Theory of Computation

Access Functions for Gödel NumbersLengths of Natural Numbers

Formal Foundations of Program Compilation & Interpretation/Execution:

Gödel Coder & Gödel Decoder

Vladimir Kulyukin

www.vkedco.blogspot.comwww.vkedco.blogspot.com

Outline● Review

– Pairing Functions

– Gödel Numbers

● Access Functions for Gödel Numbers● Lengths of Natural Numbers● Formal Foundations of Program Compilation &

Interpretation/Execution– Gödel Coder

– Gödel Decoder

Review

Review: Pairing Functions

( )

( )( )1221,

01122

1122,

+=+≥−+

−+=•

•

yyx

y

yyx

x

x

x

Review: Equation 8.1 (Ch. 3)

zyx

x, y

z

=,

:equation following the tofor solution

unique a is therenumber, natural a a is If

Review: Splitting Natural Numbers

( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) .22,219;22,219

.55,010;05,010

:Examples

.,

.,

. , and :functions twodefines 8.1Equation

====

====

==

==∈

rrrl

rrll

yyxrzr

xyxlzl

Nzzrzl

Review: l(z) and r(z) are primitive recursive

( ) [ ][ ]( ) [ ][ ] ,min)(

,min)(

Thus, .,,, then , If

yxzxzr

yxzyzl

zyzxyxzNz

zzy

zzx

=∃=

=∃=

≤≤=∈

≤≤

≤≤

Review: Gödel Numbers

( )

[ ] ∏=

=n

ii

ain

n

ppaa

aa

i

11

1

prime.th -i theis where,,..., define We

numbers. of sequence a be ,...,Let

[ ]naa ,...,1is the Gödel number (G-number) of this sequence.

Review: Example

( )

[ ] .5321,3,2

is sequence thisofnumber -G The

.2,3,1 sequence following thehave weSuppose

23123

32

11 ⋅⋅=⋅⋅= ppp

Lemma

[ ]recursive

primitive is ,..., ,each For 1 naan

Proof

3. Ch. 6.2,Corollary

by recursive, primitive isfunction recursive primitive a of

product bounded thebecause recursive, primitive is

recursive. primitive is so recursive primitive is

recursive. primitive is

1∏

=

n

i

ai

ai

y

i

i

i

p

px

p

Note 1 on Theorem 8.2 (Ch. 3)

[ ] [ ] [ ]....11......

,..., because,0,0,,...,,...,

example,For number.-G its changing without sequence the

ofright the toszero' ofnumber any addcan wegeneral,In

02

01111

111

111++ ⋅⋅⋅⋅=⋅⋅⋅⋅=⋅⋅

==

nnan

aan

aan

a

nnn

pppppppp

aaaaaannn

Note 2 on Theorem 8.2 (Ch. 3)

...75325323221

because 1, be to

defined is () sequenceempty theofnumber Godel The

0000000000 =====

Note 2 on Theorem 8.2 (Ch. 3)

[ ] [ ]

[ ] [ ][ ] [ ] .755322,1,018322,1

because,2,1,02,1

example,For

.......

because ,,...,,0,...,

general,In number.-G its changingwithout

sequence theofleft the toszero' addcannot We

21021

12011

11

11

=⋅⋅=≠=⋅=

≠

⋅⋅⋅≠⋅⋅

≠

+nn a

naa

na

nn

ppppp

aaaa

Access Function for Gödel Numbers

Motivation

● Suppose we have a Gödel number n ● By definition, n is a sequence of numbers each of which

is the power of the corresponding prime● We can think of n as an array of numbers and, as with

arrays in other programming languages, we need access to their individual elements

● In our context, we would like to access those individual prime powers

Access Function for Gödel Numbers

[ ]( )

[ ]( )( )( ) .3

;4

;2

then,3,4,2 if example,For

.1 ,function

recursive primitive a define want to We.,...,Let

3

2

1

1

===

=

≤≤==

x

x

x

x

niax

aax

ii

n

Access Function for Gödel Numbers

( ) ( )[ ]

( )( ) . allfor ,00 .2

; allfor ,0 .1

thatNote

|min

0

1

i

xx

xpx

i

ti

xti

==

¬= +

≤

Example 01

[ ]

( ) ( )[ ]( )

