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Element of the Theory of the Computation

Element of the Theory of the Computation. Lecture 4.3-4.5 Turing machines. Turing machines. Turing machines can be extended d oes not increase computational power d oes increase speed and ease of use/setup/definition r educes head movement, intermediate steps, memorization, shifting.

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Element of the Theory of the Computation

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  1. Element of the Theory of the Computation Lecture 4.3-4.5 Turing machines

  2. Turing machines • Turing machines can be extended • does not increase computational power • does increase speed and ease of use/setup/definition • reduces head movement, intermediate steps, memorization, shifting

  3. Extensions of the Turing machine • Possible extensions of the Turing machine: • multiple tapes • two-way infinite tapes • two-dimensional tapes • multiple heads • random access • nondeterministic

  4. Multiple Tape Turing machine • Multiple Tapes (see Figure 1): • each tape is connected to the finite control by means of a read/write head • for any fixed integer k  1  a k-tape Turing machine is a Turing machine equipped with k tapes and corresponding heads

  5. Multiple Tape Turing machine Figure 1

  6. Definition of the k-tape Turing machine • Def.: Let k  1 be an integer • A k-tape Turing machine is a quintuple ( K, , , s, H ), where • K: is a finite set of states, not containing the halt state h; ( h ∉K) • : is an alphabet, containing the blank symbol ⊔, but not containing the symbols L and R • s: is the initial state • : is a function (deterministic) from ( K – H ) x k to K x (   {←,→})k

  7. Definition of the k-tape Turing machine (continued) • for each state q, • each k-tuple of tape symbols: • (a1,...,ak), (q, (a1,...,ak)) = (p,(b1,...,bk)), • where p: the state of the initial configuration • bj: the action taken by M at tape j • if aj = ⊳for some j ≤ k, then bj = →

  8. Configuration of a k-tape Turing machine • Def.: • M = (K, , , s, H)

  9. Example (1) for the multiple tape TM Example (1): x + y • X on first tape, y on second tape and results written to third tape • move 1 and 2 heads to right end, move head 3 to right max (|x|,|y|) • move tape 3 head right one bit for overflow

  10. Example (1) for the multiple tape TM • add 1 and 2, bit by bit and writes each intermediate result as follows • reads bits at 1 and 2 plus carry from previous bits • if sum is 0 or 1, write it to tape 3 • if sum is 2 or 3, set carry and write 0 or 1 on tape 3 • if one string ends (beginning of tape marker) use 0 for that input and do not move that head

  11. Example (2) for the k-tape Turing machine Example (2): • The use of a k-tape Turing machine: • computing a function • deciding or semideciding a language • Adoption: • the input string is placed on the first tape, such as in case of a standard Turing machine • the other tapes are initially blank, the head is on the leftmost blank square of each • at the end of a computation, as k-tape Turing machine is to leave its output on its first tape; the contents of the other tapes are ignored

  12. Example (2) for the k-tape Turing machine • Move the heads on both tapes to the right, copying each symbol on the first tape onto the second tape, until a blank is found on the first tape, but the first square of the second tape should be left blank • Move the head on the second tape to the left until a blank is found • Move the heads on both tapes to the right, this time copying symbols from the second tape onto the first tape and halt when a blank is found on the second tape

  13. Example (2) for the k-tape Turing machine At the beginning: First tape ⊳⊔ω Second tape ⊳⊔ After (1): First tape ⊳⊔ω⊔ Second tape ⊳⊔ω⊔ After (2): First tape ⊳⊔ω⊔ Second tape ⊳⊔ω After (3): First tape ⊳⊔ω⊔ω⊔ Second tape ⊳⊔ω⊔ This sequence of actions can be pictured as follows:

  14. Example (2) for the k-tape Turing machine • The depiction of Turing machines with more than one tape is the same as single-tape Turing machines • We attach as a superscript to the symbol denoting each machine the number of the tape on wich it is to operate; all other tapes are unaffected • For example, ⊔2writes a blank on the second tape • L1⊔searches to the left for a blank on the first tape, and R1,2 moves to the right the heads of both the first and the second tape • A label a1 on an arrow denotes an action if the symbol scanned in the first tape is an a • And so on…

  15. Example (2) for the k-tape Turing machine The 2-tape version of the copying machine: Figure 2

  16. Example (3) for the k-tape Turing machine Example (3): • It is known that Turing machines can add 1 to any binary integer • Turing machines can also add arbitrary binary numbers • This task can be accomplished with two tapes (see Figure 3) • Pairs of bits such as 01 on an arrow label are a shorthand for, in this case, a1 = 0, a2 = 1

