1. Lecture 17 Oct 25, 2011 Section 2.1 (push-down automata)
Section 2.2 (pumping lemma for context-free languages)

2. Last-In First-Out pushing and popping
Pushdown Automata
Pushdown automata are for context-free languages what finite automata are for regular languages.
PDAs are recognizing automata that have a single stack (= memory):
Last-In First-Out pushing and popping
Note: PDAs are nondeterministic.

3. Informal Description PDA (1)
internal state set Q
input w =
The PDA M reads w and stack element.
Depending on - input wi  , - stack sj  , and - state qk  Q the PDA M:
- jumps to a new state, - pushes an element  (nondeterministically) x y z
stack

4. Informal Description PDA (2)
internal state set Q
input w =
After the PDA has read complete input, M will be in state  Q
If possible to end in accepting state FQ, then M accepts w x y z
stack

5. Formal Description PDA
A Pushdown Automata M is defined by a six tuple (Q,,,,q0,F), with
Q finite set of states
 finite input alphabet
 finite stack alphabet
q0 start state  Q
F set of accepting states Q
 transition function : Q      P (Q  )

6. PDA for L = { 0n1n | n0 } , $ , $ Example 2.14:
The PDA first pushes “ $ 0n ” on stack.
Then, while reading the 1n string, the zeros are popped again. If, in the end, $ is left on stack, then “accept” q1 q3 q2 q4
, $
, $
1, 0
0, 0

7. Machine Diagram for 0n1n , $ , $ 0, 0 q2 q1 1, 0 q3 q4
On w = (state; stack) evolution:
(q1; )  (q2; $)  (q2; 0$)  (q2; 00$)
(q2; 000$)  (q3; 00$)  (q3; 0$)  (q3; $)
(q4; ) This final q4 is an accepting state

8. Machine Diagram for 0n1n , $ , $ 0, 0 q2 q1 1, 0 q3 q4
On w = 0101 (state; stack) evolution:
(q1; )  (q2; $)  (q2; 0$)  (q3; $)  (q4; ) …
But we still have part of input “01”.
There is no accepting path.

9. Another Example of a PDA

10. Another example of PDA Consider the language over the alphabet {a, b}:
L = { w | #a(w) = #b(w) }
(#a(w) stands for the number of a’s in w.)
PDA design intuition: push a symbol 1 on seeing a’s, pop on seeing b’s.
Problem: what if we see a lot of b’s in the start, and a’s come later? Can change the role. Push on b, pop on a.
Need to know which one – using two different states.

11. Another example of PDA Consider the language over the alphabet {a, b}:
L = { w | #a(w) = #b(w) }

12. One more PDA – for even length palindromes
L = { w wR | w is in {0, 1}* }

13. PDAs versus CFL
Theorem 2.20: A language L is context-free if and only if there is a pushdown automata M that recognizes L.
Two step proof:
Given a CFG G, construct a PDA MG
2) Given a PDA M, make a CFG GM

14. Equivalence of PDA and CFG (0)
Part 1: For every CFG, we can build an equivalent PDA.
General construction: each rule of CFG A  w is included in the PDA’s move.

15. Equivalence of PDA and CFG (1)
Part 1: For every CFG, we can build an equivalent PDA.
Example: (page 115 of text)

16. NPDA, CFG equivalence
Proof of (): L is recognized by a NPDA implies L is described by a CFG.
harder direction
first step: convert NPDA into “normal form”:
single accept state
empties stack before accepting
each transition either pushes or pops a symbol
2011

17. NPDA, CFG equivalence
main idea: non-terminal Ap,q generates exactly the strings that take the NPDA from state p (w/ empty stack) to state q (w/ empty stack)
then Astart, accept generates all of the strings in the language recognized by the NPDA.
2011

18. NPDA, CFG equivalence Two possibilities to get from state p to q:
generated by Ap,r
generated by Ar,q
stack height p q r
abcabbacacbacbacabacabbabbabaacabbbababaacaccaccccc
input
string taking NPDA from p to q
2011

19. NPDA, CFG equivalence NPDA P = (Q, Σ, , δ, start, {accept}) CFG G:
non-terminals V = {Ap,q : p, q  Q}
start variable Astart, accept
productions:
for every p, r, q  Q, add the rule
Ap,q → Ap,rAr,q
2011

20. NPDA, CFG equivalence Two possibilities to get from state p to q:
generated by Ar,s
stack height r s p
push d
pop d q
abcabbacacbacbacabacabbabbabaacabbbababaacaccaccccc
input
string taking NPDA from p to q
2011

