1. Turing Machines The Chinese University of Hong Kong Fall 2010
CSCI 3130: Automata theory and formal languages
Turing Machines
Andrej Bogdanov

2. What is a computer? computer
program
output
input
A computer is a machine that manipulates data according to a list of instructions.

3. ? What is a computer? deep blue Hello, world world E2-E4 computer
void hello( String name) {
print( “Hello, “, name); } ?
Hello, world
world
deep blue
E2-E4
computer

4. Turing Machines Can both read from and write to the tape
control
head … a b ☐
input
blanks
Can both read from and write to the tape
Head can move both left and right
Tape is infinite
Has two special states accept and reject

5. Example Strategy: L1 = {w#w: w ∈{a, b}*} abbaa#abbaa
Read and remember the first symbol
xbbaa#abbaa
Cross it off (x)
xbbaa#abbaa
Read the first symbol past #
If they don’t match, reject
xbbaa#xbbaa
If they do, cross it off

6. Example Strategy: L1 = {w#w: w ∈{a, b}*}
Look for the first uncrossed symbol
xbbaa#xbbaa
Cross it off (x)
xxbaa#xbbaa
Read the first uncrossed symbol past #
xxbaa#xbbaa
If they match, cross it off, else reject
xxbaa#xxbaa
At the end, there should be only xs and #s
xxxxx#xxxxx
If not, reject; otherwise, accept.

7. How Turing Machines operate
current state
a/bL q1 q2 … a b a ☐
Replace a with b, and move head left
new state
a/bL q1 q2 … a b b ☐

8. Formal Definition A Turing Machine is (Q, , , , q0, qacc, qrej):
Q is a finite set of states;
 is the input alphabet not containing the blank symbol ☐
 is the tape alphabet (  ) including ☐
q0 in Q is the start state;
qacc, qrej in Q are the accepting and rejecting state
 is the transition function
d: (Q – {qacc, qrej})   → Q  G  {L, R}
Turing Machines are deterministic

9. Configurations
A configuration consists of the current state, the head position, and tape contents
configuration … a b ☐ q1
abq1a q1
qacc
a/bR … a b ☐
qacc
abbqacc

10. Configurations
We say configuration C yields C’ if the TM can go from C to C’ in a single step
The start configuration of the TM on input w is q0w
An accepting configuration is one that contains qacc; A rejecting configuration is one that contains qrej
abq1a
yields
abbqacc

11. The language of a Turing Machine
We say M accepts x if there exists a sequence of configurations C0, C1, ..., Ck where
C0 is starting
Ci yields Ci+1
Ck is accepting
The language recognized by M is the set of all strings that M accepts

12. This machine never halts
Looping
Something strange can happen in a Turing Machine:
Inputs can be divided into three types
 = {0, 1}
☐/☐R
qacc
0/0R
input: e
This machine never halts q0
1/1R
qrej
qacc
qrej
accept
reject
loop

13. Halting
We say M halts on x if there exists a sequence of configurations C0, C1, ..., Ck where
Ck is accepting
or rejecting
Ci yields Ci+1
C0 is starting
A TM M is a decider if it halts on every input
Language L is decidable if it is recognized by a TM that halts on every input

14. Programming Turing Machines
Description of Turing Machine:
L1 = {w#w: w ∈{a, b}*}
Until you reach # 1
xbbaa#xbbaa
Read and remember entry 2
xxbaa#xbbaa
Write x 3
xxbaa#xbbaa
Move right past # and past all xs 4
If this entry is different, reject 5
Otherwise
Write x
xxbaa#xxbaa 6
Move left past # and to right of first x
xxbaa#xxbaa 7
If you see only xs followed by ☐, accept 8

15. Programming Turing Machines
L1 = {w#w: w ∈{a, b}*}
a/aR
b/bR
x/xR
#/#R
a/aL
qa1
qa2
b/bL
a/aL 4
a/xL 3
b/bL
x/xR
x/xL 2
a/xR 5 6
#/#R
☐/☐R
#/#L q0 q1
qacc q2 q3
qrej 1 8
b/xR 7 6 5
b/xL 2
everything else 3
qb1 4
qb2
#/#R
a/aR
x/xR
b/bR
x/xR

