FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY. FLAC (15-453) Spring L. Blum

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1 FLAC (15-453) Spring L. lum FOAL LANGUAGES, AUOAA AND COPUAILIY ICE S HEOE, HE ECUSION HEOE, AND HE FIXED-POIN HEOE HUSDAY FE 27 FIN = { is a and L() is finite} Is FIN Decidable? Note Properties of this language: FIN is a language of uring achines If 1 2 (ie L( 1 ) = L( 2 )), then either both 1 and 2 are in FIN or both are not. here are s 1 and 2, such that 1 FIN and 2 FIN hen L is undecidable ICE S HEOE Let L be a language over uring machines. Assume that L satisfies the folloing properties: 1. For s 1 and 2, if 1 2 then 1 L 2 L 2. here are s 1 and 2, such that 1 Land 2 L EXEELY POWEFUL! hen L is undecidable ICE S HEOE Let L be a language over uring machines. Assume that L satisfies the folloing properties: 1. For s 1 and 2, if 1 2 then 1 L 2 L 2. here are s 1 and 2, such that 1 Land 2 L FIN = { is a and L() is finite} hen L is undecidable ICE S HEOE Let L be a language over uring machines. Assume that L satisfies the folloing properties: 1. For s 1 and 2, if 1 2 then 1 L 2 L 2. here are s 1 and 2, such that 1 Land 2 L E = { is a and L() = } EG = { is a and L() is regular} 1

2 FLAC (15-453) Spring L. lum hen L is undecidable ICE S HEOE Let L be a language over uring machines. Assume that L satisfies the folloing properties: 1. For s 1 and 2, if 1 2 then 1 L 2 L 2. here are s 1 and 2, such that 1 Land 2 L Will sho: A is mapping reducible to L Sho L is undecidable Sho: A is mapping reducible to L Σ* A f L Σ* (,) f (,) Sho L is undecidable Sho: A is mapping reducible to L Σ* A f L Σ* (,) 1 ICE S HEOE Define to be a that never halts Assume, WLOG, that L Why? Let 1 L (such 1 exists, by assumption) Sho A is mapping reducible to L : (,) f 2 ap (, ) here (s) = accepts if both () and 1 (s) accept loops otherise What is the language of? A is mapping reducible to L Σ* A f L Σ* (,) f 1 Problem Let S = { is a ith the property: for all, () accepts implies ( ) accepts}. S is undecidable. (,) QED 2

3 FLAC (15-453) Spring L. lum A = { (,) is a that accepts string } HAL = { (,) is a that halts on string } E = { is a and L() = } EG = { is a and L() is regular} he rest of the content of today s lecture has been a major source of headaches and misunderstandings EQ = {(, N), N are s and L() =L(N)} ALL PDA = { P P is a PDA and L(P) = Σ* } ALL UNDECIDALE Where is ice s heorm Applicable? Which are SEI-DECIDALE? he recursion theorem is just like tennis. Unless you're exposed to it at age five, you'll never become orld class. -Juris Hartmanis (uring Aard 1993) HE ECUSION HEOE heorem: Let be a uring machine that computes a function t : Σ* Σ* Σ*. hen there is a uring machine that computes a function r : Σ* Σ*, here for every string, (Note: Juris didn t see the recursion theorem until he as in his 20 s.) r() = t(<>, ) t t(<>,) ecursion heorem says: A uring machine can obtain its on description, and compute ith it. We can use the operation: Obtain your on description in pseudocode! Given a computable t, e can get a computable r such that r() = t(<>,) here <> is a description of r INSIGH: (or t) is really (or r) heorem: A is undecidable Proof (using the ecursion heorem): Assume H decides A Construct machine such that on input : 1. Obtains its on description < > 2. uns H on (<>, ) and flips the output unning on input alays does the opposite of hat H says it should! 3

