MTH401A Theory of Computation. Lecture 17

Transcription

1 MTH401A Theory of Computation Lecture 17

2 Chomsky Normal Form for CFG s

3 Chomsky Normal Form for CFG s For every context free language, L, the language L {ε} has a grammar in which every production looks like, either 1. S AB, or 2. B a

4 Chomsky Normal Form for CFG s For every context free language, L, the language L {ε} has a grammar in which every production looks like, either 1. S AB, or 2. B a Further, the grammar has no useless symbols.

5 Chomsky Normal Form for CFG s Eliminating useless symbols : A symbol is useful if it appears in some derivation of some terminal string from the start symbol. It is useless, otherwise. Eliminate all useless symbols by : 1. Eliminate symbols that derive no terminal string. 2. Eliminate unreachable symbols.

6 Chomsky Normal Form for CFG s Example : Eliminate Variables S AB C A 0A 0 B 1B C 1 Basis : A and C are identified because of A 0 and C 1. Induction : S is identified because of S C. Nothing else can be identified. S C A 0A 0 C 1

7 Chomsky Normal Form for CFG s Algorithm to eliminate unreachable symbols : 1. Basis : We can reach S (the start symbol). 2. Induction : If we can reach A and A α then we can reach all symbols in α. 3. Algorithm : Remove from the grammar all symbols not reachable from S and all productions that involve these symbols.

8 Chomsky Normal Form for CFG s Example : Eliminate Useless Symbols S AB 0 A C 0 B 0B C 1 1. If we eliminate unreachable symbols first we would find that everything is reachable. 2. A, C, 1 would never get eliminated.

9 Chomsky Normal Form for CFG s Discovering Nullable Symbols : 1. To eliminate all ε- productions, we first need to discover all nullable variables (variables A such that A * ε). 2. Basis : If there is a production A ε then A is nullable. 3. Induction : If there is a production A α and all symbols in α are nullable then A is nullable.

10 Chomsky Normal Form for CFG s Example : Eliminate ε- Productions S ABC A 0A ε B 1B ε C ε 1. A, B, C, S are nullable. 2. New grammar S ABC AB AC BC A B C A 0A 0 B 1B 1

11 Chomsky Normal Form for CFG s Example : Eliminate ε- Productions S ABC A 0A ε B 1B ε C ε 1. A, B, C, S are nullable. 2. New grammar S ABC AB AC BC A B C A 0A 0 B 1B 1 C is now useless. Eliminate its productions

12 Chomsky Normal Form for CFG s Example : Eliminate ε- Productions S ABC A 0A ε B 1B ε C ε 1. A, B, C, S are nullable. 2. New grammar S ABC AB AC BC A B C A 0A 0 B 1B 1 C is now useless. Eliminate its productions

13 Chomsky Normal Form for CFG s Unit Productions : A unit production is one whose right side consists of exactly one variable. Algorithm : Find all pairs (A,B) such that A * B by a sequence of unit productions only. Basis : Pairs (A,A) are there. Induction : If we have found (A,B) and B C is a unit production then add (A,C).

14 Chomsky Normal Form for CFG s Example : Unit Productions S A B CC DD B BOZO CC ABC

15 Chomsky Normal Form for CFG s Example : Unit Productions S A B CC DD B BOZO CC ABC Unit Productions sequences S * B

16 Chomsky Normal Form for CFG s Example : Unit Productions S A B CC DD BOZO CC ABC 0 B BOZO CC ABC S A B CC DD B BOZO CC ABC

17 Chomsky Normal Form for CFG s Example : Unit Productions S A 11 A B 1 B S 0

18 Chomsky Normal Form for CFG s Example : Unit Productions S A 11 A B 1 B S 0 Unit Productions sequences S * A, S * B; A * B, A * S; B * S, B * A;

19 Chomsky Normal Form for CFG s Example 5 : Useless Symbols S A A B B S S A 11 A B 1 B S 0

20 Chomsky Normal Form for CFG s Example 5 : Useless Symbols S A A B B S A and B are useless symbols. No way to get them from S. S A 11 A B 1 B S 0

21 Chomsky Normal Form for CFG s Cleaning Up a Grammar : Theorem : If L is a CFL, then there is a CFG for L {ε} that has No useless symbols. No ε- productions. No unit productions. In other words, every right side is either a single terminal or has length 2.

22 Chomsky Normal Form for CFG s Chomsky Normal Form Steps: Step 1 : Clean up the grammar. Step 2 : For each right side that is not a single terminal, make it all variables. For each terminal symbol a, create a new variable A a and a production A a a. Replace a by A a in right sides of length 2. Step 3 : Break right sides of length longer than 2 into a chain of productions with right sides of two variables.

