(00:01) Hello, everyone (00:02) Welcome to the next lecture (00:04) In the course and Theory of Atom Meta (00:08) In the previous lecture we studied the formal definition of a Turing machine, (00:14) and in this lecture we will see two examples of Turing machine, (00:18) and we will see how we design a Turing machine for a given language (00:22) All right So let's get started (00:26) Here we have an example for designing (00:28) of a Turing machine which recognizes the language L equal to zero, one star zero (00:35) So what this language mean? (00:37) This language mean? (00:39) It should have one0 followed by any number of ones followed by one0 at the end (00:46) So if you look, it is a regular language, (00:49) and we can easily design a finite state machine for this language (00:54) But we design theorem machine for this (00:56) language so that we can understand the machine from the very simplest example (01:02) So here I have designed the Turing machine for the language (01:06) So here we have the starting state, which is a (01:10) Then we have two States B and C, (01:13) and then we have two final States, which are the reject and the except state (01:18) Let us see how it works (01:21) So as we want to accept zero, one star zero (01:25) So at the initial state, (01:27) when we get a zero, we go to the next state B, (01:31) and in our tape we write symbol X, and we move to the right on our tape (01:38) So we are now instead B (01:41) So now we know that we have to accept any (01:45) number of ones after we have accepted the zero at the start (01:49) So instead B, if you get any number of ones, you always stay instead B (01:55) And then on the tape we write Y and we move right on the tape (02:01) And after getting any number of ones, finally, we need to get a zero (02:06) So if we get a zero, we move to the next state C, and we write X on tap (02:13) And then we move right on the tap (02:18) And after reaching this state C, we see that there is nothing more to take (02:23) So what we do, we read the blank symbol that is the spacious symbol (02:28) So when we read the blank symbol, we not write other symbol other than (02:33) the blank symbol, and we move to the accept state (02:38) So this is how it works (02:41) Now let us take an example string (02:43) from this language, which is 0110 and let us see how it works using the tape (02:51) So we come to the tape and our head is at the leftmost cell of our tape (03:02) So when we get a zero, (03:04) we write X on our tap, and we go to step B, and we move to the right under tab (03:13) Now our next input is one (03:18) So if we get a one, (03:19) we are instead B and we write a y onto our tape, and we remain in step (03:41) B But we move to the right on our tape, (03:45) and then the next input we have is a zero (03:48) So when we get zero at state B, we move to the state C and we write X onto (03:54) our tape, and we move one step to the right on the tab (04:00) Now we see that we have reached to the end of the string (04:04) So after the zero at the end, we consider it to be the blank symbol (04:11) So we are at state C, and we encounter a blank symbol (04:15) So we write blank symbol on the tab, (04:18) and we move one step to the right and we go to the accept state (04:25) So the string 0110 is accepted (04:29) And now this is the condition of our tape (04:34) Now you may ask why we write these X and y onto the tape (04:41) So as this was a very simple example language, and we actually not need (04:46) the full power of Turing machine in order to implement this (04:50) But we have to this because we are (04:53) using the Turing machine, and particularly for this language, (04:57) these tape symbols have no such specific meaning, but we just it in order (05:02) to show how Turing machine works and how Turing machine tape actually works (05:08) Also, in the first lecture on Turing (05:10) machine, I told you that Turing machines are deterministic (05:14) So here in this transition diagram, you see, we have transition for every input (05:21) Now let us discuss another example (05:25) in which we design a Turing machine for the language zero power n one power N (05:31) So it means we have to design a Turing machine for a language where number (05:35) of zeros should exactly be equal to the number of ones that follow these zeros (05:43) In the previous lecture, when we were discussing finite state (05:47) machines, we saw that we could not implement this language using finite state (05:54) machine because we need to keep count of how many zeros are there in order (06:00) to repeat the exact same number of ones so that we can implement this language (06:05) But we saw that finite state machine due to their limited memory and due to their (06:10) limited capability, it was not possible to design a finite (06:14) state machine, and therefore we declared that this was not a regular language (06:21) So we shall see using a Turing machine, how can we design this language? (06:25) So first of all, I will explain to you the algorithm of how this can be done (06:33) We first will design an algorithm (06:35) which will accomplish this task of accepting this kind of strings (06:40) And then from that algorithm, (06:42) we will see how we design the equivalent Turing machine for it (06:47) All right So here you see, we have the tape sequence (06:51) in our Turing machine, and then we have the algorithm (06:56) So the first step in algorithm says change zero to X (07:03) So we replace the first zero in our tape (07:05) to X, and then we move right until we find the first one (07:10) So if you cannot find the one, (07:12) then you have to reject means that the string will not be accepted (07:18) And if you are able to find a one, then you have to change it to Y (07:24) And after you that, you have to move left until you find (07:28) the leftmost zero, and this process will be repeated until no more zeros are there (07:34) And also you have to make sure that no more ones remain as well (07:40) Now let us apply this algorithm on our tap in the Turing machine (07:45) You notice in the tape that we have four number of zeros and four number of ones (07:50) So the number of zeros are same to the number of ones that follow (07:57) So (07:59) this string should be accepted (08:03) So first of all, our tape head is about the leftmost zero in our tab (08:12) So as in our algorithm, (08:14) the first step we perform is to change this X this zero to X (08:20) Then we move right on the tape (08:24) until we find the first one (08:27) So here I have found the first one (08:31) So according to the algorithm, we change it to a Y, (08:36) and then what we have to do, we have to move left to the leftmost zero (08:44) So I move step by step to the leftmost zero, and I find that my left most zero is (08:51) here, so we change it to X according to the algorithm (08:57) So after replacing the lift most zero (08:59) with X, I again move to the right until I find my first one (09:06) I change it to Y (09:09) Then I start moving left until I find the leftmost zero, and I change it to an X (09:17) Then again, I repeat these steps (09:22) and I start moving to the right and I find my first one (09:29) So I changed it to Y, (09:31) then start moving left, and finally I see my left most and also the final zero (09:38) I change it to X, and then I move right and I find the first and the last one (09:48) So I change it to Y (09:53) Now, if we look, there are no more zeros and also there are no more ones (09:59) So our tape  ... it in order (0 5:02) to show how Turing machine works and how Turing machine tape actually works (0 5:08) Also, in the first lecture on Turing (0 5:10) machine, I told you that Turing machines are... algorithm (0 6:35) which will accomplish this task of accepting this kind of strings (0 6:40) And then from that algorithm, (0 6:42) we will see how we design the equivalent Turing machine for it (0 6:47)... sequence (0 6:51) in our Turing machine, and then we have the algorithm (0 6:56) So the first step in algorithm says change zero to X (0 7:03) So we replace the first zero in our tape (0 7:05) to