CPSC 313 Winter, 2005 Ambiguity of Context-Free Grammars

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1 CPSC 313 Winter, 2005 Ambiguity of Context-Free Grammars This document includes several proofs concerning the ambiguity of various context-free grammars. This supplements the material on ambiguity in grammars and languages found in Section 5.4 of the textbook. 1 Proving That a Context-Free Grammar is Ambiguous Recall that a context-free grammar G = (V, P, T, S) is ambiguous if (and only if) there exists a string w in the language of G such that there are two (or more) different parse trees with root S and yield w in G. We can therefore show that a given context-free grammar G = (V, T, P, S) is ambiguous by a direct application of this definition by giving a string w L(G) along with two parse trees with root S and yield w in G. For example, consider the context-free grammar G = (V, T, P, W ), where V = {W, A, B}, T = {0, 1}, W is the start variable, and the grammar includes the following rules. W A B A 0W ɛ B 1W ɛ In order to see that this grammar is ambiguous, it is sufficient to give two different parse trees with root W and with same string, w = ɛ, as yield, such as the following. W W A B PSfrag replacements ɛ PSfrag replacements ɛ It is also true that a context-free grammar G = (V, T, P, S) is ambiguous if and only if there exists a string w in the language of G such that there are two or more leftmost derivations of w from the start variable. 1

2 Thus we can also show that the above grammar ambiguous by noting that the string w = ɛ has two leftmost derivations, namely, and W G A (using the rule W A) G ɛ (using the rule A ɛ) W G B (using the rule W B) G ɛ (using the rule B ɛ). Finally, it is true that a context-free grammar G = (V, T, P, S) is ambiguous if and only if there exists a string w in the language of G such that there are two or more rightmost derivations of w from the start variable. Thus we can prove that the above grammar G is ambiguous by giving two rightmost derivations of the empty string from the start variable W, as well. 2 Removing Ambiguity From a Grammar Sometimes we are asked to remove ambiguity from a given context-free grammar G = (V, T, P, S). In other words, we are asked to find a second context-free grammar, Ĝ = ( V, T, P, Ŝ) (with the same set of terminals) such that L(Ĝ) = L(G) and Ĝ is unambiguous. Section 5.4 of the textbook includes several heuristics that are useful when you are trying to remove ambiguity from a grammar. If you have not already done so then you should read this section of the text! However, it turns out that the current example is simple enough for ambiguity to be removed without using these rules. With a little effort it is possible to show that W G w for every string x T. This will be proved more formally later on. The variable A can be used to derive any nonnempty string in T that begins with a 0 as well as the empty string, and the variable B can be used to derive any nonempty string in T that begins with a 1 as well as the empty string. The only string that can be derived from both A and B is the empty string. It turns out that it is sufficient to remove all but one way to derive the empty string in the way that is shown below. We will be able to show that the language of the resulting grammar is T (as needed) that that the resulting grammar is unambiguous after that. With that noted, let us consider the grammar Ĝ = (V, T, P, S) that is the same as the grammar G except that Ĝ does not include the rule B ɛ: Ĝ has the rules W A B A 0W ɛ B 1W It is necessary to do two things to show that this is a correct answer that is, to show that Ĝ is a grammar that removes ambiguity from G: We must prove that L(Ĝ) = L(G), and we must also prove that Ĝ is unambiguous. However, we can see intuitively that the grammar is correct, as argued next. 2

3 2.1 Informal Argument That the Answer is Correct The only change made to the grammar G to produce Ĝ is removal of the rule allowing us to generate ɛ from B. Clearly, any string (of terminals) that could be generated from W in Ĝ can also be generated in G because G contains all the rules that Ĝ contains. If there is derivation of a string of terminals, say w,that can be generated in G but that cannot be generated in Ĝ, then the rule B ɛ must have been used in the derivation otherwise the derivation would also be a derivation in Ĝ. Part of the derivation must look as follows. uw v ubv uv w for some u, v (V T ). However, this part of the derivation can be replaced with uw v uav uv and thus the string w can also be derived in Ĝ. It follows that the language of the two grammars is the same. To see that Ĝ is unambiguous consider any string derived in Ĝ and its (leftmost) derivation. If the next symbol that we wish to derive is a 0 we must replace W with an A and then replace A with 0W. If the next symbol we wish to derive is a 1 then we must replace W with a B and replace B with 1W instead. We must continue until we have derived all terminal symbols we need, from left to right. Finally we must replace W with A which then can be replaced with ɛ. Since there is no choice during the entire derivation, there is only one derivation for each string and thus the grammar is unambiguous. The following sections prove that these intuitive arguments are indeed accurate. 2.2 Proving That Two Grammars Have the Same Language One way to prove that L(Ĝ) = L(G) is to prove that L(Ĝ) and L(G) are both equal to some given language L (in two separate proofs). In this case, L(Ĝ) and L(G) are each equal to T. This is implied by the following claims. Claim 1. W G w for every string w T. Sketch of Proof. This can be established by induction on the length of the string w that is mentioned in the claim. Basis: If the length of w is zero then w = ɛ, the empty string, and w can be derived from W as follows. W G A (using the rule W A) G ɛ (using the rule A ɛ). Inductive Step: Let n be an integer such that n 1. We must use the following Inductive Hypothesis to prove the following Inductive Claim in order to complete this step. Inductive Hypothesis: W G v for every string v T whose length is n 1. Inductive Claim: W G w for every string w T whose length is n. Completion of Inductive Step: Suppose, therefore, that w is a string in T with length n + 1. Since n 0, n and w must be nonempty. Since T = {0, 1}, w must begin with either a 0 or a 1. Each case will be considered separately, below. 3