( ) ( )[ ]( ) .2 so ,18|3 but 18;|3 ;18|3 ;18|3

and ;3 because ,218| min18

.1 so,18|2 but ;18|2 ;18|2

and 2; because ,118| min18

Then .322,118Let

3210

21

218

2

210

11

118

1

2122

11

=¬

==¬==¬

==¬=

⋅=⋅==

====

+

≤

===

+

≤

t

pp

t

pp

pp

tttt

t

t

ttt

t

t

Example 02

[ ]

( )( ) 3108

2108

Then .274323,2108Let

2

1

3222

11

==

⋅=⋅=⋅== pp

Access Function for G-Numbers: Implementation

// x is a natural number, compute (x)i

long GAccess(long x, long i) {

long p = computeIthPrime(i); // compute i-th prime pi

for t from 0 upto x {

if ( pt does not divide x )

return t – 1;

}

}

Access Function for G-Numbers: Implementation

● The previous slide assumes that we represent G-Numbers as integers

● This is a sensible choice, but only for small numbers● If we want to be generic, we can represent G-numbers as

sequences of prime powers, e.g. arrays, lists, etc. ● In this case, we can represent much larger values without

actually having to compute them● Access function becomes much easier, because we do not have

to compute primes or their powers

Lengths of Natural Numbers

Length of a Natural Number

● Since Gödel numbering and access functions give us techniques to split any natural number into its constituents

● Next question: how many constituents does a given natural number have? How long is the number?

● Example: 18=[1, 2]. So the length of 18 is 2

Length of a Natural Number: Definition

( ) ( ) ( ) ( )( )( )

( ) ( ) .010 ,definitionBy

number.-G ain power prime

zero-nonlast theofindex theis s,other wordIn

.0&0min

==

=∨≤∀≠= ≤≤

LtLt

i

xijjxxLt jxixi

Example of Lt()

( )[ ]

( )( ) ( ) ( )

( ) .320

.0...

.1

.1,0,253220

.?20

2054

3

102

=====

==⋅⋅==

=

Lt

xxx

x

x

Lt

Theorem 8.3 (Ch. 3)

[ ]( )

( ) ( )[ ] ( ) nxLtxxx

niaaa

x

n

iin

≤= ≤≤

=

>

if ,..., .2

otherwise 0

1 if ,..., .1

Then .1Let

1

1

Proof

( )[ ]( )

( ) ( ) ( )[ ] .,..., and 0 then ,

and 1 if Thus, .power prime zero-nonlast theis iff

because, divides angreater th prime nobut

, then ,0 and 1 if that,observe we2,part For

number.-G theof definition thefrom follows 1Part

1

1

xxxnnxLt

x a

n,...,aaLtx p

|xpnxLt x

n

n

nn

n

=>≤>

=>=>

Formal Foundations of

Program Compilation & Execution/Interpretation

The Big Picture

GÖDEL Coder

GÖDEL Decoder

L Program Number Universal Program

Number

???

GÖDEL Coder = CompilerGÖDEL Decoder = Reverse CompilerUniversal Program = Operating System/VM/Interpreter

The Software/Hardware Duality Principle

1) There is no point in developing a formalism unless it can be executed on an existing hardware device

2) There is no point in developing a hardware device unless there is (or there will be) a formalism for it

Formalism Hardware DeviceCo-Dependency

Gödel Coder & Decoder

● We will demonstrate that compilation (coding) and reverse compilation (decoding) are both primitive recursive

● In other words: Given an L program P, there is a primitive

recursive function that maps P to its number #(P)

Given #(P), there is a primitive recursive function that maps #(P) to its source code P

Gödel Coder

Gödel Coder

● Sub-task 1: Mapping variables and labels to numbers

● Sub-task 2: Mapping instructions to numbers

● Sub-task 3: Mapping programs to numbers

Sub-task 1: Mapping Variables & Labels

● We order the variables as follows: Y X1 Z1 X2 Z2 X3 Z3 …

● We order the labels as follows: A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 …

● Let #(V) be the position of a variable in the variable ordering

● Let #(L) be the position of a label in the label ordering ● Note that the chosen orderings are arbitrary in the

sense that other orderings are possible

Examples of #(V) & #(L)

● We order the variables as follows: Y X1 Z1 X2 Z2 X3 Z3 …

● We order the labels as follows: A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 …

● #(Y)=1, #(X1)=2, #(X3)=6, #(Z2)=5

● #(B1)=2, #(A2)=6, #(C2)=8

Sub-task 2: Mapping Instructions to Numbers

● There are four primitive instruction types in L: addition, subtraction, self-assignment (no-op), conditional dispatch

