Final Examination
Formal Language and Automata Theory
Name :________________
Reg.No.:__________________
2011/12/23
Score:______________
1. [20pts] Let L2 = { x {a,b}* | #a(x) mod 2 = 1 and #b(x) mod 2 = 0 }.
(a) Find a Strongly right linear grammar G1 to generate L2.
(b) Find a strongly left linear grammar to G2 generate L2.
Hint: You can draw a FA for L2 first and then translates the FA to G1 and G2.
2. [15 pts] Let G be the context free grammar : S   | SS | a S b | b S a
(a) Show that for all sentential form x ∈{a,b,S}*, if S * x then #a(x) = #b(x). Hence, if x ∈ L(G),
then #a(x) = #b(x).
(b) Show that for all string x ∈{a,b}*, if #a(x) = #b(x) then S * x .
Hence L(G) = {x | #a(x) = #b(x) }. Hint: Both can be proven by induction.
ANS:
(a) pf: The proof is by induction on the number k such that S k x.
Basis case: k = 0. Then x = S (not ) and #a(x) = 0 = #b(x).
Ind. case: k = t + 1 > 0. Then we must have S t S  = x) where S  is a grammar rule.
By ind.hyp., we have #a(S) = #b(S). In other words, #a() = #b()
There are 4 cases to consider depending on the last rule S applied.
case1,2:  =  or SS, then #a(x) = #a() = #a() =#b() = #b() = #b(x)
case3,4:  = aSb or bSa, then #a(x) = #a() = 1+ #a() =1+ #b() = #b() = #b(x).
(b). The proof is by induction on the length |x| of x.
Basis case: |x| = 0. Then x =  and we have S1x by applying the rule S  .
Ind. case: |x| = t + 1 > 0.
The there are 4 cases to consider:
case 0: #a(x)  #b(x). Then the implication trivially holds.
case 1: #a(x) = #b(x) and x = ayb or bya, starting and ending with different symbol.
Then we must have |y| < |x| and #a(y) = #b(y). Hence, by ind. hyp., S * y.
So we have S  a S b * ayb = x if x = ayb or
S  b S a * bya = x if x = bya.
case 2: #a(x) = #b(x) and x = aya. Now for each prefix z of y, let d(z) = #b(z) - #a(z).
1
Obviously, d(z) = 0 if z =  and d(z) = 2 if z = y. Hence there must exist a prefix u of y=
uv such that d(u) = 1. But then we have x = auva with #a(au) = #b(au), #a(vu) = #b(vu), |au|<
|x| and |va| < |x|. Hence, by ind. hyp., S * au and S * va.
As a result, we have S  SS * au S * au va = x.
case 3: #a(x) = #b(x) and x = byb. Similar to case 2. Omitted here.
3. [20 pts] Give context free grammars for each of the following languages:
(a) { x ∈ {a,b}* | x is a palindrome, i.e., x = rev(x) .}
(b) { x ∈ {a,b}* | #a(x) = 2 #b(x) } .
(c) { xy ∈ {a,b}* | |x| = |y| but x  y } [10pts] .
ANS; (a) S   | a | b | aSb | bSa
(b) S  e | b S a S a | a S b S a | a S a S b
(c) Note that a string z belongs to the set iff z = uavsbt or ubvsat for some u,v,s,t with |u|=|v| and |s|=|t|.
We thus have S  A B | B A
A  a | TAT B  b | TBT T  a | b.
4. (15 pts) We want to apply the game-theoretic version of the pumping lemma for CFL to show that the
language L = {xy ∈ {0,1}* | x ∈ {0,1}* and y is the complement of x } is not context-free. Fill in the
following procedure to derive a winning strategy for Y:
(a) Suppose D pick a number k > 0.
(b) Y choose a string z = _1K0K1K0K1K0K__. What is the restriction the selected z must obey?
_________z ∈ L and |z|  k.________
(c) Suppose D decompose z into uvwxy. What are the constraints that decomposition uvwxy must
satisfy : ___|vwx| ≤ k and |vx|  0___
(d) How should Y choose i : _______let i = 0_______
Under what condition will we say that Y wins the game?
__________uviwxiy  L.__________________________
Explain why Y always win the game according to the strategy you wrote at step (b) and (d).
ANS:
If vwx contains only one kind of symbols (all 0’s or all 1’s), then #0(uviwxiy)  #1(uviwxiy).
Hence uviwxiy  L.
Reconsider the string z = 1K A 0K B 1K C 0K D 1K E 0K__where A ~ E are boundary positions the
neighbors of which are distinct symbols. If vwx contains both 0 and 1, then it must cross one and only
one of the five boundary positions.
case 1: if vwx crosses A, then the neighboring blocks of A of 1’s and 0’s both diminish, but the
corresponding neighboring blocks of D of 0’s and 1’ does not diminish simultaneously. Hence the
resulting string uviwxiy  L.
case 2: vwx corsses D. Similar to case 1.
case 3: if vwx crosses E, then the neighboring blocks of E of 1’s and 0’s both diminish, but the
corresponding neighboring blocks of B of 0’s and 1’ does not diminish simultaneously. Hence the
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resulting string uviwxiy  L.
case 4: vwx corsses B. Similar to case 3.
case 5: vwx corsses C. Then the neighboring blocks of C of 1’s and 0’s both diminish, but the
corresponding first blocks of 0’s and corresponding last block of 1’ does not diminish. Hence the
resulting string uviwxiy  L.
Since in all cases, we have uviwxiy  L, Y always wins the game.
5. [15 pts] Let M = ({q}, {a, b}, {A, B, S}, , q, S, {q}) be a single-state NPDA which accept by empty
stack. PDA M has q as the start symbol and S as the bottom stack symbol and the transition relation  is
given in the following table:
top of stack
S
A
B
(state, input)
(q,a)
(q,B)
(q,S)
(q,),(q,BA)
(q,b)
(q,A)
(q,),(q,AB)
(q,S)
Find a context free grammar G which is equivalent to M,
ANS: S  aB | bA A aS | b | bAB B a | aBA | S.
6. [10pts] Given the following Grammar G3: SASA | aB | a
AB|S|
(a) List all symbols which are nullable: ___A,B_______________________
(b) Find a grammar G4 containing no -rule such that L(G4) = L(G3) –{}.
(c) Find a grammar G5 containing no unit rule such that L(G5) = L(G4)
ANS;
(b) S  ASA | aB | a | AS | SA | S | b
(c) S  ASA | aB | a | AS | SA | b
A  ASA | aB | a | AS | SA | b
B  b.
A B | S | b
B  b.
7. [5 pts] Let G be the CFG : S  aAc | aSc A b | bA
Find a grammar equivalent to G which is in Chomsky normal form (cnf).
aA □
c | □
aS □
c
ANS: S  □
b A
Ab | □
aA
□
aS
□
a
□
b
□
c
□
a A
 □
a S
 □
a
b
c
8. [20 pts] Consider the grammar:
SAB | BC
A  BA | a B CC | b C  Ab | a
(a) The input ‘baaba’ is a member of L(G). Draw a parse tree for it. [5pts]
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Bb|
(b) Give a left-most derivation for the input ‘baaba’. [5pts]
(c) Run the CKY algorithm on the input string ‘aabab’. Fill the results in the following table.
0a1a2b3a4b5
4
b,B
3
a,A,C
2
b,B
1
a,A,C
0
a,A,C
1
2
3
5
5
4