1. More variants of Turing Machines
The Chinese University of Hong Kong
Fall 2009
CSC 3130: Automata theory and formal languages
More variants of Turing Machines
Andrej Bogdanov

2. What does a computer look like?
CPU
instruction memory
0000 PC
arithmetic logical unit
data memory
1010
0100
0000
registers R0 R1 R2 R3
peripheral
devices

3. What does a computer look like?
CPU
instruction memory PC
0000 load 0001
0001 write R3
0010 store R5
0011 add R5
0100 jpos 0011
...
0000
arithmetic logical unit R0
0100 R1
1010
data memory
registers R2
0000
0000
0001
0010
0011
0100 R3
0000
0110
0110
0110
0110
0110

4. Instruction set load x Put the value x into R0 load Rk
Copy the value of Rk into R0
store Rk
Copy the value of R0 into Rk
read Rk
Copy the value at memory location Rk into R0
write Rk
Copy the value of R0 into memory location Rk
add Rk
Add R0 and Rk, and put result in R0
jump n
Set PC to n
jzero n
Set PC to n, if R0 is zero
jpos n
Set PC to n, if R0 is positive

5. Random access machines
instruction meaning
program counter PC 1 2 3 4 5
load -7 R0 := -7
write R2 M[R2] := R0
store R1 R1 := R0
add R1 R0 := R0 + R5
jzero 3 if R0 = 0 then PC := 3
accept
registers R0 R1 R2 … M 2 1 2 2
memory 1 2 3 4
It has registers that can store arbitrary values, a program counter, and a random-access memory

6. Random access machines
instruction meaning PC 1 2 3 1 2 3 4 5
load -7 R0 := -7
write R2 M[R2] := R0
save R1 R1 := R0
add R1 R0 := R0 + R1
jzero 3 if R0 = 0 then PC := 3
accept 4 5 R0 -7
-14 R1 -7 R2 … M -7 1 2 3 4
The instructions are indexed by the program counter

7. Random access machines
Random access machines are equivalent to Turing Machines
Simulating a Turing Machine on a RAM: M … 1 2 1 PC
head
tape R0 2 … M 1 2 1

8. Simulating a TM on a RAM … program M … … … save R1 handle for state q01 2 3 6 7 8 9 10
save R1 save head position read R1 read tape contents x
add -1
jzero 6 if x = 1 goto line 6
load 2 new value of cell
write R1 write in memory
load R1 recall head position
add 1 move head to right
jump 30 go to state q1 M … q1
1/2R … q0
qacc … … 1 2 1 30
save R1 handle for state q1 …
200
accept handle for state qacc

9. Simulating a RAM on a Turing Machine
The configuration of a RAM consists of
Program counter
Contents of registers
Indices and contents of all nonempty memory cells PC 14
configuration = (14, 3, -7, 5, (0, 2), (2, 1), (3, 2)) R0 3 R1 -7 R2 5 … M 2 1 2 1 2 3 4

10. Simulating a RAM on a Turing Machine
The TM has a simulation tape and a scratch tape
The simulation tape stores RAM configuration
The TM has a set of states corresponding to each program instruction of the RAM
The TM tape is updates according to RAM instruction M
(14,3,-7,5,(0,2),(2,1),(3,2))

11. Simulating a RAM on a Turing Machine
Example: load R1 c
(14,3,-7,5,(0,2),(2,1),(3,2))
1. Copy R1 to scratch tape -7 s
(14,-,-7,5,(0,2),(2,1),(3,2)) -7 c s
2. Write R1 to conf tape
(14,-,-7,5,(0,2),(2,1),(3,2)) c
(14,-,-7,5,(0,2),(2,1),(3,2) ) .
(14,- ,-7,5,(0,2),(2,1),(3,2))
Make more space as needed
(14,-7,-7,5,(0,2),(2,1),(3,2)) c
3. Erase scratch tape

