Transcript CS154 slides

Pushdown Automata
下推自动机
Pushdown Automata
正则语言
 正则文法
 有限状态自动机
上下文无关语言
 上下文无关文法
 下推自动机(PA)
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Pushdown Automata
正则语言
 正则文法
 NFA = DFA
上下文无关语言
 上下文无关文法
 NPA (NPA  DPA)
DPA 仅定义了上下文无关语言一个真子集
 大部分程序设计语言可以用DPA描述
 DPA can model parsers
如果没有特殊说明，本节课中提及的PA一般都是
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非确定PA
PA的直观思考
 下推自动机可以看作是一个具有栈操作
能力的ε-NFA
 PA = ε-NFA + 栈(及其操作)
 PA迁移取决于:
1. 当前状态
2. 当前输入符号 (可以为 ε)
3. 当前栈顶符号
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PA的直观思考
 在每次迁移中，PA做以下动作：
1. 改变状态
2. 将栈顶符号替换为一个长度可以为0符号序
列
 长度为零时 = “pop” 出栈
 长度大于0时 = sequence of “pushes”
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PA 的形式化定义
 A PA is defined by a seven tuple
(Q, Σ, Γ, δ, q0, Z0, F):
1.
2.
3.
4.
5.
6.
7.
Q, a finite set of states
与有限状态自动机
Σ, an input alphabet
的区别是什么？
Γ, a stack alphabet
δ, a transition function 栈！
q0 in Q, a start state
Z0 in Γ, a start stack symbol
F ⊆ Q, a set of final states
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常用符号
a, b, … 输入符号.
 But sometimes we allow ε as a possible
value.
…, X, Y, Z 栈符号
…, w, x, y, z 输入符号串
, ,… 栈符号串
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迁移函数
 表达为：
δ(q, a, Z)={(p1, 1), (p2, 2), ...}
 函数δ有3个参数:
1. q, a state in Q
2. a, an input, which is either a symbol in Σ or ε
3. A stack symbol in Γ
 函数δ(q, a, Z) 的结果为 a set of actions of
the form (p, )


p is a state;  is a string of stack symbols.
为什么是a set of actions?
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迁移函数
δ(q, a, Z)={(p1, 1), (p2, 2), ...}
 直观含义：
 当前，状态为q, 栈顶符号为Z,
 遇到了输入符号a,
 该下推自动机迁移到状态pi,
 从输入符号串首去掉符号a,
 栈顶符号Z退出,
 i进入
 (pi, i) in {(p1, 1), (p2, 2), ...}
 如果是确定的, δ(q, a, Z)=(p, )
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迁移函数
δ(q, a, Z)={(p1, 1), (p2, 2), ...}
a, a1,a2,...
a1,a2,...
q
pi
Z,Z1,..
i, Z1, ...
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PA的图形表示
是否可以像FSA那样，可以有一个PA的
图形表示？
q1
1,Z0/1
0,Z0/0
1,Z0/1
q2
e,Z0/e
q3
0,0/e
1,1/e
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Example
设计一个下推自动机接收语言 {0n1n | n
> 1}.
状态:
 q = 初始状态.
• 到目前为止，只看到了0
 p = we’ve seen at least one 1 and may
now proceed only if the inputs are 1’s.
 f = final state
• Accept.
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Example
栈符号:
 Z0 = start symbol
• 标记栈底
• so we know when we have counted the same
number of 1’s as 0’s.
 X = marker
• used to count the number of 0’s seen on the
input.
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Example
迁移：
 δ(q, 0, Z0) = {(q, XZ0)}.
 δ(q, 0, X) = {(q, XX)}.
 These two rules cause one X to be pushed
into the stack for each 0 read from the
input.
 δ(q, 1, X) = {(p, ε)}. When we see a 1,
go to state p and pop one X.
 δ(p, 1, X) = {(p, ε)}. Pop one X per 1.
 δ(p, ε, Z0) = {(f, Z0)}. Accept at bottom.
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Actions of the Example PA
000111
q
Z0
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Actions of the Example PA
00111
q
X
Z0
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Actions of the Example PA
0111
q
X
X
Z0
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Actions of the Example PA
111
q
X
X
X
Z0
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Actions of the Example PA
11
p
X
X
Z0
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Actions of the Example PA
1
p
X
Z0
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Actions of the Example PA
p
Z0
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Actions of the Example PA
f
Z0
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PA的瞬时描述
 瞬时描述(instantaneous description,
ID) is a 3-triple (q, w, )
 表达了PA当前的情况:
1. the current state, q.
