Homework One Solution CSE 355

Transcription

1 Homework One Solution CSE 355 Due: 3 Jnury 2 Plese note tht there is more thn one wy to nswer most of these questions. The following only represents smple solution. Prolem : Linz 2..7()(c)(g), nd : Find df s for the following lnguges on Σ = {, } (): L = {w : w mod 5 } A df for L is given y the following trnsition grph:,,,,, (c): L = {w : n (w) mod 3 > } A df for L is given y the following trnsition grph:

2 (g): L = {w : w mod 3 =, w 6} A df for L is given y the following trnsition grph:,,,,,,,,,, 2.2.7: Design n nf with no more thn five sttes for the set { n : n } { n : n }. An nf for the set is given y the following trnsition grph: λ 2.2.: Find n nf with foour sttes for L = { n : n } { n : n }. An nf for L is given y the following trnsition grph: λ 2

3 Prolem 2: Linz 2.39 nd : Let L e regulr lnguge tht does not contin λ. Show tht there exists n nf without λ-trnsitions nd with single finl stte tht ccept L. Since L is regulr there exists df, D = (Q, Σ, δ, q, F ), with n ssocited trnsition grph, G D, such tht L(D) = L. We will construct n nf N = (Q {q f }, Σ, δ, q, {q f }) where q f / Q y giving its trnsition grph G N s follows:. From G D, remove the finl lel from every finl stte (mking them nonfinl sttes). 2. Add new stte q f nd lel it s finl stte. 3. For every stte q i, if there is trnsition from q i to stte in F on input Σ, then dd trnsition from q i to q f on input. Clerly, N hs single ccept stte, q f, nd no λ-trnsitions (since D is df nd we did not dd ny λ-trnsitions in our construction of N). We will now show tht L(N) = L. First note tht since λ / L, every w L cn e written s w = v for some v Σ nd n Σ. Now, w = v L iff there is wlk on G D leled with w from q to q i with q i F iff there is wlk on G D leled with v from q to q j nd trnsition from q j to q i on input iff there is wlk on G N leled with v from q to q j nd trnsition from q j to q f on input (since every trnsition in G D is trnsition in G N nd from step (3) in the construction of G N ) iff there is wlk on G N leled with w from q to q f iff w L(N). Thus, w L iff w L(N). Therefore we conclude tht L(N) = L nd tht for ny regulr lnguge tht does not contin λ, there exists n nf without λ-trnsitions nd with single finl stte tht ccept L : Show tht if L is regulr, so is L R. Since L is regulr lnguge, we cn construct corresponding df, N, such tht L(N) = L (For every regulr lnguge, there is corresponding df, y definition, nd for every df, there is n equivlent nf). By definition, L R consists of ll strings in lnguge L in reverse order. We will construct nf, N R, representing L R such tht L(N R ) = L R. N R will contin n dditionl strt stte with λ-trnsitions to the finl sttes of N. The direction of every trnsition in N is reversed. Also, the strt stte of N will e the finl stte of N R. The construction of nf N R is s follows: Let N = (Q, Σ, δ, q n, F ) N R = (Q {q }, Σ, δ r, q r, {q n }) Set of sttes of N R = set of sttes of N long with q = Q {q r } Σ = lphet of N R = sme s N q r = strt stte of N R {q n } = set of finl sttes of N R = strt stte of N Trnsition function: δ r (q, ) = {q : δ(q, ) = q} δ r (q r, λ) = F 3

4 δ r (q r, ) =, if λ Now we will show tht L R = L(N R ). w L R iff w R L iff there is wlk on the trnsition grph of N with lel w R from q n to some q i F iff there is wlk on the trnsition grph of N R from q r to q i with lel λ nd wlk from q i to q n with lel w (Following the reverse of every trnsition in the originl grph) iff w L(N R ). Since L R cn e represented y nf, it is regulr (y equivlence of nf to df, nd df to regulr lnguge). Prolem 3: Linz : A run in string is sustring of length t lest two, s long s possile nd consisting entirely of the sme symol. For instnce, the string contins run of s of length three nd run of s of length two. Find df s for the following lnguges on {, }. (): L = {w : w contins no runs of length less thn four}. 4

5 (): L = {w : every run of s hs length either two or three}. (c): L = {w : there re t most two runs of s of length three}., 5

6 (d): L = {w : there re exctly two runs of s of length 3}., Prolem 4: Linz : Let L e regulr lnguge on some lphet Σ, nd let Σ Σ e smller lphet. Consider L, the suset of L whose elements re mde up only of symols from Σ, tht is, L = L Σ. Show tht L is lso regulr. Since L is regulr lnguge, there should e df, N, representing L such tht L(N) = L, where N = (Q, Σ, δ, q, F ). Since L is mde up of strings with lphets from Σ, Σ Σ, nd L is suset of L, L contins only strings tht re ccepted y L s well. We cn construct df, M, for L s follows:. From the trnsition grph of N, remove every trnsition tht is leled with some / Σ. Now we will show tht L(M) = L. w = 2... n L iff there is wlk on the trnsition grph of N with lel w from q to some q i F nd every i Σ iff there is wlk on the trnsition grph of M from q to q i with lel w (it will e the exct sme pth s it ws in N) iff w L(M). Since L cn e represented y df, it is regulr. 6

7 Prolem 5: Linz nd : Convert the following nf into n equivlent df (see textook for the digrm). {,} {,2} {} {,2} {,,2} {} {2} Φ 2.3.8: Find n nf without λ-trnsitions nd with single finl stte tht ccepts L = {} { n : n }. Noting tht λ / L, we cn use the technique given in (Prolem 2) nd we get the nf given y the following trnsition grph:, 7