  17. Example (3) for the k-tape Turing machine Figure 3

  18. Example (3) for the k-tape Turing machine • The operation of this machine: • it copies the first binary integer in its second tape, writing zeros in its place (and in the place of the ”;” separating the two integers) in the first tape  the first tape contains the second integer zeros added in front • it performs binary addition by the „school method,” starting from the least significant bit of both integers • it adds the corresponding bits writing the result in the first tape

  19. Extensions of the Turing machine Theorem: M= (K, , , s, H) is a k-tape Turing machine for some k 1 Standard Turing machine M’ = (K’, ’, ’, s’, H), where ’  for any input string x *, M on input x halts with output y on the first tape if and only if M’ on input x halts at the same halting state, and with the same output y on its tape. If M halts on input x after t steps, then M’ halts on input x after a number of steps which is O (tּ ( x +t)).

  20. Theorem: M= (K, , , s, H) is a k-tape Turing machine for some k 1. Then there is a standard Turing machine M’ where ’, and such that: • For any input string x, M on input x halts with output y on the first tape if M’ on input x halts at the same halting state and with the same output on its tape • If on input x, M halts after t steps, then M’ halts after a number of steps which is O (tּ ( x +t))

  21. Corollary from the k-tape Turing machine • Corollary: Any function that is computed or language that is decided or semidecided by a k-tape Turing machine is also computed, decided, or semidecided, respectively, by a standard Turing machine

  22. Equivalence of multiple tapes TM and ordinary TM • Theorem: • every multiple tapes Turing machine has an equivalent single-tape Turing machine • Proof ideas: • convert a multi-tape TM M to an equivalent single-tape TM S

  23. Two-way infinite Turing machine • Two-way infinite Tape: • the tape is infinite in both directions • all squares are (exception: those containing the input) • it can be simulated by a 2-tape machine: • tape 1: contains the part of the tape to the right of the square containing the first input symbol • tape 2: contains the part of the tape to the left of this in reverse

  24. Multiple heads Turing machine • Multiple Heads : • uses a single tape and multiple heads • in any state only one head can write or move • the heads all sense the scanned symbols and move or write independently • simulations in case of k-tape machines can be carried out for Turing machines with several heads, too • he tape is divided into tracks • the number of steps needed is quadriatic

  25. Multiple heads Turing machine • L = ( anbncn | n = 0, 1, 2, …) • given string w, position first head at beginning of input • position second head past all a’s to the first b • position third head past all a’s and all b’s to first c • enter loop verifying that, on each iteration, head 1 reads an a, head 2 reads a b and head 3 reads a c • if third head reaches end of input string at the same time head 1 reads the first b and head 2 reads the first c, machine erases the input string and writes a 1 into cell 1 to signify acceptance

  26. Two-dimensional tape Turing machine • Two-Dimensional Tape : • the input string is placed on the first tape, such as in case of a standard Turing machine • tape: an infinite two-dimensional grid • one head on a two dimensional grid that could expand indefinitely down and to the right • head can move in four different directions • end of tape markers on left and top sides

  27. Theorem for extensions of the Turing machine • Theorem: Any language decided or semidecided, and any function computed by Turing machines with several tapes, heads, two-way infinite tapes, or multi-dimensional tapes, can be decided, or computed, respectively, by a standard Turing machine

  28. Random access Turing machines Figure 4

  29. Random access Turing machines • A random access Turing machine has: • a fixed number of registers • a one-way infinite tape • a program counter • program contains a finite sequence of instructions (see Figure 5)

  30. Sequence of instructions Instruction Operand Sematics read j R0 := T [Rj] write j T [Rj] := R0 store j Rj := R0 load j R0 := Rj load =c R0 := c add j R0 := R0+Rj add =c R0 := R0+c sub j R0 := max {R0 – Rj, 0} sub =c R0 := max {R0 – c, 0} half R0 := [R0 / 2] jump s k := s jpos s if R0 > 0 then k := s jzero s if R0 = 0 then k := s halt k := 0 Figure 5

  31. Notes for the sequence of instructions • j stands for a register number, 0 ≤ j < k • T [i] denotes the current contents of tape square i • Rj denotes the current contents of Register j • s ≤ p denotes any instruction number in the program • c is any natural number • All instructions change k to k+1, unless explicitly stated otherwise

  32. Definition of the random access Turing machine • Def.: A random access TM is a pair M = (k, П), where: • k >0 is the number of registers • П = (π1, π2,…, πp), the program, is a finite sequence of instructions ( see Figure 5)

  33. Configuration of a random access Turing machine • A configuration of a random access Turing machine (k, П) is a k + 2-tuple C=(k, R0, R1,…, Rk-1, T), where: • k  N is the program counter, an integer between 0 and p • if k = 0, the configuration is a halted configuration • for each j, 0 ≤ j < k, Rj є N is the current value of register j • T, the tape contents, is a finite set of pairs of positive integers

