(00:00) Exactly the same tasks that we have written in our algorithm (00:04) So let's see here we have the starting state, which is A (00:08) Then we have stats B, C, and D, and then we have the accept state (00:14) here we we have not mentioned the reject state, but we presume that any missing (00:19) input in each state leads us to the reject state (00:23) And also, as you know that your machine is deterministic (00:27) So there should be transition for each input symbol (00:31) But we have not mentioned the missing transitions, and we have presumed these (00:36) missing transitions in each state will lead us to the reject state (00:42) So now let us see how it works (00:45) So we are in the start state (00:49) and we get the input zero (00:51) So we replace zero with an X (00:56) Now we are in state B and we move to the right until we find the first one (01:05) So instead B the state B help us to find the first one (01:14) How (01:18) instead B, if we find a zero, so zero will remain zero and we have to move right (01:26) And also if we encounter a Y, so Y will remain Y and we have to move right (01:33) But if we find a one, (01:36) then we replace that one with Y and we move to the next state, which is C (01:42) So in transition diagram, you see that the transition one tends to Y L (01:49) So this transition mean that I replace one with a Y (01:53) and I move to the left (01:56) So now (01:58) we are instead C state C help us (02:01) in finding the leftmost zero, because when we encounter zero or Y (02:06) instead C, we change nothing until we have find an X (02:12) So when we encounter X, we come to know that following this X is (02:18) my leftmost zero, so X remains X, but we move one step to the right (02:24) Now my tab head is over the leftmost zero, and I am at state A again, (02:30) so I change zero with an X, and this whole process will continue again (02:37) So on encountering Y, (02:39) I have to change it to encountering one, I have to change it to Y (02:46) and moving left to find the leftmost zero and change it to X (02:54) So we move left, replacing each zero with an X, (02:58) and then move right and replacing each one with a Y, (03:05) and we repeat these steps again and again (03:10) until we (03:13) replace all zeros with X and all ones with Y's (03:20) So unchanging the last one with Y (03:25) I am instead C and the tape head will move to the left (03:30) So if I encounter Y, (03:32) I remain in state C and I don't change nothing but keep moving left (03:38) And when I encounter the first X, (03:41) I go to state A and here as I don't have no more zeros (03:46) Also, the tab head moves to the right where I encounter with a Y (03:53) So now I make transition from state A (03:55) to state D and my tab head moves to the right (04:00) The loop over state D indicates that unencountering each y (04:05) I don't replace it, but I keep moving to the right until I find a blank symbol (04:12) and unencountering a blank symbol (04:16) I go to the accept state that is, my string is accepted and my tape head (04:22) moves to left so this string 0011 will be accepted (04:30) So this was how the transition diagram and tape and Tutoring machine works (04:35) So hopefully (04:36) So you now understand how Turing machine is designed and how the tape works (04:43) So thank you for watching and see you in the next lecture (04:46) Bye