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Nondeterminism

CSC 4170 Theory of Computation. Nondeterminism. Section 1.2. 1.2.a. An NFA without  -transitions. q 1. 1. q 2. 0,1. q 3. 0,1. 0 1 0 1 0. What language does this NFA recognize?. 1.2.b. An NFA with  -transitions. 1. b. . a. a. 3. a,b. 2. b a b a b. Does this NFA accept:

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Nondeterminism

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  1. CSC 4170 Theory of Computation Nondeterminism Section 1.2

  2. 1.2.a An NFA without -transitions q1 1 q2 0,1 q3 0,1 0 1 0 1 0 What language does this NFA recognize?

  3. 1.2.b An NFA with -transitions 1 b  a a 3 a,b 2 b a b a b • Does this NFA accept: •  ? • a ? • b ? • aaa…a ?

  4. 1.2.c Formal definition of a nondeterministic finite automaton Let  =   {} • A NFA is a 5-tuple (Q, , , s, F), where: • Q is a finite set called the states, •  is a finite set called the alphabet, •  is a function of the type Q  P(Q) called the transition function, • s is an element of Q called the start state, • F is a subset of Q called the set of accept states.

  5. 1.2.d Our automaton formalized 1 Q: : : s: F: b  a a a b  1 2 3 3 a,b 2 A = (Q, , , s, F)

  6. 1.2.e Formal definition of accepting by NFA M = (Q, , , s, F) 1 b  a a 3 a,b 2 • M accepts the string x iff x can be written as • u1 u2 … un • where each ui is in , andthere is a sequence • r1, r2,…, rn, rn+1 • of states such that: • r1= s • ri+1(ri,ui), for each i with 1in • rn+1  F Example: aa u1 u2 … un aa = r1, r2,…, rn, rn+1

  7. 1.2.f What language does this NFA recognize? 0 0   0 0 0 0 0 0 0 0 0 0 0

  8. 1.2.g What language does this DFA recognize? 1 2 0 0 0 0 3 0 0 5 4 0

  9. 1.2.h Equivalence of NFAs and DFAs Two machines are said to beequivalent, if they recognize the same language. Theorem 1.39Every NFA has an equivalent DFA. Proof. Consider an NFA N = (Q, , , s, F) We need construct an equivalent DFA D = (Q’, , ’, s’, F’) using a procedure called the subset construction (next slide). • Notation: • For RQ, let • R = {q | q can be reached from R by traveling • along 0 or more -arrows} a a • For RQ and a, let • (R,a) = {q | q can be reached from R by traveling • along an a-arrow} a

  10. 1.2.i The Subset Construction Constructing DFA D = (Q’, , ’, s’, F’) from NFA N = (Q, , , s, F) • Q’ = P (Q) • ’(R,a) = (R,a) • s’ = {s} • F’= {R | R is a subset of Q containing an accept state of N} • D obviously works correctly: • at every step in the computation, it clearly enters a state that • corresponds to the subset of states that N could be in at that point. a  a  a

  11. 1.2.j Q’: : ’: s’: F’: Applying the subset construction to our NFA N = (Q, , , s, F) 1 a b  {1} {2} {3} {1,2} {1,3} {2,3} {1,2,3} b  a a 3 a,b 2 • Q’ = P (Q) • ’(R,a) = (R,a) • s’ = {s} • F’= {R | R is a subset of Q containing an • accept state of N}

  12. 1.2.k {,{1},{2},{3}, {1,2},{1,3},{2,3},{1,2,3}} D a,b Q’: : ’: s’: F’: The resulting DFA {3} b  {a,b} a a a b  {1} {2} {3} {1,2} {1,3} {2,3} {1,2,3} b   {1,3} {1}  {2} b {2,3} {3} a b b {1,3}  {2,3} {2,3} {1,3} {2} {2,3} a {2} {1,2,3} {3} {1,2,3} {2,3} a b a,b {1,2,3} {1,2} {1,3} {{1},{1,2},{1,3},{1,2,3}} a

  13. 1.2.l {,{1},{2},{3}, {1,2},{1,3},{2,3},{1,2,3}} D a,b Q’: : ’: s’: F’: Removing unreachable states {3} b  {a,b} a a b  {1} {2} {3} {1,2} {1,3} {2,3} {1,2,3} b   {1,3}  {2} b {2,3} {3} a b {1,3}  {2,3} {2,3} {1,3} {2} {2,3} a {2} {1,2,3} {3} {1,2,3} {2,3} a b {1,2,3} {1,3} {{1},{1,2},{1,3},{1,2,3}} a

  14. 1.2.m N D a,b Testing in work {3} b  1 a b  a b {1,3} b a a b 3 a,b 2 {2,3} a {2} a b b a a {1,2,3} a

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