23 Chomsky Normal Form for CFG s Example : Terminals S 0A1B DEF A CC

24 Chomsky Normal Form for CFG s Example : Terminals S 0A1B DEF A CC Z 0 O 1

25 Chomsky Normal Form for CFG s Example : Terminals S ZAOB DEF A CC Z 0 O 1

26 Chomsky Normal Form for CFG s Example : Long productions S ZAOB DEF A CC Z 0 O 1

27 Chomsky Normal Form for CFG s Example : Long productions S ZAOB DEF A CC Z 0 O 1 Too long.

28 Chomsky Normal Form for CFG s Example : Long productions S ZAOB DEF A CC Z 0 O 1 Too long. New group of symbols.

29 Chomsky Normal Form for CFG s Example : Long productions S ZM DEF A CC Z 0 O 1 M AOB

30 Chomsky Normal Form for CFG s Example : Long productions S ZM DEF A CC Z 0 O 1 M AOB New group of symbols.

31 Chomsky Normal Form for CFG s Example : Long productions S ZM DH A CC Z 0 O 1 M AN N OB H EF

32 Chomsky Normal Form for CFG s Chomsky Normal Form : CFG is said to be in a Chomsky Normal Form if the grammar has no useless symbols and every production is of one of these two forms : 1. A BC (right side is two variables), 2. A a (right side is a single terminal). Theorem : If L is a CFL, then L {ε} has a CFG in Chomsky normal form.

33 Pushdown Machines

34 Pushdown Machines

35 Pushdown Machines Finite State Machine

36 Pushdown Machines Finite State Machine

37 Pushdown Machines

38 Pushdown Machines Finite Pushdown State Machine

39 Formalism : Pushdown Machines A pushdown machine is described by: 1. A finite set of states (Q, typically). 2. An input alphabet (Σ, typically). 3. A stack alphabet (Γ, typically). 4. A transition function (δ, typically). 5. A start state (q 0, in Q, typically). 6. A start symbol (Z 0 in Γ, typically). 7. A set of final states (F Q, typically).

40 Convention : Pushdown Machines a, b, c, are input symbols. Sometimes, we also allow ε as a possible value., X, Y, Z are stack symbols., w, x, y, z are strings of input symbols. α, β, γ, are strings of stack symbols.

41 Pushdown Machines The Transition Function :

42 Pushdown Machines The Transition Function : Takes three input arguments. 1. A state, in Q. 2. An input, which is either a symbol in Σ or ε. 3. A stack symbol in Γ.

43 Pushdown Machines The Transition Function : Takes three input arguments. 1. A state, in Q. 2. An input, which is either a symbol in Σ or ε. 3. A stack symbol in Γ. δ(q, a, Z) is a set of zero or more actions of the form (p, α). p is a state and α is a string of stack symbols.

44 Pushdown Machines Actions of The Pushdown Machine : If δ(q, a, Z) contains (p, α) among its actions, then one thing the pushdown machine can do in state q, with a at the front of the input, and Z on top of the stack is : 1. Change the state to p. 2. Remove a from the front of the input (a may be ε). 3. Replace Z on the top of the stack by α.

45 Example : Pushdown Machines Construct a pushdown machine to accept {0 n 1 n : n 1}.

46 Example : Pushdown Machines Construct a pushdown machine to accept {0 n 1 n : n 1}. The states :

47 Example : Pushdown Machines Construct a pushdown machine to accept {0 n 1 n : n 1}. The states : 1. q = start state. We are in state q if we have seen only 0 s so far.

48 Example : Pushdown Machines Construct a pushdown machine to accept {0 n 1 n : n 1}. The states : 1. q = start state. We are in state q if we have seen only 0 s so far. 2. p = we have seen at least one 1 and may now proceed only if the inputs are 1 s.

49 Example : Pushdown Machines Construct a pushdown machine to accept {0 n 1 n : n 1}. The states : 1. q = start state. We are in state q if we have seen only 0 s so far. 2. p = we have seen at least one 1 and may now proceed only if the inputs are 1 s. 3. f = final state; accept.

50 Pushdown Machines Example (continued) : The stack symbols :

51 Pushdown Machines Example (continued) : The stack symbols : 1. Z 0 = start symbol. Also marks the bottom of the stack.

52 Pushdown Machines Example (continued) : The stack symbols : 1. Z 0 = start symbol. Also marks the bottom of the stack. 2. X = marker, used to count the number of 0 s seen on the input.

