Formal Language, chapter 6, slide 2 Copyright © 2007 by Adam Webber The problem with implementing NFAs is that, being nondeterministic, they do not really define computational procedures for testing language membership. To implement an NFA we must give a computational procedure that can look at a string and decide whether the NFA has at least one sequence of legal transitions on that string leading to an accepting state. This seems to require searching through all legal sequences for the given input string—but how? One approach is to implement a direct backtracking search. Another is to convert the NFA into a DFA and implement that instead. This conversion is both useful and theoretically interesting: the fact that it is always possible shows that in spite of their extra flexibility, NFAs have exactly the same power as DFAs. They can define exactly the regular languages.

Outline:

Formal Language, chapter 6, slide 3 Copyright © 2007 by Adam Webber Outline 6.1 NFA Implemented With Backtracking Search 6.2 NFA Implemented With Bit-Mapped Parallel Search 6.3 The Subset Construction 6.4 NFAs Are Exactly As Powerful As DFAs 6.5 DFA Or NFA?

An NFA Example:

Formal Language, chapter 6, slide 4 Copyright © 2007 by Adam Webber An NFA Example L ( N ) is the language strings over the alphabet {0,1} that have a 1 as the next-to-last symbol We will implement it with backtracking search in Java We will use a three-dimensional transition array delta[s,c-'0'] will be an array of 0 or more possible next states

Slide 5:

Formal Language, chapter 6, slide 5 Copyright © 2007 by Adam Webber /** * A nondeterministic finite-state automaton that * recognizes strings of 0s and 1s with 1 as the * next-to-last character. */ public class NFA1 { /* * The transition function represented as an array. * The entry at delta[s,c-'0'] is an array of 0 or * more ints, one for each possible move from * state s on character c. */ private static int[][][] delta = {{{0},{0,1}}, // delta[q0,0], delta[q0,1] {{2},{2}}, // delta[q1,0], delta[q1,1] {{},{}}}; // delta[q2,0], delta[q2,1]

Slide 6:

Formal Language, chapter 6, slide 6 Copyright © 2007 by Adam Webber /** * Test whether there is some path for the NFA to * reach an accepting state from the given state, * reading the given string at the given character * position. * @param s the current state * @param in the input string * @param pos index of the next char in the string * @return true iff the NFA accepts on some path */ private static boolean accepts (int s, String in, int pos) { if (pos==in.length()) { // if no more to read return (s==2); // accept iff final state is q2 }

Slide 7:

Formal Language, chapter 6, slide 7 Copyright © 2007 by Adam Webber char c = in.charAt(pos++); // get char and advance int[] nextStates; try { nextStates = delta[s][c-'0']; } catch (ArrayIndexOutOfBoundsException ex) { return false; // no transition, just reject } // At this point, nextStates is an array of 0 or // more next states. Try each move recursively; // if it leads to an accepting state return true. for (int i=0; i < nextStates.length; i++) { if (accepts(nextStates[i], in, pos)) return true; } return false; // all moves fail, return false }

Slide 8:

Formal Language, chapter 6, slide 8 Copyright © 2007 by Adam Webber /** * Test whether the NFA accepts the string. * @param in the String to test * @return true iff the NFA accepts on some path */ public static boolean accepts(String in) { return accepts(0, in, 0); // start in q0 at char 0 } } Not object-oriented: all static methods All recursive search information is carried in the parameters Usage example: if (NFA1.accepts(s)) ...

Outline:

Formal Language, chapter 6, slide 9 Copyright © 2007 by Adam Webber Outline 6.1 NFA Implemented With Backtracking Search 6.2 NFA Implemented With Bit-Mapped Parallel Search 6.3 The Subset Construction 6.4 NFAs Are Exactly As Powerful As DFAs 6.5 DFA Or NFA?

Parallel Search:

Formal Language, chapter 6, slide 10 Copyright © 2007 by Adam Webber Parallel Search The previous implementation was a backtracking search Try one sequence of moves If that fails, back up a try another Keep going until you find an accepting sequence, or run out of sequences to try You can also search all sequences at once Instead of keeping track of one current state, keep track of the set of all possible states

Bit-Coded Sets:

Formal Language, chapter 6, slide 11 Copyright © 2007 by Adam Webber Bit-Coded Sets We'll use machine words to represent sets One bit position for each state, with a 1 at that position if the state is in the set The set { q 1 , q 2 } The set { q 0 , q 2 }

Bit-Coded Sets in Java:

Formal Language, chapter 6, slide 12 Copyright © 2007 by Adam Webber Bit-Coded Sets in Java The << operator in Java shifts an integer to the left We're using 1<<i for state i The | operator combines integers using logical OR The set { q 1 , q 2 } Java: 1<<1 | 1<<2 The set { q 0 , q 2 } Java: 1<<0 | 1<<2

Slide 13:

Formal Language, chapter 6, slide 13 Copyright © 2007 by Adam Webber /** * A nondeterministic finite-state automaton that * recognizes strings with 0 as the next-to-last * character. */ public class NFA2 { /* * The current set of states, encoded bitwise: * state i is represented by the bit 1<<i. */ private int stateSet; /** * Reset the current state set to {the start state}. */ public void reset() { stateSet = 1<<0; // {q0} }

Slide 14:

Formal Language, chapter 6, slide 14 Copyright © 2007 by Adam Webber /* * The transition function represented as an array. * The set of next states from a given state s and * character c is at delta[s,c-'0']. */ static private int[][] delta = {{1<<0, 1<<0|1<<1}, // delta[q0,0] = {q0} // delta[q0,1] = {q0,q1} {1<<2, 1<<2}, // delta[q1,0] = {q2} // delta[q1,1] = {q2} {0, 0}}; // delta[q2,0] = {} // delta[q2,1] = {}

Slide 15:

Formal Language, chapter 6, slide 15 Copyright © 2007 by Adam Webber /** * Make one state-set to state-set transition on * each char in the given string. * @param in the String to use */ public void process(String in) { for (int i = 0; i < in.length(); i++) { char c = in.charAt(i); int nextSS = 0; // next state set, initially empty for (int s = 0; s <= 2; s++) { // for each state s if ((stateSet&(1<<s)) != 0) { // if maybe in s try { nextSS |= delta[s][c-'0']; } catch (ArrayIndexOutOfBoundsException ex) { // in effect, nextSS |= 0 } } } stateSet = nextSS; // new state set after c } }

Slide 16:

Formal Language, chapter 6, slide 16 Copyright © 2007 by Adam Webber /** * Test whether the NFA accepted the string. * @return true iff the final set includes * an accepting state */ public boolean accepted() { return (stateSet&(1<<2))!=0; // true if q2 in set } } Usage example: NFA2 m = new NFA2(); m.reset(); m.process(s); if (m.accepted()) ...

Larger NFAs:

Formal Language, chapter 6, slide 17 Copyright © 2007 by Adam Webber Larger NFAs Generalizes easily for NFAs of up to 32 states Easy to push it up to 64 (using long instead of int ) Harder to implement above 64 states Could use an array of  n /32  int variables to represent n states That would make process slower and more complicated

Outline:

Formal Language, chapter 6, slide 18 Copyright © 2007 by Adam Webber Outline 6.1 NFA Implemented With Backtracking Search 6.2 NFA Implemented With Bit-Mapped Parallel Search 6.3 The Subset Construction 6.4 NFAs Are Exactly As Powerful As DFAs 6.5 DFA Or NFA?

From NFA To DFA:

Formal Language, chapter 6, slide 19 Copyright © 2007 by Adam Webber From NFA To DFA For any NFA, there is a DFA that recognizes the same language Proof is by construction: a DFA that keeps track of the set of states the NFA might be in This is called the subset construction First, an example starting from this NFA:

Slide 20:

Formal Language, chapter 6, slide 20 Copyright © 2007 by Adam Webber Initially, the set of states the NFA could be in is just { q 0 } So our DFA will keep track of that using a start state labeled { q 0 }:

Slide 21:

Formal Language, chapter 6, slide 21 Copyright © 2007 by Adam Webber Now suppose the set of states the NFA could be in is { q 0 }, and it reads a 0 The set of possible states after reading the 0 is { q 0 }, so we can show that transition:

Slide 22:

Formal Language, chapter 6, slide 22 Copyright © 2007 by Adam Webber Suppose the set of states the NFA could be in is { q 0 }, and it reads a 1 The set of possible states after reading the 1 is { q 0 , q 1 }, so we need another state:

Slide 23:

Formal Language, chapter 6, slide 23 Copyright © 2007 by Adam Webber From { q 0 , q 1 } on a 0, the next set of possible states is  ( q 0 , 0)   ( q 1 , 0) = { q 0 , q 2 } From { q 0 , q 1 } on a 1, the next set of possible states is  ( q 0 , 1)   ( q 1 , 1) = { q 0 , q 1 , q 2 } Adding these transitions and states, we get…

Slide 24:

And So On:

Formal Language, chapter 6, slide 25 Copyright © 2007 by Adam Webber And So On The DFA construction continues Eventually, we find that no further states are generated That's because there are only finitely many possible sets of states: P ( Q ) In our example, we have already found all sets of states reachable from { q 0 }…

Slide 26:

Accepting States:

Formal Language, chapter 6, slide 27 Copyright © 2007 by Adam Webber Accepting States It only remains to choose the accepting states An NFA accepts x if its set of possible states after reading x includes at least one accepting state So our DFA should accept in all sets that contain at least one NFA accepting state

Slide 28:

Lemma 6.3:

Formal Language, chapter 6, slide 29 Copyright © 2007 by Adam Webber Lemma 6.3 Suppose L is L ( N ) for some NFA N = ( Q N ,  ,  N , q N , F N ) Construct a new DFA D = ( Q D ,  ,  D , q D , F D ), where: If L = L ( N ) for some NFA N , then L is a regular language.