12. Nondeterministic Turing Machine… 1
The transition function is nondeterministic:
The language recognized by N are those strings that can lead N to end in qacc
d: (Q – {qacc, qrej})   → subsets of (Q  G  {L, R})

13. Equivalence of NTM and TM
Nondeterministic Turing Machines recognize the same languages as Turing Machines
Let us look more deeply into NTMs
Nondeterministic choices can be numbered
1/2R 1 q5 q3
The first m steps of the computation
are then fully specified by a sequence of m numbers q6
1/1L 2
1/1R 3

14. Simulating a nondeterministic TM
input tape N M … 1
simulation tape … 1 … 1
address tape … 1 3 2 2
k =
maximum number of nondeterministic choices in any state
G’ = G ∪ {1,..., k}
Address tape specifies nondeterministic choices of N
First, input is copied from input tape to simulation tape
Then, M simulates N using address tape data

15. How to use the address tape
input tape N M … 1
simulation tape … 1 … 1
address tape … 1 3 2 2
Suppose N accepts x when nondeterminism =
Then we want to make sure the address tape contains the string at some point in its execution
To ensure this we try all possibilities for the address tape

16. Simulating a nondeterministic TM
input tape x
Description of M: … 1
Initially:
x = input of N a is empty
For all possible strings a:
Copy x to z.
Simulate N on input z using a as the nondeterminism
If a specifies an invalid choice or simulation runs out of choices, abort simulation.
If N enters its accept state, accept and halt.
If N enters its reject state, abort simulation.
simulation tape z … 1
address tape a … 1 3 2 2

17. Simulating a nondeterministic TM
Try all possible choices of the address tape:
address tape
abort simulation runs out of choices at q0 …
abort simulation runs out of choices at q1 … 1
abort impossible choice at q0 … 2 M
1/1R q1
abort out of choices at q0 … 1 1 1 1 q0 2
0/1R
0/0R
qacc
accept … 1 2
input: 10

18. Simulating a nondeterministic TM
Try all possible choices of the address tape:
address tape
abort simulation runs out of choices at q0 …
abort simulation runs out of choices at q1 … 1
abort impossible choice at q0
0/0L … M 2 2
1/1R q1
abort out of choices at q0 … 1 1 1 1 q0 1
2/0R
0/0L
abort out of choices at q1 …
qacc 1 2
input: 10
Simulation will loop forever

19. Correctness of simulation
If N accepts x:
Eventually, M will encounter the correct a
Description of M:
For all possible strings a:
Copy x to z.
Simulate N on input z using a as the nondeterminism
If a specifies an invalid choice or simulation runs out of choices, abort simulation.
If N enters its accept state, accept and halt.
If N enters its reject state, abort simulation.
So M will accept
Provided that all previous simulations halt!

20. Correctness of simulation
Claim: Simulation step always halts (never loops)
Description of M:
For all possible strings a:
Copy x to z.
Simulate N on input z using a as the nondeterminism
If a specifies an invalid choice or simulation runs out of choices, abort simulation.
If N enters its accept state, accept and halt.
If N enters its reject state, abort simulation.
Since a is finite, the simulation will eventually run out of choices

21. Correctness of simulation
If N does not accept x, then for every a:
Description of M:
Either N will loop, so simulation runs out of choices
For all possible strings a:
Copy x to z.
Simulate N on input z using a as the nondeterminism
If a specifies an invalid choice or simulation runs out of choices, abandon simulation.
If N enters its accept state, accept and halt.
If N enters its reject state, abort simulation.
Or N rejects, so simulation is abandoned anyway
In either case, simulation fails, so M loops forever!

22. Context-free languages are recognizable
Since Nondeterministic Turing Machines are more general than Pushdown Automata, it follows that
In fact
Context-free languages are recognizable (by TMs)
Context-free languages are decidable (by TMs)

23. Recap of computational models
NTM
NFA
DFA
PDA TM
toy computers
RAM