Computation Theory

1
Computation Theory

• Introduction to Turing Machine

2
Turing Machines

• Proposed by Alan Turing, 1936. England
• Turing Award

3
Differences between finite automata and Turing
machine

• A Turing machine can both write on the tape and
  read from it.
• The read-write head can move both to the left and
  to the right.
• The tape is infinite.
• The special states for rejecting and accepting
  take immediate effect.

4
eg.
Consider a Turing machine M1 for testing
membership in the language Bww w?0,1
? 0 1 1 0 0 0 0 1 1 0 0 0 B ? X 1 1 0 0 0
0 1 1 0 0 0 B
? X 1 1 0 0 0 X 1 1 0 0 0 B ? X 1 1 0 0 0
X 1 1 0 0 0 B ? X X 1 0 0 0 X 1 1 0 0
0 B
? X X X X X X X X X X X X
B
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• M1 On input string w
• Scan the input to be sure that it contains a
  single symbol. If not, reject.
• Zig-zap across on either side to check on
  whether these positions contain the same symbol.
  If not, reject. Cross off symbols as they
  checked.
• If any symbols remain, reject otherwise accept.

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Formal definition of a Turing machine

• d QG? QGL,R
• transition function.
• d(q,a) (r,b,L)

r
b
7
Def

• A Turing Machine is a 7-tuple (Q, S, G, S, q0,
  qaccept, qreject), where Q, S, G are all finite
  sets and
• Q set of states,
• S the input alphabet NOT containing the special
  blank symbol B,
• G the tape alphabet, where B? G and S? G,
• dQG?QGL,R transition func.
• q0?Q is the start state,
• qaccept?Q is the accept state, and
• qreject?Q is the reject state, where qreject ?
  qaccept

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• As a Turing machine computes, it may halt with
  accept or reject, or it may never halt!
• During computation, changes occur in
• the current state,
• the current tape contents, and
• the current head location.
• The above three items form a configuration of
  the Turing machine.

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• Write uqv for the configuration where q is the
  current state, uv is the current contents, the
  current head location is the first symbol of v.
• 1011qT0111
• Let C1 and C2 be two configurations. Say that C1
  yields C2, if the Turing machine can legally go
  from C1 to C2 in a single step.

qT
B
1
0
1
1
0
1
1
1
1
B
uaqibv yeilds uqjacv if d(qi,b)(qj,c,L) a,b?G,
u,v?G
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• uaqibv yields uacqjv if d(qi,b)(qj,c,R)
• qibv yields qjcv if the transition is right
  moving.
• We dont move the head out of bound!
• We dont have configuration uaqiB.
• Starting configuration q0w
• Accepting configuration the state is qaccept
• Rejecting configuration the state is qreject

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halting configurations

• A Turing machine M accepts input w if a seq. of
  configurations C1, C2, , Ck exists where
• C1 is the start conf. of M on input w,
• each Ci yields Ci1, and
• Ck is an accepting configuration.
• The collection of strings that M accepts is the
  language of M, denoted L(M).

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• Def A language is Turing-recognizable if some
  Turing machine recognizes it.
• (or recursively enumerable)
• r.e.
• Def A language is Turing-decidable or decidable
  if some Turing machine decides it.
• (or recursive)
• r.

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eg.
Show a TM M that recognizes the language
A02nn?0

• MOn input string w
• Sweep left to right across the tape, crossing off
  every other 0.
• If in stage 1 the tape contained a single 0,
  accept.
• If in stage 1 the tape contained more than a
  single 0 and the number of 0s was odd, reject.
• Return the head to the left-hand and of the tape.
• Goto 1.

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• eg. (Element Distinctness problem)
• MOn input w
• 1. Place a mark on top of the leftmost tape
  symbol. If
• that symbol was not a , reject.
• 2. Scan right to the next and place a second
  mark
• on top of it. If no is encountered before a
  blank
• symbol, only x1 was present, so accept.

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• 3. By zig-zagging,compare the two strings to the
• right of the marked s .If they are equal
  reject.
• 4. Move the rightmost of the two marks to the
  next
• symbol to the right. If no symbol is
• encountered before a blank symbol, move the
• leftmost mark to next to its right.
• 5. Goto step 3.

