::ics 804:: theory of computation - ibrahim otieno - [email protected] +254-0722-429297

::ICS 804::Theory of

Computation

- Ibrahim Otieno -

[email protected]

+254-0722-429297

SCI/ICT Building Rm. G15

mailto:[email protected]

Course OutlineCourse Outline Mathematical Preliminaries Turing Machines Recursion Theory Markov Algorithms Register Machines Regular Languages and finite-state

automata Aspects of Computability

Last Week: Additional Varieties of Last Week: Additional Varieties of Turing MachinesTuring Machines

Turing Machines with One-Way-Infinite tape

Turing Machines that accept by terminal state

Multitape Turing Machines Encoding of Turing Machines Universal Turing Machines Nondeterministic Turing Machines Turing-Computability Turing Machines and Artificial Intelligence Turing Machines and Cognitive Science

Course OutlineCourse Outline

Mathematical Preliminaries Turing Machines

◦ Additional Varieties of Turing Machines Recursion Theory Markov Algorithms Register Machines Regular Languages and finite-state

automata Aspects of Computability

Recursion TheoryRecursion Theory

The primitive recursive functions The partial recursive functions The class of partial recursive

functions = class of Turing-computable functions

The Primitive Recursive The Primitive Recursive FunctionsFunctions

Initial FunctionsInitial Functions

Recall function forming operations:◦ Unary inversion operator( f-1) – must be

injective◦ Binary composition operator – h (n) = f°g

(n)

We look at 2 other function forming operators:

◦ Comp – generalization of binary composition

◦ Pr – Primitive Recursion

Initial FunctionsInitial Functions The class of initial functions contains the

following functions and no others:◦ The unary successor function succ

◦ The k-ary constant-0 function Ck0 for k 0

k: number of arguments 0: indicates the value

◦ The projection function pkj for each k 1

and 1 j kk: number of argumentsj: argument unto which the function projects

Function compositionFunction composition

Given function◦ f: Nm N with m 1 ◦ functions g1, g2, … gm each of which is Nk N

with k 0. Then applying function composition to the m+1

functions f,g1,g2,…,gm yields a new function h: Nk N defined by

h(n1,…,nk) = f(g1(n1,…,nk), g2(n1,…,nk),…,gm(n1,…,nk))

Comp[f,g1,g2,…,gm]: composition operator for function h

Function compositionFunction composition

Result is another k-ary function Comp[f,g1,g2,…,gm](n1,…,nk)

= Comp[f,g1,g2,…,gm] (n)

= f(g1(n),g2(n),…,gm(n))

IllustratiIllustrationon

The class is considered as a single, multipart computer of a k-ary number-theoretic function

h = Comp[f,gh = Comp[f,g11,g,g22,…,g,…,gmm]]

ExampleExample

j(n,m) = n3+m2+7Comp[j,p3

3,p32](57849,6,8)

= j(p33(57849,6,8),p3

2(57849,6,8))

= j(8,6)= 83 + 62 + 7= 555

CompositionComposition

h = f g= Comp[f,g] with m = k = 1

SummaryGiven function f: Nm N with m 1 and functions g1, g2, … gm each of which is Nk N with k 0. Then applying function composition to the m+1 functions f, g1,g2,…,gm yields a new function h: Nk N defined by


Comp[f,g1,g2,…,gm]: composition operator for function h

ExerciseExercise

j(n,m) = n3+m2+7Calculate

Comp[p21,p2

2,j](8,6) = ???

Comp[p32,p2

1,j,p22](8,6) = ???

ExerciseExercise

j(n,m) = n3+m2+7Calculate

Comp[p21,p2

2,j](8,6) = p21(6,555) = 6

Comp[p32,p2

1,j,p22](8,6) = p3

2(8,555,6) = 555

ExerciseExercise

If j(n1,n2,n3) = n1.n2+n3

(a)What is j(6,0,7) (b)What is the value of m and k in case

of h = Comp[plus,j,p33]


(c) How many arguments does h have (d) What is f, g1 and g2? (e) What is h(6,0,7)

h(6,1,14)

ExerciseExerciseIf j(n1,n2,n3) = n1.n2+n3

(a)What is j(6,0,7) 7(b)What is the value of m and k in case of h =

Comp[plus,j,p33] m=2 k=3


(c) How many arguments does h have 3(d) What is f, g1 and g2? f=plus g1=j

g2=p33

(e) What is h(6,0,7) 14 h(6,1,14) 34

Primitive RecursionPrimitive RecursionPlus function (recap)

