Foundations of Computing Discrete Mathematics Solutions to exercises for week 7

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1 Foundations of Computing Discrete Mathematics olutions to exercises for week 7 Agata Murawska (agmu@itu.dk) October 25, 2013 xercise (13.1.4). Let G = (V, T,, P ) be the phrase-structure grammar with V = {0, 1, A, }, T = {0, 1} and the set of productions P consisting of 1, 00A, A 0A, and A 0. a) how that belongs to the language generated by G A b) how that does not belong to the language generated by G. When we look at the productions from initial state we notice that we can only produce 1s at the front. Once we decide to use the rule 00A we have no rule that allows us to add 1 to the word we are creating. Therefore cannot be in the language generated by G, as it contains 1 at the end. c) What is the language generated by G? We have already noticed that we can produce an arbitrary number of 1s in front of the word (possibly none, as we can can use the second rule right away). Once we stop producing 1s, we have to add 00A to the word. Then, the rule for A allows us to produce an arbitrarily long, non-empty string of 0s, so: L(G) = {1 m 000 n m N, n N + } = {1 m 0000 n m, n N} = {1 m 0 n+3 m, n N} xercise (13.1.8). how that the grammar given in xample 5 generated the set {0 n 1 n n N}. The grammar G from xample 5 looks like this: 01, λ. We need to prove that L(G) = {0 n 1 n n N}. As a remainder, L(G) = {w {0, 1} w}. 1

2 Looking at the grammar we can see that the only way to derive a word is to use the production 01 some number of times and then end the word with the rule λ. Using 01 k times results in a string 0 k 1 k, so: L(G) = {w {0, 1} w} = {w {0, 1} for some n, n 0 n 1 n 0 n 1 n = w} = {w {0, 1} for some n, w = 0 n 1 n } = {0 n 1 n n N} xercise ( ). how that the grammar given in xample 7 generates the set {0 n 1 n 2 n n N}. The grammar G from xample 7 looks like this 1 : V = {0, 1, 2,, A, B, C}, T = {0, 1, 2}, starting symbol and productions C, C 0CAB, C λ, BA AB, 0A 01, 1A 11, 1B 12 and 2B 22. We need to prove that L(G) = {0 n 1 n 2 n n N}. First let us look at the grammar and see how does it work. From starting symbol we can only go to C and we can t ever go back. From C we can either produce an empty word (which allows us e.g. to produce = λ which is in the language {0 n 1 n 2 n n N}), or use C 0CAB. If you look carefully you ll notice that no other production generates new 0s the 0A 01 simply preserves the existing ones. o this gives us some hint we will have to use this rule k times in total to produce the leading 0s in 0 k 1 k 2 k. What else can we do before we remove the C using C λ? Only BA AB is an option. What we can use it for is to reorder the ABABABAB..AB sequence at the end of the string in fact, we can sort this sequence into A...AB...B using only this rule! Let us try to see exactly how could we generate = using this grammar: C 0CAB 00CABAB 000CABABAB 000ABABAB 000AABBAB 000AABABB 000AAABBB 0001AABBB 00011ABBB BBB BB B This is not the only way in which this word can be generated, but it is one that can be generalized into an algorithm for generating 0 k 1 k 2 k for any k: 1. Use C 2. Use C 0CAB k times to generate k leading 0s (now the string is 0 k C(AB) k ) 3. Use C λ to remove C and mark the end of generation of 0s 4. Use BA AB to reorder (AB) k into A k B k (now the string is 0 k A k B k ) 1 Actually, the grammar originally had λ instead of C λ, but that seems to be a typo. 2

3 5. Use 0A 01 once 6. Use 1A 11 k 1 times (now the string is 0 k 1 k B k ) 7. Use 1B 12 once 8. Use 2B 22 k 1 times to finally obtain 0 k 1 k 2 k This shows that {0 n 1 n 2 n n N} L(G), as we can use the procedure above to generate any member of the set {0 n 1 n 2 n n N}. What remains to be shown is that the reverse is also true that is that L(G) {0 n 1 n 2 n n N}. Let us take any w L(G) and show that it is of the form 0 n 1 n 2 n for some n. Let us assume that the number of 0s in w is equal to k. If k = 0 then trivially w is of the correct form, as any word from L(G) either contains some 0s or is equal to λ. Let us consider the case with k > 0. There are a couple of observations to keep in mind. Fact 1. very word in language L(G) has equal number of 0s, 1s and 2s. To see this, notice that C C0AB produces one 0, one A and one B. Looking at productions we can notice that each A can only be transformed to 1, each B to 2. Fact 2. All the 0s are produced in front of the string and before any 1s or 2s. In order to use 0A 01 we need 0 next to A. As long as we have not used C λ rule we won t have that, afterwards we cannot go back to produce any more 0s. Moreover all the 0s are at the front, as that is how the rule C 0CAB is placing them and there is no rule to move them around. Fact 3. There is a single point of transforming nonterminals A and B to terminals 1 and 2. After all the 0s have been produced, all four rules exchanging A with 1 and B with 2 require <terminal><nonterminal> left-hand side. But we have terminals only on the far left side of the word, so there is always only one point where we can use one of the rules 0A 01, 1A 11, 1B 12, 2B 22. Fact 4. In a derivation of a valid word, all the 1s are produced before (and in front of) any 2s. If not all the 1s are produced before the first B is exchanged to 2 (using 1B 12) that means there is some A to the left. This A will never be used in any production, so it will stay in the word. A valid word contains no nonterminals, so no As, therefore such derivation will not end with a valid word. Combining this with the previous fact we get that all the 1s are before any 2 in a valid word. 3