4 FLAC (15-453) Spring L. lum heorem: A is undecidable Proof (using the ecursion theorem): Assume H decides A eject if H (x, ) accepts Let H (x, ) = Accept if H (x, ) rejects (Here x is vieed as a code for a ) y the ecursion heorem, there is a such that: () = H (<>, ) = Contradiction! eject if H (<>, ) accepts Accept if H (<>, ) rejects IN = { is a minimal, rt } heorem: IN is not E. Proof (using the ecursion heorem): Assume E enumerates IN Construct machine such that on input : 1. Obtains its on description <> 2. uns E until a machine D appears ith a longer description than of 3. Simulate D on Contradiction. Why? IN = { is a minimal, rt } heorem: IN is not E. Proof (using the ecursion heorem): Assume E enumerates IN Let E (x, ) = D() here <D> is first in E s enumeration s.t. <D> > x y the ecursion heorem, there is a such that: () = E (<>, ) = D() here <D> is first in E s enumeration s.t. <D> > <> Contradiction. Why? HE FIXED-POIN HEOE heorem: Let f : Σ* Σ* be a computrable function. here is a such that f(<>) describes a that is equivalent to. Pseudocode for the : On input : 1. Obtain the description <> 2. Let g = f(<>) and interpret g as a code for a G 3. Accept iff G() accepts HE FIXED-POIN HEOE heorem: Let f : Σ* Σ* be a computrable function. here is a such that f(<>) describes a that is equivalent to. HE FIXED-POIN HEOE heorem: Let f : Σ* Σ* be a computrable function. here is a such that f(<>) describes a that is equivalent to. Let f (x, ) = G() here <G> = f (x) (Here f(x) is vieed as a code for a ) Let f (x, ) = G() here <G> = f (x) (Here f(x) is vieed as a code for a ) y the ecursion heorem, there is a such that () = f (<>, ) = G() here <G> = f (<>) Hence G here <G> = f (<>), ie <> f (<>) So is a fixed point of f! y the ecursion heorem, there is a such that () = f (<>, ) = G() here <G> = f (<>) Hence G here <G> = f (<>), ie <> f (<>) So is a fixed point of f! 4

5 FLAC (15-453) Spring L. lum HE FIXED-POIN HEOE heorem: Let f : Σ* Σ* be a computrable function. here is a such that f(<>) describes a that is equivalent to. Examples: 1. For any 1-1 computable enumeration of Σ* (or Gödel numbering of s), there ill alays be a that is equivalent to its successor in the enumeration 2. Let a virus flip the first bit of each ord in Σ* (or in each ). hen there is a that remains uninfected. HE ECUSION HEOE heorem: Let be a uring machine that computes a function t : Σ* Σ* Σ*. hen there is a uring machine that computes a function r : Σ* Σ*, here for every string, r() = t(<>, ) t t(<>,) HE ECUSION HEOE heorem: Let be a uring machine that computes a function t : Σ* Σ* Σ*. hen there is a uring machine that computes a function r : Σ* Σ*, here for every string, r() = t(<>, ) Lemma: here is a computable function q : Σ* Σ*, here for any string, q() is the description of a P that on any input, prints out and then accepts Q <P > s P o Start: Need to sho ho to construct a that computes its on description. Q computes q A SELF HA PINS <SELF> A SELF HA PINS <SELF> P () P (), on any input, prints the code for a that on any input outputs the result of ith input What about on input <>? () = < P > here P () = () So, (<>) = < P <> > here P <> () = (<>) No, P <> () = (<>) = <P <> > > So, let SELF = P <> 5

6 FLAC (15-453) Spring L. lum A SELF HA PINS <SELF> A SELF HA PINS <SELF> P () P SELF SELF <> P <> P <> (<>) <> P <> P <> HE ECUSION HEOE heorem: Let be a uring machine that computes a function t : Σ* Σ* Σ*. here is a uring machine that computes a function r : Σ* Σ*, here for every string, t r() = t(<>, ) t t(<>,) t t P P < P <> > P <> < P <> > 6

7 FLAC (15-453) Spring L. lum t t P P <> =??? <> =??? < P <> > < P <> > P <> t t P P <> (= <P <> >) <> (= <P <> >) < P <> > P <> < P <> > P <> t t P P <> (= <P <> >) <> (= <P <> >) < P <> > < P <> > t(<>,) 7

8 FLAC (15-453) Spring L. lum HE ECUSION HEOE heorem: Let be a uring machine that computes a function t : Σ* Σ* Σ*. here is a uring machine that computes a function r : Σ* Σ*, here for every string, r() = t(<>, ) ead Chapter 6.1 and 6.3 for next time t t(<>,) 8