15 AUGUST 2002 SOLUTIONS

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1 CD58 AUTOMATE, BEÄKNINGA OCH FOMELLA METODE, 5 p. DFA nd egulr Expressions 5 AUGUST 22 SOLUTIONS () Build the miniml DFA tht ccepts the strings over Σ ={,} whose ll proper prefixes (i.e the prefix is not the string) do not codify, in inry, multiple of 3. (lmd, the empty string, is not considered multiple of 3). We hve to check ll different prefixes. At the moment we find one cse where it s multiple of 3, nd it s not proper (i.e there re more symols to red) we stop nd reject the string. Such n utomton is (I consider tht is not multiple of 3). 3 2, 4 Sttes, nd 2 represent multiple of 3, multiple of 3 plus nd multiple of 3 plus 2 respectively. Now we hve to check tht this utomton is miniml. We pply the itertive lgorithm, seprting ccepting nd rejecting sttes. Iter I II II I I I II II I I I II Now, we seprte stte from the others. Iter I II III III II II II III III I II II III We hve to seprte now stte from 2,3. Iter 2 I II III IV IV III II III IV IV I III II IV Finlly, we hve to split 2 nd 3, resulting with the utomton we lredy hve. So it s miniml.

2 CD58 AUTOMATE, BEÄKNINGA OCH FOMELLA METODE, 5 p 5 AUGUST 22 SOLUTIONS () Let e e the regulr expression e=(+(+ * ) * ) *. Prove tht L(e)={w (+) * in(w)= o 6 }. If we wnt to prove tht, one wy is to crete the miniml utomt for oth, the lnguge nd the regulr expression, nd then compre them. If we see tht oth of them re the sme, we re done. First, I crete the utomton for the lnguge {w (+) * in(w)= o 6 }. The ide is tht ech stte represents one of the possile sttes of numer: multiple of 6, multiple of 6 plus, multiple of 6 plus 2... nd so one Now, it s time to mke sure tht the utomton is miniml. If we tke look to the sttes, it seems tht 5 nd 2 do the sme, nd tht 4 nd should e together. We pply the itertive lgorithm s in the previous exercise. Iter I II I I II I I II I I I I I I We see tht stte 3 is different, so we split up the set I. Iter I II III I I I I III III II I II I I I Agin, we hve to split set I into two sets, one for -4 nd the other for 2-5. Iter I II III IV II II I II IV IV III III II II I I 2

3 CD58 AUTOMATE, BEÄKNINGA OCH FOMELLA METODE, 5 p 5 AUGUST 22 SOLUTIONS We re done. This is miniml! Now, it s time to clculte the utomton tht represents the regulr expression. This utomton is: There s no need to minimize it, since we cn see tht it s exctly the sme one tht we minimized just ove. (c) Write (simple) grmmr for the following regulr expression: (+) * () + ( * * ) *. We cn generte the grmmr from the regulr expression, lthough it s it tricky nd it cn e error prone. Thus, we first mke the utomton, nd we get the grmmr from it. I use NFA since it s esier to represent this regulr expression. The NFA is:, The following step is determinize it. [] STAT --> S [,2] [] [,2] --> A [,2] [,3] [,3] --> B [,2,4] [] [,2,4] ACC [,2,4] [,3,5] [,3,5] ACC [,2,4] [,5] [,5] ACC [,4,2] [,5] Now, if we wnt to get very simple grmmr, we cn minimize it. If not, we cn use this utomton to get grmmr. 3

4 CD58 AUTOMATE, BEÄKNINGA OCH FOMELLA METODE, 5 p 5 AUGUST 22 SOLUTIONS If we minimize it, we ll see tht we never get out of the cceptnce sttes once we rech one, so we sustitute ll of them y one (clled C). Thus, the grmmr is (following the trnsitions from the determiniztion tle): S S A A A B B C S C C C λ (d) Build the miniml DFA tht recognizes the strings tht hve the sustring ut do not hve the sustring. (Ide: you my wnt to mke one utomton for ech cse nd then clculte the intersection of oth of them.) This cn e done stright from the mind. But nice nd methodic wy of doing it is creting first one utomton tht recognizes the strings tht hve the sustring (which is very esy). After tht, nother one tht recognizes the strings tht don t hve the sustring (which is very esy too). And finlly, you mke the intersection (nd minimize it). The utomton tht recognizes the strings tht hve the sustring is:, A B C D The other utomton is:, E F G H 4

