Lecture 12 Examples of CFL

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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}.
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