Regular Expression Clarity

Question

I have a question

Write down a regular expression for all binary strings that end with an even (nonzero) number of 0's.

So far I have come up with L = { (0/1)* (00)+ }

But it doesn't seem to work all the time. The (0/1)* can generate 0110 and the (00)+ will add 0110 00 - which doesn't make an ending of even number of 0's.

Can anyone help me with this?

Thanks

score 2 · Accepted Answer · answered Aug 13 '14 at 14:13

2

I see you are using + to denote a superscript + for 1 or many, and a forward slash to denote a normal plus for a union.

This would work:

L = { (0/1)* 1/E (00)+ }

answered Aug 13 '14 at 14:13

RK2015

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score 1 · Answer 2 · answered Aug 10 '14 at 11:24

1

I've found the solution.

Its L = { (0/1)* 1/E (00)+ }

answered Aug 10 '14 at 11:24

Leo

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is this regex? `0/1` `1/E`? – Kent Aug 10 '14 at 11:27
According to my lecturer, he suggests to use a slash for OR instead of + to avoid ambiguity when using + for 1 or more – Leo Aug 10 '14 at 11:29

score 0 · Answer 3 · edited May 23 '17 at 12:20

0

This will work:

\b(([01]*1(0{2})+)|((00)+))\b

You can test it here

You can also check this similar problem:

Regex: Odd number of occurrences of a char

edited May 23 '17 at 12:20

Community

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answered Aug 10 '14 at 11:28

Acsisr

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score 0 · Answer 4 · answered Aug 10 '14 at 11:30

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Per condition you stated in your problem. Shouldn't it be { (0/1)* 1(00)+ } ? * denotes 0 or more occurrences. Whereas + denotes one or more occurrences.

answered Aug 10 '14 at 11:30

havish

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Regular Expression Clarity

4 Answers4