正则表达式和自动机

Question

I'm studying Regular Expressions by reading Aho's book. I don't understand two of the statements in the book:

Question A:

1(0+1)*1 + 1 : denotes the set of all strings beginning and ending with a 1.

My question why is +1 added at end of the regular expression? Shouldn't 1(0+1)*1 be sufficient?

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I'm also having trouble with the following:

Question B:

The set of strings containing only 0's and 1's that have atmost one 1 as below

    0*+0*10* 

Can you explain how the solution 0*+0*10* is arrived at, step by step?

Answer 1

As to question a: 1(0+1)*1 does not match the one-character string 1, which begins and ends with 1. One needs a special case for it, which the example does.

As to question b: I cannot speak for the author. However... Any string that contains at most one 1 is a string that either has no 1s or has exactly one 1. Assuming that the alphabet is {0,1}, the former means any string that contains zero or more 0s, that is, 0*. The latter, with the same assumption, means any string that contains zero or more 0s followed by one 1 followed by zero or mpre 0s, that is, 0*10*. Combining these yields the example.

Answer 2

For Question a : 1(0+1)*1 denotes set of all strings beginning and ending with one but does not contain string 1 which has length one and starts and ends with one.

For Question b : Set of strings containing atmost one 1 = A + B where A is set of all strings containing zero 1 s and B is the set of all strings containing exactly one 1

So A is 0* and B is 0*10* Hence we get the answer as 0* + 0*10*

Answer 3

For the first example, the string that is accepted by the + 1 but not by the rest is 1 . The rest of the expressions can handle 11, but not a string where the first and last character are the same.

It's similar reasoning for the second string - 0* handles strings of all zeroes, 0*10* handles strings of 1 one.