Slide 1

CSci 4011

INHERENT LIMITATIONS
OF COMPUTER PROGRAMS


Slide 2

THE PUMPING LEMMA
Let L be a regular language with |L| = 
Then there exists a length P such that
∀w  L, if |w| ≥ P
then there exist xyz = w where:
1. |y| > 0
2. |xy| ≤ P
3. xyiz  L for any i ≥ 0


Slide 3

PUMPING DOWN
C = { 0i1j | i > j ≥ 0}
Assume … pumping length P
Find a w ∈ C longer than P
0P+11P
Show that … w cannot be pumped:
P+1

P

w= 00…0011…11
If w = xyz with |xy| ≤ P then
y = 0J for
y some J ≥ 1.
0z = xz = 0P+1-J1P  C
Then
xy
xz = 0…0011…11
xyyz = 00…00011…11
> P+1

P

≤P

P


Slide 4

CHOOSING WISELY
Let BALANCED = { w | w has an equal # of 1s and 0s }

Assume … there is a P
Find a w ∈ BALANCED longer than P
(01)P

0P1P

Show that w cannot be pumped:
If w = xyz with |xy| ≤ P then
y = 0J for some J>0.
Then xyyz = 0P+J1P  BALANCED


Slide 5

REUSING A PROOF
Pumping a language can be lots of work…
Let’s try to reuse that work!
{0n1n : n ≥ 0} = BALANCED ∩ 0*1*
If BALANCED is regular then so is {0n1n: n ≥ 0}


Slide 6

USING CLOSURE
Prove: A is not regular
A

∩

C

=

(regular)

B
(Not regular)

or any of {¬, ∪, ∩, ･, R, *}
If A is regular, then A ∩ C (= B) is regular.

But B is not regular so neither is A.


Slide 7

Prove A = {0i1j : i  j} is not regular
using B = {0n1n : n ≥ 0}

¬A = B ∪{ strings that mix 0s and 1s }

B = ¬A ∩ 0*1*


Slide 8

PUMPING
NON-REGULAR LANGUAGES
Let F = {aibjck | i, j, k ≥ 0 and i=1 ⇒ j=k}
F has pumping length 2!
i=0
i=1
i=2

F ∩ ab*c* = {abncn | n ≥ 0}
A non-regular language may still satisfy the pumping lemma.


Slide 9

4011 SO FAR…
MODEL OF A PROBLEM: LANGUAGE
MODEL OF A PROGRAM: DFA
EQUIVALENT MODELS: NFA, REGEXP

PROBLEMS THAT A DFA CAN’T SOLVE

ARE WE DONE?


Slide 10

NONE OF THESE ARE REGULAR
Σ = {0, 1}, L = { 0n1n | n ≥ 0 }
Σ = {a, b, c, …, z}, L = { w | w = wR }
Σ = { (, ) }, L = { balanced strings of parens }

We can write a SCHEME or JAVA program for any of them!


Slide 11

PUSHDOWN AUTOMATA
FINITE
STATE
CONTROL

STACK
(Last in,
first out)

INPUT


Slide 12

string

pop
ε,ε → $

0011

push
11
0011
011

0,ε → 0

1,0 → ε
ε,$ → ε

STACK

$
0 0
$ $

1

1,0 → ε


Slide 13

string

pop
ε,ε → $

001

push
1
001
01

0,ε → 0

1,0 → ε
ε,$ → ε

STACK

1,0 → ε

$
0 0
$ $

PDA to recognize L = { 0n1n | n ≥ 0 }


Slide 14

Definition: A (non-deterministic) PDA is a tuple
P = (Q, Σ, Γ, , q0, F), where:
Q is a finite set of states

Σ is the input alphabet
Γ is the stack alphabet

 : Q  Σε  Γε → 𝒫(Q  Γε)
q0  Q is the start state
F  Q is the set of accept states
𝒫(Q) is the set of subsets of Q and Σε = Σ  {ε}


Slide 15

q0

ε,ε → $

0,ε → 0

q1

1,0 → ε

q3

Q = {q0, q1, q2, q3}

ε,$ → ε

1,0 → ε

q2

Σ = {0,1}

Γ = {$,0,1}

 : Q  Σε  Γε → 𝒫(Q  Γε)
(q1,1,0) = { (q2,ε) }

(q2,1,1) = 


Slide 16

EVEN-LENGTH PALINDROMES
Σ = {a, b, c, …, z}

q0

ε,ε → $

q1

,ε → 

ε,ε → ε
q3

ε,$ → ε

q2

 , → ε


Slide 17

Build a PDA to recognize
L = { aibjck | i, j, k ≥ 0 and (i = j or i = k) }
c,ε → ε

b,a → ε

q0

ε,$ → ε

q2

q3

ε,ε → $

q1

ε,ε → ε

a,ε → a

q4

b,ε → ε

ε,ε → ε

q5

ε,$ → ε

c,a → ε

q6


Slide 18

Build a PDA to recognize…
L = { x0y | |x| = |y| }


Slide 19

CONTEXT-FREE GRAMMARS
start variable

A → 0A1
A→B
B→#
variables

production
rules

terminals

A  0A1  00A11  00B11  00#11


Slide 20

VALLEY GIRL GRAMMAR
→

→
→ LIKE
→ UMM
→ YOU KNOW
→ ε
→ GAG ME WITH A SPOON

→ AS IF
→ WHATEVER
→ LOSER


Slide 21

VALLEY GIRL GRAMMAR
→ |
→ LIKE | UMM
→ YOU KNOW | ε
→ GAG ME WITH A SPOON | AS IF | WHATEVER
| LOSER


Slide 22

CONTEXT-FREE GRAMMARS
A context-free grammar (CFG) is a tuple G = (V,
Σ, R, S), where:
V is a finite set of variables
Σ is a finite set of terminals (disjoint from V)
R is set of production rules of the form A → W,
where A  V and W  (VΣ)*
S  V is the start variable
G = { {S}, {0,1}, R, S }

R = { S → 0S1, S → ε }

L(G) = { 0n1n | n ≥ 0 }


Slide 23

CONTEXT-FREE GRAMMARS
start variable

A → 0A1
A→B
B→#
variables

production
rules

terminals

A  0A1  00A11  00B11  00#11
uVw yields uvw if (V → v) ∈ R.
A derives 00#11 in 4 steps.


Slide 24

WRITE A CFG FOR EVEN-LENGTH
PALINDROMES
S → S for all   Σ
S→ε


Slide 25

WRITE A CFG FOR THE EMPTY SET

G = { {S}, Σ, , S }


Slide 26

GIVE A CFG FOR…

L3 = { strings of balanced parens }

L4 = { aibjck | i, j, k ≥ 0 and (i = j or j = k) }


Slide 27

VERY INTERESTING CFGs…
http://java.sun.com/docs/books/jls/third_edition/html/j3TOC.html
http://unununium.org/darcs/stackless-python/Grammar/Grammar