Recursive Sets and Relations. Computability and Logic

Transcription

1 Rersive Sets and Relations Comptability and Logi

2 The Plan Eventally I will show that any Tring-omptable * fntion is a rersive fntion thereby losing the loop : All Tring-omptable * fntions are rersive All rersive fntions are Abas-omptable * (already shown) All Abas-omptable * are Tring-omptable * (already shown) Ths we will have shown that these three sets are eatly the same providing evidene in favor of the Chrh-Tring Thesis. OK bt to show that any Tring-omptable * fntion is a rersive fntion I will need a whole lot more mahinery: I need to prove a bnh more fntions to be rersive. I will define rersive sets and relations whih will be a great help in showing ertain fntions to be rersive and vie versa

3 Rersive Sets The harateristi fntion S of a set S N is defined as follows: S () = 1 if S S () = 0 if S A set S is a rersive set iff its harateristi fntion S is a rersive fntion Eamples of rersive sets The empty set ( S = z) The set of all natral nmbers ( S = onst 1 ) The set of even nmbers ( S =?)

4 Rersive Relations The harateristi fntion R of a relation R N k is defined as follows: S ( 1 k ) = 1 if < 1 k > S S ( 1 k ) = 0 if < 1 k > S A relation R is a rersive set iff its harateristi fntion R is a rersive fntion Eamples of rersive relations: < > = ( y) = sg( y ) < ( y) = sg( y) > ( y) = sg( y) ( y) = sg( y) sg( y ) =

5 Finding new Rersive Fntions and Relations In the net slides we ll go over a bnh of different methods to define new fntions and relations (and sets bt they an be seen as 1-plae relations) from eisting ones. In eah ase we an show that if the eisting fntions and relations are rersive then the reslting fntions and relations will be rersive as well.

6 Proesses From fntions to fntions: Composition Rersion Minimization (we saw this!) From fntions and relations to fntions: Definition by Cases From fntions and relations to relations: Sbstittion From fntions to relations: Graph From relations to relations: Logial operations From relations to fntions: Bonded Minimization and Maimization

7 Definition by Cases Sppose f( 1 n ) is defined by: f( 1 n ) = g 1 ( 1 n ) if R 1 ( 1 n ) f( 1 n ) = g m ( 1 n ) if R m ( 1 n ) Where: If: R 1 R m are mtally elsive i.e. there is no 1 n i j: R i ( 1 n ) and R j ( 1 n ) R 1 R m are olletively ehastive i.e. for all 1 n there is a i: R i ( 1 n ) g 1 g m are all rersive fntions R 1 R m are all rersive relations Then: f is a rersive fntion Proof: f( 1 n ) = g 1 ( 1 n ) R1 ( 1 n ) + + g m ( 1 n ) Rm ( 1 n )

8 Eample: min and ma min(y) is a rersive fntion Proof: min(y) an be defined by ases: min(y) = if y min(y) = y if > y ma(y) is a rersive fntion as well: ma(y) = if > y ma(y) = y if y

9 Sbstittion Given: Relation R(y 1 y m ) Fntions f 1 ( 1 n ) f m ( 1 n ) We an define relation R ( 1 n ) as follows: R ( 1 n ) iff R(f 1 ( 1 n ) f m ( 1 n )) If: R(y 1 y m ) is a rersive relation f 1 ( 1 n ) f m ( 1 n ) are rersive fntions Then: R is a rersive relation Proof: R ( 1 n ) = R (f 1 ( 1 n ) f m ( 1 n ))

10 Eample Consider relation R(yz) defined as follows: R(yz) iff y z We see that R is the reslt of sbstitting the rersive fntion into rersive relation Ths R is rersive (Tehnially R is the reslt of sbstitting the fntions f 1 (yz) = y z and f 2 (yz) = into and we need to show that f 1 (yz) = y z and f 2 (yz) = are rersive bt that s trivial sing the identity fntions)

11 Graph Remember that any fntion f:x Y an be seen as a relation defined over X Y The Graph operation will obtain a relationship from a fntion in eatly this manner. Given fntion f( 1 n ) Define R f ( 1 n y) iff f( 1 n ) = y If f is rersive then R f is rersive. Proof: R f is the reslt of sbstitting rersive fntion f into rersive relation =