● There are three pieces of information we have to known in order to map an instruction to a number:

label or no label variable used in the instruction instruction type

Sub-Task 2: Assigning Numbers to Instruction Types

( ) 2# n typeinstructio be willThis // 0

2 n typeinstructio be willThis // 1

1 n typeinstructio be willThis // 1

0 n typeinstructio be willThis //

+≠←

+←←

L GOTO LIF V

-VV

VV

VV

Sub-Task 2: Mapping Instructions to Numbers

( )

ninstructio in the used variable theofnumber theis .3

number n typeinstructio theis 2.

number label theis 1.

where,,,#Then n.instructio primitive a be Let

c

b

a

cbaILI =

Sub-Task 2: Mapping Instructions to Numbers

( )

( )( )

( )( ) .22

case in this that Note .2then ,0! is if 7.

;2 then ,1 is if 6.

;1 then ,1 is if 5.

;0 then , is if .4

;1 then ,in used is variablea if .3

; then , label some with labeled is if 2.

;0 then unlabeled, is if 1.

where,,,#Then n.instructio primitive a beLet

>+=+==

=←=+←

=←−=

==

=

L#b

L#b GOTO LIF VI

b-VVI

bVVI

bVV I

V#cIV

L#aLI

aI

cbaILI

Instruction Coding Example 01

Let I be X1 X1 + 1← . What is #(I)?

Instruction Coding Example 01

( )

( ) ( )

( ) ( )( )[ ]( )

[ ]( ) .101111113221

111112222

111,1221,1,0#

Thus, 4.

;21 because ,111 3.

addition; is because ,1 .2

unlabeled; is because ,0 .1

where,,,# so ;111 is

10

0

=−=−+−⋅⋅⋅=−+−+⋅⋅

=−+==

==−===

=+←

I

X#X#c

Ib

Ia

cbaIXXI

Instruction Coding Example 02

Let I be [A1] X1 X1+1← . What is #(I)?

Instruction Coding Example 02

[ ]( ) ( ) ( )

( )( )[ ]( ) .211112111112222

111,1221,1,1

11#,1,1##

,111 1 is Since

111

1

=−⋅=−+−+⋅

=−+⋅=

=−=

+←

XAI

XXAI

Sub-task 3: Mapping Programs to Numbers

● We know how to do sub-task 1 of mapping labels and variables into numbers through label and variable orderings

● We know how to do sub-task 2 of mapping individual primitive L instructions into numbers

● What is left? We need to do sub-task 3, i.e., mapping finite sequences of primitive L instructions into numbers

Sub-task 3: Definition of #(P)

( ) ( ) ( ) ( )[ ] .1#,...,#,##

thenns,instructio primitive are ,...,, and

0 where,,...,, of consists that program a be Let

21

21

21

−=

≥

k

k

k

IIIP

LIII

kIIIP

Example of #(P)

[ ]1 GOTO 0 1 IF

111 1

program? thisofnumber theisWhat

AX

XXA

≠+←

Example of #(P)

[ ]

( ) ( ) ( )[ ] 1#,##

Thus,

1 GOTO 0 1 IF is

111 1 is

21

2

1

−=

≠+←

IIP

AXI

XXAI

Example of #(P)

( ) [ ]( )

( )

( )( ) 21111,1221,1,1 So,

11211# .3

1 2.

11# .1

where

,,,111 1## example, previous aBy

1

1

=−+=

=−=−==

==

=+←=

Xc

b

Aa

cbaXXAI

Example of #(P)

( ) ( )

( )( )

( ) ( )

( )

( ) .46147112321

111112222

111,3221,3,0# Thus,

.111# .3

;32121# .2

;0 .1

where,,,1 GOTO 0! 1 IF##

30

02

2

=−=−+⋅

=−

+

−+⋅

=−+==

=−==+=+=

=

===

I

Xc

Ab

a

cbaAXI

Example of #(P)

( )( )

( ) ( ) ( )[ ] [ ]32969420735054819781858692840132

146,211#,##

46#

21#

4621

21

2

1

=−⋅

=−=−=

==

IIP

I

I

Back to The Big Picture

GÖDEL Coder

GÖDEL Decoder

L Program Number Universal Program

Number

???

The red arrows show the implemented tasks (program compilation): we now know how to map L programs to numbers.

