Turing Machines and Equivalent Models

1. Turing Machines and Equivalent Models Section 13.2 The Church-Turing Thesis

2. Section 13.2 The Church-Turing Thesis • The Church-Turing Thesis: Anything that is intuitively computable can be computed by a Turing machine. • It is a thesis rather than a theorem because it relates the informal notion of intuitively computable to the formal notion of a Turing machine.

3. Computational Models • A computational model is a characterization of a computing process that describes the form a program and describes how the instructions are executed. • Example. The Turing machine computational model describes the form of TM instructions and how to execute them. • Example. If X is a programming language, the X computational model describes the form of a program and how each instruction is executed.

4. Equivalence of Computational Models • Two computational models are equivalent in power if they solve the same class of problems. • Any piece of data for a program can be represented by a string of symbols and any string of symbols can be represented by a natural number. • So even though computational models may process different kinds of data, they can still be compared with respect to how they process natural numbers. • We’ll make the assumption that there is an unlimited amount of memory available. So we can represent any natural number or any finite string. • Each of the following models of computation is equal in power to the TM model.

5. The Simple Programming Language • This imperative programming model processes natural numbers. The language is defined as follows: • Variables have type N. • Assignment statements: X := 0; X := succ(Y); X := pred(Y). (assume pred(0) = 0) • Composition of statements: S1; S2. • while X ≠ 0 do S od. • This simple language model has the same power as the Turing machine model. • For input and output use the values of the variables before and after execution.

6. Some Macros X := Y X := 2 Z := Z + X Z := X + Y Z := X monus Y while X < Y do S od if X ≠ 0 then S1 else S2 fi Macro Expansion X := succ(Y); X := pred(X). X := 0; X := succ(X); X := succ(X). C := X; while C ≠ 0 do Z := succ(Z); C := pred(C) od. Z := X; C := Y; while C ≠ 0 do Z := succ(Z); C := pred(C) od. Z := X; C := Y; while C ≠ 0 do Z := pred(Z); C := pred(C) od. T := Y monus X; while T ≠ 0 do S; T := Y monus X od. U := X; V := 1; while U ≠ 0 do S1; V := 0; U := 0 od; while V ≠ 0 do S2; V := 0 od. Example/Quiz To demonstrate the power of this language, define the following macros.

7. Partial Recursive Functions • This model consists of a set of functions that take natural numbers as arguments and as values. The functions are defined as follows, where x can represent zero or more arguments. • Initial functions: zero(x) = 0, succ(n) = n + 1, and projections (e.g., p2(a, b, c) = b). • Composition: e.g., ƒ(x) = h(g1(x), …, gm(x)), where h, g1, …, gm are partial recursive. • Primitive recursion: ƒ(x, 0) = h(x) (h is partial recursive) ƒ(x, succ(y)) = g(x, y, ƒ(x, y)) (g is partial recursive). • Unbounded search (minimalization): ƒ(x) = min(y, g(x, y) = 0) (g is total partial recursive). • This means that ƒ(x) = y is the minimum y such that g(x, y) = 0, if such a y exists. • This model for constructing functions has the same power as the Turing machine model.

8. Example • The predecessor and monus functions are partial recursive as follows: • pred(0) = 0 • pred(succ(y)) = y, which can be written in the form p1(y, pred(y)). • monus(x, 0) = x • monus(x, succ(y)) = pred(monus(x, y)), which can be written pred(p3(x, y, monus(x, y))).

9. Quiz • Show that pred(monus(x, y)) = monus(pred(x), y)). • Proof: The equation is true if x = 0 or y = 0. So assume x > 0 and y > 0. Then we have: • pred(monus(x, y)) = if x > y then x – y – 1 else 0 • monus(pred(x), y)) = if x – 1 > y then x – 1 – y else 0. Although x > y and x – 1 > y are different, the if-then-else values are equal. QED.

10. Quiz • Let p and q be partial recursive with p(x), q(x) ∊ {0, 1}, for false and true. Show that the logical operations p(x) Λq(x), ¬ p(x), and p(x) Vq(x), are partial recursive functions. • Solutions: p(x) Λq(x) = p(x)q(x) ¬ p(x) = monus(1, p(x)) p(x) Vq(x) = p(x) + monus(1, p(x))q(x).

11. Examples Unbounded Search, Minimalization • ƒ(x) = min(y, xy = 0) defines ƒ(x) = 0. • ƒ(x) = min(y, x + y = 0) defines ƒ(x) = if x = 0 then 0 else undefined. • ƒ(x) = min(y, monus(x, y) = 0) defines ƒ(x) = x. • ƒ(x) = min(y, monus(y, x) = 0) defines ƒ(x) = 0. • ƒ(x, y) = min(z, monus(x + z, y) = 0) defines ƒ(x, y) = if x ≤ y then 0 else undefined

12. Example/Quiz • For y ≠ 0, let ƒ(x, y) = min(z, monus(x, yz) = 0). What function does ƒ compute? • Answer: ƒ(x, y) = ⌈x / y ⌉. To see this, notice that the definition ƒ(x, y) = min(z, monus(x, yz) = 0) means that ƒ(x, y) = z is the smallest natural number such that monus(x, yz) = 0. The definition of monus tells us that monus(x, yz) = 0 means that x ≤ yz. So z is the smallest natural number such that x ≤ yz. In other words, we have y(z – 1) < x ≤ yz. Divide by y to obtain z – 1 < x/y ≤ z. Therefore z = ⌈x / y ⌉.

