Department of Computer Science & Engineering CSC3130 Mid-term Review bridger.zhao@gmail.com Department of Computer Science & Engineering

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

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

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

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

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

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

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

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

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

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

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

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

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:0+1+0+ and 0+1+.So, 0n-k1n 0n1n  L 0n1n0n1n CSC3130 Tutorial One 1/13/2019

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

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

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

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