00:01
We are given sets and we are asked to find a deterministic finite state automaton that recognizes each set.
00:10
So in part a, we're given the set 0.
00:21
Now, since this is an automaton, we have to have a start state as zero.
00:36
Now, because lambda, the empty string should not be recognized, it follows that s .0.
00:47
Is not a final state.
00:56
Now suppose that the string starts with 0.
01:09
You see that the string should be recognized, so we'll want to move from the non -final state as 0 to the final state s1.
01:24
And if the string contains any more bits, then we want to move from the final state to a non -final state.
01:35
So suppose there are any more bits.
01:43
Then we'll move to the non -final state.
01:49
In this case, we don't want to move back to s -0, because if we were to do that, then we would have that, say, the string 0 -0 would be accepted, which we don't want.
02:01
So move to a new non -final state, s2.
02:10
And once we're at s2, we should not be able to escape because the string can never be recognized.
02:23
So we will remain.
02:25
Remain at s2 for any input.
02:39
Now on the other hand, suppose that the string starts with the 1.
02:52
Well, in this case, we have that 1 should not be recognized, so we should move from s0 to the non -final state s2, and as before, we're going to remain at s2 for any input.
03:07
And this concludes the description for the deterministic machine in part b, we're given the set 1 .00.
03:29
Now once again, let s0 be a start state, which we know must exist.
03:37
Since the empty string is not contained in this set, it follows the s0 is a non -final state.
03:53
Now, suppose that the first bit is a zero.
04:02
We'll always see that this string should not be accepted so far.
04:09
But it has potential to be accepted.
04:14
So we want to move to a new non -final state, otherwise the string 0 -0 would not be accepted when it should be.
04:21
So we'll move to non -final state s1.
04:30
And then once we're at s1, suppose that the second bit is a 0.
04:39
Well, we'll want to move to a final state, s2, and we see that if that's a final state, there are any more bits, then the string should not be accepted, and we should move to a non -final state.
05:13
And this can't be the same at s -0 because this would imply that a string like 0 -0 -0 -0 would be accepted when it's not.
05:27
And we can't make this the same as s -1 because this would mean that a string like 0 -0 -0 -0 -0.
05:37
Would be accepted, which is also not true.
05:41
So move to a new non -final state s3.
05:49
And we see that once we're at s3, the strain can never be recognized.
05:53
So we shall remain at s3 for any input.
06:04
So that's one case.
06:06
Now suppose that the first bit is a 1...
