CSCI 2670Introduction to Theory of

Computing

October 7, 2004

October 7, 2004 2

Agenda• Yesterday

– Test• Today

– Continue Turing Machines

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Announcements• Homework due Wednesday

– 3.2 b,d 3.5 all, 3.7, 3.8 a,c (high-level descriptions)

• Reminder: tutorial sessions are suspended until further notice– Extended office hours while tutorials

are suspended• Monday 11:00 – 12:00• Tuesday 3:00 – 4:00• Wednesday 3:00 – 5:00

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Formal definition of a TMDefinition: A Turing machine is a 7-

tuple (Q,,,,q0,qaccept,qreject), where Q, , and are finite sets and

1. Q is the set of states,2. is the input alphabet not

containing the special blank symbol ~

3. is the tape alphabet, where ~ and ,

4. : QQ{L,R} is the transition function,

October 7, 2004 5

Formal definition of a TMDefinition: A Turing machine is a 7-

tuple (Q,,,,q0,qaccept,qreject), where Q, , and are finite sets and

5. q0Q is the start state,6. qacceptQ is the accept state, and7. qrejectQ is the reject state, where

qrejectqaccept

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Computing with a TM• M receives input w = w1w2…wn*

on leftmost n squares of tape– Rest of tape is blank (all ~ symbols)

• Head position begins at leftmost square of tape

• Computation follows rules of • Head never moves left of leftmost

square of the tape– If says to move L, head stays put!

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Completing computation• Continue following transition

rules until M reaches qaccept or qreject– Halt at these states

• May never halt if the machine never transitions to one of these states!

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TM configurations• The configuration of a Turing

machine is the current setting– Current state– Current tape contents– Current tape location

• Notation uqv– Current state = q– Current tape contents = uv

• Only ~ symbols after last symbol of v– Current tape location = first symbol of v

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Configuration C1 yields C2

• C1 yields C2 if the TM can legally go from C1 to C2 in one step– Assume a, b and u, v *

– uaqibv yields uqkacv if (qi,b)=(qk,c,L)– uaqibv yields uacqkv if (qi,b)=(qk,c,R)

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Configuration C1 yields C2

• Special cases if head is at beginning or end of tape– qibv yields qkcv if (qi,b)=(qk,c,L)– qibv yields cqkv if (qi,b)=(qk,c,R)– uaqi is the same as uaqi~

• “handle this case as before”• uaqi~ yields uqkac~ if (qi,b)=(qk,c,L)• uaqi~ yields uacqk~ if (qi,b)=(qk,c,R)

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Special configurations• Start configuration

– q0w• Halting configurations

– Accepting configuration: uqacceptv– Rejecting configuration: uqrejectv

• u, v *

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Strings accepted by a TM• A Turing machine M accepts input

sequence w if a sequence of configurations C1, C2, …, Ck exist, where

1. C1 is the start configuration of M on input w

2. each Ci yields Ci+1 for i = 1, 2, …, k-13. Ck is an accepting configuration

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Language of a TM• The language of M, denoted L(M),

is– L(M) = {w | M accepts w}

• A language is called Turing-recognizable if some Turing machine recognizes it

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Deciders• A Turing machine is called a

decider if every string in * is either accepted or rejected

• A language is called Turing-decidable if some Turing machine decides it– These languages are often just called

decidable

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Turing machine notation(qi,b)=(qk,c,D), where D = L or R,

is represented by the following transition

qi qkb c, D

• In the special case where (qi,b)=(qk,b,D), i.e., the tape is unchanged, the right-hand side will just display the direction

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Example• Write a TM that accepts all strings

of the form 101001000100001 …– Start with a 1– End with a 1– Progressively more 0’s between

consecutive 1’s

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Design• Check first symbol is a 1

– If not reject• Move right and check if second

symbol is a 0– If not reject– If so, replace with X and begin

recursion

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Recursion (high level)• Go back and forth on either side of

each 1– Replace 0’s on right side of 1 with an X– Replace X’s on left side of 1 with a Y

• After all X’s on left side of 1 are replaced with Y’s, there should be exactly one on the right side that has not been X’ed– If not, reject– If so, repeat process (recursion step)

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Exit condition• If you begin to look for the next

group of 0’s and reach a ~ then accept

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October 7, 2004 21

Group project 1• Design a Turing machine to accept

any string in {a,b}* after making a copy of it on the tape– The tape will start with w – After TM processes the string, the

tape should read ww

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Group project 2• Write a Turing machine that

accepts the language {w {a,b}* | |w| is even}

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Group project 3• Write a Turing machine that

accepts the language {anbm | nm and nm}

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Group project 4• Write a Turing machine that

accepts the language {anbman+m | n0 and m1}

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Group project 5• Write a Turing machine that

accepts the language {wwR | w{a,b}*}

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Group project 6• Design a Turing machine that

accepts the language {w{a,b}* | w has more a’s than

b’s}