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Theory of Computing Lecture 24 MAS 714 Hartmut Klauck 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 MyhillNerode Index • Questions: – How can we find a minimal DFA for L? – Is it unique (up to renaming states)? – Are NFA smaller than DFA? – Are two-way DFA smaller than DFA? Question 4 • Consider the language L={xi : x2{0,1}n and i2{0,1} log n and xi=1} – n is fix, this is a finite language • It is easy to see that there are 2n rows in the communication matrix – Rows labeled x contain the string on columns i=1…n • Hence any DFA for L has 2n states. • Can give a 2-way DFA with O(n2) states: – go to the right end of the input, read i into the state, move n-i steps left, accept if there is a 1 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(n2) states for L • DFA-size: the matrix has more than 2n 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 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 comm. matrix • Edges are also determined uniquely 1) Optimal DFA • Theorem: For every regular L there is a unique DFA A that has minimum size. Given a DFA M for L we can find A in polynomial time. • Corollary: Given DFA A,B we can decide in polynomial time whether they compute the same language – Minimize DFA, compare graphs (note that this is not an instance of the (hard) graph isomorphism problem, since we know the starting state and edge labels) • Corollary: Given DFA A, we can decide in polynomial time whether LA is empty Note • The corresponding questions for Turing machines are undecidable • These properties (easy to find unique representation, easy to find small representation, easy to check for emptiness and easy to compare) make DFA a good datastructure for languages/Boolean functions DFA Minimization • All algorithms try to recover equivalence classes in the Myhill-Nerode characterization • I.e., try to find the smallest partition of the rows of the communication matrix into equal rows • Input: DFA with n states and alphabet size k • Hopcroft: O(nk log n) • Moore: O(n2k) • Idea in both: – Start from partition of states into accepting/rejecting – Refine partition until no longer possible Proof • We are given a DFA M with n states and alphabet size k • Want to find an optimal DFA A • First: remove unreachable states – unreachable means in the graph, cannot be reached from q0 on any input – Identify them with DFS from the starting state • Equivalent states: q,q’ are equivalent, if all strings y, starting from q or q’ lead to the same result (acc/rej) – can assume |y|< n, otherwise y contains a loop in M – equivalent states have the same row in the communication matrix A simple algorithm • Algorithm: – Remove unreachable states – Build a n£n matrix D that containing blank entries – For all rows q, and columns q’, where q2 F and q’ is not, set D[q,q’]=D[q’,q]=² – Iterate [until nothing to do, at most n times] • Loop over all letters a and all pairs q,q’ with D[q,q’] blank • If D[±(q,a),±(q’,a)] is not blank, set D[q,q’]=a – Join states q,q’ that have D[q,q’] still blank A simple algorithm • D allows us to find a witness that q,q’ are not equivalent: D[q,q’]=a, then find recursively the witness y for ±(q,a) and ±(q’,a), witness for q,q’ is ay – Follow ay from q and from q’, one leads to accept, the other to reject • Witnesses need not be longer than n-1 • If no witness exists, then q,q’ are equivalent – by definition of equivalence Time • Each iteration takes time O(n2k) • If q and q’ are not equivalent, a witness of length no more than n-1 can be found, i.e., number of iterations is at most n-1 – typically much less: longest simple path in DFA Correctness • All q,q’ with D[q,q’] not blank are not equivalent • Claim: algo will find a witness for non-equivalent q,q’ Induction over length of witness – length 0: in first iteration – Assume length k witnesses are found in k iterations – If q,q’ are not equivalent, and there is a string ay of length k such that ±(q,ay) accepts, ±(q’,ay) rejects, then some witness y’ for ±(q,a) and ±(q’,a) of length k-1 is found earlier – ay’ is a witness and found in iteration k • Claim: all equivalent states will never be declared not equivalent – Clear from the witness computed Minimizing NFA? • NFA are frequently smaller than DFA • Can we minimize their size? • Theorem: – NFA-minimization is PSPACE-hard – This remains true if the input is a DFA but we search the smallest NFA – Even approximation is hard Conclusion • DFA can be transformed into unique minimal DFA in polynomial time – Can then compare, decide emptiness etc. • NFA and 2-way DFA/NFA can be exponentially smaller, but cannot be minimized efficiently • Limits of DFA: – Communication method Regular Languages (Definition 2) • Simple recursive description of languages • Definition: Regular expression over alphabet § – a2§ is a r.e. – The empty word ² is a r.e. – ; is a r.e. [empty set] – R1[ R2 is regular [union] Notation: r1 | r2 – R1R2 is regular [concatenation] – R1* is regular [Kleene] Examples • • • • • • 0*10*: strings with one 1 §* 1§*: strings with at least one 1 (§§)*: even length 1*(011*)*: every 0 followed by a 1 (0*10*10*)*: even number of 1’s 0*|1*: contains only 0’s or only 1’s Regular vs. DFA • Definition: a language describable by a regular expression is called regular • Theorem: The set of regular languages is the same set as the set of languages decidable by DFA • Proof: • We show – NFA can simulate regular expression – Regular expression can be constructed from DFA Regular to NFA • Consider an NFA with ² transitions – Edges can be labeled with ² (empty word) • Exercise: NFA with ² transitions can be transformed into NFA, same number of states • R a regular expression • Inductively find an NFA for R • Basis cases: R=², R=a2§, R=; – NFA is trivial (at most 3 states) • Induction: Find NFA for Union, Concatenation, Kleene Hull Regular to NFA • Union R1[ R2: – NFA M1 and M2, starting state ²-move to start in M1 and to start in M2 • Concatenation: • Kleene: DFA to regular • Generalized DFA: edges can carry regular expressions instead of letters • Idea 1: start with a normal DFA, shrink to a generalized DFA with 1 edge • Idea 2: dynamic programming type argument • Ri,j,k regular expression for the set of strings that lead from qi to qj using q0,…, qk only DFA to regular • R0,0,0=(a|b|c)* if there are edges labeled a,b,c from q0 to q0 • Ri,i,0 = ² otherwise • Ri,j,0 = a|b etc., if there are edges labeled a and b from qi to qj • Ri,j,k = Ri,j,k-1 | Ri,k,k-1 (Rk,k,k-1)* Rk,j,k-1 • Finally take union over all R0,i,n, where qi is accepting • Result: a reg. expr. for the DFA • Works even for NFA Conclusion • Theoretical Computer Science – Algorithms – Computability – Complexity – Machine Models – Formal Languages – More: Cryptography, Information/Communication Theory … Topics: • Algorithms: – – – – – Basic Algorithms [Sorting/Searching] Graph Algorithms Paradigms [Greedy, Dynamic Programming] Linear Programming Algorithms for hard problems [Approximation] • Other theory: – – – – Hardness and reductions [eg. NP-completeness] Computability [eg. Halting problem, Universality] Time-Hierarchy [etc.] Finite Automata