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

Example/Quiz

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.

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) Λ q(x), ¬ p(x), and p(x) V q(x), are partial recursive functions.

• Solutions: p(x) Λ q(x) = p(x)q(x)¬ p(x) = monus(1, p(x))p(x) V q(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@ → @a5. 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# → @05. 1@ → @06. 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.

• ExecutionGiven 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#X2. bX → @b#X3. X#aY → Xaa#Y4. X#bY → Xb#Y5. @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#aY4. 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