(i) plus(n,0) = n(ii) plus(n,m+1) = succ(plus(n,m))

plus(5,2) = plus(5,1+1)= succ(plus(5,1))= succ(plus(5,0+1))= succ(succ(plus(5,0))))= succ(succ(5))= succ(6)= 7

Primitive RecursionPrimitive Recursion Definition of plus is recursive: applying plus to

decreasing values (5,1) and (5,0) Function forming operation to construct plus from

the function succ

Given Function f: NkN with k 0 and function g: Nk+2N, we can form a new function h: Nk+1N defined by

(i) h(n1,…,nk,0) = f(n1,…,nk)

(ii) h(n1,…,nk,m+1) = g(n1,…,nk,m,h(n1,…,nk,m))

We write Pr[f,g] for the function h formed by applying primitive recursion to given functions f and g

Primitive RecursionPrimitive Recursion

Given Function f: NkN with k 0 and function g: Nk+2N, we can form a new function h: Nk+1N defined by

(i) h(n1,…,nk,0) = f(n1,…,nk)

(ii) h(n1,…,nk,m+1) = g(n1,…,nk,m,h(n1,…,nk,m)) We write Pr[f,g] for the function h formed by applying

primitive recursion to given functions f and g

or:

(i) h(n,0) = f(n)

(ii) h(n,m+1) = g(n,m,h(n,m))

illustrationillustration

Classroom with two students◦ StudentA is good at computing (k+2)-ary function g ◦ StudentB is good at computing k-ary function f

◦ Instructor provides (k+1)-tuple <n1,…,nk,m>

(i) h(n,0) = f(n)(ii) h(n,m+1) = g(n,m,h(n,m))


Classroom with two students◦ Instructor provides (k+1)-tuple <n1,…,nk,m>

◦ If m=0, StudentB reports the value of f right away, because that means clause (i) applies



Classroom with two students◦ Instructor provides (k+1)-tuple <n1,…,nk,m>

◦ If m≠0, StudentB must still first compute the value of f for k-tuple <n1,…,nk> before Student A can start

◦ This is passed on to StudentA who computes g(n1,…,nk,0,h(n1,…,nk,0))



◦ This is passed on to StudentA who computes g(n1,…,nk,0,h(n1,…,nk,0))

= h(n1,…,nk,1)◦ StudentA then computes the value of g(n1,…,nk,1,h(n1,

…,nk,1))◦ Until the 2nd argument of his function equals the one on the

blackboard


DefinitionDefinition The class of primitive recursive

functions: An initial function is a primitive recursive

function If g1,…,gm are each k-ary primitive recursive

functions for some k 0 and f is an m-ary primitive recursive function for some m 1 then Comp[f,g1,…,gm] is a k-ary primitive recursive function

If f is a k-ary primitive recursive function and g is a (k+2)-ary primitive function for some k 0, then Pr[f,g] is a (k+1)-ary primitive recursive function

No other number-theoretic functions are primitive recursive except those listed above


(i) plus(n,0) = n(ii) plus(n,m+1) = succ(plus(n,m))

(i) plus(n,0) = p11(n)

(ii) plus(n,m+1) = succ(p33(n,m,plus(n,m)))

= Comp[succ,p33](n,m,plus(n,m))

So plus: Pr[p11,Comp[succ,p3

3]]



(i) plus(n,0) = p11(n)



Classroom scenario:+ teacher writes k+1 tuple on the blackboard, in this case a

2-tuple <n,m>, e.g. <5,2>+ m≠0 so the calculation will include StudentA+ StudentB calculates the value for 1-tuple <n>p1

1(5) = 5+ and passes the answer on to StudentA


illustrationillustration(i) plus(n,0) = p1

1(n)



Classroom scenario:+ StudentA then calculates the g-function, in this case

Comp[succ,p33](5,1,plus(5,1)) ……(i)

Expanding plus(5,1)

= succ(p33(5,0,plus(5,0))) = succ(p3

3(5,0,5)) plus(5,0) = p1

1(5)= succ(5)= 6 Replacing in (i)

Comp[succ,p33](5,1,6)

