Question

Name: Solve problems 1-3. Problem 4 counts for extra credit. Each problem counts for 3 points. 1. Construct a non-ambiguous grammar generating the language consisting of all strings over the alphabet = {0,1,2), which contain no adjacent 1's. Provide a justification of correctness of your construction. 2. A Huffman tree constructed out of characters aj, az, az, ..., an, occurring with frequencies fi f2, f3, ..., fn, is given. Explain the terms: the maximal, the minimal, the average code length in terms of the features of the Huffman tree. Can it happen that the lengths are equal to 1, 2, 3,...,n-2, n-1, n-1? Can it happen that all code lengths are equal? Provide illustrating examples. 3. Construct non-ambiguous grammars properly interpreting operator precedence, equivalent to G:E → E + EE* E|( E id, and satisfying the following properties (i) both + and * are left associative, (ii) both + and * are right associative, (iii) operation + is left associative and operation * is right associative, (iv) operation + is right associative and operation * is left associative. Justify your constructions. In each case draw a parse tree of the string 1 * 2 * 3+4+5+6+7* 8*9. You are allowed to reuse portions of already drawn parse trees. 4. Explain how one may construct an equivalent non-ambiguous grammar out of a given deterministic automaton. Provide illustrating examples. Hint: Express the reading process of the automaton as a left-most derivation. Construct variables out of states.

If you could please help with 1-3. 4 if you can but it is not necessary. Thanks

Answer 1

1.

Regular Expression: (0+2+10+12)*(1+epsilon)

Explanation
(0+2+10+12)*
Any number of 0s or 2s can come without restriction (0+2)
If 1 comes then there has to be either 0 or 2 followed since there shouldn't be 1.

(1+epsilon)
If string is gonna end with 1, it can be there because previously 10 or 12 or 0 or 2 came.
If string is not gonna end with 1, then it can be epsilon

Unambiguous Grammar
S -> 0S | 2S | 10S | 12S | 1 | epsilon

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