16. Programming Turing Machines
input: aab#aab
a/aR
b/bR
x/xR
configurations:
#/#R
a/aL
qa1
qa2
q0aab#aab
b/bL
a/aL 4
a/xL 3
b/bL
xqa1ab#aab
x/xR
x/xL 2
a/xR 5 6
xaqa1b#aab
#/#R
☐/☐R
#/#L q0 q1
qacc q2 q3 1
xabqa1#aab 8
b/xR 7 6 5
xab#qa2aab
b/xL 2 3
xabq2#xab
qb1 4
qb2
#/#R
xaq3b#xab
a/aR
x/xR
xq3ab#xab
b/bR
q3xab#xab
xq0ab#xab
x/xR

17. Programming Turing Machines
L2 = {aibjck: i × j = k and i, j, k > 0 }
aabbcccc
High-level description of TM:
For every a:
aabbcccc
Cross off the same number of bs and cs
aabbcccc
Restore the crossed of bs (but not the cs)
aabbcccc
Cross off this a
aabbcccc
If all as and cs are crossed off, accept.
aabbcccc
aabbcccc
S = {a, b, c}
G = {a, b, c, a, b, c, ☐}
aabbcccc

18. Programming Turing Machines
L2 = {aibjck: i × j = k and i, j, k > 0 }
Low-level description of TM:
Scan input from left to right to check it looks like aa*bb*cc*
Move the head to the first symbol of the tape
For every a:
Cross off the same number of bs and cs
Restore the crossed of bs (but not the cs)
Cross off this a
If all as and cs are crossed off, accept.
how do we know?
how to do this?

19. Programming Turing Machines
aabbcccc
Implementation details:
Move the head to the first symbol of the tape
aabbcccc 
Put a special marker on top of first a
Cross off the same number of bs and cs:
aaqbbcccc 
aabqbcccc 
Replace b by b
Move right until you see a c
Move left just past the last b
If any bs are left, repeat
Replace c by c
aabbqcccc 
aabqbcccc 
aabbqcccc 
aabbcqccc 
aabbqcccc 
S = {a, b, c}
G = {a, b, c, a, b, c, a, a, ☐} 

20. Programming Turing Machines
L3 = {#x1#x2...#xl : xi ∈ {0, 1}* and xi ≠ xj for each i ≠ j}
High-level description of TM:
#01#0011#1
On input w,
For every pair of blocks xi and xj in w,
Compare the blocks xi and xj
If they are the same, reject.
Otherwise, accept.

21. Programming Turing Machines
L3 = {#x1#x2...#xl : xi ∈ {0, 1}* and xi ≠ xj for each i ≠ j}
Low-level description:
#01#0011#1 0.
If input is e, or has exactly one #, accept.
Place a mark on the leftmost # (i.e. replace # by #) and move right  1.
#01#0011#1 
Place another mark on next unmarked # (If there are no more #, accept) 2.
#01#0011#1 

22. Programming Turing Machines
L3 = {#x1#x2...#xl : xi ∈ {0, 1}* and xi ≠ xj for each i ≠ j}
current state:
#01#0011#1  
Compare the two strings to the right of marked #. If there are same, reject 3. 
#01#0011#1  4.
Move the right # to the right
If not possible, move the left # to the next # and put the right # on the next If not possible, accept 
#01#0011#1 
Repeat Step 3 5.
#01#0011#1 

23. Programming Turing Machines
L3 = {#x1#x2...#xl : xi ∈ {0, 1}* and xi ≠ xj for each i ≠ j} 3.
Compare the two strings to the right of marked #. If there are same, reject
#01#0011#1  
#01#0011#1  4.
Move the right # to the right 
If not possible, move the left # to the next # and put the right # on the next If not possible, accept 
accept 

24. How to describe Turing Machines?
Unlike for DFAs, NFAs, PDAs, we rarely give complete state diagrams of Turing Machines
Usually we give a high-level description: A recipe about the workings of the Turing Machine
Just like in cooking, practice makes perfect!