13. Markov Algorithms • This model processes strings. An algorithm consists of a finite, ordered, sequence of productions of the form x → y, where x, y ∊A* for some alphabet A. • Any production can be suffixed with (halt) although this is not required. • Execution Given an input string w ∊A*, perform the following execution step repeatedly. • Scan the productions x → y sequentially to see whether x occurs as a substring of w. If so, • replace the leftmost occurrence of x in w by y and reset w to this string. Otherwise halt. • If the x → y is labeled with (halt), then halt. • Assumption: w = Λw. So a production of the form Λ → y would transform w to yw. • The Markov algorithm model has the same power as the Turing machine model.

14. Examples • Example. The Markov algorithm consisting of the single production a → Λ will delete all a’s from any string. • Example. A more instructive Markov algorithm to delete all a’s from any string over {a, b} can be written as the following sequence of productions (# is an extra symbol). 1. #a → # 2. #b → b# 3. # → Λ (halt) 4. Λ → #. An example trace: abab (input) #abab (by 4) #bab (by 1) b#ab (by 2) b#b (by 1) bb# (by 2) bb (by 3, halt).

15. Quiz • Find a Markov algorithm to replace each a with aa in strings over {a, b}. • Answer. Modify the previous example: 1. #a → aa# 2. #b → b# 3. # → Λ (halt) 4. Λ → #.

16. Example/Quiz • Find a Markov algorithm to delete the rightmost b from strings over {a, b}. • Solution: 1. #a → a# 2. #b → b# 3. # → @ 4. a@ → @a 5. b@ → Λ (halt) 6. @ → Λ (halt) 7. Λ → #.

17. Example/Quiz • Find a Markov algorithm to implement succ(x), where x is a natural number represented in binary. Assume no leading zeros (except 0 itself). • Solution: 1. #0 → 0# 2. #1 → 1# 3. 0# → 1 (halt) 4. 1# → @0 5. 1@ → @0 6. 0@ → 1 (halt) 7. @ → 1 (halt) 8. Λ → #.

18. Post Algorithms • This model processes strings. An algorithm consists of a finite set of productions of the form s → t where s and t are strings over the union of an input alphabet A with a set of variables and other symbols. Restriction: If a variable X occurs in t then X occurs in s. A production can be suffixed with (halt) although this is not required. • Execution Given an input string w ∊A*, perform the following execution step repeatedly. • Find a production x → y such that w matches x. • If so, use the match and y to construct a new string w. Otherwise halt. • If the x → y is labeled with (halt), then halt. • Assumption: A variable may match Λ.

19. Examples • Example. If the input string is 1, then 1 matches 1X. So a production like 1X → 1X0 would transform 1 into 10. • The Post algorithm model has the same power as the Turing machine model. • Example. A Post algorithm with the single production XaY → XY will delete all a’s from any string. Notice the nondeterminism.

20. Example • A Post algorithm to replace each a with aa in strings over {a, b}. • Solution: 1. aX → @aa#X 2. bX → @b#X 3. X#aY → Xaa#Y 4. X#bY → Xb#Y 5. @X# → X (halt).

21. Example • A Post algorithm to delete the rightmost b from any string over {a, b}. • Solution: 1. Xb → X (halt) 2. Xa → X#a@ 3. Xa#Y → X#aY 4. Xb#Y@ → XY (halt) 5. #X@ → X (halt).

22. Example/Quiz • Find a Post algorithm to implement succ(x), where x is a natural number represented in binary. Assume no leading zeros (except 0 itself). • Solution: 1. X0 → X1 (halt) 2. X1 → X#0# 3. X1#Y# → X#0Y# 4. X0#Y# → X1Y (halt) 5. #Y# → 1Y (halt).

23. Post Systems • This model generates a set of strings from axioms (a given set of strings) and inference rules (a given set of productions that map strings to strings by matching). • Execution: Match the left side of a production with an axiom string or a string that has already been constructed. Then use the match and the right side to construct a new string.

24. Example • A Post system to generate the binary representations of natural numbers: • Axioms: 0, 1 • Inference Rules: 1X → 1X0 1X → 1X1. • The system generates the set {0, 1, 10, 11, 100, 101, 110, 111, … }.

25. Quizzes • Quiz. Find a Post system to generate {a}*. • Solution: • Axiom: Λ • Inference rule: X → aX. • Quiz. Find a Post system to generate the set {anbncn| n ∊N}. • One of Many Solutions: • Axioms: Λ, abc • Inference rule: aXbcY→ aaXbbccY.

26. Post system and Turing machine The Post system model has the same power as the Turing machine model in the following sense: A function ƒ : A* → A* is Post-computable if there is a Post system to compute the function as a set of ordered pairs in the form {x#ƒ(x) | x ∊A*}. Post-computable functions coincide with Turing-computable functions.

27. Example/Quiz • Generate the function ƒ : {a}* → {a}* where ƒ(an) = a2n. • Solution: • Axiom: Λ#Λ • Inference rule: X#Y → Xa#Yaa (or the simpler X → aXaa) • This system generates the set {Λ#Λ, a#aa, aa#aaaa, …, an#a2n, … }.

28. Example/Quiz • Generate the function ƒ : N → N defined by ƒ(n) = n2, where n is represented as a string over {a} of length n. • A Solution: • Axiom: Λ#Λ • Inference rule: X#Y → Xa#YXXa. • Proof: • In terms of natural numbers, from the pair n#n2 we must construct (n + 1)#(n + 1)2. • We can write (n + 1)2 = n2 + 2n + 1 = n2 + n + n + 1. • So from n#n2 we must get (n + 1)#(n2 + n + n + 1). • In terms of strings over {a}, we transform X#Y into Xa#YXXa. QED.

29. The End of Chapter 13 Section 2