21. NPDA P = (Q, Σ, , δ, start, {accept}) CFG G:
NPDA, CFG equivalence
NPDA P = (Q, Σ, , δ, start, {accept})
CFG G:
non-terminals V = {Ap,q : p, q  Q}
start variable Astart, accept
productions:
for every p, r, s, q  Q, d  , and a, b  (Σ  {ε})
if (r, d)  δ(p, a, ε), and
(q, ε)  δ(s, b, d), add the rule
Ap,q → aAr,sb
from state p, read a, push d, move to state r
from state s, read b, pop d, move to state q
2011

22. NPDA P = (Q, Σ, , δ, start, {accept}) CFG G:
NPDA, CFG equivalence
NPDA P = (Q, Σ, , δ, start, {accept})
CFG G:
non-terminals V = {Ap,q : p, q  Q}
start variable Astart, accept
productions:
for every p  Q, add the rule
Ap,p → ε

23. NPDA, CFG equivalence two claims to verify correctness:
if Ap,q generates string x, then x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack)
if x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack), then Ap,q generates string x
2011

24. NPDA, CFG equivalence
1. if Ap,q generates string x, then x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack)
induction on length of derivation of x.
base case: 1 step derivation. must have only terminals on rhs. In G, must be production of form Ap,p → ε.
2011

25. NPDA, CFG equivalence
1. if Ap,q generates string x, then x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack)
assume true for derivations of length at most k, prove for length k+1.
verify case: Ap,q → Ap,rAr,q →k x = yz
verify case: Ap,q → aAr,sb →k x = ayb
2011

26. NPDA, CFG equivalence
2. if x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack), then Ap,q generates string x
induction on # of steps in P’s computation
base case: 0 steps. starts and ends at same state p. only has time to read empty string ε.
G contains Ap,p → ε.
2011

27. NPDA, CFG equivalence
2. if x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack), then Ap,q generates string x
induction step. assume true for computations of length at most k, prove for length k+1.
if stack becomes empty sometime in the middle of the computation (at state r)
y is read going from state p to r (Ap,r→* y)
z is read going from state r to q (Ar,q→* z)
conclude: Ap,q → Ap,rAr,q →* yz = x
2011

28. NPDA, CFG equivalence
2. if x can take NPDA P from state p (w/ empty stack) to q (w/ empty stack), then Ap,q generates string x
if stack becomes empty only at beginning and end of computation.
first step: state p to r, read a, push d
go from state r to s, read string y (Ar,s→* y)
last step: state s to q, read b, pop d
conclude: Ap,q → aAr,sb →* ayb = x
2011

29. PDACFG conversion
Summary of the construction:

30. Non-CF Languages
The language L = { anbncn | n0 } does not appear to be context-free.
Informal: A PDA can compare #a’s with #b’s. But by the time b’s are processed, the stack is empty. Not possible to count a’s with c’s.
The problem of A * vAy : If S * uAz * uvAyz * uvxyz  L,
then S * uAz * uvAyz * … * uviAyiz * uvixyiz  L as well, for all i=0,1,2,…

31. Pumping Lemma for CFLs
Idea: If we can prove the existence of derivations for elements of the CFL L that use the step
A * vAy, then a new form of ‘v-y pumping’
holds: A * vAy * v2Ay2 * v3Ay3 * …)
Observation: We can prove this existence if the parse-tree is tall enough.

32. Recall Parse Trees
Parse tree for S  AbbcBa * cbbccccaBca  cbbccccacca S b a c A B

33. Pumping a Parse Tree S A A u v x y z
If s = uvxyz  L is long, then its parse-tree is tall. Hence, there is a path on which a variable A repeats itself. We can pump this A–A part.

34. A Tree Tall Enough
Let L be a context-free language, and let G be its grammar with maximal b symbols on the right side of the rules: A  X1…Xb
A parse tree of depth h produces a string with maximum length of bh. Long strings implies tall trees.
Let |V| be the number of variables of G. If h = |V|+2 or bigger, then there is a variable on a ‘top-down path’ that occurs more than once.

35. uvxyz L S A A u v x y z
By repeating the A–A part we get…

36. uv2xy2z L S A A A R y u v x z y v x
… while removing the A–-A gives…

37. uxz  L S A x u z
In general uvixyiz  L for all i=0,1,2,…

38. Pumping Lemma for CFL
For every context-free language L, there is a pumping length p, such that for every string sL and |s|p, we can write s = uvxyz with 1) uvixyiz  L for every i{0,1,2,…} 2) |vy|  1 3) |vxy|  p
Note that
1) implies that uxz  L
2) says that v and y cannot be both empty strings  Condition 3) is not always used. (It is not crucial part of pumping lemma, but helps to reduce the number of cases.)

