Regexp for exceptions 101 and 110

Question

What is a regular expression that accepts everything in {0,1} but does not have substring 110 or 101?

Accept:

• 111111
• 000011111
• 100001000001001
• 010
• 1

Reject

• 100110
• 010100
• 123

Edit: In the comments on the answers below, this question requires formal regular expression.

Answer 1

Limited to the formal designation of regular expressions:

((1|0*|0*1)(000*))*0*(10*|1*)

Answer 2

This solution (even without viewing):

/^0*(11*$|10$|100+)*$/

• Start with any number of zeros.
• Loop (I know: the line processed so far does not end with "1" or "10")
  • "$ 1" in order (& stop)
  • "11", , , .
  • "10 $" .
  • "10" , . .

Answer 3

, /101|110/

Answer 4

, , , , regex lookahead.

/^(1(?!01|10)|0)*$/

Answer 5

:

/^([01])\1*$/

Answer 6

DFA .

, "" ( "", "xor", "not", )

,

(0 | 100 | (1 | 10 | 11 *) $) *

This can be solved with a possessive comparison. (111 + $) is 111 ++