53 Pushdown Machines Example (continued) : The transitions :

54 Pushdown Machines Example (continued) : The transitions : 1. δ(q, 0, Z 0 ) = {(q, XZ 0 )}.

55 Pushdown Machines Example (continued) : The transitions : 1. δ(q, 0, Z 0 ) = {(q, XZ 0 )}. 2. δ(q, 0, X) = {(q, XX)}. These two rules make sure that one X is pushed into the stack for each 0 read from the input.

56 Pushdown Machines Example (continued) : The transitions : 1. δ(q, 0, Z 0 ) = {(q, XZ 0 )}. 2. δ(q, 0, X) = {(q, XX)}. These two rules make sure that one X is pushed into the stack for each 0 read from the input. 3. δ(q, 1, X) = {(p, ε)}. When we see a 1, go to state p and pop one X.

57 Pushdown Machines Example (continued) : The transitions : 1. δ(q, 0, Z 0 ) = {(q, XZ 0 )}. 2. δ(q, 0, X) = {(q, XX)}. These two rules make sure that one X is pushed into the stack for each 0 read from the input. 3. δ(q, 1, X) = {(p, ε)}. When we see a 1, go to state p and pop one X. 4. δ(p, 1, X) = {(p, ε)}. Pop one X per 1.

58 Pushdown Machines Example (continued) : The transitions : 1. δ(q, 0, Z 0 ) = {(q, XZ 0 )}. 2. δ(q, 0, X) = {(q, XX)}. These two rules make sure that one X is pushed into the stack for each 0 read from the input. 3. δ(q, 1, X) = {(p, ε)}. When we see a 1, go to state p and pop one X. 4. δ(p, 1, X) = {(p, ε)}. Pop one X per δ(p, ε, Z 0 ) = {(f, Z 0 )}. Accept at bottom.

59 Pushdown Machines Example (continued) : In action Z q

60 Pushdown Machines Example (continued) : In action Z q δ(q, 0, Z 0 )

61 Pushdown Machines Example (continued) : In action Z q δ(q, 0, Z 0 ) = {(q, XZ 0 )}

62 Pushdown Machines Example (continued) : In action X Z 0 q

63 Pushdown Machines Example (continued) : In action X Z 0 q δ(q, 0, X)

64 Pushdown Machines Example (continued) : In action X Z 0 q δ(q, 0, X) = {(q, XX)}

65 Pushdown Machines Example (continued) : In action X X Z 0 q

66 Pushdown Machines Example (continued) : In action X X Z 0 q δ(q, 0, X)

67 Pushdown Machines Example (continued) : In action X X Z 0 q δ(q, 0, X) = {(q, XX)}

68 Pushdown Machines Example (continued) : In action X X X Z 0 q

69 Pushdown Machines Example (continued) : In action X X X Z 0 q δ(q, 1, X)

70 Pushdown Machines Example (continued) : In action X X X Z 0 q δ(q, 1, X) = {(p, ε)}

71 Pushdown Machines Example (continued) : In action X X 1 1 Z 0 p

72 Pushdown Machines Example (continued) : In action X X 1 1 Z 0 p δ(p, 1, X)

73 Pushdown Machines Example (continued) : In action X X 1 1 Z 0 p δ(p, 1, X) = {(p, ε)}

74 Pushdown Machines Example (continued) : In action X 1 Z 0 p

75 Pushdown Machines Example (continued) : In action X 1 Z 0 p δ(p, 1, X)

76 Pushdown Machines Example (continued) : In action X 1 Z 0 p δ(p, 1, X) = {(p, ε)}

77 Pushdown Machines Example (continued) : In action Z 0 p

78 Pushdown Machines Example (continued) : In action Z 0 p δ(p, ε, Z 0 )

79 Pushdown Machines Example (continued) : In action Z 0 p δ(p, ε, Z 0 ) = {(f, Z 0 )}

80 Pushdown Machines Example (continued) : In action Z 0 f

81 Pushdown Machines Instantaneous Descriptions : We can formalize the pictures just seen with an instantaneous description (ID). An ID is a 3-tuple (q, w, α), where: 1. q is the current state. 2. w is the remaining input. 3. α is the stack contents, top at the left.

82 Pushdown Machines The Goes-To relation: To say that ID I can become ID J in one move of the pushdown machine, we write I J.

83 Pushdown Machines The Goes-To relation: To say that ID I can become ID J in one move of the pushdown machine, we write I J. Formally, (q, aw, Xα) (p, w, βα), for any w and α, if δ(q, a, X) contains (p, β).

84 Pushdown Machines The Goes-To relation: To say that ID I can become ID J in one move of the pushdown machine, we write I J. Formally, (q, aw, Xα) (p, w, βα), for any w and α, if δ(q, a, X) contains (p, β). Extend to *, meaning zero or more moves by Basis : I * I. Induction : If I * J and J K, then I * K.