1. Text search and closure properties
The Chinese University of Hong Kong
Fall 2010
CSCI 3130: Automata theory and formal languages
Text search and closure properties
Andrej Bogdanov

2. Text search

3. The program grep grep -E regexp file
Searches for the occurrence of patterns matching a regular expression
cat|12 {cat, 12} union
[abc] {a, b, c} shorthand for a|b|c
[ab][12] {a1, a2, b1, b2} concatenation
(ab)* {e, ab, abab, ...} star
[ab]? {e, a, b} zero or one
(cat)+ {cat, catcat, ...} one or more
[ab]{2} {aa, ab, ba, bb} {n} copies

4. Searching with grep Words containing savor or savour
Words with 5 consecutive a or b
grep –E `savou?r` words
grep –E `[ab]{5}` words
outsavor
savor
savored
savorer
savorily
savoriness
savoringly
savorless
savorous
savorsome
savory
savour
unsavored
unsavoredly
unsavoredness
unsavorily
unsavoriness
unsavory
grabbable
grep –E `zo+zo+` words
zoozoo

5. More grep commands a n f a s u f f e r l r e g u l a r e n s t a r
. any symbol 1 2 3 4 5
[a-z] anything in a range a l e n
\< beginning of line
n f a
$ end of line
s u f f e r
r e g u l a r
1 If a DFA is too hard, I do an ...
2 CSCI 3130 homework makes me ...
s t a r
3 If a DFA likes it, it is ...
4 $ = $10?
5 I study 3130 hard because it will make me ...
grep –E `\<.ff.u..t` words

6. how do you look for... Words that start in cat and have another cat
grep –E `\<cat.*cat` words
Words with at least ten vowels?
grep –E `([aeiouy].*){10}` words
Words without any vowels?
[^R] does not contain R
grep –E `\<[^AEIOUYaeiouy]*$` words
Words with exactly ten vowels?
grep –E `\<[^AEIOUYaeiouy]* ([aeiouy][^AEIOUYaeiouy]*){10}$` words

7. How grep (could) work regular expression NFA NFA without e DFA input
text file
differences
in class
in grep
[ab]?, a+, (cat){3}
not allowed
allowed
input handling
matches whole
looks for pattern
output
accept/reject
finds pattern

8. Implementation of grep
How do you handle expressions like:
[ab]? zero or one
→ ()|[ab]
R? → e|R
(cat)+ one or more
→ (cat)(cat)*
R+ → RR*
a{3} {n} copies
→ aaa
R{n} → RR...R
n times ?
[^aeiouy] not containing any

9. Closure properties

10. Example The language L of strings that end in 101 is regular
How about the language L of strings that do not end in 101?
(0+1)*101

11. Example
Hint: w does not end in 101 if and only if it ends in: or it has length 0, 1, or 2
So L can be described by the regular expression
000, 001, 010, 011, 100, 110 or 111
(0+1)*( )
+ e + (0 + 1) + (0 + 1)(0 + 1)

12. Complement
The complement L of a language L is the set of all strings that are not in L
Examples (S = {0, 1})
L1 = all strings that end in 101
L1 = all strings that do not end in = all strings end in 000, …, 111 or have length 0, 1, or 2
L2 = 1* = {e, 1, 11, 111, …}
L2 = all strings that contain at least one = (0 + 1)*0(0 + 1)*

13. Example The language L of strings that contain 101 is regular
How about the language L of strings that do not contain 101?
(0+1)*101(0+1)*
You can write a regular expression,
but it is a lot of work!

14. Closure under complement
If L is a regular language, so is L.
To argue this, we can use any of the equivalent definitions for regular languages:
The DFA definition will be most convenient
We assume L has a DFA, and show L also has a DFA
regular expression
NFA
DFA

15. Arguing closure under complement
Suppose L is regular, then it has a DFA M
Now consider the DFA M’ with the accepting and rejecting states of M reversed
accepts L
accepts strings not in L
this is exactly L

16. NO! Food for thought Can we do the same thing with an NFA? (0+1)*10
0, 1 q0 q1 q2
(0+1)*10 1
0, 1 q0 q1 q2
(0+1)*

17. Intersection
The intersection L  L’ is the set of strings that are in both L and L’
Examples:
If L, L’ are regular, is L  L’ also regular?
L = (0 + 1)*11
L’ = 1*
L  L’ =
1*11
L = (0 + 1)*10
L’ = 1*
L  L’ = ∅

18. Closure under intersection
If L and L’ are regular languages, so is L  L’.
To argue this, we can use any of the equivalent definitions for regular languages:
regular expression
NFA
DFA
Suppose L and L’ have DFAs, call them M and M’
Goal: Construct a DFA (or NFA) for L  L’

19. An example M M’ L = even number of 0s L’ = odd number of 1sM s0 s1 1 M’
L = even number of 0s
L’ = odd number of 1s
r0, s1 1
r0, s0
r1, s1
r1, s0
L  L’ = even number of 0s and odd number of 1s

20. Closure under intersection
M and M’
DFA for L  L’
states
Q = {r1, ..., rn}
Q’ = {s1, ..., sn’}
Q × Q’ = {(r1, s1), (r1, s2), ...,
(r2, s1), ..., (rn, sn’)}
start state
ri for M
sj for M’
(ri, sj)
accepting
states
F for M
F’ for M’
F × F’ = {(ri, sj): ri ∈ F, sj ∈ F’}
Whenever M is in state ri and M’ is in state sj,
the DFA for L  L’ will be in state (ri, sj)

21. Closure under intersection
M and M’
DFA for L  L’ ri rj a
transitions
in M
ri, si’
rj, sj’ a
si’
sj’ a
in M’

22. Reversal The reversal wR of a string w is w written backwards
The reversal LR of a language L is the language obtained by reversing all its strings
w = cave
wR = evac
L = {cat, dog}
LR = {tac, god}

23. Reversal of regular languages
L = all strings that end in 101 is regular
How about LR?
This is the language of all strings beginning in 101
Yes, because it is represented by
(0+1)*101
101(0+1)*

24. Closure under reversal
If L is a regular language, so is LR.
How do we argue?
regular expression
NFA
DFA

25. Arguing closure under reversal
Take a regular expression E for L
We will show how to reverse E
A regular expression can be of the following types:
The special symbols  and e
Alphabet symbols like a, b
The union, concatenation, or star of simpler expressions

26. Proof of closure under reversal
regular expression E
reversal ER Æ Æ e e
a (alphabet symbol) a
E1 + E2
E1R + E2R
E1E2
E2RE1R
E1*
(E1R)*

27. ? A question LDUP = {ww: w  L} L = {cat, dog} LDUP = {catcat, dogdog}
Ex.
LDUP = {catcat, dogdog}
If L is regular, is LDUP also regular? ?
regular expression
NFA
DFA

28. LDUP = LL A question Let’s try with regular expression: L = {a, b}
Let’s try with NFA:
L = {a, b}
LDUP = {aa, bb}
LL = {aa, ab, ba, bb}
LDUP = LL q0 q1 
NFA for L

29. An example L = 0*1 is regular LDUP = {1, 01, 001, 0001, ...}
= {0n10n1: n ≥ 0}
Let’s try to design an NFA for LDUP

30. An example LDUP = {11, 0101, 001001, 00010001, ...} = {0n10n1: n ≥ 0}1 01 1
001 1
0001 1
LDUP = {11, 0101, , , ...}
= {0n10n1: n ≥ 0}