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自動機理論與正規語言
期中考試
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一、 是非題(40%)
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1. Compared to an accepter, a transducer is a more general automation, capable of
producing strings of symbols as output.
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2. Every language accepted by an nfa is regular.
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3. The transition function of a deterministic finite accepter (dfa) should be a total
function.
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4. A linear grammar is a grammar in which at most one alphabet can occur on the right
side of any production, without restriction on the position of this alphabet.
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5. A language L is regular if and only if there exists a left-linear grammar G such that L
= L(G).
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6. The pumping lemma for regular languages can be defined as follows:
Let L be a regular language. Then these exists some positive integer m such that any
w L with | w | m can be decomposed as w  xyz , with | xy | m , and | y | 1 ,
such theat wi  xy i z is also in L for all i  0,1,2,3,... .
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7. In the exhaustive search parsing, if we restrict ourselves to leftmost derivations, we
can have no more than |P| sentential forms after one round, no more than |P|2
sentential forms after second round, and so on. (here P is the set of production rules
of a context-free grammar G  (V , T , S , P) )
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8. A regular language can be inherently ambiguous.
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9. If G is an s-grammar, then any string w in L(G) can be parsed with an effort
proportional to |w|.
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10. L  ww R : w  {a, b}*
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11.
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12. Language L  {a n bm c k : k  n  m , n, m, k  0} is not context-free.
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13. Let L1 and L2 be regular languages, then L  w : w  L1 , w R  L2
language.
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14. Two finite accepters are said to be equivalent if they both accept the same language.
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15.  ,  , and a   are all regular expressions. These are called primitive regular
expressions.
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16. Let G  (V , T , S , P) be a context-free grammar. In its derivation tree, every leaf
has a label from V T  .
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17. The family of regular languages is closed under union, intersection, concatenation,
complementation, reversal, difference, and star-closure.

L  a b a
n
l
k

is a regular language.
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: n  l  k  5 is a context-free language.

1
 is also a regular
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18. Let L1 and L2 be a language on the same alphabet, then right quotient of L1 with L2
is defined as L1 L2  {x : xy  L2 for some y  L1} .
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19. A context-free grammar G  (V , T , S , P) is said to be a simple grammar if all its
productions are of the form A  ax , where A V , a  T , and x  T * .
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20. If L is a context-free language for which there exists an ambiguous grammar, then L
is said to be ambiguous.
二、 (9%)請寫出下列 grammars 所產生出的 languages。
1. (3%)The grammar G  ({S},{a, b}, S , P) , with productions S  aSa , S  bSb , S  a .
Ans:
2. (3%)The grammar G  ({S, S1 , S2 },{a, b}, S , P) , with productions S  S1ab , S1  S1ab | S2 ,
S2  a .
Ans:
3. (3%) The grammar G  ({S , A, B},{a, b}, S , P) , with productions
S  aB | bA , A  a | aS | bAA , B  b | bS | aBB .
Ans:
三、 (6%)針對下列各 regular languages，請寫出適當的 right-linear grammars。
1.(3%) L  {w : na ( w) and nb ( w) are both odd}
Ans:
2.(3%) L(( aab * ab)*)
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四、 (3%)請寫出下列的 finite automata 所能接受的 language。
a
b
q1
q0
b
q2
a
a
Ans:
2
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五、 (6%)請寫出下列 regular languages 的 regular expressions。
1. (3%) L  {vwv : v, w  {a, b}*, | v | 2}
Ans:
2. (3%) L(r)  w {0,1}* : w has at least one pair of consequiti ve zeros

Ans:
六、 (8%)請畫出可接受下列語言的 transition graph。(不可以畫 generalized transition graph)
1. (4%) L( ab * a*)  L( a * b * a )
Ans:
2. (4%) L  w  {a, b}* : na ( w) is even and nb ( w) is odd
Ans:
七、 (4%) Let L1  L(bba * baa*) and L2  L(ab*) . Find L1 L2 .
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八、 (8%) 給定一 grammar G  ({S , A, B},{a, b}, S , P) , with productions S  aAB , A  aBb ,
B  A| 。
1. (4%)請畫出此一 grammar 的 dependency graph.
Ans:
2. (4%)請畫出 string aaabb 的 derivation tree.
Ans:
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九、 (8%)針對下列各 languages，請寫出適當的 context-free grammars。
1. (4%) L  {a n b m | n  m  3, n, m  0}
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2. (4%) L  {w  {a, b}* : na (w)  nb (w) and na (v)  nb (v), where v is any prefix of w}
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十、 (8%) Simplification of contest-free grammars.
1.(4%) Eliminate useless productions from the grammar
S  aS | AB
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A  bA | Bc
B  AA
C  cSD | c
D  ddd
what language does this grammar generate?
2.(4%) Eliminate all  -productions from
Ans:
S  ABaCD
A  BC
B  b|
C  D |
Dd
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