00:01
In part a, we are asked to find the k equivalence classes of the automaton m, which we are given in exercise 62 for the values of k equaling 0, 1, 2, and 3, and to find the star equivalence classes of m.
00:43
So looking at problem 62, you'll see the automaton, m that they're referring to.
00:54
Now, looking at this machine, we see that it has three states, s0, s1, and s2, and we see that first i'll construct the zero equivalence classes.
01:28
These are the set of all final states and the set of all non -final states.
02:08
Now, the final states we see in the given figure are generated by two circles, while the non -final states.
02:15
Final states have a single circle.
02:18
So we see that the set of non -final states, this is equal to s0, s1, s2, and s4, and the set of final states, which has s3 as a representative, is s3, s5, and s6.
03:15
Now consider the one equivalence classes.
03:21
We have that s and t are in the same name what equivalence class if s and t are both non -final states or are both final states.
04:17
And on top of that, if we arrive at both final states when the input is a and when starting from states s and t.
05:00
So we see that s zero is the only non -final state that moves to a non -final state when the input is zero.
05:36
Therefore it belongs in its own equivalence class.
05:38
So we have s0, the class containing s0 r1 is equal to simply the set s0.
05:54
Now we have that s1 is the only remaining non -final state that moves to a non -final state when the input is 1.
06:30
Therefore, it follows that it belongs to its own equivalence class, and we have the equivalence class with representative s1 is simply s1.
06:49
Now we see that s2 and s4 are the remaining non -final states, and we see that both s2 and s4 move to final states when the input is both 0 and 1.
07:40
Therefore, it follows that s2 and s4 belong to the same equivalence class.
07:54
So the equivalence class with representative s2 is the set s2, s4.
08:03
Now we see that s6 is the only final state that remains at a final state when the input is one.
08:33
Therefore, it belongs to its own equivalence class.
08:36
So the equivalence class with representative s6 is simply the set s6.
08:48
Now we see that s3 and s5 are remaining final states that move to a final state when the input is both zero.
09:00
And it moved to a non -final state when the input is 1.
10:01
So it follows the s -1 is in its own 1 -no, no, sorry.
10:11
Therefore, it follows the s -3 and s -5 are in the same equivalence class.
10:18
So the equivalence class with the representative s -3 is the set s -3 -s -5.
10:33
Okay, so going through, i'm going to circle the equivalence classes that we identified.
10:39
This was our equivalence class from, keep going up.
10:51
Here's for k equals 0.
10:53
These are the two equivalence classes.
10:56
And for k equals 1, we have this one, we have this one, we have this one, we have this one, and we have this last one...