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Lecture 12 Examples of CFL n n Example 1. L={0 1 | n > 0} is a CFL. To show that L is a CFL, we need to construct a CFG G generating L. Let G = (V, Σ, R, S) where V = {S} Σ = {0, 1} R ={S → ε|0S1 } Example 2. Construct CFG to generate R L={xx | x is in (0+1)*}. G = (V, Σ, R, S) where V = {S} Σ = {0, 1} R = { S → ε | 0S0 | 1S1} Example 3 L={ x (0+1)* | #0(x) = #1(x)} Idea: what relations exist between longer strings and shorter strings in this language? Consider four cases: Case 1. x = 0w1. Then x is in L iff w is in L. If we use S to represent a string in L, then the relation in Case 1 can be represented As a rule S → 0S1 Case 2. x=1w0. Similar to case 1. This case gives rule S → 1S0 Case 3. x=0w0. Suppose w = w1w2···wn. Consider the following sequence: #0(0) - #1(0) > 0, #0(0w1) - #1(0w1), …, #0(0w1···wn) - #1 (0w1···wn) < 0. Note that in this sequence, two adjacent numbers Have difference 1. Therefore, there exists I such that #0(0w1···wi) - #1 (0w1···wi) = 0 This means that x is the concatenation of two shorter strings in L. So, we have a rule S → SS Case 4. x=1w1. Similar to Case 3. Based on the above analysis, we have CFG G=(V, Σ, R, S) where V = {S} Σ = {0, 1} R = {S → ε | 0S1 | 1S0 | SS} Each nonterminal symbol represents a language. Each rule represents a relationship between languages represented by nonterminal symbols. CFL is closed under union. Proof. Suppose A = L(GA) and B = L(GB) where GA = (VA, ΣA, RA, SA) GB = (VB, ΣB, RB, SB) Without loss of generality, assume VA ∩ VB= Ǿ. (Otherwise, we may change some nonterminal symbols.) Then A U B = L(G) for G=(V, Σ, R, S) where V = VA U VB U {S} Σ = ΣA U ΣB R = RA U RB U {S → SA | SB } Example 4 L = { 0 1 | m ≠ n, m, n > 0} m m n n m n L = {0 1 | m > n > 0} U {0 1 | n > m > 0} Hence, L = L(G) for G = ({S, SA, SB}, {0,1}, R, S) where R = {S → SA | SB, SA → 0 | 0SA | 0SA1, SB →1 | SB1 | 0SB1} CFL is closed under concatenation. Proof. Suppose A = L(GA) and B = L(GB) where GA = (VA, ΣA, RA, SA) GB = (VB, ΣB, RB, SB) Without loss of generality, assume VA ∩ VB= Ǿ. (Otherwise, we may change some nonterminal symbols.) Then AB = L(G) for G=(V, Σ, R, S) where V = VA U VB U {S} Σ = ΣA U ΣB R = RA U RB U {S → SASB } R + Example 5 L = {xx w | x (0+1), w (0+1)*} + L = {xx | x (0+1) }{0,1}* R L=L(G) for G = ({S, SA, SB}, {0, 1}, R, S) where R = { S → SASB, SA→ 00 | 11 | 0SA0 | 1SA1, SB→ ε | 0SB | 1SB } CFL is closed under star-closure. Proof. Suppose L = (G) for G=(V, Σ, R, S). Then L* = L(G*) for G* =(V, Σ, R*, S) where R* = R U { S → ε | SS}. S represents L and S * represents L * . Then S * | S * S . So, S * S . Example 6 L=(0+1)*00 L=L(G) for G=({S, A}, {0,1}, R, S) where R={S → A00, A → ε | AA | 0 | 1 } The role of nonterminat symbol. Every nonterminal symbol A represents a language which can be generated by using A as start symbol. Example 7. m n p q L={a b c d | m+n = p+q, m, n, p, q > 0} Let S represent L, A represent {b n c n| n > 0} n B represent {am b c p | m+n = p, m, n, p > 0} n p q C represent {b c d | n = p+q, n, p, q > 0} Then we can find relations S → aSd | B | C, B → aBc | A, C → bCd | A, A → bAc | ε. Example 8 L ={x (0+1)*| x ≠ ww for any w (0+1)*} Let us analyze what would happens for x in L. Case 1. |x| = odd. In this case, x is either in language A = {u0v | |u|=|v|, u, v (0+1)*} or B = {u1v | |u|=|v|, u, v (0+1)*}. Case 2. |x| = even. Write x = x1 x2 … xm y1 y2 …ym . There exists i such that xi ≠ yi . Subcase 2.1 x = x1… xi-10xi+1 …xmy1…yi-11yi+1…ym. x is in AB Subcase 2.2 x = x1… xi-11xi+1 …xmy1…yi-10yi+1…ym. x is in BA Thus, L = A+B+AB+BA. So, L=L(G) for G=({L,A,B}, {0,1}, R, L) where R ={ L → A | B | AB | BA, A → 0 | 0A0 | 0A1 | 1A0 | 1A1, B → 1 | 0B0 | 0B1 | 1B0 | 1B1}.