00:01
We are asked to construct a deterministic finite state automaton that recognizes the set of all bit strings that do not contain three consecutive zeros.
00:18
So the set of strings we're interested in, if it doesn't contain three consecutive zeros, let s be the set of all bit strings not containing three consecutive zeros.
01:14
And let's pick a start state s -0, and let's consider four different states, s -0, s -1, s -2, and s -3.
01:42
So s -1 represented the last digit was zero, and on top of that, there was no zero -zero preceding the last digit.
02:12
S -2 represented the last two digits were zero -zero, but there was no zero preceding it, the bit preceding it wasn't zero.
02:41
And finally, s3 represents the string contained 0 -00.
02:52
S -3 represents, string contains 0 -00.
03:10
And now, to describe the transitions, we're going to, of course, start at s -0, since we designated this will be our start.
03:19
And we have the s -0, s -1, and s -2 will all be final states.
03:23
Since all the strings containing three consecutive zeros and the state s3.
03:42
Okay, so suppose that the string contains a zero, then we're going to move from s0 to s1, and then suppose the string contains a second zero...