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Homework #5 Solutions

Homework #5 Solutions. 1. True or False. a) Regular Languages are always Context-Free Languages True False b) Context Free Languages are always Regular Languages

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Homework #5 Solutions

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  1. Homework #5 Solutions

  2. 1. True or False a) Regular Languages are always Context-Free Languages True False b) Context Free Languages are always Regular Languages True False c) The grammar S 0S |0S1S |e is ambiguous d) The language {anbncn} is regular e) The language {anbncn} is context-free True False True False True False

  3. 2. Minimize the following dfa (1) Dividing into Final and Non-Final States: Partition I  Partition 2 

  4. #2 cont. (2) In Partition 1, all states “do the same thing” on an a. But on a b, states 1 and 6 both go to Partition II. We’ll move them to their own partition: (3) We cannot partition further: L(M) = (a*ba*ba*)*

  5. #3. a) Create a grammar thatgenerates the set of all strings over {0,1} with an equal number of 0’s and 1’s Also b) construct a parse tree and c) leftmost derivation of 0011. d) Is your grammar ambiguous? Why or why not? • S-> 0 S 1, S  1 S 0, S  S S, S e b) c) Yes. There is more than 1 parse tree for e as well as for other strings. S0 S 1  0 0 S 1 1  0 0 1 1

  6. #4. Find the Start symbol for the Java grammar shown at: http://www.cse.psu.edu/~saraswat/cg428/lecture_notes/LJava2.html • The start symbol is CompilationUnit. • It doesn’t appear on the left-hand-side. It is good technique to write a programming language grammar so that the Start symbol does not occur on the right-hand-side, and all grammars can be changed to an equivalent grammar having this property (how?)

  7. #5. For the grammar G: S  X Z Z X X x X e Z zZ ea) What is L(G)? { e, x, z, xz, xx, zz, zx, xzz, xzx. zzx, xzzx }, a finite language • b) ProofLet X = { e, x, z, xz, xx, zz, zx, xzz, xzx. zzx, xzzx }. • To show X = L(G) requires 2 proof parts: • if w e X, then w e L(G) • if w e L(G), then w e X • 1. Given: w e X • Prove: w e L(G) • * • To show w e L(G) means we have to show S  w • Since the language is finite, we can show this for each string: • w = e: • SXZZXeZZXeeZXeeeXeeee = e Similarly for the other elements of L.

  8. #5 cont 2. if w e L(G), then w e X Derivations of strings of length 0 in L(G): SXZZXeZZXeeZXeeeXeeee = e and e is in X Derivations of strings of length 1 in L(G): SXZZXxZZXxeZXxeeXxeee=x (can be derived another way also) And x is in X SXZZXeZZXezZXezeXezee=z (can be derived another way also And z is in X No other derivations result in strings of length 1 Similarly for derivations of strings of length 3 and 4

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