Regular expression for checking every substring

Question

I had this question in my end semester exam, unfortunately I couldn't solve it and tried it for several days and no luck.

Condition- For every substring of length 4 in a string of length n, write a RE to force the rule- there must be exactly three 1's.

My solution kind of looked like (1+11+111+€)(0111)*(0+€).

But this is obviously wrong, string 11011 is a valid solution too.

Update- my new solution is (1+11+111+€)(0111)*(0+01+011+€).

Update- the plus operator is actually 'OR'

Update- € is empty string

Update - the string length has no requirements. A string of length 5 will have 2 substrings of length 4, the first 4 chars and the last 4 chars

What happens if the string has 10 characters? How do we handle the final two characters? — Tim Biegeleisen, Feb 07 '17 at 10:59
We can handle only 4 consecutive characters at a time, the last substring would be the last 4 characters — Ajinkya Jumbad, Feb 07 '17 at 11:02
Nope, the string length has no requirements. A string of length 5 will have 2 substrings of length 4, the first 4 chars and the last 4 chars — Ajinkya Jumbad, Feb 07 '17 at 11:05
You seem to misunderstand the repetition operator. `11+11+1` matches five or more 1s, where the second and the fourth may be repeated more than once. (This doesn't make any sense; because of greediness, the second will repeat as many times as possible, and the fourth will never get more than one.) — tripleee, Feb 07 '17 at 11:19
You might want to document the other oddities in your notation, too. What's €? — tripleee, Feb 07 '17 at 11:32
There's a character for that, obviously. [ε aka GREEK SMALL LETTER EPSILON (U+03B5)](http://www.fileformat.info/info/unicode/char/3b5/index.htm). Instead of plus I guess you mean [∪ aka UNION (U+222A)](http://www.fileformat.info/info/unicode/char/222a/index.htm) — tripleee, Feb 07 '17 at 12:01

score 0 · Answer 1 · answered Feb 07 '17 at 11:32

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I think the professor is looking for the realization that in order for the condition to hold, a 1 may never be bracketed by zeros on both sides.

(0?11)*

To complete the picture, we also need to include the case for n=1 where either a 0 or a 1 is allowed (I presume).

^[01]$|(0?11)*

I'm using traditional regex here, where [...] denotes a character class, | is "or", parentheses do grouping, and * specifies zero or more repetitions (Kleene star).

answered Feb 07 '17 at 11:32

tripleee

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Wouldn't 011011 be in your RE language? But consider the first substring of length 4, 0110 - it has two 1's but we need exactly three 1's. – Ajinkya Jumbad Feb 07 '17 at 11:42
Oh sorry, I missed the "exactly" and assumed "at least". Do you want me to delete this answer or should I leave it as a clue for someone else to perhaps develop to a working answer? – tripleee Feb 07 '17 at 11:53
It's okay, you can leave it here. If you get the answer, post it here ! – Ajinkya Jumbad Feb 07 '17 at 11:56

Rohit · Answer 2 · 2017-02-11T10:36:34.377

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Edit: Consider the following regular expression where € refers to the empty string and assuming the alphabet consists of only {0,1,€}:

€ + (0111)* + (1110)* + (1101)* + (1011)*

edited Feb 11 '17 at 10:36

answered Feb 11 '17 at 10:08

Rohit

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01110 is not in your RE but 01110 is a valid string – Ajinkya Jumbad Feb 11 '17 at 12:13

Omar Sherif · Answer 3 · 2017-02-17T14:24:31.470

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Python 2

import re

# regular expression
# '^' in the start of the expression means from the very start of the string
# "[^1.]*" means match any character unless it is '1'
# '$' in the end of the expression means till the very end of the string
# Summary:
# from the start of the string till the end of it check if we have
# '1' exactly 3 times, 
# and before or after them you might find any type of characters
# and these characters must not be equal '1' ...
expression = '^[^1.]*1[^1.]*1[^1.]*1[^1.]*$'

# testing strings
tests = ["01110", "11100", "00111", "010101", "00100100111", "00100"]


prog = re.compile(expression)

for test in tests:
    print 'matched' if prog.match(test) else 'not matched'

edited Feb 17 '17 at 14:24

answered Feb 13 '17 at 18:23

Omar Sherif

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could you explain how you came up with this ? – Ajinkya Jumbad Feb 17 '17 at 10:40
I have updated my answer adding more explanation on how the regular expression works, this is a link that may help you figure out how regular expressions work http://www.rexegg.com/regex-quickstart.html – Omar Sherif Feb 17 '17 at 14:26

Regular expression for checking every substring

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