计算理论答案ch2

Undergraduate Course ELEMENTS OF COMPUTATION THEORY College of Computer Science Chapter2ZHEJIANG UNIVERSITYFall-Winter,2006P602.1.1Let M be a deterministicfinite automaton.Under exactly what cir-cumstances is e∈L(M)?Prove your answer.Solution:e∈L(M)if and only if s∈F.Suppose e∈L(M).Then,by definition of L(M),(s,e) ∗M(q,e),where q∈F.Because it is not the case that(s,e) M(q,w)for any configuration(q,w)(w=e).(s,e) ∗M(q,e)must be in the reflexive transitive closure of M by virtue of reflexivity −that is,(s,e)=(q,e).Therefore,s=q and thus s∈F.Suppose s∈F.Because ∗M is reflexive,(s,e) ∗M(s,e).Because s∈F,we havee∈L(M)by definition of L(M).2.1.2Describe informally the languages accepted by the following DF A.Solution:(c)All strings with the same number of a s and b s and in which no prefix has more thantwo b s than a s,or a s than b s.(d)All strings with the same number of a s and b s and in which no prefix has more thanone more a than b,or vice-versa.2.1.3Construct DF A accepting each of the following languages.(c){w∈{a,b}∗:w has neither aa nor bb as a substring}.(e){w∈{a,b}∗:w has both ab and ba as a substring}.Solution:(c)M=(K,Σ,δ,sF),whereK={q0,q1,q2,q3},Σ={a,b},s=q0,F={q0,q1,q2}q aδ(q,a)q0a q1q0b q2q1a q3q1b q2q2a q1q2b q3q3a q3q3b q3(e)M=(K,Σ,δ,sF),whereK={q0,q1,q2,q3,q4,q5},Σ={a,b},s=q0,F={q5}q aδ(q,a)q0a q1q0b q2q1a q1q1b q3q2a q4q2b q2q3a q5q3b q3q4a q4q4b q5q5a q5q5b q5P742.2.2Which regular expression for the languages accepted by the NF A of Problem2.2.1.Solution:a)a∗b)a(ba∪baa)∗(b∪ba)2.2.6(a)Find a simple NF A accepting(ab∪aab∪aba)∗.(b)Convert the NF A of part(a)to a DF A by the method in section2.2.Solution:(a)M=(K,Σ,∆,sF),where K={q0,q1,q2,q3},Σ={a,b},s=q0,F={q0}(qσq i)(q0a q1)(q1a q2)(q1b q0)(q1b q3)(q2a q0)(q3b q0)(b)Determinizing the above machine results in the following DFA:K={{q0},{q1},{q3},{q0,q1},{q0,q2},{q1,q3},∅},Σ={a,b},s={q0},F={{q0},{q0,q1},{q0,q2}}{q}σ{δ(q,σ)}{q0}a{q1}{q0}b∅{q1}a{q3}{q1}b{q0,q2}{q0,q2}a{q0,q1}{q0,q2}b∅{q0,q1}a{q1,q3}{q0,q1}b{q0,q2}{q3}a∅{q3}b{q0}{q1,q3}a{q3}{q1,q3}b{q0,q2}∅a∅∅b∅2.2.10Describe exactly what happens when the construction of this sectionapplied to a F A that is already deterministic.Solution:Only |K |of the 2|K |states of the new automaton will be reachable.Each of these states will have {q }for some q ∈K .If we identify {q }with q ,we have a bijection between the states of the old automata and the reachable states of the new one.With respect to this bijection,δ,s ,and F will be identical between the old mach-ine and the new.Sinceis the same,there is a natural isomorphism between the old and the automaton formed from the new one by discarding unreachable states.P 83-842.3.4Using the construction in the proof of theorem 2.3.1,construct F A accepting these languages.(b )((a ∪b )∗(e ∪c )∗)∗.Solution:(b )Ommited.2.3.7Apply the construction in Example 2.3.2to obtain regular expressions responding to each of the F A above.Simplify the resulting regular expres-sions as much as you can.Solution:(a )a ∗b (ba ∗b ∪a )∗(b )((a ∪b )(a ∪b ))∗(c )(a ∪b )∗abaa (a ∪b )∗(d )(a ∪∅∗)(ba ∗a )∗b (b ∪a )∗P 902.4.5Using the pumping theorem and closure under intersection,show that the following are not regular.(a ){ww R :w ∈{a,b }∗}Solution:Assume L is regular,by the closure property under intersection so is L 1=L ∩a ∗bba ∗.Consider language L 1.Let k be the constant whose existence the pumping theorem guarantees.Choose string w =a k bba k ∈L 1.Clearly |w |≥k .So the pumping theorem must hold.Let w =xyz such that |xy |≤k and y =e ,then y =a i where i >0.But then xy n z =a k +(n −1)i bba k ,which is clearly asymmetric for any n =1.The theorem fails,and thus that L 1is not regular,therefore L is not regular.2.4.8Are the following statements true od false?Explain you answer in each case.(a )Every subset of a regular language is regular.(b )Every regular language has a regular proper subset.(c )If L is regular,then so is {xy |x ∈L and y ∈L }.(d ){w |w =w R }is regular.(e )If L is regular,then so is {w |w ∈L and w R ∈L }.(f )If C is any set of regular languages,then ∪C is a regular language.(g ){xyx R |x,y ∈Σ∗}is regular.Solution:(a )False.Every language,including those we know not to be regular,is a subset of theregular language∗.(b )False.The empty set,which is a regular language,has no proper subsets at all,so it certainly cannot have a proper subset which is also a regular.(c )True.{xy |x ∈L and y ∈L }=L ◦¯L.Since L is regular,so is its complement,and thus their concatenation is regular.(d )False.This can be shown by trying to pump the string a k ba k .y will have to consist only of a s and the resulting xy 2z will unbalanced.Note,however,that this language is regular over an alphabet of one symbol.(It is true when C is required to be finite).(e )True.This language is L ∩L R .If L is regular,then so is L R .Since both L and L R are regular,so is their intersection.Solution:(f)False.Any language can be written as the(possibly infinite)union of the singleton sets containing its individual elements.Since not every language is regular,this claim is false.(g)True.{xyx R|x,y∈Σ∗}=Σ∗.By letting x=e,y can vary over all the stringsofΣ∗.)。