39. Formal Proof of Pumping Lemma
Let G=(V,,R,S) be the grammar of a CFL. Maximum size of rules is b2: A  X1…Xb
A string s requires a minimum tree-depth  logb|s|.
If |s|  p=b|V|+2, then tree-depth  |V|+2, hence there is a path and variable A where A repeats itself: S * uAz * uvAyz * uvxyz
It follows that uvixyiz  L for all i=0,1,2,…
Furthermore: |vy|  1 because tree is minimal |vxy|  p because bottom tree with  p leaves has a ‘repeating path’

40. Pumping lemma for {anbncn | n >= 0}
Assume that B = {anbncn | n0} is CFL
Let p be the pumping length, and s = apbpcp  B
P.L.: s = uvxyz = apbpcp, with uvixyiz  B for all i0
Options for vxy: 1) The strings v and y are uniform (v=a…a and y=c…c, for example). Then uv2xy2z will not contain the same number of a’s, b’s and c’s, hence uv2xy2zB
2) At least one of v or y is not uniform. (i.e., it has at least two different symbols occurring in it).
Then uv2xy2z will not be a…ab…bc…c Hence uv2xy2zB

41. Pumping lemma applied to {anbncn} continued
Assume that B = {anbncn | n0} is CFL Let p be the pumping length, and s = apbpcp  B P.L.: s = uvxyz = apbpcp, with uvixyiz  B for all i0
We showed: For every way of partitioning s into uvxyz, there is an i such that uvixyiz is not in B. Contradiction.
B is not a context-free language.

42. Another example
Proof that C = {aibjck | 0ijk } is not context-free.
Let p be the pumping length, and s = apbpcp  C
P.L.: s = uvxyz, such that uvixyiz  C for every i  0
vxy can’t have a’s and c’s. Why?
So only two options for vxy:
1) vxy belongs to a*b*, then the string uv2xy2z has not enough c’s, hence uv2xy2zC
2) vxy belongs to b*c*, then the string uv0xy0z = uxz has too many a’s, hence uv0xy0zC
Contradiction: C is not a context-free language.

43. D = { ww | w{0,1}* } (Ex. 2.22)
Carefully take the strings sD. Let p be the pumping length, take s=0p1p0p1p.
Three options for s=uvxyz with 1  |vxy|  p:
If a part of y is to the left of | in 0p1p|0p1p, then second half of uv2xy2z starts with “1”
2) Same reasoning if a part of v is to the right of middle of 0p1p|0p1p, hence uv2xy2z  D
3) If x is in the middle of 0p1p|0p1p, then uxz equals 0p1i 0j1p  D (because i or j < p)
Contradiction: D is not context-free.

44. Pumping lemma for CFG - remarks
Using the CFL pumping lemma is more difficult than the pumping lemma for regular languages.
You have to choose the string s carefully, and divide the options efficiently. Additional CFL properties would be helpful (like we had for regular languages).
What about closure under standard operations?

45. Union Closure Properties
Lemma: Let A1 and A2 be two CF languages, then the union A1A2 is context free as well.
Proof: Assume that the two grammars are G1=(V1,,R1,S1) and G2=(V2,,R2,S2).
Construct a third grammar G3=(V3,,R3,S3) by:
V3 = V1  V2  { S3 } (new start variable) with
R3 = R1  R2  { S3  S1 | S2 }.
It follows that L(G3) = L(G1)  L(G2).

46. Intersection, Complement?
Let again A1 and A2 be two CF languages.
One can prove that, in general,
the intersection A1  A2 ,
and
the complement Ā1= * \ A1 are not context free languages.

47. Intersection, Complement?
Proof for complement:
Recall that a problem in HW 5 shows that
L = { x#y | x, y are in {a, b}*, x != y} IS context-free.
Complement of this language is
L’ = { w | w has no # symbol} U
{ w | w has two or more # symbols} U
{ w#w | w is in {a,b}* }.
We can show that L’ is NOT context-free.

48. Context-free languages are NOT closed under intersection
Proof by counterexample: Recall that in an earlier slide in this lecture, we showed that
L = {anbncn | n >= 0} is NOT context-free.
Let
A = {anbncm | n, m >= 0} and
B = L = {anbmcm | n, m >= 0}. It is easy to see that both A and B are context-free. (Design CFG’s.)
This shows that CFG’s are not closed under intersection.