1. Decidable Languages
A language L is decidable if there is a Turing machine ML such that given any word w  0*, then:
Input of ML: a  b  … w
Output of ML: 1  …
If w  L
If w  L 

2. Complement of Decidable Languades
Suppose that L is decidable, is LC = {w : w  L} decidable?
YES
Proof: Let ML be the Turing machine that decides L: …

3. Semidecidable Languages
Definition. A language L is called semidecidable if there exists a Turing machine, ML, such that:
If w  L then ML(w) halts
If w  L then ML(w)
Why do you think they are called “semidecidable”?
Because if we run ML on a word w and it hasn’t stop, there are two possibilities:
Maybe it just needs to run more time before it finally ends, or
Maybe it will never stop (w  L)

4. Decidable vs Semidecidable
If L Decidable then L is semidecidable:
Use the Turing machine ML that decides L a follows: …
The other way around is not true.
If L is not even semidecidable then L cannot be decidable

5. There are Non-Decidable and Even Non-Semidecidable Languages
That is, there are languages for which no algorithm exists that decides if a given word is in the language or not
Example: Consider the following set:
Halting = {M(w) : M is a Turing machine that halts on w}
Halting is not decidable
HaltingC is not even semidecidable

6. Halting is Not Decidable but it is Semidecidable
Proof that Halting is semidecidable: Trivial.
What this means is that you cannot write a program (in any language, for example in C) that receives as input another program P (for example in C) and an input w for P and correctly decides if P(w) halts or not

7. Multi-tape Turing Machines: Informal Description
Control … a1 a2 
Tape1
head1 … a1 a2 
Tape2
head2
We add a finite number of tapes …

8. Multi-tape Turing Machines: Informal Description (II)
Each tape is bounded to the left by a cell containing the symbol 
Each tape has a unique header
Transitions have the form (for a 2-tape Turing machine):
( (p,(x1, x2), (q, (y1, y2) )
Such that each xi is in  and each yi is in  or is  or 

9. Multi-tape Turing Machines: Formal Definition
Definition. A 2-tape Turing machine is a 5-tuple (S, , , s,H), where:
S is a set of states
 is an alphabet. It must contain  and . It cannot contain  or 
s  S is the initial state
H  S is the set of halting states

10. Multi-tape Turing Machines: Formal Definition (II)
 is a collection of transitions defined by the function:
: (S-H)     
S  (  (, })  (  (, })
such that:
1. For any q  S – H, p  S with
(q,(x1, x2)) = (p, (y1, y2)) ), if xi is  then
2. There cannot be any p, q  with
(q,(x1, x2)) = (p, (y1, y2)) ), with xi ≠  such that yi = 
yi must be either  or 

11. Multi-tape Turing Machines: Example
Lets construct a 3-tape Turing machine that receives as input 2 binary numbers of the same lenght in the first two tapes and outputs the OR of the two numbers on the third tape
Example:
Tape 1: 010111
Tape 2: 100011
Tape 3: 110111

12. Multi-tape Turing Machines
Construct a 2-tape Turing machine that recognizes the language:
L = {anbn : n = 0, 1, 2, …}
Tape1: w
Input:
Tape2: 
Tape1: 1… if w  L or
Tape1: 0… if w  L
Output:
Hints:
use the second tape as an stack
Use the machines M1 and M0

13. Multi-tape Turing Machines vs Turing Machines a1 a2 … ai …  b1 b2 … … bj
We can simulate a 2-tape Turing machine M2 in a Turing machine M:
we can represent the contents of the 2 tapes in the single tape by using special symbols
We can simulate one transition from the M2 by constructing multiple transitions on M
We introduce several (finite) new states into M

14. Using States to “Remember” Information
Configuration in a 2-tape Turing Machine M2:
Tape1  a b a b
State: s
Tape2  b b a
State in the Turing machine M:
“s+b+1+a+2”
Which represents:
M2 is in state s
Cell pointed by first header in M2 contains b
Cell pointed by second header in M2 contains an a

15. Using States to “Remember” Information (2)
State in the Turing machine M:
“s+b+1+a+2”
(# states in M2) * | or  or | * | or  or |
How many states are there in M?
Yes, we need large number of states for M but it is finite!

16. Configuration in a 2-tape Turing Machine M2: a b a b
State in M2: s
Tape2  b b a
Equivalent configuration in a Turing Machine M: a b 1  2  4 3 e 
State in M: s+b+1+a+2

17. Simulating M2 with M
The alphabet  of the Turing machine M extends the alphabet 2 from the M2 by adding the separator symbols: 1, 2, 3 , 4 and e, and adding the mark symbols:  and 
We introduce more states for M, one for each 5-tuple p++1+ +2 where p in an state in M2 and +1+ +2 indicates that the head of the first tape points to  and the second one to 
We also need states of the form p++1++2 for control purposes

18. Simulating transitions in M2 with M
State in M: s+b+1+a+2 a b 1  2  4 3 e 
At the beginning of each iteration of M2, the head starts at e and both M and M2 are in an state s
We traverse the whole tape do determine the state p++1+ +2, Thus, the transition in M2 that is applicable must have the form:
( (p,(, )), (q,(,)) ) in M2
p+ +1+ +2
q+ +1++2
in M

19. Simulating transitions in M2 with M (2)
To apply the transformation (q,(,)), we go forwards from the first cell.
If the  (or ) is  (or ) we move the marker to the right (left): 1  …  i 1  … i 
If the  (or ) is a character, we first determine the correct position and then overwrite

20. state: s  1 a b a b 2      3 b b a 4     e a b   2  4 3 e 1 a b  1 2   4 3 e  b  a 1 2  4 3 e    a b 1 2 4 3 e a b   2  4 3 e
Output: 1
state: s+b+1

21. Multi-tape Turing Machines vs Turing Machines (6)
We conclude that 2-tape Turing machines can be simulated by Turing machines. Thus, they don’t add computational power!
Using a similar construction we can show that 3-tape Turing machines can be simulated by 2-tape Turing machines (and thus, by Turing machines).
Thus, k-tape Turing machines can be simulated by Turing machines

22. Implications
If we show that a function can be computed by a k-tape Turing machine, then the function is Turing-computable
In particular, if a language can be decided by a k-tape Turing machine, then the language is decidable
Example: Since we constructed a 2-tape TM that decides
L = {anbn : n = 0, 1, 2, …}, then L is Turing-computable.