00:01
We're asked to construct a touring machine with tape symbols 0 ,1, and b, such that if we're given a bit string as an input, this can't have any blanks in it.
00:13
Then the machine is going to replace the first two consecutive ones on the tape with zeros, and will not change any of the other symbols on the tape.
00:24
So we know that our turing machine t is a four tuple with first component s, second component i, third component f, and fourth component s0, where s is the set of states, i is the set of tape symbols, which is 0 1b, f is the partial function representing the transitions, and s0 is the start state.
01:01
So for this problem, we want the machine to run past all zeros until it reaches its first one.
01:43
And we see that once t encounters its first one, so in state s zero, if t finds a one, then what we have is that t should enter state s1 and nothing else should happen really.
02:08
Should just move to the right because we aren't changing the one yet.
02:15
So i'll just say, and it will move right.
02:28
Actually, instead of changing s1, all right, now in s1, if t finds a zero, in other words, if the next term is a zero, then t is going to enter s0 again, we're essentially starting over, and it's going to keep moving right.
03:27
However, if we're an s1 and if t finds a 1, then this means we found a pair of ones.
03:39
And so t is going to enter s2.
03:51
And we also have that t is going to change the 1 to a 0.
04:01
And because this is the second one in the first pair of consecutive ones, this means that t should then move left after this.
04:17
So now we're in front of that first one in the pair of consecutive ones.
04:25
And therefore it follows that if we're an s2, if t finds, maybe i should, if we're an s2 and when t finds a one, we actually want to enter a new state s3 and we're going to change the one to a zero...