4 Case: w begins with 0. In this case, w = 0x for some string x T whose length is n. It follows by the inductive hypothesis that W G x. This implies that 0W G 0x as well (see the online notes on Correctness of Context-Free Grammars if this is not clear). Therefore W G A (using the rule W A) G 0W (using the rule A 0W ) G 0x = w Case: w begins with 1. In this case, w = 1x (using the derivation described above). for some string x T whose length is n. It follows by the inductive hypothesis that W G x. This implies that 1W G 1x as well. Therefore W G B (using the rule W B) G 1W (using the rule B 1W ) G 1x = w (using the derivation described above). It has now been established that W G w in every possible case. Since w was an arbitrarily chosen string with length n + 1, this implies that W G w for every string w T with length n + 1, as required to complete the inductive step. It follow that W G w for every string w T, as claimed. Claim 2. W b G w for every string w T. The two grammars, G and Ĝ are similar enough for it to be easy to modify the proof of the first claim in order to prove the second claim. This is left as an exercise. Claim 3. L(Ĝ) = L(G). Proof. It follows by the two claims, given above, that L(Ĝ) = T = L(G). Another way to prove that two grammars G 1 and G 2 have the same language that will be helpful when Normal Forms for Context-Free Grammars are discussed is to describe a construction that can be be used to transform a derivation of a string w from the start variable of G 1, in the grammar G 1, into a derivation of the same string w from the start variable of G 2, in the grammar G 2, and then to prove that this transformation is correct establishing that L(G 1 ) L(G 2 ). Repeating the process, with the roles of the grammars G 1 and G 2 reversed, establishes that L(G 2 ) L(G 1 ) as well. Exercise: Try to do something like this. Transforming a derivation in Ĝ into one in G does not require any work at all. On the other hand, you should be able to use the ideas in Section 2.1 to prove that a derivation in G can also be transformed into a derivation in Ĝ either using induction on the number of steps in the original derivation or (even better) using induction on the number of times that the rule B ɛ is used in the derivation you start with. 4

5 2.3 Proving That a Context-Free Grammar is Unambiguous It is considerably more challenging to prove that a given context-free grammar is unambiguous than it is to prove that a grammar is ambiguous. In particular, in order to prove that a grammar G = (V, T, P, S) is unambiguous, it is necessary to prove that there is only one leftmost derivation (or rightmost derivation, or parse tree) of any given string w T from the start variable S. In the example, the grammar Ĝ is simple enough for it to be possible to prove this using mathematical induction on the length of w. A sketch of a proof that establishes this is given below. Claim 4. There is exactly one leftmost derivation of the string w from the variable W in the grammar Ĝ, for every string w T. Proof. This will be proved by induction on the length of w. Basis: Suppose that w has length zero: Then w is the empty string. Notice that W bg A (using the rule W A) bg ɛ (using the rule A ɛ) is a leftmost derivation of the empty string from W in Ĝ. We must show there is no other leftmost derivation of the empty string from W in this grammar. Suppose, to obtain a contradiction, that some other leftmost derivation of the empty string from W in Ĝ. Either this leftmost derivation begins with an application of the rule W A (just like the leftmost derivation that is listed above) or it does not. These cases will be considered separately. Case: The other leftmost derivation also begins with the rule W A. Consider what happens after this rule is applied: Since we are left with the variable A we must continue by using some rule with A on its left hand side. Now, if we used the rule A ɛ then we would be done but the resulting leftmost derivation would not be different from the above one, after all! So, this cannot be the case. It follows that we must use some other rule with A as its head. There is only one such rule, namely, the rule A 0W, so our leftmost derivation must begin as follows: W bg A (using the rule W A) bg 0W (using the rule A 0W ) However, now we are stuck! It is impossible to get rid of the leftmost 0 that is generated, so that the only strings in T that can possibly be generated have the form 0v for v T. Question: How could you prove the above claim in a more formal way? In particular, it is not possible to derive the empty string, because the empty string does not begin with a 0. Thus this case cannot arise. Case: The other leftmost derivation does not begin with the rule W B. In this case, we must begin by using the rule W B. Since there is only one rule with B as its head, 5