Lemma 6.3, Proof Continued:

Formal Language, chapter 6, slide 30 Copyright © 2007 by Adam Webber Lemma 6.3, Proof Continued By construction we have, for all x   *, D simulates N ’s behavior on each input x D accepts if and only if N accepts L = L ( N ) = L ( D ) L is a regular language jump can be bridged by routine induction

Implementation Note:

Formal Language, chapter 6, slide 31 Copyright © 2007 by Adam Webber Implementation Note The subset construction defined the DFA transition function by This is exactly what process computed in its inner loop, in our bit-mapped implementation in Java So that implementation really just computes the subset construction, one step for each input symbol

Start State Note:

Formal Language, chapter 6, slide 32 Copyright © 2007 by Adam Webber Start State Note In the subset construction, the start state for the new DFA is Often this is the same as q D = { q N }, as in our earlier example But the difference is important if there are -transitions from the NFA's start state

Unreachable State Note:

Formal Language, chapter 6, slide 33 Copyright © 2007 by Adam Webber Unreachable State Note The formal subset construction generates all states Q D = P ( Q N ) These may not all be reachable from the DFA's start state In our earlier example, only 4 states were reachable, but | P ( Q N )| = 8 Unreachable states don't affect L ( D ) When doing the construction by hand, it is usually better to include only the reachable states

Empty-Set State Note:

Formal Language, chapter 6, slide 34 Copyright © 2007 by Adam Webber Empty-Set State Note The empty set is a subset of every set So the full subset construction always produces a DFA state for {} This is reachable from the start state if there is some string x for which the NFA has no legal sequence of moves:  N *( q N , x ) = {} For example, this NFA, with L ( N ) = {}

Slide 35:

Trap State Provided:

Formal Language, chapter 6, slide 36 Copyright © 2007 by Adam Webber Trap State Provided The subset construction always provides a state for {} And it is always the case that so the {} state always has transitions back to itself for every symbol a in the alphabet It is a non-accepting trap state

Outline:

Formal Language, chapter 6, slide 37 Copyright © 2007 by Adam Webber Outline 6.1 NFA Implemented With Backtracking Search 6.2 NFA Implemented With Bit-Mapped Parallel Search 6.3 The Subset Construction 6.4 NFAs Are Exactly As Powerful As DFAs 6.5 DFA Or NFA?

From DFA To NFA:

Formal Language, chapter 6, slide 38 Copyright © 2007 by Adam Webber From DFA To NFA This direction is much easier A DFA is like a special case of an NFA, with exactly one transition from every state on every symbol So it is easy to show that whenever there is a DFA M with L ( M ) = L (so L is regular), there is an NFA N with L ( N ) = L There's just a little technicality involved in changing the type of the  function

Lemma 6.4:

Formal Language, chapter 6, slide 39 Copyright © 2007 by Adam Webber Lemma 6.4 Let L be any regular language By definition there must be some DFA M = ( Q ,  ,  , q 0 , F ) with L ( M ) = L Define a new NFA N = ( Q ,  , ' , q 0 , F ), where '( q , a ) = {( q , a )} for all q  Q and a  , and '( q ,) = {} for all q  Q Now '*( q , x ) = {*( q , x )}, for all q  Q and x  * Thus L ( N ) = L ( M ) = L If L is any regular language, there is some NFA N for which L(N) = L . jump can be bridged by routine induction

Theorem 6.4:

Formal Language, chapter 6, slide 40 Copyright © 2007 by Adam Webber Theorem 6.4 Follows immediately from the previous lemmas Allowing nondeterminism in finite automata can make them more compact and easier to construct But in the sense of Theorem 6.4, it neither weakens nor strengthens them A language L is L ( N ) for some NFA N if and only if L is a regular language.

DFA, Pro And Con:

Formal Language, chapter 6, slide 41 Copyright © 2007 by Adam Webber DFA, Pro And Con Pro Faster and simpler Con There are languages for which DFA-based implementation takes exponentially more space than NFA-based Harder to extend for non-regular constructs Example: scanner in a compiler Speed is critical Token languages do not usually bring out the exponential-size pathology of DFAs

NFA, Pro And Con:

Formal Language, chapter 6, slide 42 Copyright © 2007 by Adam Webber NFA, Pro And Con Pro Easier to extend for non-regular language constructs No exponential-space pathologies Con Slower and trickier Example: regular-expression programming language features (Perl, Python, Ruby, etc.) Need to handle non-regular constructs as well as regular ones More about these when we look at regular expressions

Hybrids:

Formal Language, chapter 6, slide 43 Copyright © 2007 by Adam Webber Hybrids Some applications use both techniques in combination Use DFA techniques for purely regular parts, but switch to NFA techniques when non-regular extensions are needed Use NFA techniques but cache frequently-used state sets and transitions A spectrum of techniques, not just two points