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• eg. For ?a,b, design a Turing Machine that
  accepts Lanbn n?1
• Sol
• Qq0,q1,q2,q3,q4,
• ?a,b
• ?a,b,x,y,B,
• q4accept state

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• ?(q0,a)(q1,x,R) ?(q1,a)(q1,a,R)
• ?(q1,y)(q1,y,R) ?(q1,b)(q2,y,L)
• ?(q2,y)(q2,y,L) ?(q2,a)(q2,a,L)
• ?(q2,x)(q0,x,R)
• ?(q0,y)(q3,y,R) ?(q3,y)(q3,y,R)
• ?(q3,B)(q4,B,R)
• q0aabb xq1abb xaq1bb xq2ayb q2xayb
• xq0ayb xxq1yb xxyq1b xxq2yy
• xq2xyy xxq0yy xxyq3y xxyyq3B
• xxyyBq4

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Variants of Turing machine

• 1. Multi-tape Turing Machines
• a Turing Machine with several tapes.

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• Thm Every multi-tape Turing machine has an
  equivalent single tape Turing machine.
• Pf
• Idea

0
1
0
1
0
B
M
a
a
a
B
b
a
B
S

0
1
0
1
0

a
a
a

b
a

B
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• Non-deterministic Turing Machines
• Thm Every non-deterministic TM has an equivalent
  deterministic TM.
• Pf

0
0
1
D
x
x

0
1
x
B
q
0
1
0
0
0
q
q
0
q
0
0
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• Initially tape 1 contains the input w, and tapes
  2 and 3 are empty.
• Copy tape 1 to tape 2.
• Use tape 2 to simulate N with input w on one
  branch of its non-det. computation. Before each
  step of N consult the next symbol on tape 3 to
  decide which branch to move. If no symbol remains
  or this choice is invalid goto step 4. If reject
  also goto 4.
• Increase the count on tape 3. go to step 2.

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• CorollaryA language is Turing-recognizable if
  and only if some non-deterministic TM recognizes
  it.
• Corollary A language is decidable if and only
  if some non-deterministic TM decides it

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Enumerators

• A TM with an attached printer.
• The language enumerated by an enumerator is the
  collection of all the strings that it eventually
  prints out.

Control
0
0
1
0

Work tape
Enumerator
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• Theorem A language is Turing-recognizable if and
  only if some enumerator enumerates it.
• Pf
• If an enumerator E that enumerates a language A,
  a TM M recognizes A.
• Mon input w
• Run E. Every time that E outputs an string,
  Compare it with w.
• If w ever appears in the output of E, accept.

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• If TM M recognizes a language A, we can construct
  an enumerator E for A.
• Let S1, S2, be a list of all possible strings
  in ?.
• EIgnore the input.
• Repeat the following for i 1, 2, 3,
• Run M for i steps on each input, S1, S2, , Si,
• If any computations accept, print out the
  corresponding Si.

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Definition of algorithm

• Hilberts problems
• In 1900, mathematician David Hilbert delivered a
  famous address at the International congress of
  Mathematicians in Paris.
• He proposed 23 mathematical problems.
• Hilberts tenth problem
• Design an algorithm that tests whether a
  multi-variable polynomial has an integral root.

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The Church-Turing Thesis

• In 1970, Yuri Matijasevic, building on work of
  Martin Davis, Hilary Putnam, and Julia Robinson,
  Showed that no algorithm exists for testing
  whether a poly. Has integral roots.

Intuitive notion of algorithms
Turing machine algorithms

Necessary to resolve Hilberts tenth problem
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• D P P is a polynomial with an integral root
• Hilberts tenth problem
• Is D decidable?
• A special case
• D1P P is a polynomial over x with an integral
  root
• D1 is decidable!

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• M1 The input is a polynomial P over x
  Evaluate P with x to the values 0, 1, -1, 2,
  -2, 3, -3, . If at any point the polynomial
  evaluates to 0, accepts!
• Problem 3.18 All the roots of P is between
• Thus, Di is decidable!