1. Department of Computer Science & Engineering
CSC3130 Mid-term Review
Department of Computer Science & Engineering

2. Scope: regular languages
DFA, NFA and conversion from NFA to DFA.
Regular expression.
Equivalence of DFA, NFA and regular expression.
Closure properties of regular languages.
Pumping lemma for regular languages.
DFA minimization algorithm, DFA equivalence.
CSC3130 Tutorial One
1/13/2019

3. Scope: context free languages
Context free grammars.
Derivations, parse trees, and ambiguity.
Removal of ε and unit productions, Chomsky Normal Form.
The CYK parsing algorithm.
Pushdown automata.
Pumping lemma for context-free languages
CSC3130 Tutorial One
1/13/2019

4. NFA to DFA & State minimization
CSC3130 Tutorial One
1/13/2019

5. DFA Hacking
Let L be a regular language, L’={u|uv is in L and |u|=|v|}. That is, L’ is the first halves of strings in L. Prove that L’ is also regular.
Hints: Let M be the DFA for L, How do we characterize a middle state?
CSC3130 Tutorial One
1/13/2019

6. Closure Properties
For regular languages: union, intersection, concatenation, Kleen closure, reverse, complement, etc.
For context free languages, exclude intersection and complement.
Usage: is L L regular? Is non palindromes regular?
CSC3130 Tutorial One
1/13/2019

7. Context Free Language S → |aaSbbb CFG to PDA see 08L10,17/23
CSC3130 Tutorial One
1/13/2019

8. CNF Consider the following simple grammar:
A  cA | a
B  ABC | b
C  c
How to convert this grammar to CNF?
CSC3130 Tutorial One
1/13/2019

9. Conversion into CNF A B1B2…Bn or
Step 1: Convert every production into either:
A B1B2…Bn or
A a
e.g. A  bCDeF becomes:
A  BCDEF
B  b
E  e
CSC3130 Tutorial One
1/13/2019

10. Conversion into CNF e.g. A  BCDEF becomes:
Step 2: Convert production of the form A B1B2…Bn into A  C1C2 :
e.g. A  BCDEF becomes:
A  BX
X  CY
Y  DZ
Z  EF
CSC3130 Tutorial One
1/13/2019

11. Example Convert the following CFG into Chomsky Normal Form: S  
S  ABBA
B  bCb
A  a
C  c
CSC3130 Tutorial One
1/13/2019

12. Solution Result: S   S   S  ABBA S AE B  bCb E BF A  a F BA
Step 1:
B bCb becomes
B DCD
D b
Step 2:
S  ABBA becomes
S AE
E BF
F BA
B DCD becomes
B DG
G CD
Result:
S  
S AE
E BF
F BA
B DG
G CD
A  a
C  c
D b
S  
S  ABBA
B  bCb
A  a
C  c
CSC3130 Tutorial One
1/13/2019

13. Pumping Lemma for CFL L = { ww | w is in {0,1}* } Assume L is a CFL
Let n be the Pumping Lemma constant
Consider z = 0n1n0n1n, |z| = 4n > n
We want write z = uvwxy with |vwx|  n and |vx| 1
There are several cases for uvwxy
CSC3130 Tutorial One
1/13/2019

14. Pumping Lemma for CFL Case 1: vwx is within the first block 0’s
v and x must contain 0’s
Let vx consist of k 0’s, where k > 0
Choose i=0, uviwxiy = 0n-k1n 0n1n
If 0n-k1n 0n1n in form of ww, the end of first w must lie in the second block of 0’s
Then, the two w’s are in the different pattern: and 0+1+.So, 0n-k1n 0n1n  L
0n1n0n1n
CSC3130 Tutorial One
1/13/2019

15. Pumping Lemma for CFL
0n1n0n1n
Case 2: vwx spread over the first block of 0’s and first block of 1’s
vx has at least one 0’s or one 1’s
Let vx consist of k 0’s and p 1’s, where k0 ,p>0
or k>0,p 0
Choose i=0, uviwxiy = 0n-k1n-p 0n1n
If 0n-k1n-p 0n1n in form of ww, the end of first w must lie after the first block of 1’s
Then, the number of 1’s is different in two w. So, 0n-k1n-p 0n1n  L
CSC3130 Tutorial One
1/13/2019

16. Pumping Lemma for CFL Case 3: vwx is within the first block of 1’s
v and x must contain 1’s
Let vx consist of k 1’s, where k > 0
Choose i=0, uviwxiy = 0n1n-k 0n1n
If 0n1n-k 0n1n in form of ww, the end of first w must lie in the second block of 0’s
Then, the number of 1’s is different in two w. So, 0n1n-k 0n1n  L.
0n1n0n1n
CSC3130 Tutorial One
1/13/2019

17. Pumping Lemma – Example
0n1n0n1n
Case 4: vwx spread over the first block of 1’s and second block of 0’s
vx has at least one 1’s or one 0’s.
Let vx consist of k 1’s and p 0’s, where k 0,p>0
or k>0,p 0
Choose i=0, uviwxiy = 0n1n-k 0n-p1n
For 0n1n-k 0n-p1n in form of ww, if the end of first w lie in the first block of 1’s, then the number of 0’s is different in two w.
if the end of first w lie in the second block of 0’s, then the number of 1’s is different in two w. So, 0n1n-k 0n-p1n  L
CSC3130 Tutorial One
1/13/2019

18. Pumping Lemma for CFL Other cases: vwx is within the second half of z
the argument is symmetric to the cases where vwx is within the first half of z
We can conclude that L = { ww | w is in {0,1}* } is not context free.
0n1n0n1n
0n1n0n1n
0n1n0n1n
CSC3130 Tutorial One
1/13/2019