25. Programming Turing Machines
L4 = {〈G〉: G is a connected undirected graph}
Q: How do we feed a graph into a Turing Machine?
A: We represent it by a string, e.g.
(1,2,3,4)((1,4),(2,3),(3,4)(4,2)) 1 2
Convention for describing graphs:
(nodes)(edges) 3 4
no node must repeat
edges are pairs (node1,node2)

26. Programming Turing Machines
L4 = {〈G〉: G is a connected undirected graph}
On input 〈G〉, 0.
Verify that 〈G〉 is the description of a graph
(no vertex repeats; edges only go between nodes) 1.
Mark the first node of G x x 1 2 2.
Repeat until no new nodes are marked:
For each node, mark it if it is attached to an already marked node x x 3 4 3.
If all nodes are marked accept, otherwise reject.

27. Programming Turing Machines
L4 = {〈G〉: G is a connected undirected graph}
(1,2,3,4)((1,4)(2,3)(3,4)(4,2)) 
(1,2,3,4)((1,4)(2,3)(3,4)(4,2)) 
(1,2,3,4)((1,4)(2,3)(3,4)(4,2)) 
(1,2,3,4)((1,4)(2,3)(3,4)(4,2))  1 2
(1,2,3,4)((1,4)(2,3)(3,4)(4,2)) 
(1,2,3,4)((1,4)(2,3)(3,4)(4,2))  3 4
(1,2,3,4)((1,4)(2,3)(3,4)(4,2)) 
(1,2,3,4)((1,4)(2,3)(3,4)(4,2)) 
(1,2,3,4)((1,4)(2,3)(3,4)(4,2)) 
etc.
(1,2,3,4)((1,4)(2,3)(3,4)(4,2)) 

28. What’s so great about
Turing Machines?

29. Hilbert’s list of 23 problems
Leading mathematician in the 1900s
At the 1900 International Congress of Mathematicians, proposed a list of 23 problems for the 20th century
Hilbert’s 10th problem:
David Hilbert
( )
Find a method to tell if an equation like xyz – 3xy + 6xz + 2 = 0 has integer solutions

30. A brief history of computing devices
Z3 (Germany, 1941)
ENIAC (Pennsylvania, 1945)
PC (1980)
MacBook Air (2008)

31. Computation is universal
In principle, all these
have the same problem-solving ability

32. The Church-Turing Thesis
Turing Machine
If an algorithm can be implemented on any realistic computer, then it can be implemented on a Turing Machine.

33. The Church-Turing Thesis
Turing Machine
quantum computing
cosmic computing
DNA computing

34. Alan Turing
Invented the Turing Test to tell humans and computers apart
Broke German encryption machines in World War II
The Turing Award is the “Nobel prize of Computer Science”
Alan Turing
( )
Turing’s motivation: By studying Turing Machines, we can understand the limitations of real computers.

35. There is no such Turing Machine!
Undecidability
What Hilbert meant to ask was:
Building on work of Gödel, Church, Turing, Davis, Putnam, Robinson, in 1970 Matijasievič showed:
Find a Turing Machine to tell if an equation like xyz – 3xy + 6xz + 2 = 0 has integer solutions
There is no such Turing Machine!

36. Computer program analysis
public static void main(String args[]) { System.out.println("Hello World!"); }
What does this program do?
public static void main(String args[]) {
int i = 0;
for (j = 1; j < 10; j++) {
i += j;
if (i == 28) {
System.out.println("Hello World!"); }
How about this one?

37. Computers cannot analyze programs!
No Turing Machine can do this:
input: The code of a java program P
Accept if P prints “hello, world”
Reject if not
Significance: It is impossible for computer to predict what a computer program will do!

38. How do you argue things like that?
To argue what computers cannot do,
we need to have a precise definition of what a computer is.
Turing’s answer: A computer is just
a Turing Machine.
1936:
“On Computable Numbers, with an Application to the Entscheidungsproblem”
Section 1. Computing Machines

39. The Church-Turing Thesis
All arguments [for the CT Thesis] which can be given are bound to be, fundamentally, appeals to intuition, and for this reason rather unsatisfactory mathematically.
The arguments which I shall use are of three kinds:
1. A direct appeal to intuition A proof of the equivalence of two definitions (In case the new definition has greater intuitive appeal) Giving examples of large classes of numbers which are computable.
1936:
“On Computable Numbers, with an Application to the Entscheidungsproblem”
Section 9. The extent of the computable numbers