= succ(6)= 7

Other primitive recursive Other primitive recursive functionsfunctions

Comment: every primitive recursive function is total

Multiplication and Exponentiation are primitive recursive functions

Monus(n,m) = n – m if n m0 otherwise

Sg(n) = 1 if n = 00 otherwise

Partial Recursive FunctionsPartial Recursive Functions

Partial recursive Partial recursive functionsfunctions Compared to Turing-computable functions

New definition of a class: partial recursive functions◦ The smallest class of number-theoretic

functions containing certain basic functions and closing under certain function-forming operations (cf. class of primitive recursive functions)

Primitive recursive functions constitute an infinite class, but are effectively enumerable

Partial recursive functionsPartial recursive functionsTheorem: there exists a computable function that is total, but not

primitive recursiveProofLet f0,f1,f2,… be the enumeration of unary primitive recursive functions.

Consider f*f*(n) = fn(n) + 1◦ f* is unary and total◦ If f* is primitive recursive, it occurs in the enumeration (above)◦ Assume f* is fk for some k◦ Ask for the value of f*(k)

f*(k) = fk(k) Substituting n for k (above) this means f*(k) = fk(k) + 1 as well (Proof by) Contradiction!!!!

◦ f* is computable: for any n, find the nth function in the enumerable list of primitive recursive functions, compute the value of fn and add 1

◦ f* is Turing-computable

Number-theoretic Predicates Number-theoretic Predicates (recap)(recap)

Properties of objects, relations, holding between pairs of objects, relations, holding among triples of objects are expressed using predicates. If the objects are members of N, then one speaks of number-theoretic predicates

e.g. prime(n) is a unary number-theoretic predicate that is satisfied by 2 and 3, but not by 4 or 6

Least number operatorLeast number operator : least number operator m[C(n,m)] the least natural number m

such that predicate C(n,m) holds m[C(n,m)] = 0: the least natural number

such that f(n,m) is equal to 0

MinimizationMinimization New function-forming operation, Mn

suppose that f:Nk+1N with k 0 is given. We write Mn[f] for the function g:NkN defined as:

g(n) = m[Cf(n,m) = 0 and such that

for all j < m, f(n,j) is defined and ≠ 0

MinimizationMinimization

e.g. f(3,0) = 7f(3,1) = 4f(3,2) = 4f(3,3) = 2f(3,4) = 9f(3,5) = 0f(3,6) = 4…

Then g(3) = Mn[f](3) = m[f(3,m) and such that for all j < m, f(3,j) is defined and ≠0] = 5


e.g. f(3,0) = 7f(3,1) is undefinedf(3,2) = 4f(3,3) = 2f(3,4) = 9f(3,5) = 0f(3,6) = 4…

m[f(3,m)= 0] = 5 but g(3) = Mn[f](3) is undefined


e.g. f(3,0) = 7f(3,1) = 4f(3,2) = 4f(3,3) = 2f(3,4) = 9f(3,5) = 5f(3,6) = 4…

g(3) = Mn[f](3) is undefined

ExerciseExercise

f(0,0) = 6 f(1,0) = 8 f(2,0) = 3 f(3,0) = 9 f(4,0) = 6f(0,1) = 4 f(1,1) = 8 f(2,1) = 1 f(3,1) = 6 f(4,1) = 6f(0,2) = 6 f(1,2) = 6 f(2,2) = 6 f(3,2) = 4 f(4,2) = 6f(0,3) = 0 f(1,3) = 6 f(2,3) is undefined f(3,3) = 6 f(4,3) = 6f(0,4) = 6 f(1,4) = 5 f(2,4) = 6 f(3,4) = 0 f(4,4) = 6f(0,5) = 8 f(1,5) = 0 f(2,5) = 0 f(3,5) = 6 f(4,5) = 6

with f(4,m) = 6

for all m

Calculate g(0) = Mn[f](0)g(1) = Mn[f](1) g(2) = Mn[f](2)g(3) = Mn[f](3) g(4) = Mn[f](4)

