CSCI 2670 Introduction to Theory of Computing October 7, 2004

2 Agenda Yesterday –Test Today –Continue Turing Machines

October 7, 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

October 7, Formal definition of a TM Definition: A Turing machine is a 7- tuple (Q, , , ,q 0,q accept,q reject ), 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.  : Q  Q  {L,R} is the transition function,

October 7, Formal definition of a TM Definition: A Turing machine is a 7- tuple (Q, , , ,q 0,q accept,q reject ), where Q, , and  are finite sets and 5.q 0  Q is the start state, 6.q accept  Q is the accept state, and 7.q reject  Q is the reject state, where q reject  q accept

October 7, Computing with a TM M receives input w = w 1 w 2 …w n  * 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!

October 7, Completing computation Continue following  transition rules until M reaches q accept or q reject –Halt at these states May never halt if the machine never transitions to one of these states!

October 7, 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

October 7, Configuration C 1 yields C 2 C 1 yields C 2 if the TM can legally go from C 1 to C 2 in one step –Assume a, b   and u, v   * –uaq i bv yields uq k acv if  (q i,b)=(q k,c,L) –uaq i bv yields uacq k v if  (q i,b)=(q k,c,R)

October 7, Configuration C 1 yields C 2 Special cases if head is at beginning or end of tape –q i bv yields q k cv if  (q i,b)=(q k,c,L) –q i bv yields cq k v if  (q i,b)=(q k,c,R) –uaq i is the same as uaq i ~ “handle this case as before” uaq i ~ yields uq k ac~ if  (q i,b)=(q k,c,L) uaq i ~ yields uacq k ~ if  (q i,b)=(q k,c,R)

October 7, Special configurations Start configuration –q0w–q0w Halting configurations –Accepting configuration: uq accept v –Rejecting configuration: uq reject v u, v   *

October 7, Strings accepted by a TM A Turing machine M accepts input sequence w if a sequence of configurations C 1, C 2, …, C k exist, where 1.C 1 is the start configuration of M on input w 2.each C i yields C i+1 for i = 1, 2, …, k-1 3.C k is an accepting configuration

October 7, 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

October 7, 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

October 7, Turing machine notation  (q i,b)=(q k,c,D), where D = L or R, is represented by the following transition qiqi qkqk b  c, D In the special case where  (q i,b)=(q k,b,D), i.e., the tape is unchanged, the right-hand side will just display the direction

October 7, Example Write a TM that accepts all strings of the form … –Start with a 1 –End with a 1 –Progressively more 0’s between consecutive 1’s

October 7, 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

October 7, 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)

October 7, Exit condition If you begin to look for the next group of 0’s and reach a ~ then accept

October 7,

October 7, 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

October 7, Group project 2 Write a Turing machine that accepts the language {w  {a,b} * | |w| is even}

October 7, Group project 3 Write a Turing machine that accepts the language {a n b m | n  m and n  m}

October 7, Group project 4 Write a Turing machine that accepts the language {a n b m a n+m | n  0 and m  1}

October 7, Group project 5 Write a Turing machine that accepts the language {ww R | w  {a,b} * }

October 7, Group project 6 Design a Turing machine that accepts the language {w  {a,b} * | w has more a’s than b’s}