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CSC 3130: Automata theory and formal languages Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130 The Chinese University of Hong Kong Variants of Turing Machines Fall 2008

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Recognizing versus deciding Looping is an undesirable behavior so we say –Such TM are called deciders The language recognized by a TM is the set of all inputs that make it reach q acc A TM decides language L if it recognizes L and does not loop on any input acceptrejectloop

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Turing Machines state control … infinite tape 010 input Why is this a universal model of computation?

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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.

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Variant 1: The multitape Turing Machine The transition may depend on the contents of all the cells Different tape heads can be moved independently state control tape 1 … 010 tape 2 … 01 tape 3 … 100

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Variant I: The multitape Turing Machine Multitape Turing Machines recognize the same languages as single-tape Turing Machines M … 010 … 01 … 100 = {0, 1, ☐ } S … 01010##0#10 ’ = {0, 1, ☐, 0, 1, ☐, #} #

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Variant I: The multitape Turing Machine For every transition in M, do following in S : Until you find the last #, Find the next dotted symbol Update this symbol according to transition in M Move the dot according to transition in M If there is no space to move the dot, Move everything to the right of the dot to make space first Move head of S to beginning of tape S … 01010##0#10 ’ = {0, 1, ☐, 0, 1, ☐, #} #

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Variant 2: The random access machine It has registers that can store arbitrary values, a program counter, and a random-access memory load -7 R 0 := -7 write R3 M[R 3 ] := R 0 store R5 R 5 := R 0 add R5 R 0 := R 0 + R 5 jpos 3 if R 0 > 0 then PC := 3 accept instructionmeaning 012345012345 0 PC 0 R0R0 0 R1R1 0 R2R2 2 M 1220 0 1234 program counter registers memory …

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Variant 2: The random access machine load -7 R 0 := -7 write R2 M[R 2 ] := R 0 save R1 R 1 := R 0 add R1 R 0 := R 0 + R 1 jzero 3 if R 0 = 0 then PC := 3 accept instructionmeaning 012345012345 0 PC 0 R0R0 0 R1R1 0 R2R2 0 M 0000 0 1234 … -7 -14 1 2 3 4 5 The instructions are indexed by the program counter

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Variant 2: The random access machine Simulating a Turing Machine on a RAM: Random assess machines recognize the same languages as Turing Machines PC 2 R0R0 1 M 2100 … M … 211 head tape

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save R1 handle for state q 0 0 Simulating a TM on a RAM M q0q0 q1q1 q acc 1/2R … save R1 save head position read R1 read tape contents x add -1 jzero 6 if x = 1 goto line 6 load 2 new value of cell write R1 write in memory load R1 recall head position add 1 move head to right jump 30 go to state q 1 program 0 1 2 3 6 7 8 9 10 … 211 save R1 handle for state q 1 … … … accept handle for state q acc 30 200

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Simulating a RAM on a Turing Machine The configuration of a RAM consists of –Program counter –Contents of registers –Indices and contents of all nonempty memory cells 14 PC 3 R0R0 -7 R1R1 5 R2R2 2 M 0120 1234 … configuration = (14, 3, -7, 5, (0, 2), (2, 1), (3, 2)) 0

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Simulating a RAM on a Turing Machine The TM has a simulation tape and a scratch tape The simulation tape stores RAM configuration The TM has a set of states corresponding to each program instruction of the RAM The TM tape is updates according to RAM instruction (14,3,-7,5,(0,2),(2,1),(3,2)) M

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Simulating a RAM on a Turing Machine Example: load R1 (14,3,-7,5,(0,2),(2,1),(3,2)) 2. Write R 1 to conf tape c 1. Copy R 1 to scratch tape -7 s (14,-,-7,5,(0,2),(2,1),(3,2)) -7 c s (14,-,-7,5,(0,2),(2,1),(3,2)) c c. c.. Make more space as needed (14,-7,-7,5,(0,2),(2,1),(3,2)) c 3. Erase scratch tape

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Variant 3: Nondeterministic Turing Machine The transition function is nondeterministic: The language recognized by N are those strings that can lead N to end in q acc N … 010 : (Q – {q acc, q rej }) → subsets of (Q {L, R})

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Equivalence of NTM and TM Let us look more deeply into NTMs Nondeterministic Turing Machines recognize the same languages as Turing Machines q3q3 q5q5 1/2R q6q6 1/1L 1/1R Nondeterministic choices can be numbered ➀ ➁ ➂ The first m steps of the computation are then fully specified by a sequence of m numbers

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Simulating a nondeterministic TM N … 100 k = maximum number of nondeterministic choices in any state M … 100 … 100 … 321 input tape 2 simulation tape address tape Address tape specifies nondeterministic choices of N First, input is copied from input tape to simulation tape Then, M simulates N using address tape data ’ = ∪ {1,..., k}

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How to use the address tape Suppose N accepts x when nondeterminism = 11221321 Then we want to make sure the address tape contains the string 11221321 at some point in its execution N … 100 M … 100 … 100 … 321 input tape 2 simulation tape address tape To ensure this we try all possibilities for the address tape

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Simulating a nondeterministic TM … 100 … 100 … 321 input tape x 2 simulation tape z address tape a Initially: x = input of N a is empty For all possible strings a : Copy x to z. Simulate N on input z using a as the nondeterminism If a specifies an invalid choice or simulation runs out of choices, abandon simulation. If N enters its accept state, accept and halt. If N enters its reject state, abandon simulation. Description of M:

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Correctness of simulation If N accepts x : For all possible strings a : Copy x to z. Simulate N on input z using a as the nondeterminism If a specifies an invalid choice or simulation runs out of choices, abandon simulation. If N enters its accept state, accept and halt. If N enters its reject state, abandon simulation. Eventually, M will encounter the correct a Description of M: So M will accept Provided that all previous simulations halt!

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Correctness of simulation Claim: Simulation step always halts (never loops) For all possible strings a : Copy x to z. Simulate N on input z using a as the nondeterminism If a specifies an invalid choice or simulation runs out of choices, abandon simulation. If N enters its accept state, accept and halt. If N enters its reject state, abandon simulation. Description of M: Since a is finite, the simulation will eventually run out of choices

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Correctness of simulation If N does not accept x, then for every a : For all possible strings a : Copy x to z. Simulate N on input z using a as the nondeterminism If a specifies an invalid choice or simulation runs out of choices, abandon simulation. If N enters its accept state, accept and halt. If N enters its reject state, abandon simulation. Description of M: Either N will loop, so simulation runs out of choices Or N rejects, so simulation is abandoned anyway In either case, simulation fails, so M loops forever!

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Context-free languages are recognizable Corollary In fact If L is context-free, then it is recognizable by a TM If L is context-free, then it is decided by some TM

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