1 Concept Review 11 Undecidability By a cardinality argument we know that almost all languages are undecidable This argu ment however does not give us an explicit construction The following theorem does just that Theorem 11 The language Mw accepts ID: 74165
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1.3Rice'stheoremTheorem1.2(Rice'stheorem).LetPbeanysubsetoftheclassofr.e.languagessuchthatPanditscomplementarebothnonempty.ThenthelanguageLP=fhMi:L(M)2Pgisundecidable.Intuitively,Rice'stheoremstatesthatTuringmachinescannottestwhetheranotherTuringmachinesatisesa(nontrivial)property.Forexample,letPbethesubsetoftherecursivelyenumerablelanguageswhichcontainsthestringa.ThenRice'stheoremclaimsthatthereisnoTuringmachinewhichcandecidewhetheraTuringmachineacceptsa.2ExercisesExercise2.1.Reductionscanbetrickytogetthehangof,andyouwanttoavoid\goingthewrongway"withthem.InwhichofthesescenariosdoesL1mL2provideusefulinformation(andinthosecases,whatmayweconclude)?(a)L1'sdecidabilityisunknownandL2isundecidableNothing(b)L1'sdecidabilityisunknownandL2isdecidableL1isdecidable.ThisisinSipser.(c)L1isundecidableandL2'sdecidabilityisunknownUndecidable.Corollarytotheabovequestion.(d)L1isdecidableandL2'sdecidabilityisunknownNothingExercise2.2.Arguethatmisatransitiverelation.LetfbethefunctionthatreducesAtoB,i.e.,AmBbyf,andletgbethefunctionthatreducesBtoC,i.e.,BmCbyg.Thenw2A()f(w)2B.Furthermore,x2B()g(x)2C.Itfollowsthatforeveryw2A,g(f(w))2C,andfurthermore,foreveryx2C,thereexistsay2Bsuchthatg(y)=x,andforeveryy2B,thereexistsaw2Asuchthatf(w)=x,sothatforeveryx2C,thereexistsaw2Awithg(f(w))=x2C.Thusw2Ai g(f(w))2C,sothatAmC.Exercise2.3.Determine,withproof,whetherthefollowinglanguagesaredecidable.2