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Theory of Computing Lecture 23 MAS 714 Hartmut Klauck

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The game Two players, Alice and Bob Alice receives an input x 2 § * Bob receives y 2 § * Alice sends a message to Bob, who decides whether x ± y 2 L – ± : concatenation, we will also just write xy Cost of the game: number of messages used

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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

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Application How to show that we need many messages? Consider the infinite matrix M, rows labeled with x 2 § * and columns labeled with y 2 § * and M[x,y]=1 iff xy 2 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

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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’ Bob cannot distinguish x and x’, so there is an error on xy or on x’y On the other hand it is enough for Alice to send the same message for rows that are not distinct

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Examples Parity: M has two distinct rows – one for x with odd number of 1’s and for even {0 n 1 n }: – Consider the rows labeled 0 k for all k=1,…, 1 – The rows are all different Conclusion: {0 n 1 n : n natural} is not a regular language Note: We can consider subsets of the rows and columns of M and show that this submatrix has infinitely many distinct rows. Then M also has infinitely many distinct rows

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Example L={xx: x 2 {0,1}* } is not regular Proof: we need to show the number of distinct rows is infinite Consider the rows and columns labeled by 0 n 1 for all natural n Take two rows labeled by x=0 p 1 and x’=0 r 1 Clearly there is a 1 in M[x,y] for y=0 p 1 but there is no 1 in M[x’,1] Again M contains an infinite identity matrix

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Proof [DFA to game] Assume L is regular. There is a DFA M for L. Alice simulates M on x, and sends the state of M reached after reading x to Bob, who continues the simulation Bob accepts iff M accepts xy Clearly this is a correct protocol, and the number of messages is the size of the set of states

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Proof [game to DFA] Now assume that there is a protocol with m messages We can assume that Alice uses one message for each distinct row of the matrix, otherwise the protocol is not optimal – i.e., for two rows that are not distinct she uses the same message We construct a DFA with m states corresponding to the m messages q 0 [the starting state] is the message that Alice sends on the empty string We use m-1 other states, corresponding to messages Set of accepting states: messages/states, such that Bob accepts on the message when y is empty string Remains to define the transition function

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Proof Assume that on some input x Alice sends message q, and on input xa she sends message q’ Then we include the transition q,a q’ Clearly this defines transitions, but are they consistent? I.e., If Alice sends q on x and q’ on xa, and Alice also sends q on x’, then she should send q’ on x’a But x and x’ have the same row, hence xa and x’a also have the same row [if not, there is some y such that xay 2 L and x’ay not in L, hence rows of x and x’ not the same] Hence the transition function is well-defined. We get a DFA with m states for L

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Note The minimum number of messages in our communication game is usually called the Myhill-Nerode index

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2-way DFA Theorem: 2-way DFA can recognize only regular languages Note: – Sometimes it is easier to show that L is regular by using a 2-way DFA – Sometimes 2-way DFA need fewer states – Proof: We show that the Myhill-Nerode index is finite.

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Examples L={w: third last symbol of w is 1} 2-way DFA: move to the right end of the input, move 3 steps left, accept if there is a 1 L rev for a regular language L is the set of w, such that w read in reverse is in L – Move to the right end and then simulate the DFA for L while reading the input backwards

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Proof (theorem) Proof: We show that the Myhill-Nerode index is finite. Language L is decided by a 2-way DFA M, where M has n states Consider words w=xy Suppose the tape-head of M comes back from y into x while M is in state q, and the same happens later in state q’ – q q’ otherwise there is an infinite loop Hence the states on which M crosses from y to x are distinct, and there are at most n such crossings Simulation in the communication game:

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Proof We describe the communication protocol Let S=[q 1,...,q m ] be a sequence of any m · n distinct states of M Alice simulates M on x, assuming that when M first leaves x it later returns to x in state q 1 – Etc: If M leaves x for the k-th time it returns to x in state q k Alice’s message: for every possible S Alice sends the m states in which M leaves x (for even m), and additionally if she would accept, for odd m Bob finds the sequence S that is consistent with M’s behavior on y and Alice’s message (S is the correct sequence occurring on M and xy) Bob accepts if M accepts It is easy to see that this protocol is correct, and uses no more than O(n 2n ) messages, which is finite

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2-way NFA Definition: a two-way NFA is a nondeterministic Turing machine that cannot write and cannot leave the tape area that contains the input. Theorem: Two-way NFA can decide only regular languages Proof: Same as for two-way DFA, Alice and Bob check consistency with respect to the nondeterministic A – Is there a computation that leads to the communicated sequence of states and accepts?

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Open Problem Is there a language, for which 2-way NFA need much fewer states than 2-way DFA?

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Size of Automata? We know L is regular, if there is a DFA with a finite number of states for L Minimum number of states is the Myhill-Nerode Index Questions: 1. How can we find a minimal DFA for L? 2. Is the minimal DFA unique (up to renaming states)? 3.Are NFA sometimes smaller than minimal DFA? 4.Are two-way DFA sometimes smaller than DFA?

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Question 4 Consider the language L={xi : x 2 {0,1} n and i 2 {0,1} log n and x n-i =1} – n is fix, this is a finite language, hence regular It is easy to see that there are at least 2 n rows in the communication matrix – Consider only columns for the different i=n-1,…,0 – Rows labeled x contains the string x at entries i=n-1…0 Hence any DFA for L has at least 2 n states Can give a 2-way DFA with O(n 2 ) states: – go to the right end of the input, read i into the state, move i steps left, accept if there is a 1

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Question 3 Consider L={xy: x,y 2 {0,1} n, x y} – This is a FINITE language, n is fixed Exercise: there is an NFA with O(n 2 ) states for L DFA-size: the matrix has more than 2 n distinct rows – DFA size is exponential Hence NFA can be exponentially smaller than DFA for some languages Example where they are not: complement of L

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2) Uniqueness Uniqueness of the minimal DFA (up to vertex names) follows from the Myhill-Nerode characterization The optimal DFA has exactly one state for each set of equal rows of the communication matrix Edges/the transition function are also determined uniquely, i.e., if x is in set q and xa is in set q’ then there must be an edge labeled a from q to q’

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