CS 461 – Sept. 23

1. CS 461 – Sept. 23 Context-free grammars • Derivations • Ambiguity • Proving correctness

2. Derivations • A sequence of steps showing how a word is generated using grammar rules. • Drawing a diagram or tree can help. • Example: derivation of 1110 from this grammar: S  AB S A  ε | 1A A B B  ε | 1B0 1A 1B0 11A 1ε0 11ε 10 11

3. Derivations, con’d • There are several ways to express a derivation • Tree  • Leftmost derivation = as you create string, replace leftmost variable. Example: S  AB  1AB  11AB  11B  111B0  1110 • Rightmost derivation • Often, grammars are ambiguous • For some string,  2+ derivation trees. Or, equivalently:  2+ leftmost derivations. • Example?

4. More examples • Even number of 0’s • Words of form { 0i 1j } where j is constrained. i  j  2i (not to be confused with { 0n 12n } ) i  j  2i + 3 i  j  1.5i • Next topic: how do we show that our grammar is correct?

5. Correctness • Given a language L and a grammar G, how do we know if L = L(G)? Must show: • L  L(G). All words in L can be derived by grammar. Tell how to derive the words. • L(G)  L. All words derived by grammar are in L. Use induction.

6. First example • Show that S  ε | 0S1 is a CFG for the language { 0n1n } • Step #1: Any word in 0n1n can be generated by the grammar. We explain the derivation. • Apply the rule 0S1 n times • Finally, apply the ε rule. • Step #2: All words generated by the grammar are of the form 0n1n. • Induction on the number of times we use a rule. • S  ε for the basis, and S  0S1 for the inductive step.

7. Second example L is { even number of 0s } and G is S  ε | S1 | S0S0 Step #1. Suppose w  L. How can G generate w? If w = ε, we’re done. Do the following until w is ε: • If w ends with 1s, take them off by virtue of “S1”. • Now w ends with 0, so take 2 0’s away by virtue of “S0S0”. We know there must be 2 0’s in w because w had an even number to start with.

8. Illustration S  ε | S1 | S0S0 Let’s see how we can derive “010100”: • The last 2 0s: S  S0S0  ε0ε0 = 00 • 0101 is 010 + 1 010 is S  S0S0  ε0S10  ε0ε10 = 010 1 is S  S1  ε1 = 1 • We can reconstruct derivation from beginning in form of a tree. 

9. L(G) is in L S  ε | S1 | S0S0 Step #2: Need to show all generated words are in L. Base case: ε is in L. It has even number of 0s.  Recursive case. Let w be a word generated by the grammar. Assume it has even # 0s. We can create a larger word by applying rule “S1” or “S0S0”. S1 adds no more 0s. S0S0 adds two 0s. Either way the # of 0s stays even.

10. Extra example • Let’s show that this grammar G S  ε | 0S1 | 00S111 generates the language L = { 0i 1j | i  j  1.5i } • Step #1: any word in L can be generated • This is the hard part, so let’s motivate the proof with an example or two. • Step #2: all words generated are in L • Just induction again.

11. L is in L(G) S  ε | 0S1 | 00S111 • How could we generate 011 116? • We have 5 extra 1s. So we use rule 00S111 five times. • Undoing these steps, # 0s = 11-5*2 = 1 and # 1s = 16-5*3 = 1. Then we use 0S1 once and we are left with ε. • Okay, how about 012 116? • We have 4 extra 1s, so use rule 00S111 four times. • Undoing these steps, # 0s = 12-4*2 = 4 and # 1s = 16-4*3 = 4. They match! So use “0S1” 4 times.

12. Thinking out loud… S  ε | 0S1 | 00S111 Let w = 0i 1j L. In other words, i  j  1.5i Consider the number j – i. This is the number of times to apply rule #3 (00S111). Note that using rule #3 (j – i) times will account for 2(j – i) zeros. Then apply rule #2 (0S1) the “appropriate number” of times. How many? Well, we want i 0’s and we’ve already got 2(j – i), so we need the difference: i – 2(j – i) = 3i – 2j.

13. Finishing step 1 S  ε | 0S1 | 00S111 Let w = 0i 1j L. In other words, i  j  1.5i Use rule #1 once to start with empty word. Use rule #3 (j – i) times. Use rule #2 (3i – 2j) times. Total # 0s = 3i – 2j + 2(j – i) = i Total # 1s = 3i – 2j + 3(j – i) = j Thus, the word can be generated.

14. L(G) is in L S  ε | 0S1 | 00S111 Now the easy part. Need to show that when we apply any rule, we preserve: i  j  1.5i Base case: ε has no 0s or 1s. 0  0  1.5*0 √ Recursive case. Let w be generated by the grammar with i 0s and j 1s satisfying i  j  1.5i . If we apply either rule 0S1 or 00S111, we can show (i + 1)  (j + 1)  1.5(i + 1) (i + 2)  (j + 3)  1.5(i + 2) (Need to work out arithmetic.)

15. Think about… • Can you write a CFG for { 0i 1j | 0.5i  j  1.5i } ? in other words, the ratio of 1s to 0s must be between ½ and 3/2.