Report

Theory of Computing Lecture 22 MAS 714 Hartmut Klauck Nondeterministic Finite Automata • Definition: – A nondeterministic finite automaton (NFA) is defined like a DFA, except that there are possibly several transitions from a pair (state, letter) [but at least one] – For the graph this means that there are several edges from state q labeled with letter a – Acceptance if there is a path to an accepting state • Note: variants include: allow ² transitions (transition without reading) and incomplete edge sets (input gets stuck/leads to no state) Example • Language that contains all strings having a 1 in the third position from the end A note • There is no good `guess and check’ definition of NFA • One can prove that NFA cannot do all the nondeterministic guesses at the beginning of the computation and then just proceed deterministically – Reason: can store only O(1) guesses in the internal state, which is not enough DFA vs. NFA • Theorem: The set of languages that can be decided by NFA is equal to the set of regular languages • Proof: • Take any NFA A. We have to find a DFA that decides the same language as A • Idea: power-set construction: – use one state for every nonempty subset of states of A – After reading a string x the DFA will be in the state that the set of reachable states of A (on x) Proof • A has s states, simulating DFA M has 2s -1 states • Starting state: {q0} • Accepting states of M: those that contain an accepting state of A • Edges: Consider a state {q1,…, qk} of M. There is an edge labeled with letter a to the state which is the union of the ±(qi,a) for i=1…k • Claim: The computation in M on input x ends in the state s that is the set of states in A that can be reached on x – Proof by induction • Then: M accepts x iff A accepts x DFA vs. NFA • Hence every language L decided by an NFA is regular, and there is a DFA for L that is at most exponentially larger than the smallest NFA • This is also best possible • Note that the size (number of states) of the NFA/DFA is a constant (independent of the input length) Languages that are not regular • Intuitively, DFA cannot count • We will see that languages like {0n1n :n is a natural number} are not regular • To show this we simulate DFA by a communication game • Game is equivalent to the Myhill-Nerode method • Most common approach: pumping lemma – much less general The game • • • • Two players, Alice and Bob Alice receives an input x2§* Bob receives y2§* Alice sends a message to Bob, who decides whether x±y2 L – ±: concatenation, we will also just write xy • Cost of the game: number of messages used Example • • • • L={0n1n: n is natural} How many messages are needed? Alice has an input x If x contains a 1 before a 0, or x=0n1n+k she sends the message `reject’ • Otherwise 2 cases: – Alice has 0k: she sends the message `0,k’ – Alice has 0n1k: she sends the message `1,n-k’ • Bob can decide if xy is in L from the message and y • These are infinitely many different messages! • We can still count the number of messages as a function of the input length m, number of different messages is O(m) Example 2 • Parity={w: w has odd number of 1’s} • Alice sends the number of 1’s in her part of the input modulo 2 • 2 messages: 0/1 • Bob accepts if the message bit is different from the number of 1’s in his input modulo2 Example 3 • L={w: third last symbol of w is 1}, alphabet={0,1} • Alice sends last 3 symbols of her input x (if possible) • Otherwise: – If x=ab, she sends the message for 0ab – If x=a she sends message for 00a – If x is the empty string she sends 000 • Total: 8 messages Example 3 • L={w: third last symbol of w is 1}, alphabet={0,1} • Correctness of the protocol: – Given message abc Bob can append his input y and check if the third last symbol of abcy is 1 – If Alice’s input has length 1 or 2 then ab=00 resp. a=0 • DFA with 8 states exists (see next theorem) – Will show later: no DFA for L has less than 8 states DFA and the game • Theorem [DFA vs. game]: – If L is regular, then the optimal number of messages in the game is equal to the smallest number of states in a DFA. – If L is not regular, then the smallest number of messages is infinite Application • How to show that we need many messages? • Consider the infinite matrix M, rows labeled with all possible x and columns labeled with all possible y and M[x,y]=1 iff xy2 L • Call this matrix the communication matrix of L • Lemma: The minimum number of messages in the game is equal to the number of distinct rows in the matrix Proof • Two rows labeled x, x’ are distinct if there is a column y where M[x,y]=1 and M[x’,y]=0 – or vice versa • If Alice sends the same message on x and x’ then Bob cannot distinguish them, so there is an error either on xy or on x’y • On the other hand it is enough for Alice to send the same message on x, x’ if they their rows are not distinct Examples • Parity: M has only two distinct rows • {0n 1n}: – Consider the rows of M labeled 0k for all k=1,…,1 – The rows are all different [the row for 0k has a 1 only in column 1k] • Conclusion: {0n 1n: n natural} is not a regular language Example • {w: third last symbol of w is 1} • Consider the rows labelled by all strings of length 3 over {0,1} – 8 such rows • Easy to see that these rows are all distinct • Hence no DFA for the language can have less than 8 states