6 our leftmost derivation must begin as follows. W bg B (using the rule W B) bg 1W (using the rule B 1W ) However, now we are stuck once again: The only strings in T that can possibly be derived are strings that begin with 1. Thus this case cannot arise either. We have now obtained a contradiction, so the only assumption that we made must be false: There is only one leftmost derivation of the empty string from W in the grammar Ĝ. Inductive Step: Let n be an integer such that n 1. We must use the following Inductive Hypothesis to prove the following Inductive Claim: Inductive Hypothesis: If v T such that v = n 1 then there is exactly one leftmost derivation of v from W in the grammar Ĝ. Inductive Claim: If w T such that w = n then there is exactly one leftmost derivation of w from W in the grammar Ĝ. Completion of the Inductive Step: Let w be a string in T with length n. Since n 1 w is not the empty string, and T = {0, 1}, either w begins with a 0 or w begins with 1. These cases are considered separately, below. Case: w begins with 0. In this case, w = 0v for some string v T such that the length of v is n 1. Notice that it follows by the inductive hypothesis that there is exactly one leftmost derivation of v from W in Ĝ. To complete the proof in this case, you should give an argument like the one in the basis to show that every leftmost derivation of w from W in Ĝ must begin as follows. W bg A (using the rule W A) bg 0W (using the rule A 0W ) You can prove this in much the same way as a similar result in the basis: Notice that there is only a finite (and small!) number of other ways to begin a leftmost derivation. Consider every one of these other ways and obtain a contradiction in every case. Suppose, to obtain a contradiction, that there are at least two leftmost derivations of w from W in Ĝ that begin in the above way. Notice that if you deleted the first two lines of these, and then removed the leftmost 0 after that, you would produce two different leftmost derivations of v from W in Ĝ. Since it follows by the inductive hypothesis that there is only one leftmost derivation of v, this results in a contraction. It follows that there is only one leftmost derivation of w from W in Ĝ in this case. Case: w begins with 1. This case is very, very similar and its completion is left as an exercise. It follows (after the above exercise is completed) that there is exactly one leftmost derivation of w from W in each case. Since w was an arbitrarily chosen string with length n in T, this completes the inductive step. The desired result now follows by induction on the length of w. 6

7 About This and Other Proofs As the above proof should suggest, you should often consider a proof by contradiction: Assume that two leftmost derivations of some string (from some variable) exist and then establish a contradiction. In more complicated examples it can be necessary to prove that several other context-free grammars namely, the grammars that you get by choosing different start variables (and changing nothing else) are unambiguous too. Mutual inductions on the length of the string being derived, or the length the (shorter) derivation being considered, can be useful when you do this. In other cases, you might not need to use mathematical induction at all just as you do not necessarily need to use mathematical induction in some cases when you are trying to show that a given context-free grammar has a given language. Exercise: Modify the above proof to show that the following context-free grammars are unambiguous as well: 1. G A : This is the same context-free grammar as Ĝ, except that A is the start variable instead of W. 2. G B : This is the same context-free grammar as Ĝ, except that B is the start variable instead of W. 3 Proving That a Context-Free Language is Not Inherently Ambiguous Suppose that you are given a context-free language L probably, represented by a given context-free grammar G such that L(G) = L. Suppose, as well, that you wish to prove that L is not inherently ambiguous. One way to do this is to prove that the given context-free grammar, G, is unambiguous. Another way to do this (which will work if G is ambiguous, instead), is to give another context-free grammar Ĝ, and then to show that L(Ĝ) = L(G) (so that L(Ĝ) = L) and that Ĝ is unambiguous. In other words, another way to do this, that works when G is an ambiguous context-free grammar, is to remove ambiguity from the grammar G, as described in the previous section of this document. For example, if you were given the context-free language L = {0, 1}, represented by the context-free grammar G that was presented at the beginning of this document, then you could prove that L is not inherently ambiguous by giving the grammar Ĝ that has been presented, above, along with the claims and proofs. that are used to show that L(Ĝ) = L(G) and Ĝ is unambiguous, and that are found in the previous section. 4 Proving That a Context-Free Language is Inherently Ambiguous As noted in lectures and the textbook, the following language L is inherently ambiguous. L = {a i b j c k d l either i = j 0 and k = l 0, or i = l 0 and j = k 0} An informal argument that this is true can be found in Section of the text. A more formal proof of this can be found in the first edition of the textbook. However, this is for interest only: Proving that a given context-free language is inherently ambiguous is beyond the scope of this course, and students will not be asked to do this, or to describe how to do this, in CPSC

8 Instead, it will be sufficient for students to understand (and accept) that inherently ambiguous contextfree languages really do exist. 8