ExerciseExercise

f(0,0) = 6 f(1,0) = 8 f(2,0) = 3 f(3,0) = 9 f(4,0) = 6f(0,1) = 4 f(1,1) = 8 f(2,1) = 1 f(3,1) = 6 f(4,1) = 6f(0,2) = 6 f(1,2) = 6 f(2,2) = 6 f(3,2) = 4 f(4,2) = 6f(0,3) = 0 f(1,3) = 6 f(2,3) is undefined f(3,3) = 6 f(4,3) = 6f(0,4) = 6 f(1,4) = 5 f(2,4) = 6 f(3,4) = 0 f(4,4) = 6f(0,5) = 8 f(1,5) = 0 f(2,5) = 0 f(3,5) = 6 f(4,5) = 6

with f(4,m) = 6

for all m

Calculate g(0) = Mn[f](0) = 3g(1) = Mn[f](1) = 5g(2) = Mn[f](2) is undefinedg(3) = Mn[f](3) = 4g(4) = Mn[f](4) is undefined

Even if f(2,3) were defined,

Mn[f](n) = g(n) would still

be partial

PartialPartial

The exercise shows that applying minimization operator Mn to a total function may result in a partial function

Partial recursive functions Partial recursive functions defineddefined

The class of partial recursive functions is the smallest class containing all initial functions and closed under the operations Comp, Pr, Mn

So: every primitive recursive function is also a partial recursive function

A total partial recursive function = recursive function

World of number-theoretic World of number-theoretic functionsfunctions

The class of partial recursive The class of partial recursive functions = class of Turing-functions = class of Turing-

computable functionscomputable functions

TheoremsTheorems A number-theoretic function is partial

recursive if and only if it is Turing Computable

If number-theoretic function h is partial recursive, then h is Turing-computable

Forward Proof (partial recursive Turing Computable):◦ Initial functions◦ Comp functions◦ Pr functions◦ Mn functions

Proof: initial functionsProof: initial functions

Successor function

0 1 21 : L B : 1

Figure 1.5.1

Proof: initial functionsProof: initial functions

Constant-k function ◦ Erases k groups of 1s off to the right,

writes a single 1 and halts◦ e.g. C0

0

B:10 1

Proof: initial functionsProof: initial functions Projection function. Erase all

sequences of 1s on the tape, except the one that is projected unto, e.g. p3

2

B:R0 1

1:BB:R

2 1:R

B:R3

1:BB:R

4

5B:L

B:L

61:L

1:L

7

B:R

Proof: CompProof: Comph = Comp[f,g1,…gm]f is an m-ary partial recursive

functiong1,…,gm are k-ary partial recursive

functionsWe have Turing Machines Mf,Mg,…

Mm

Mcomp will simulate the composition of Turing Machines Mf,Mg,…Mm

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B BInput tape

Argument n1 Argument n2Argument nk

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B BWorktape-1


1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B BWorktape-m


B B …Output tape

Copy k arguments from input tape to m worktapes

…

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B BInput tape


1 1 1 … 1 B …Worktape-1

g1(n1,n2,…,nk)

1 1 1 … 1 B ...Worktape-m

B B …Output tape

Mcomp simulates Mg,…Mm

on m worktapes

gm(n1,n2,…,nk)

…

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B BInput tape


Worktape-1

g1(n1,n2,…,nk)

1 1 1 … 1 B ...Worktape-m

B B …Output tape

Mcomp copies the contents of m worktapes

unto worktape-1

gm(n1,n2,…,nk)

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B B

g2(n1,n2,…,nk) gm(n1,n2,…,nk)

…

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B BInput tape


Worktape-1

g1(n1,n2,…,nk)

Output tape

Mcomp simulate Mf on worktape-1

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B B

g2(n1,n2,…,nk) gm(n1,n2,…,nk)

B B …

f(g1(n1,n2,…,nk), g2(n1,n2,…,nk),…, gm(n1,n2,…,nk))

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B BInput tape


Worktape-1

Output tape

Mcomp copies the contents to the output tape

1 1 1 … 1 B …

f(g1(n1,n2,…,nk), g2(n1,n2,…,nk),…, gm(n1,n2,…,nk))

1 1 1 … 1 B …

Proof: Pr[f,g]Proof: Pr[f,g]f is a k-ary partial recursive

functiong is a k+2-ary partial recursive

functionBoth are Turing ComputableWe have Mf and Mg that compute

f and gMPr: 5 tape Turing Machine that

will simulate primitive recursion

Proof: Pr[f,g]Proof: Pr[f,g]

