring will be the unsplit remainder.
n == 0: the result is nil (zero substrings)
n < 0: all substrings
I came up with the following, but that seems a bit too verbose:
import "regexp"
r := regexp.MustCompile("[^\\s]+")
r.FindAllString(" word1 word2 word3 word4 ", -1)
which will evaluate to:
[]string{"word1", "word2", "word3", "word4"}
Is there a more compact or more idiomatic expression?
You can use package strings function split
strings.Split(someString, " ")
strings.Split
Using regular expressions (in any language) is there a way to match a pattern that wraps around from the end to the beginning of a string? For example, if i want the match the pattern:
"street"
against the string:
m = "et stre"
it would match m[3:] + m[:2]
You can't do that directly in the regexp. What you can do is some arithmetic. Append the string to itself:
m = "et stre"
n = m + m //n = "et street stre"
If there is an odd number of matches in n (in this case, 1), the match was 'circular'. If not, there were no circular matches, and the number of matches in n is the double of the number of matches in m.