Gödel Decoder

Gödel Decoder

● Sub-Task 1: Obtain the source code number● Sub-Task 2: Extract from the source code

number the numbers for individual instructions● Sub-Task 3: Extract the source from each

individual instruction number

GÖDEL Decoder: Sub-Task 1

( ) ( ) ( ) ( )[ ]

( ) ( ) ( )[ ] ( ) .1##,...,#,#

isnumber code source theThus,

.1#,...,#,##

21

21

+=

−=

PIII

IIIP

k

k

Side Note: A Prime Factorization Algorithm

// n is a natural number > 0, i is the prime counter, initially// set to 1. We need this algorithm to solve sub-task 2 of // Gödel DecodingFactors(n, i)

If n is prime then n is the factorization;If n is composite thenIf pi divides n then

add pi to the list of factors;Factors((n / pi), i);

If pi does not divide n ThenFactors(n, i+1)

Prime Factorization Algorithm: Example

( )

( )

2] 2, [2,return

2] [2,return

[2]return so prime a is 2

1) factors(2, to2 add so

4|2

1) factors(4, to2 add so

8|2

1)factors(8,

1

1

=

=

p

p

GÖDEL Decoder: Sub-Task 2

( ) ( ) ( ) ( )[ ]( ) ( ) ( )[ ]

( )

number-G the

i.e., powers, prime oflist obtain the factors, prime oflist theFrom .3

factors prime oflist obtain the to,1,1nfactorsRun 2.

#,...,#,#1 isnumber code source theSo,

1#,...,#,# be Let 1.

21

21

+=+

−=

k

k

IIIn

IIIP#n

Examples of Source Code Extraction

lyrespective 1, and 1,0,0, into compile that nsinstructiofour of consists P .3

1475321] 0, 0, [1,number -G the toconverted is [2,7] 2.

[2,7] returns ,1)factors(14 then 13, into compiles P If .1

3number into compilesn that instructio one of consists code source sP' .5

82[3]number -G the toconverted is ]2,2,2[ 4.

2] 2, [2, returns 1)factors(8, 3.

8 is Pfor number code source The 2.

7number into compiled is P program some Suppose 1.

1001

3

=⋅⋅⋅=

==

GÖDEL Decoder: Sub-Task 3

( )( )

( )( )( )

( )( ) ordering. variablein the 1 isposition

whose variableis in used variable theThus, . 3.

ordering. label in the 2 isposition

whoselabel is L whereL GOTO 0 ! V IF typeof is 2.4.

;2 if 1 - V V typeof is 2.3.

;1 if 1 V V typeof is 2.2.

;0 if V V typeof is 2.1.

Then . 2.

ordering. label in the

number label has Otherwise, unlabeled. is ,0 If . 1.

.,,#Let

number.-G in then instructioeach of code source eextract th

toneed we2,task -sub fromnumber -G thehave that weNow

c

Iqrrc

b-

I

b I

b I

b I

qrlb

ql

IIaqla

cbaqI

+=

==←=+←=←

=

==

==

#(P) P: GÖDEL Decoding Algorithm

( ) ( ) ( ) ( )[ ]( ) ( ) ( )[ ]

( )

recursive primitive are operations All 5.

3 stepin obtained

number-G theofelement each from code source eExtract th 4.

powers prime of sequence the

get i.e. factors, prime oflist thefromnumber -G Obtain the .3

1,1nFactorsRun 2.

#,...,#,#1 isnumber code source theSo,

;1#,...,#,# be Let 1.

21

21

+=+

−=

k

k

IIIn

IIIP#n

GÖDEL Decoding Algorithm: Example1. Let #(P) = 199.2. The source code number = 199+1=200.3. The prime factorization of 200 is 233052. Thus, the G-number = [3,0,2].4. 3 = <2, <0, 0>>. So the label is B1, the instruction variable is Y and the type of instruction is self-assignment. Thus, first instruction is [B1] Y Y.←5. 0 = <0, <0, 0>>. Thus, second instruction is Y Y.←6. 2 = <0, <1, 0>>. Thus, third instruction is Y Y + 1.←

GÖDEL Decoding Algorithm: Example

Let #(P) = 199. The source code number = 199+1=200. So, the source code of P is

[B1] Y ← Y

Y ← Y

Y ← Y + 1

Back To The Big Picture

GÖDEL Coder

GÖDEL Decoder

L Program Number Universal Program

Number

???

GÖDEL Coder = CompilerGÖDEL Decoder = Reverse CompilerUniversal Program = Operating System/VM/Interpreter

Reading Suggestions

● Chapters 3 & 4, Computability, Complexity, and Languages, 2nd Edition, by Davis, Weyuker, Sigal

● http://en.wikipedia.org/wiki/Kurt_Gödel