In case (i) (i.e. n1,…,nk,0), <n,0> are represented on input tape. MPr copies arguments n to worktape-1 and simulates the operation of Mf

on worktape-1. When it halts in a value-representing configuration, the value is copied onto the output tape

Proof: Pr[f,g]Proof: Pr[f,g]In case (ii) (i.e. n1,…,nk,m+1), <n,m+1> are

represented on the input tape. MPr copies arguments n to worktape-1 and simulates the operation of Mf on worktape-1. When it halts in a value-representing configuration, MPr writes a single 1 on worktape-3 (indicating iteration0 has been completed). Then several items are copied onto worktape-2:

◦arguments n from the input tape◦The contents of worktape-3◦Contents of worktape-1, currently

representing h(n0) = f(n)

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B 1 B 1 … 1 B …

Argument n1 Argument n2 Argument nk Argument m+1

1 1 1 … 1 B …

f(n1,…,nk) = h(n1,…,nk,0)

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B 1 B 1 … 1 B …

n1 n2 nk

f(n1,…,nk) = h(n1,…,nk,0)0

Copied from input tape Copied from worktape3

Copied from worktape1

1 B …

B B …

input

work1

work2

work3

output

After 1 iteration

Proof: Pr[f,g]Proof: Pr[f,g]

Mpr then simulates operation of Mg on worktape-2, obtaining a representation of h(n,1) on that tape. This value is copied onto worktape-1 and the representation on worktape-3 is incremented by 1. Then the contents of worktape-2 are overwritten:◦arguments n from the input tape◦The contents of worktape-3◦Contents of worktape-1, currently

representing h(n,1) = f(g(n,0,h(n,0))

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B 1 B 1 … 1 B …

Argument n1 Argument n2 Argument nk Argument m+1

1 1 1 … 1 B …

h(n1,…,nk,1)

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B 1 1 B 1 … 1 B …

n1 n2 nk h(n1,…,nk,1)1

Copied from input tape Copied from worktape3

Copied from worktape1

1 1 B …

B B …

input

work1

work2

work3

output

After 2 iterations

Proof: Pr[f,g]Proof: Pr[f,g]Mpr then simulates operation of Mg

on worktape-2, obtaining a representation of h(n,2) on that tape.

◦This process is iterated until the representation on worktape-3 exceeds the rightmost argument on the input tape. At that point, the representation on worktape-2 represents h(n,m+1)

Pr[f,g] for plus(n,m)Pr[f,g] for plus(n,m)

The plus function:(i) plus(n,0) = n(ii) plus(n,m+1) = succ(plus(n,m))

The schema for Pr[f,g]:(i) h(n,0) = f(n)(ii) h(n,m+1) = g(n,m,h(n,m))

Applied to plus (see Slide 28):

(i) plus(n,0) = f(n) = p11(n) = n

(ii) plus(n,m+1) = g(n,m,h(n,m)) = succ(p33(n,m,plus(n,m)))

Case-study: plus(3,2)


(i) plus(n,0) = f(n) = p11(n) = n

(ii) plus(n,m+1) = g(n,m,h(n,m)) = succ(p33(n,m,plus(n,m)))

Case-study: plus(3,2)The 2nd argument is not 0, so:

◦ MPr copies argument n to worktape-1 and simulates the operation of Mf on worktape-1 (f(n)=n=plus(n,0))

◦ MPr writes a single 1 on worktape-3 (indicating 1st iteration has been completed).

◦ Then several items are copied onto worktape-2: argument n from the input tape The contents of worktape-3 Contents of worktape-1, currently representing plus(3,0) = 3

1 1 1 1 B 1 1 1 1 B

Argument n Argument m+1

1 1 1 1 B

f(n) = n

1 1 1 1 B 1 B 1 1 1 1 B

n f(n)=n0

1 B …

B B …

input

work1

work2

work3

output

After 1 iteration


Mpr then simulates operation of Mg on worktape-2◦ Worktape-2 contains <3,0,3>, the 3 arguments for function g

1st argument: n (=3) 2nd argument: m (=0) 3rd argument: plus(3,0) =3

◦ Function g is defined as succ(p33(n,m,plus(n,m)))

