Post on 14-Jan-2016
description
CSCI 3130: Formal languages and automata theory
Andrej Bogdanov
http://www.cse.cuhk.edu.hk/~andrejb/csc3130
The Chinese University of Hong Kong
Turing Machines
Fall 2011
What is a computer?
A computer is a machine that manipulates data according to a list of instructions.
computerinput
programoutput
What is a computer?
world
void hello( String name){ print( “Hello, “, name);} Hello, world
E2-E4
computer
deep blue
?
?
?
Turing Machines
…a b b ☐ ☐
input blanks
controlhead
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
Example
L1 = {w#w: w ∈{a, b}*}
Strategy:
Read and remember the first symbol
Cross it off (x)
Read the first symbol past #
If they don’t match, reject
abbaa#abbaa
xbbaa#abbaa
xbbaa#abbaa
If they do, cross it off xbbaa#xbbaa
Example
L1 = {w#w: w ∈ {a, b}*}
Strategy:
Look for the first uncrossed symbol
Cross it off (x)
Read the first uncrossed symbol past #
xbbaa#xbbaa
xxbaa#xbbaa
xxbaa#xbbaa
If they match, cross it off, else reject xxbaa#xxbaa
At the end, there should be only xs and #sxxxxx#xxxxx
If not, reject; otherwise, accept.
How Turing Machines operate
…a b a ☐q1 q2
a/bL
Replace a with b, and move head left
current state
…a b b ☐q1 q2
a/bL
new state
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
: (Q – {qacc, qrej}) → Q {L, R}
Turing Machines are deterministic
Configurations
• A configuration consists of the current state, the head position, and tape contents
…a b a ☐q1 abq1a
q1 qacca/bR
abbqacc…a b b ☐
qacc
configuration
Configurations
• We say configuration C yields C ’ if the TM can go from C to C ’ in one 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 abbqaccyields
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+1Ck is accepting
The language recognized by M is the set of all strings that M accepts
Looping
• Something strange can happen in a Turing Machine:
• Inputs can be divided into three types
q0
qacc
qrej
= {0, 1}
input: ☐/☐R
0/0R
1/1RThis machinenever halts
qacc qrej
accept reject loop
Halting
• We say M halts on x if there exists a sequence of configurations C0, C1, ..., Ck where
C0 is starting Ci yields Ci+1
Ck is acceptingor rejecting
A TM M is a decider if it halts on every inputLanguage L is decidable if it is recognized by a TM that halts on every input
Programming Turing Machines
L1 = {w#w: w ∈ {a, b}*}Description of Turing Machine:
Until you reach #
Read and remember entry
Move right past # and past all xsIf this entry is different, reject
Otherwise
Move left past # and to right of first x
If you see only xs followed by ☐, accept
xbbaa#xbbaa
xxbaa#xxbaa
Write x xxbaa#xbbaaxxbaa#xbbaa
Write x xxbaa#xxbaa
1
2
3
4
5
6
7
8
Programming Turing Machines
L1 = {w#w: w ∈ {a, b}*}
q0
qa1 qa2
q2
qb1 qb2
q3 qrejq1 qacc
a/xR
#/#R
#/#R
b/xL
b/xR
a/xL
a/aRb/bR
a/aRb/bR
x/xR
x/xR
x/xR
a/aLb/bLx/xL
☐/☐R#/#R #/#L
a/aLb/bL
x/xR
everything else
1
23
23
4
4
56
56
78
Programming Turing Machines
q0
qa1 qa2
q2
qb1 qb2
q3q1 qacc
a/xR
#/#R
#/#R
b/xL
b/xR
a/xL
a/aRb/bR
a/aRb/bR
x/xR
x/xR
x/xR
a/aLb/bLx/xL
☐/☐R#/#R #/#L
a/aLb/bL
x/xR
1
23
23
4
4
56
56
78
input: aab#aab
configurations:
q0aab#aabxqa1ab#aabxaqa1b#aabxabqa1#aabxab#qa2aabxabq2#xabxaq3b#xabxq3ab#xabq3xab#xabxq0ab#xab
Programming Turing Machines
L2 = {aibjck: i j = k and i, j, k > 0 }
High-level description of TM:
For every a:
Cross off the same number of bs and cs
Uncross the crossed bs (but not the cs)
Cross off this a
If all as and cs are crossed off, accept.
aabbcccc
aabbcccc
= {a, b, c}
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?