5 CD58 AUTOMATE, BEÄKNINGA OCH FOMELLA METODE, 5 p 5 AUGUST 22 SOLUTIONS Now, we hve to compute the intersection. We strt joining the strting sttes, AE, nd we compute ll different trnsitions. AE () BE AF BE (2) CE AF AF (3) BE AG CE (4) DE AF AG (5) BE AH DE (6) --> ACC DE DF AH (7) BH AH DF (8) --> ACC DE DG BH (9) CH AH DG () --> ACC DE DH CH () DH AH DH (2) DH DH Now, we cn minimize it. ejecting sttes={,2,3,4,5,7,9,,2}, ccepting sttes={6,8,} After minimizing (using the sme lgorithm s lwys), we cn join sttes {7,9,,2}, which is sink stte. 2. egulr or not? Prove! Prove whether the following lnguges L re regulr or not. () L={u#v u,v (+) * u is sustring of v} We choose N. Then, we tke string, tht cn e s simple s N # N. Now, we choose decomposition. All possile decompositions re x= k, y= j, z= N-k-j # N, xy <N. Now, if we pump the string otining w =xy i z= N - j(i-) # N. Choosing i= is cler tht w doesn t fll into the lnguge, since j>. () L={xy x,y (+) + x = y +} The lnguge is regulr, since it s the lnguge of the strings tht hve odd length. A regulr expression is (+) 2i+, i (c) L={ 2... n n } (see exm June 22). 3. Context-free lnguges () Build the NPDA tht recognizes { n n2... nk k i ni=i } (Explntion: you hve pckets of... So you hve the strings such tht one of the pckets hs the sme numer of s s the position where it is. You d ccept since the third pcket hs 3 s, ut you shouldn t ccept since the first pcket hs two s it hd to hve - nd the second one only one.) 5

6 CD58 AUTOMATE, BEÄKNINGA OCH FOMELLA METODE, 5 p 5 AUGUST 22 SOLUTIONS The ide is to use indeterminism. On one hnd, when we red n, we push n A to the stck so we know how mny pckets we hve red, which should e the numer of s we wnt to red. On the other hnd, we just move nd strt to check the s with the current stck. So, it s like hving two copies of the stck ll the time, nd ll this is possile due to indeterminism. The utomton is: Z ZA A AA A A A λ Z Z Z Z Z ZA A AA Zλ Z Z Z () Which lnguge does this grmmr generte? S-> A A A A-> BB BB B-> B B B B λ If you tke look to the different rules, you cn produce ny comintion of strings. Thus, you re generting either (,) * or (,) +. Since the first rules doesn t llow to mke the empty string, we re tlking out (,) +. (c) Generte (simple) grmmr tht descries the lnguge of ll the strings tht hve more s tht s. If we tke look to exm 22, it s lmost the sme chnging s for s. The only difference is tht we should mke sure tht we lwys hve more s. Such grmmr is: S BSSB SBSB B B B 6

7 CD58 AUTOMATE, BEÄKNINGA OCH FOMELLA METODE, 5 p 5 AUGUST 22 SOLUTIONS 4. Primitive recursion Descrie the mpping defined y dd o ( mult o ( id, id), dd o ( id, id)) Solution f ( n) = dd o ( mult o ( id, id), dd o ( id, id))( n) = dd o ( mult o ( id( n), id( n)), dd o ( id( n), id( n))) = dd o ( mult( n, n), dd( n, n)) 2 = dd( n,2n) = n 2 + 2n 5. Turing Mchine Wht is the lnguge the following Turing mchine recognizes? Demonstrte how it works., L x x, L x x, # #, q 4 # #, L q # #, # #, x, q #, q2 # #,, x x, x, q reject q ccept q 3 x x, # #, Solution The lnguge consisting of ll strings of s whose length is power of 2, n L = { 2 n }. This is clssroom exmple. (Se Senste nytt, 9/5 TM, steg-för-steg). 7

8 CD58 AUTOMATE, BEÄKNINGA OCH FOMELLA METODE, 5 p 5 AUGUST 22 SOLUTIONS 6. Decidle? Motivte! (i) L = {<M > L(M) hs exctly 3557 elements}. (ii) L = { <M> L(M) is not recognizle}. (iii) L = { <M> L(M) hs fewer thn sttes nd hlts on input }. Solution (i) (ii) (iii) Undecidle. This is non-trivil property of lnguges, so ice's Theorem pplies. Decidle. This is tricky question. L(M) is lwys recognizle; in prticulr, it is the lnguge recognized y Turing mchine M. Therefore, this is the empty lnguge, which is recognizle. Decidle. There re only finitely mny progrms with less thn sttes. Any suset of them, like the ones hlting on input, will lso e finite set. Every finite set is Turing computle since list of ll memers cn e mde, nd n lgorithm only need to check the list. 8

Homework 3 Solutions

CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.