◦ g(3,0,3) = succ(p33(3,0,plus(3,0))) = succ(3) = 4

Worktape-2 now contains the representation of plus(3,1)◦ This value is copied onto worktape-1 ◦ The representation on worktape-3 is incremented by 1◦ Then the contents of worktape-2 are overwritten:

arguments n from the input tape The contents of worktape-3 Contents of worktape-1, currently representing plus(3,1)

1 1 1 1 B 1 1 1 1 B


1 1 1 1 1 B

g(3,0,3) = succ(3) = 4

1 1 1 1 B 1 1 B 1 1 1 1 1 B

n g(3,0,3) = succ(3) = 41

1 1 B …

B B …

input

work1

work2

work3

output

After 2 iterations


Mpr then simulates operation of Mg on worktape-2◦ Worktape-2 contains <3,1,4>◦ g(3,1,4) = succ(p3

3(3,1,plus(3,1))) = succ(4) = 5



1 1 1 1 B 1 1 1 1 B


1 1 1 1 1 1 B

g(3,1,4) = succ(4) = 5

1 1 1 1 B 1 1 1 B 1 1 1 1 1 1 B

n 2

1 1 1 B …

B B …

input

work1

work2

work3

output

After 3 iterations

g(3,1,4) = succ(4) = 5



3(3,2,plus(3,2))) = succ(5) = 6



1 1 1 1 B 1 1 1 1 B


1 1 1 1 1 1 1 B

g(3,2,5) = succ(5) = 6

1 1 1 1 B 1 1 1 1 B 1 1 1 1 1 1 1 B

n 3

1 1 1 1 B …

B B …

input

work1

work2

work3

output

After 4 iterations

g(3,2,5) = succ(5) = 6



3(3,3,plus(3,3))) = succ(6) = 7



1 1 1 1 B 1 1 1 1 B


1 1 1 1 1 1 1 1 B

g(3,3,6) = succ(6) = 7

1 1 1 1 B 1 1 1 1 1 B 1 1 1 1 1 1 1 1 B

n 4

1 1 1 1 1 B …

B B …

input

work1

work2

work3

output

After 5 iterations

g(3,3,6) = succ(6) = 7

Pr[f,g] for plus(n,m)Pr[f,g] for plus(n,m)Now the representation on worktape-3

exceeds the rightmost argument on the input tape. At that point, the representation on worktape-2 represents plus(3,3+1) and it can be copied to the output tape

Proof: Mn[f]Proof: Mn[f]Suppose Mn[f] where f is a (k+1)-ary

partial recursive function. Let Mf be a single-tape Turing Machine that computes f.

MMn will compute h = Mn[f]◦Four-tape Turing machine◦Arguments n appear on input tape◦MMn will write a single 1 on worktape-1

(=0)◦Copies arguments n and the single 1

from worktape-1 onto worktape-2

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B BInput tape


1 B …Worktape-1

Worktape-2

B B …Output tape

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B 1 B …

n1 n2nk

From worktape-1

MMn simulates Mf on worktape-2

Proof: Mn[f]Proof: Mn[f]

After execution of Mf on worktape-2, MMn checks whether the value representing configuration equals to 0 (i.e. scanning a single 1 on an otherwise blank tape).

If this is the case:◦ The contents of worktape-1 will represent the least

number operator◦ The contents of worktape-1 are then copied unto

the output tape

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B BInput tape


1 B …Worktape-1

Worktape-2

B B …Output tape

1 1 B …

f(n,0)

In this example f(n,0) = 1, so 0 is not the least number operator

Proof: Mn[f]Proof: Mn[f]

If that is not the case MMn will◦ add a 1 to worktape-1 ◦ Copy the k arguments of the input tape unto

worktape-2◦ Copy the contents of worktape-1 unto worktape-2

◦ Simulate Mf on worktape-2

This process is repeated until worktape-2’s value representing configuration equals to 0.

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B BInput tape


1 1 B …Worktape-1

Worktape-2

B B …Output tape

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B 1 1 B …

n1 n2nk

From worktape-1

MMn simulates Mf on worktape-2

1 1 1 … 1 B 1 1 1 … 1 B … 1 1 1 … 1 B BInput tape


Worktape-1

Worktape-2

Output tape

1 B …

f(n,1)

In this example f(n,1) = 0, so 1 is the least number operator and contents of worktape-1 can be copied to output tape

1 1 B …

1 1 B …

::ics 804:: theory of computation - ibrahim otieno - [email protected] +254-0722-429297

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