1) The document discusses Turing machines and their properties such as having a finite set of states and read/write tape memory. The output depends only on the input and previous output based on definite transition rules. 2) Reducibility is introduced as a primary method for proving problems are computationally unsolvable by converting one problem into another problem such that solving the second solves the first. 3) Decidability and undecidability of languages are defined. Undecidable problems have no algorithm to determine membership regardless of whether a Turing machine halts or not on all inputs.

1. PREPARED FOR:
Dr. Gede Pramudya Ananta
PREPARED BY:
THAMER J.ABBAS
SAIF MOHAMMED MAKKI
SAIF ZUHAIR ABDULMAJEED
M031020009
M031020010
M031110012

2. Turing machines
a device with a finite amount of read-only “hard” memory
(states), and an unbounded amount of read/write tape-memory
 The output depends only on input
and the previous output
Black box reads a
sequence of 0’s and
1’s
 The main thing is ????
That the changes from one output state to
the next
 Given by definite rules, called the
TRANSITION rules

3. Reducibility
Definition
 Primary method for Proving
computationally unsolvable.
that problems are
Reducibility also occurs in mathematical problems .
A Reduction : is a way of converting one problem
to another problem in such a way that a solution to the
second problem can be used to solve the first problem.
Such reducibilities come up often in everyday life, even
if we don't usually refer to them in this.

4. EXAMPLE: use problem B to
solve problem A
A
CAN'T TAKE IT DIRECT.. XX
B

5. DECIDABLE
Recall that:
A language
machine
A
and
M
A
is decidable, if there is a Turing
(decider) that accepts the language
halts on every input string.
Turing Machine
Input
string
Decider for
M
A
Decision
On Halt:
YES
Accept
NO
Reject

6. UNDECIDABLE
• Undecidable
problems
have
no
algorithm, regardless of whether or not
they are accepted by a TM that fails to
halt on some inputs
• Undecidability: undecidable languages
that cannot be decided by any Turing
Machine

7. Example for: Decidable And Undecidable
x=2
y=1
z =5
Assume, we have a program which assigns all possible
combination of 3 integers to variables x, y and z. For the
first case there is at least one solution (x = 2, y = 1, z
=5). Thus, the program will eventually stops. But for the
second case we don’t know if this system has a solution.
If there is no solution for the second system, then the
program never stops.

8. Examples: Halting Problem
halts(“2+2”) True
halts(“def f(n):
if n==0: return 1
else: return n * f(n-1)
f(5)”) True
halts(“def f(n):
if n==0: return 1
else: return n * f(n-1)
f(5.5)”) false

9. 2+
X
10
2+
Y
2
Z
11
12
365
6
=
13
=
8
7
149
2+
A
14
365
10
9
≠
2
B
182

10. Post’s Correspondence Problems (PCP)
Definition
An instance of PCP consists of two lists of strings
over some alphabet S.
The two lists are of equal length, denoted as A and
B.
The instance is denoted as (A, B).
We write them as
A = w1, w2, …, wk
B = x1, x2, …, xk
for some integer k.
For each i, the pair (wi, xi) is said a corresponding
pair.

11. PCP Instances
An instance of PCP is a list of pairs of
nonempty strings over some alphabet Σ
Say (w1, x1), (w2, x2), …, (wn, xn).
The answer to this instance of PCP is
“yes” if and only if there exists a
nonempty sequence of indices
i1,…,ik, such that wi1…win = xi1…xin.

12. Post’s Correspondence Problem
 (PCP) is an example of a problem
that does not mention TM’s in its
statement, yet is undecidable.
 From
PCP, we can prove many
other non-TM problems undecidable.

13. Example: PCP
• Let the alphabet be {0, 1}.
• Let the PCP instance consist of the two
pairs (0, 01) and (100, 001).
• We claim there is no solution.
• You can’t start with (100, 001), because the
first characters don’t match.

14. Example: PCP
Recall: pairs are (0, 01) and (100, 001)
0 100 100
01 001 001
Must start
with first
pair
Can add the
second pair
for a match
But we can never make
the first string as long
as the second.
As many
times as
we like

15. Example: PCP – (3)
 Suppose
we add a third
pair, so the instance becomes:
1 = (0, 01); 2 = (100, 001); 3
= (110, 10).
Now 1,3 is a solution; both
strings are 0110.
In fact, any sequence of
indexes in 12*3 is a solution.

16. A Simple Undecidable Problem
We say this instance of PCP has a
solution, if there is a sequence of
integers, i1, i2, …, im, that, when
interpreted as indexes for strings in the A
and B lists, yields the same string, that
is, wi1wi2…wim = xi1xi2…xim.
We say the sequence is a solution to this
instance of PCP.

17. A Simple Undecidable Problem
The Post’s corresponding problem is:
given an instance of PCP, tell whether this
instance has a solution.
The solution to an instance of PCP
sometimes is not unique.
Also, an instance of PCP might have no
solution.