  34. Example (1) for a random access TM Example (1): • program of a random access Turing machine, deciding the language {anbncn : n  0}. acount := bcount := ccount := 0, n :=1 while T[n] = 1 do : n := n +1, acount :=acount +1 while T[n] = 2 do : n := n +1, bcount :=bcount +1 while T[n] = 3 do : n := n +1, acount :=acount +1 if acount = bcount = ccount and T[n] = 0 then accept else reject

  35. Example (1) for a random access TM • We are assuming here that E(a) = 1, E(b) = 2, E(c) = 3 • We are using the variables acount, bcount, and ccount to stand for the number of a’s, b’s, and c’s • We are also using the abbreviation accept for ”load =1, halt” and reject for ”load =0, halt”

  36. Theorem for a random access Turing machines • Theorem: Any recursive or recursively enumerable language, and any recursive function, can be decided, semidecided, and computed, respectively, by a random access Turing machine.

  37. Theorem for a random access Turing machines • Theorem: Any language decided or semidecided by a random access Turing machine, and any function computable by a random access Turing machine, can be decided, semidecided, and computed, respectively, by a standard Turing machine. • If the machines halt on an input, then the number of steps taken by the standard Turing machine is bounded by a polynomial in the number of steps of the random access Turing machine on the same input.

  38. Nondeterministic Turing machines • Def.:A nondeterministic Turing machine is a quintuple M = (K, ,  , s, H) , where is a transitive relation. (( K – H ) x ) x (K x (   {←,→})) ❑abab ❑abab ❑abab ❑abab ❑bbab

  39. Nondeterministic semideciding • M = (K, ,  , s, H) a nondeterministic Turing machine • M accepts an input w (Σ - {, ❑ })* if • (s, ❑w ) yields a least one accepting configuration • M semidecides a language • L ⊆ (Σ-{, ❑ })* if • for all w (Σ - {, ❑ })* : w  L if • (s, ❑w ) yields at least one halting configuration

  40. An example of nondeterministic semideciding • L = {w {a, b, c, d}* :there are two of at least one letter } Figure 6

  41. Nondeterministic deciding and computing • M decides a language L if, for all w (Σ - {, ❑ })* : • all of M’s computations on w halt, and • w  L if at least one of M’s computations accepts • M computes a function f if, for all w (Σ - {, ❑ })* : • all of M’s computations halt, and • all of M’s computations result in f(w)

  42. An example of nondeterministic deciding • L = {w {0, 1}* : w is the binary encoding of a composite number } • M decides L by doing the following on input w: • nondeterministically choose two binary numbers 1< p, q, where |p| and |q|≤ |w|, and write them on the tape, after w, separated by ❑110011;111;1111❑❑ • multiply p and q and put the answer, A, on the tape, in place of p and q ❑110011;1011111❑❑ • compare A and w, if equal, go to y, else go to n

  43. Equivalence of nondeterministic TM and deterministic TM • Theorem: • every nondeterministic Turing machine has an equivalent deterministic Turing machine • Proof Ideas: • simulate any nondeterministic TM N with a deterministic TM D ( see Figure 7 ) • let D try all possible branches of N’s computation tree. If D finds an accept state on one of these branches, D accepts

  44. N 0 0 1 0 0 0 1 0 D x x # 0 1 x 1 2 3 3 2 3 1 2 1 1 3 Proof of Theorem Deterministic TM D simulating nondeterministic TM N input tape simulation tape address tape Figure 7

  45. Proof of theorem • D has three tapes: • tape 1 always contains the input string and is never altered • tape 2 maintains a copy of N’s tape on some branch of its nondeterministic computation • tape 3 keeps track of D’s location in N’s nondeterministic computation tree • address string 231

  46. Proof of theorem ( continued ) • 1. initially tape 1 contains the input w, and tape 2 and 3 are empty • 2. copy tape 1 to tape 2 • 3. use tape 2 to simulate N with input w on one branch of its nondeterministic computation;before each step of N, consult the next symbol on tape 3 to determine which choice to make among those allowed by N’s transition • if an accepting configuration is encountered, accept the input • if no more symbols remain on tape 3 or if this nondeterministic choice is invalid, abort this branch and go to stage 4

  47. Proof of theorem ( continued ) • if a rejecting configuration is encountered, go to state 4 • 4. replace the string on tape 3 with the lexicographically next string;simulate the next branch of N’s computation by going to stage 2

  48. Nondeterministic Turing machines • Theorem: Any language accepted by a nondeterministic Turing machine is also accepted by a deterministic Turing machine.

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