Transcript Document
CSCI 2670 Introduction to Theory of Computing
September 28, 2005
Agenda
• This week – Turing machines – Read section 3.1
September 28, 2005
Announcements
• Tests will be returned tomorrow • 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 September 28, 2005
Recap to date
• Finite automata (both deterministic and nondeterministic) machines accept regular languages – Weakness : no memory • Pushdown automata accept context free grammars – Add memory in the form of a stack – Weakness : stack is restrictive September 28, 2005
Turing machines
• Similar to a finite automaton – Unrestricted memory in the form of a tape • Can do anything a real computer can do! – Still cannot solve some problems • Church-Turing thesis some Turing machine : any effective computation can be carried out by September 28, 2005
Touring machine schematic
Control a b a ~ • Initially tape contains the input string – Blanks everywhere else (denoted ~ in class … different symbol in book) • Machine may write information on the tape September 28, 2005
Touring machine schematic
Control a b a ~ • Can move tape head to read information written to tape • Continues computing until output produced – Output values accept or reject September 28, 2005
Touring machine schematic
Control a b a ~ • Turing machine results – Accept – Reject – Never halts • We may not be able to tell result by observation September 28, 2005
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Differences between TM and FA
TM has tape you can read from and write to Read-write head can be moved in either direction Tape is infinite Accept and reject states take immediate effect September 28, 2005
Example
• How can we design a Turing machine to find the middle of a string?
– If string length is odd, return middle symbol – If string length is even, reject string • Make multiple passes over string Xing out symbols at end until only middle remains September 28, 2005
Processing input
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Check if string is empty If so, return reject Write X over first and last non-X symbols After this, the head will be at the second X Move left one symbol If symbol is an X, return reject even in length) Move left one symbol If symbol is an X, return accept even in length) Go to step 2 (string is (string is September 28, 2005
Example
• 00110~ – First check if string is empty – X first and last non-X symbols • X011X~ – Move left one symbol • X011X~ – Is symbol an X? No – Move left one symbol • X011X~ – Is symbol an X? No – Write X over first and last non-X symbols September 28, 2005
Example
• XX1XX~ – Move left one symbol • XX1XX~ – Is symbol an X? No – Move left one symbol • XX1XX~ – Is symbol an X? Yes – Return accept September 28, 2005
Formal definition of a TM
Definition: A Turing machine is a 7 1.
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tuple (Q, Q, , and , , ,q 0 ,q accept Q is the set of states, ,q reject ), where is the input alphabet not containing the special blank symbol ~ is the tape alphabet, where ~ and , : Q Q {L,R} is the transition function, September 28, 2005
What is the Transition Function??
Q = set of states, : Q Q {L,R} • Or = tape alphabet • • Given: • The current internal state • The symbol on the current tape cell Then tells us what the TM does: Changes to new internal state Either writes new symbol Q Q
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.
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q 0 Q is the start state, q accept Q is the accept state, and q reject Q is the reject state, where q reject q accept September 28, 2005
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 square of the tape – If • Head never moves left of leftmost says to move L, head stays put!
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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!
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Another TM example
• We want to create a TM to add two numbers • Use a simple tape alphabet {0,1} plus the blank symbol • Represent a number n by a string of n+1 1’s terminated by a zero • Input to compute 3+4 looks like this: September 28, 2005
Result of the TM addition example
• Note that the TM is initially positioned on the leftmost cell of the input.
• When the TM halts in the accept state, it must also be on the leftmost cell of the output: September 28, 2005
Breaking down the addition problem
• Good computer scientists like to simplify • A successor TM appends a 1 to the right end of a string of 1’s September 28, 2005
The successor subroutine
• The TM starts in the initial state s and stays in state s 0 0 , positioned on the leftmost of a string of 1’s • If it sees a 1, it writes a 1, moves right, • If it sees a 0, it writes a 1 and moves to September 28, 2005
Successor subroutine state transitions
• • < s 0 , 0, s 1 , 1 > • < s 1 , 1, s 1 , « > • < s 1 , 0, s 2 , » > (original state, input, new state, action) September 28, 2005
TM state interpretation
• S and is scanning right • S • S 0 1 and is scanning left 2 – the TM has seen only 1’s so far – the TM has seen its first zero – the TM has returned to the leftmost 1 and halts.
View movie of this TM September 28, 2005
From successor TM to addition TM
• The successor TM will join the two blocks of n+1 1’s and m+1 1’s into a single block of n+m+3 1’s • To complete the computation, knock off two 1’s from left end (states s 2 and s 3 ) September 28, 2005
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 September 28, 2005
Acknowledgements
• TM addition example from Stanford, http://plato.stanford.edu/entries/turing-machine/ September 28, 2005