Download presentation

Presentation is loading. Please wait.

Published byRohan Francy Modified over 3 years ago

1
INHERENT LIMITATIONS OF COMPUTER PROGRAMS CSci 4011

2
SET THEORY 101 A function ƒ : A → B is: 1-1 (or injective) if onto (or surjective) if bijective if it is 1-1 and onto. ƒ(x)=ƒ(y) ⇔ x=y ∀ y Ǝ x: y = ƒ(x) 1-1 then |A| |B| onto then |A| |B| bijective then |B| |A|. ƒ can help us count. If ƒ is: ≤ ≥ =

3
Let S be any set and (S) be the power set of S Theorem: There is no onto map from S to P(S) Proof: Assume, for a contradiction, that there is an onto map ƒ : S (S) Let D ƒ = { x S | x ƒ(x) } If D ƒ = ƒ(y) then y D ƒ if and only if y D ƒ We construct an D ƒ ⊆ S that cannot be the output, ƒ(y), for any y S. D ƒ is called the diagonal set for ƒ.

4
x y 1 ∈ ƒ(x)?y 2 ∈ ƒ(x)?y 3 ∈ ƒ(x)?y 4 ∈ ƒ(x)? … y1y1 YNYY y2y2 NYNY y3y3 NNNN y4y4 YNNY Y Y Y N (y i ∈ D) ≡ (y i ∉ ƒ(y i ))

5
The process of constructing a counterexample by “contradicting the diagonal” is called DIAGONALIZATION

6
No matter what, (S) always has more elements than S

7
THE CHURCH-TURING THESIS L is recognized by a program for some computer* ⇔ L is recognized by a TM * The computer must be “reasonable”

8
There are languages over {0,1} that are not decidable If we believe the Church-Turing Thesis, this means there are problems that computers inherently cannot solve. We proved this using a “simple” counting argument: There are more Languages than Turing Machines.

9
Turing Machines Languages over {0,1}

10
Not all languages over {0,1} are decidable Turing Machines Strings of 0s and 1s Sets of strings of 0s and 1s Languages over {0,1} S(S) OK, but what does an undecidable language look like?

11
THE DIAGONAL LANGUAGE D TM = { 〈 M 〉 | M does not accept 〈 M 〉 } is undecidable. T M 1 ∈ L(T)?M 2 ∈ L(T)?M 3 ∈ L(T)?M 4 ∈ L(T)? … M1M1 YNYY M2M2 NYNY M3M3 NNNN M4M4 YNNY Y Y Y N *Note: Hypothetical. May not correspond to an actual encoding *

12
Programs can “accept” their code as input: $ grep grep /tmp/example.pl <

13
A TM = { 〈 M,w 〉 | M is a TM that accepts string w } THE ACCEPTANCE PROBLEM Theorem: A TM is Turing-recognizable but NOT decidable A TM is Turing-recognizable: Define TM U as follows: On input 〈 M,w 〉, U runs M on w. If M ever accepts: accept. If M ever rejects: reject. If M(w) loops forever: U( 〈 M,w 〉 ) loops forever.

14
A TM = { 〈 M,w 〉 | M is a TM that accepts string w } A TM is undecidable:(proof by contradiction) Assume TM acceptsInput decides A TM acceptsInput(M,w) = Acceptif M(w) accepts Reject otherwise Construct a new TM LLPF as follows: on input 〈 M 〉 run acceptsInput (M, 〈 M 〉 ) and output the opposite LLPF( M ) = Rejectif M accepts M Accept if M does not accept M LLPF

15
M1M1 M2M2 M3M3 M4M4 : M1M1 M2M2 M3M3 M4M4 … accept reject OUTPUT OF acceptsInput(M,N)* accept reject LLPF reject accept reject accept ? *Note: Hypothetical. May not correspond to an actual encoding

16
Theorem: A TM is Turing-recognizable (r.e.) but NOT decidable Theorem: A TM is not even Turing-recognizable! Proof: Suppose ¬A TM is recognized by TM V. We know A TM is recognized by U. Build Z to decide A TM : run V(M,w) and U(M,w) in parallel. If V accepts, reject; if U accepts, accept.

17
A language is called Turing-recognizable, semi-decidable, or recursively enumerable if some TM recognizes it A language is called decidable or recursive if some TM decides it recursive languages r.e. languages all languages

18
HALT TM = { 〈 M,w 〉 | M is a TM that halts on string w } Theorem: HALT TM is undecidable THE HALTING PROBLEM Proof: Assume, for a contradiction, that TM haltsOnInput decides HALT TM haltsOnInput(M,w) = Acceptif M(w) halts Reject if M(w) loops We construct the machine liarLiar as follows: liarLiar( 〈 M 〉 ): if haltsOnInput(M, 〈 M 〉 ): return liarLiar( 〈 M 〉 ) else: return true. liarLiar( 〈 liarLiar 〉 ) halts iff liarLiar( 〈 liarLiar 〉 ) loops!

19
We can also prove that A TM is undecidable by “reusing” the proof that D TM is undecidable. Theorem. If A TM is decidable, then so is D TM. Proof: Suppose we have a program, acceptsInput, that decides A TM, e.g. acceptsInput(M,w) returns true if 〈 M,w 〉 A TM and returns false otherwise. We can build a TM notDiagonal to decide ¬D TM : Theorem. If HALT TM is decidable, then so is A TM. Proof: acceptsInput(M,w) = let M’ = M with q reject replaced by q loop. return haltsOnInput(M’,w) notDiagonal( 〈 M 〉 ) = return acceptsInput(M, 〈 M 〉 )

Similar presentations

OK

Donghyun (David) Kim Department of Mathematics and Computer Science North Carolina Central University 1 Chapter 4 Decidability Some slides are in courtesy.

Donghyun (David) Kim Department of Mathematics and Computer Science North Carolina Central University 1 Chapter 4 Decidability Some slides are in courtesy.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on series and parallel circuits practice Ppt on tokyo stock exchange Ppt on computer malwares free Best ppt on save water Tcp fast open ppt on mac Ppt on natural resources and conservation major Ppt on diffusion of innovation Ppt on save water and electricity Conceptual architecture view ppt on android Ppt on idiopathic thrombocytopenia purpura in pregnancy