18. Post’s (Domino) Correspondence
Problem
PCP as a game
Usually dominoes is played as follows:
18

19. Post’s (Domino) Correspondence
Problem
Usually dominoes is played as follows:
•
•
•
19

20. Post’s (Domino) Correspondence
Problem
Usually dominoes is played as follows:
•
•
•
•
•
•
•
•
•
20

21. Post’s (Domino) Correspondence
Problem
Usually dominoes is played as follows:
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
21

22. Post’s (Domino) Correspondence
Problem
Usually dominoes is played as follows:
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
22

23. Post’s (Domino) Correspondence
Problem
Usually dominoes is played as follows:
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
23

24. Post’s (Domino) Correspondence
Problem
We’ll play horizontally instead of vertically.
Furthermore, dominoes will not be allowed to
be flipped so each half will be a different color:
•
•
•
24

25. Post’s (Domino) Correspondence
Problem
Aim of the game is to have same total
number of dots on the top as on the
bottom. Player is given a set of domino
prototypes to choose from, and can
choose as many of a given prototype as
necessary.
Let’s play with the following 2 prototypes:
•
•
•
• •
• •
• •
•
25

26. Post’s (Domino) Correspondence
Problem
Let’s play with the following 2 prototypes:
•
•
•
• •
• •
• •
•
Total
Total
26

27. Post’s (Domino) Correspondence
Problem
Let’s play with the following 2 prototypes:
•
•
•
•
•
•
• •
• •
• •
•
Total
1
Total
2
27

28. Post’s (Domino) Correspondence
Problem
Let’s play with the following 2 prototypes:
•
•
•
•
•
•
•
•
•
• •
• •
• •
•
Total
2
Total
4
28

29. Post’s (Domino) Correspondence
Problem
Let’s play with the following 2 prototypes:
•
•
•
•
•
•
•
•
•
• •
• •
• •
•
• •
• •
• •
•
Total
8
Total
5
29

30. Post’s (Domino) Correspondence
Problem
Let’s play with the following 2 prototypes:
•
•
•
•
•
•
•
•
•
• •
• •
• •
•
•
•
•
• •
• •
• •
•
Total
9
Total
7
30

31. Post’s (Domino) Correspondence
Problem
Let’s play with the following 2 prototypes:
• •
• •
• •
•
•
•
•
•
•
•
•
•
•
• •
• •
• •
•
•
•
•
•
•
•
Total
10
Total
9
31

32. Post’s (Domino) Correspondence
Problem
Let’s play with the following 2 prototypes:
• •
• •
• •
•
•
•
•
•
•
•
•
•
•
• •
• •
• •
•
•
•
•
•
•
WINNER!
•
•
•
•
Total
11
Total
11
32

33. Post’s (Domino) Correspondence
Problem
Could have represented dominos using unary
strings:
•
•
•
• •
• •
• •
1
111111
•
11
1
Point of game is to get the same string to be
written on top as bottom.
33

34. Post’s (Domino) Correspondence
Problem
In general, could use arbitrary strings. EG:
a
c
ba
acb
ac
ba
a
b
Aim: Get the same string on top as bottom.
PCP: Given an alphabet S and finite set of string
pairs (u1,v1), (u2,v2), … , (un,vn) with ui ,vi S*, can
a non-empty sequence of indices i1, i2, i3, … , it be
chosen so that
ui1ui2ui3…uit =
vi1vi2vi3…vit
?
34

35. Post’s (Domino) Correspondence
Problem
Let’s play with the following 4 prototypes:
1:
a
ac
, 2:
c
ba
, 3:
ba
a
, 4:
acb
b
Total
Total
Indices
35

36. Post’s (Domino) Correspondence
Problem
Let’s play with the following 4 prototypes:
1:
a
ac
, 2:
c
ba
, 3:
ba
a
, 4:
acb
b
Total
a
a
ac
Total
ac
Indices
1
36

37. Post’s (Domino) Correspondence
Problem
Let’s play with the following 4 prototypes:
1:
a
ac
, 2:
c
, 3:
ba
ba
a
, 4:
acb
b
Total
ac
a
ac
c
ba
Total
acba
Indices
12
37

38. Post’s (Domino) Correspondence
Problem
Let’s play with the following 4 prototypes:
1:
a
ac
, 2:
c
, 3:
ba
a
ac
ba
, 4:
a
c
ba
acb
b
ba
a
Total
acba
Total
acbaa
Indices
123
38

39. Post’s (Domino) Correspondence
Problem
Let’s play with the following 4 prototypes:
1:
a
ac
, 2:
c
, 3:
ba
a
ac
ba
, 4:
a
c
ba
acb
b
ba
a
a
ac
Total
acbaa
Total
acbaaac
Indices
1231
39

40. Post’s (Domino) Correspondence
Problem
Let’s play with the following 4 prototypes:
1:
a
ac
, 2:
c
, 3:
ba
a
ac
ba
, 4:
a
c
ba
acb
b
ba
a
a
ac
Answer: YES! (solution is 12314)
acb
b
Total
acbaaacb
Total
acbaaacb
Indices
12314
40

41. PCP is undecidable
We can try all lists i1, i2, … , ik in
order of k. if we find a solution,
the answer is “yes”
But if we never find a solution,
how can we be sure no longer
solution exists?
So, we can never say no

42. A Simple Undecidable Problem
• Example 1
Two lists of an instance of PCP are shown
• A solution is 2, 1, 1, 3 (strings may be
repeated)
because
List A
List B
w2, w1, w1, w3
= 101111 110
i
wi
xi
= 10111 1110
1
1
111
= x2, x1, x1, x3
• Another solution is
2
10111
10
2, 1, 1, 3, 2, 1, 1, 3
3
10
0

43. There is much
more I haven’t
told you about It