2. the remaining input, w.
3. stack contents,  (top at the left)
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The “Goes-To” Relation
Let I and J be two different IDs
I⊦J: ID I can become ID J in one move of
the PA
Formally：
(q, aw, X)⊦(p, w, )
for any w and , if δ(q, a, X) contains (p, ).
Extend ⊦ to ⊦*, meaning “zero or more
moves” .
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Example: Goes-To
The previous PA of {0n1n | n > 1}
We can describe the sequence of moves by:
(q, 000111, Z0)⊦
(q, 00111, XZ0)⊦
(q, 0111, XXZ0)⊦
(q, 111, XXXZ0)⊦
(p, 11, XXZ0)⊦
(p, 1, XZ0)⊦
(p, ε, Z0)⊦
(f, ε, Z0)
(q, 000111, Z0)
⊦*
(f, ε, Z0)
What would happen on input 0001111?
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Answer
Legal because a PA can use
ε input even if input remains.
δ(p, ε, Z0) = {(f, Z0)}
(q, 0001111, Z0)⊦(q, 001111, XZ0)⊦
(q, 01111, XXZ0)⊦(q, 1111, XXXZ0)⊦
(p, 111, XXZ0)⊦(p, 11, XZ0)⊦
(p, 1, Z0)⊦(f, 1, Z0)
Note the last ID has no move.
0001111 is not accepted, because the
input is not completely consumed.
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Language of a PA
The common way to define the
language of a PA is by final state.
If P is a PA, then L(P) is the set of
strings w such that (q0, w, Z0) ⊦*
ε, ) for final state f and any .
(f,
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Language of a PA – (2)
Another language defined by the same
PA is by empty stack.
If P is a PA, then N(P) is the set of
strings w such that (q0, w, Z0) ⊦*
(q, ε, ε) for any state q.
(q0, w, Z0) ⊦* (f, ε, )
(q0, w, Z0) ⊦* (q, ε, ε)
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Equivalence of Language
Definitions
1. For any L(P), there exists a PA P' such
that N(P')=L(P).
2. For any N(P'), there exists a PA P''
such that L(P'')=N(P').
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Proof: L(P) -> N(P’) 直观思考
 用P’模拟 P.
 Idea:
 当P接收时，将P'的栈置空.
 特殊情况：栈空且没有接收的情况
 用一个特殊的栈底标记来表示栈空且没有接
收的情况
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Proof: L(P) -> N(P’)
 P’ 包含P的所有状态、符号、以及迁移，
另外:
1. 栈符号X0 (used to identify the stack bottom
against accidental emptying)
2. New start state s and “erase” state e.
3. δ(s, ε, X0) = {(q0, Z0X0)}. Get P started.
4. δ(f, ε, X) = δ(e, ε, X) = {(e, ε)} for any
final state f of P and any stack symbol X.
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Proof: L(P) -> N(P’)
start
e, X/e
s
q0
e, X0/Z0X0
P
e
e, X/e
e, X/e
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Proof: N(P) -> L(P’’) 直观思考
 P” simulates P.
 P” 通过一个特殊的栈底标志来表示P的
栈为空的情况
 If so, P” accepts.
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Proof: N(P) -> L(P’’)
 P’’ 包含P的所有状态、符号、以及迁移
，另外:
1. Stack symbol X0, used to guard the stack
bottom.
2. New start state s and final state f.
3. δ(s, ε, X0) = {(q0, Z0X0)}. Get P started.
4. δ(q, ε, X0) = {(f, ε)} for any state q of P.
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Proof: N(P) -> L(P’’)
e, X0/e
start
e, X0/e
s
q0
e, X0/Z0X0
P
f
e, X0/e
e, X/e
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Aside: FA and PDA Notations
We represented moves of a FA by an
extended δ, which did not mention the
input yet to be read.
We could have chosen a similar
notation for PDA’s, where the FA state is
replaced by a state-stack combination,
like the pictures just shown.
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FA and PDA Notations – (2)
Similarly, we could have chosen a FA
notation with ID’s.
 Just drop the stack component.
 (q,w)=p iff (q,wx)⊦*(p,x).
Why do not adopt ID for FA?
 Can you answer it?
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Deterministic PAs
DPA 的表达能力严格的大于正则语言
DPA的表达能力严格的小于上下文无关
语言
DPA表达的语言并不恰好等于CFL的非固
有歧义的子集
More about DPAs, please read the
book!
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