4 The proof that w is of the correct form follows directly from the observations above. Namely, w must necessarily have the same number of 0s, 1s and 2s, so the only thing that can go wrong is that they are not sorted. But we know that all the 0s are in the front and all the 1s are created before 2s, so the word w is in fact sorted that is, it is of the form 0 k 1 k 2 k (where k is the number of times the rule C 0CAB has been used). xercise ( ). Construct phrase-structure grammars to generate each of these sets. a) {01 2n n N} G = ({0, 1, }, {0, 1},, { 0, 11}) b) {0 n 1 2n n N} G = ({0, 1, }, {0, 1},, { λ, 011}) c) {0 n 1 m 0 n n, m N} G = ({0, 1,, A}, {0, 1},, { 00, A, λ, A 1A, A λ}) xercise ( ). A palindrome is a string that reads the same backward as it does forward that is, a string w, where w = w R, where w R is the reversal of the string w. Find a context-free grammar that generates the set of all palindromes over the alphabet {0, 1}. G = ({0, 1, }, {0, 1},, { 00, 11, 0, 1, λ}) Let us informally motivate this construction. Productions 00 and 11 are the only ones that allow you to continue building a word (as they contain a nonterminal on the right). The both preserve the feature of being a palindrome that is, if w was a palindrome, then 1w1 and 0w0 are also palindromes. Notice that they produce even-length words. A letter in the middle of a palindrome, if it has odd number of letters, can be anything so we have productions 0 and 1 to put the middle letter. A palindrome can also have an even number of letters; that we can obtain by finishing the derivation with λ. xercise ( ). Let G be the grammar with V = {a, b, c, }, T = {a, b, c}; starting symbol ; and productions ab, bc,, a, cb. Construct derivation trees for a) bca 4

5 bc a b) bca bc a c) bcabcb bc ab cb xercise ( ). a) xplain what the productions are in a grammar if the Backus-Naur form for productions is as follows: <expression> ::= ( <expression> ) <expression> + <expression> <expression> * <expression> <variable> <variable> ::= x y Productions for this grammar are the following: (), +,, V, V x, V y 5

6 with terminal symbols T = {x, y,, +, (, )}, a starting symbol and another nonterminal V. To see why that is we have to look at the BNF form of the grammar. ach left-hand side corresponds to a nonterminal object; <expression> is represented by and <variable> by V. ach element of the barseparated right-hand side is a production. Terminal objects are parts of the right-hand side that are not enclosed in brackets < >. b) Find a derivation tree for (x y) + x in this grammar. + ( ) V * x V V x y xercise ( ). Describe how productions for a grammar in extended Backus-Naur form can be translated into a set of productions for the grammar in Backus-Naur form. xtended Backus-Naur form (BNF) differs from BNF in several places: It (usually) does not use brackets to denote nonterminals. We can still tell which symbols are nonterminal by looking at all the left-hand sides (the names defined there are nonterminals, the rest of the symbols are terminal). It is then easy to recover the < > for nonterminals? symbol can be used to indicate that some part (a symbol or group of symbols enclosed in parentheses) of the production is optional. For example ::= a?bcde? is a simple BNF style grammar that can produce strings abcde, abcd, bcde, bcd as taking a and e is optional. The way to capture that in BNF is to have 2 productions for each?x: one with x and one without. In our example, the BNF equivalent is then: <> ::= abcde bcde abcd bcd this is obtained in 2 steps, first we remove?a to get 6

7 <> ::= abcd?e bcd?e and then we remove?e. * symbol is used to indicate that something occurs zero or more times. For example ::= a*b describes all strings with arbitrary number of a s in front and ending with b (so the language described by this grammar is {a n b n N}). We can simulate x* in BNF by creating a new nonterminal object, let us call it <Xstar>. Now, production for <Xstar> will be <Xstar> ::= x<xstar> λ and we will use <Xstar> where x* was in the original grammar. ::= a*b becomes then <> ::= <Astar>b <Astar> ::= a<astar> λ + symbol is used to indicate a symbol occurs one or more times. But notice that x+ is the same as xx* so we can exchange them and we already know how to transform x*. How to transform a more complicated BNF expression? We just have to keep transforming the grammar step by step until all the + and * symbols are gone. The result may not be optimal in terms of size (number of rules), but the two grammars will describe the same language. 7