00:01
We are asked to construct a finite state automaton with four states that recognizes the set of bit strings containing an even number of ones and an odd number of zeros.
00:25
So let s be a set of all bit strings with an even number of ones and an odd number of zeros.
00:45
And let s0 be a start state.
01:07
Consider the empty string.
01:11
The empty string does not have an odd number of zeros.
01:16
It has an even number of zeros.
01:18
So it follows that the empty string is not recognized.
01:21
And therefore, s0 is not a final state.
01:34
And now, as the problem suggests, let's consider the four states.
01:39
S0, s1, s2, and s3.
01:42
So s0 represents that there were an...
01:46
Even number of zeros and an even number of ones in the previous digits.
02:12
S1 represents there were an even number of zeros and an odd number of ones in the previous bits.
02:31
So far neither of these are final states.
02:36
S2 will represent that there were an odd number of zeros and an even number of ones in the previous bits.
02:53
We see that this is the same criteria as our set s.
02:58
So this is going to be a final state.
03:19
And finally, s3 represents the fact that there were an odd number of zeros and an odd number of ones in previous bits.
03:37
And this is also not a final state.
03:47
Let's consider the transitions between the states.
03:51
So suppose the input is 1 at s0, or i guess instead suppose we are at s0, then if the input is a 1, well, at s0, we know that the number of zeros is even, the number of 1s is even.
04:20
So if the input is 1, we now have an even number of zeros and an odd numbers of 1...