TM_Universal

Transcript

1. Turing Machines and Regular Languages We’ll do this on a 4 tape TM. Our TM’s first task is to move things onto the right tapes: Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 0 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t 5 / 40
2. Turing Machines and Regular Languages Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 0 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t We search tape 0 for the current state (as recorded on tape 2) Here, we already found it. 6 / 40
3. Turing Machines and Regular Languages Tapes: Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 0 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t Then we search tape 0 for the input symbol (as found on tape 3) We use the TM’s state to keep track of whether we are after a , or an =. 7 / 40
4. Turing Machines and Regular Languages Tapes: Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 0 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t Then we move across the = on tape 0, find the state number and update Tape 2 (no change on tape 2: already in state 0) 8 / 40
5. Turing Machines and Regular Languages Tapes: Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 0 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t Then we scan back to the start of Tape 0, and advance Tape 3 by one position Tape 3’s movement is the DFA going to the next character of input. 9 / 40
6. Turing Machines and Regular Languages Tapes: Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 0 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t Already at state 0’s description, so find 1 10 / 40
7. Turing Machines and Regular Languages Tapes: Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 1 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t Go after the =, and update state number (tape 2) 11 / 40
8. Turing Machines and Regular Languages Tapes: Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 1 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t Scan to left of Tape 0, advance on Tape 3. 12 / 40
9. Turing Machines and Regular Languages Tapes: Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 1 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t Now, we search for state 1’s description on Tape 0 Use TM state to remember if we just crossed a ; 13 / 40
10. Turing Machines and Regular Languages Tapes: Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 1 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t Find input symbol (1 from Tape 3) 14 / 40
11. Turing Machines and Regular Languages Tapes: Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 0 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t Find destination state (0) and update tape 2. 15 / 40
12. Turing Machines and Regular Languages Tapes: Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 0 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t Scan to left on Tape 0 and move right one on Tape 3. We’ll skip to the end now... 16 / 40
13. Turing Machines and Regular Languages Tapes: Tape 0: Transition function Tape 1: Accepting states Tape 2: Current state (starts at 0) Tape 3: Input to DFA t 0 : 1 = 1 , 0 = 0 ; 1 : 1 = 0 , 0 = 1 t t 1 t t t t t t t t t t t t t t t t t t t t 1 t t t t t t t t t t t t t t t t t t t t 0 1 1 0 1 t t t t t t t t t t t t t t t When Tape 3 sees t, we have reached the end of the DFAs input We need to see if the current state (recorded on Tape 2) is accepting (listed on tape 1) If we find it, we accept, if not, we reject (here we accept) 17 / 40