12 Logial Operations Given n-plae relations R R 1 and R 2 we an define: R( 1 n ) iff not R( 1 n ) (i.e. < 1 n > R) R 1 R 2 ( 1 n ) iff R 1 ( 1 n ) and R 1 ( 1 n ) R 1 R 2 ( 1 n ) iff R 1 ( 1 n ) or R 1 ( 1 n ) If R R 1 and R 2 are rersive then R R 1 R 2 and R 1 R 2 are rersive: R = 1 R R1 R2 = R1 R2 (or: R1 R2 = min( R1 R2 ) ) R1 R2 = sg( R1 + R2 ) (or: R1 R2 = ma( R1 R2 ) )

13 Bonded Qantifiation Given n+1-plae relation R( 1 n y) we an define: v [R]( 1 n ) iff there eists some v sh that R( 1 n v) We ll simply write this as v R( 1 n v) v [R]( 1 n ) iff for all v : R( 1 n v) We ll simply write this as v R( 1 n v) If R is rersive then v [R] and v [R] are rersive as well: = = v n R n R v v sg ] [ )) ( ( ) ( = = v n R n R v v ] [ ) ( ) (

14 Using Strit Bonds = < = < if if pred n R v n R v )) ( ( ) ( 1 ] [ 1 ] [ = < = < if if pred n R v n R v )) ( ( ) ( 1 ] [ 1 ] [

15 Eample: Prime Consider the 1-plae relation P() where P() iff is prime (alternatively onsider the set P of all primes) P() is rersive (P is rersive) sine: P() iff 1< y< z< y z = That is: P() an be defined by applying the proesses of logial operators ( and bonded qantifiation) sbstittion and omposition to other rersive fntions (onst 1 and ) and rersive relations (< and =).

16 Bonded Minimization and Maimization Given n+1-plae relation R( 1 n y) define n+1-plae fntions Min[R] and Ma[R]: Min[R]( 1 n w) = smallest y<=w for whih R( 1 n y) if sh a y eists Min[R]( 1 n w) = w + 1 if no sh y eists Ma[R]( 1 n w) = largest y<=w for whih R( 1 n y) if sh a y eists Ma[R]( 1 n w) = 0 if no sh y eists

17 Proof that Min[R] is Rersive if R is Rersive If R is rersive then Min[R] is rersive as well: w = 1 n w) S ( 1 n i) i= 0 Min[ R]( where S is the harateristi fntion of the relation S defined as t i[ R]( 1 n i)

18 Why This Works Sppose we want to know Min[R](w) where R is defined as below: w R(w) (e.g.) R(w) S(w) = t w[ R](w) = t w R(t) S (w) 0 F T T F T T T F F F T F T F F F T F 0 2 w i= 0 S ( i) Verify that this is indeed Min[R](w)

19 Ma[R] is Rersive too Left as HW qestion Make sre to demonstrate that yor fntion works by providing a similar table (and note that for the speifi R relation as defined in that table the Ma olmn entries shold be )

20 Eample: qo and rem are Rersive Define qo(tient) and rem(ainder) fntions as follows: qo(y) = the largest z sh that z*y if y > 0 qo(y) = 0 if y = 0 rem(y) = y*qo(y) qo is rersive sine it is a definition by ases where one of the ases ses the bonded maimization of a rersive relation (and every other fntion and relation sed is rersive) rem is rersive sine - * and qo are rersive.

21 Eample: The Net Prime Let π () = the least y sh that < y and y is prime π () is rersive sine it an be defined as the bonded minimization of a rersive relation: π () = Min[<y Prime(y)](!+1) Eplanation:!+1 is not divisible by any nmber so either!+ 1 is prime itself or it has a prime fator greater than in either ase there eists a prime nmber greater than bt smaller or eqal to!+1

22 Eample: Modified Logarithms Consider the following two modified logarithm fntions lo(y) and lg(y): lo(y) = the largest z sh that y z divides if and y > 1 where divides y iff for some z: z * = y lo(y) = 0 otherwise lg(y) = the largest z sh that y z if > 1 and y > 1 lg(y) = 0 otherwise lo and lg are rersive sine they an be defined sing bonded maimization (se as pper bond) and other rersive operations over rersive fntions and relations (divides an be defined as bonded eistential qantifiation (again se as bond))

23 Prime Coding and Deoding Fntions A seqene 1 k an be enoded sing the following prime oding : ode( 1 n ) = 2 n *3 1 *5 2 * π(n) n where π(n) is the n-th prime and where 2 is the 0-th prime. π(n) is rersive sine π(0) = 2 and π(n+1) = π (π(n)) ode( 1 n ) is therefore rersive as well Given some ode nmber s the seqene an be deoded sing the following (rersive) fntion: ent(si) = the i-th entry (the 0-th entry gives the length) = lo(s π(i))