Programming Turing Machines
Implementation details:
Cross off the same number of bs and cs:
Move right until you see a c
If any bs are left, repeat
Put a special marker on top of first a
Move the head to the first symbol of the tape
aabbcccc
aabbcccc
aaqbbcccc
= {a, b, c}
Programming Turing Machines
L3 = {#x1#x2...#xl : xi ∈ {0, 1}* and xi ≠ xj for each i ≠ j}
#01#0011#1High-level description of TM:
On input w,
For every pair of blocks xi and xj in w,
Compare the blocks xi and xjIf they are the same, reject.Otherwise, accept.
Programming Turing Machines
L3 = {#x1#x2...#xl : xi ∈ {0, 1}* and xi ≠ xj for each i ≠ j}
#01#0011#1Low-level description:
Place a mark on the leftmost #(i.e. replace # by #) and move right
1.
0.If input is , or has exactly one #, accept.
#01#0011#1
Place another mark on next unmarked #(If there are no more #, accept)2. #01#0011#1
Programming Turing Machines
L3 = {#x1#x2...#xl : xi ∈ {0, 1}* and xi ≠ xj for each i ≠ j}
current state:
Repeat Step 35.
#01#0011#1
#01#0011#1
#01#0011#1
#01#0011#1
Compare the two strings to the rightof marked #. If there are same, reject3.
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
Programming Turing Machines
L3 = {#x1#x2...#xl : xi ∈ {0, 1}* and xi ≠ xj for each i ≠ j}
#01#0011#1
#01#0011#1
accept4.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
Compare the two strings to the rightof marked #. If there are same, reject3.
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!
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
(1,2,3,4)((1,4),(2,3),(3,4)(4,2))
Convention for describing graphs:
(nodes)(edges)
no node must repeatedges are pairs (node1,node2)
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
2.Repeat until no new nodes are marked:
For each node, mark it if it is attached to an already marked node
3.If all nodes are marked accept, otherwise reject.
1 2
3 4
x
x
x
x
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))
(1,2,3,4)((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)) etc.
1 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))
What’s so great aboutTuring Machines?
Hilbert’s list of 23 problems
• Leading mathematician in the 1900s
• At the 1900 International Congressof Mathematicians, proposed a list of23 problems for the 20th century
• Hilbert’s 10th problem:David Hilbert
(1862-1943)
Find a method to tell if an equation like xyz – 3xy + 6xz + 2 = 0 has integer solutions
A brief history of computing devices
Z3 (Germany, 1941) ENIAC (Pennsylvania, 1945)
PC (1980) MacBook Air (2008)
Computation is universal
In principle, all thesehave the same problem-solving ability
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.
The Church-Turing Thesis
Turing Machinequantum computing
DNA computing
cosmic computing
Alan Turing
• Invented the Turing Test to tellhumans and computers apart
• Broke German encryption machines in World War II
• The Turing Award is the “Nobel prize of Computer Science”
Alan Turing (1912-1954)
Turing’s motivation: By studying Turing Machines, we can understand the limitations of real computers.
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!
Computer program analysis
public static void main(String args[]) { System.out.println("Hello World!"); }
What does this program do?
How about this one?
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!"); } } }
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!
How do you argue things like that?
To argue what computers cannot do,we need to have a precise definitionof what a computer is.
“On Computable Numbers, with an Application to the Entscheidungsproblem”
1936:
Section 1. Computing Machines
Turing’s answer: A computer is just a Turing Machine.
The Church-Turing Thesis
“On Computable Numbers, with an Application to the Entscheidungsproblem”
1936:
Section 9. The extent of the computable numbers
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 2. A proof of the equivalence of two definitions (In case the new definition has greater intuitive appeal) 3. Giving examples of large classes of numbers which are computable.