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CSE 135: Introduction to Theory of Computation Equivalence of DFA and NFA Sungjin Im University of California, Merced 02-05-2014 Expressive Power of NFAs and DFAs I Is there a language that is recognized by a DFA but not by any NFAs? Expressive Power of NFAs and DFAs I Is there a language that is recognized by a DFA but not by any NFAs? No! Expressive Power of NFAs and DFAs I Is there a language that is recognized by a DFA but not by any NFAs? No! I Is there a language that is recognized by an NFA but not by any DFAs? Expressive Power of NFAs and DFAs I Is there a language that is recognized by a DFA but not by any NFAs? No! I Is there a language that is recognized by an NFA but not by any DFAs? No! Main Theorem Theorem A language L is regular if and only if there is an NFA N such that L(N) = L. Main Theorem Theorem A language L is regular if and only if there is an NFA N such that L(N) = L. In other words: I For any DFA D, there is an NFA N such that L(N) = L(D), and I for any NFA N, there is a DFA D such that L(D) = L(N). Converting DFAs to NFAs Proposition For any DFA D, there is an NFA N such that L(N) = L(D). Converting DFAs to NFAs Proposition For any DFA D, there is an NFA N such that L(N) = L(D). Proof. Is a DFA an NFA? . Converting DFAs to NFAs Proposition For any DFA D, there is an NFA N such that L(N) = L(D). Proof. Is a DFA an NFA? Essentially yes! Syntactically, not quite. The formal definition of DFA has δDFA : Q × Σ → Q whereas δNFA : Q × (Σ ∪ {}) → P(Q). . Converting DFAs to NFAs Proposition For any DFA D, there is an NFA N such that L(N) = L(D). Proof. Is a DFA an NFA? Essentially yes! Syntactically, not quite. The formal definition of DFA has δDFA : Q × Σ → Q whereas δNFA : Q × (Σ ∪ {}) → P(Q). For DFA D = (Q, Σ, δD , q0 , F ), define an “equivalent” NFA N = (Q, Σ, δN , q0 , F ) that has the exact same set of states, initial state and final states. Only difference is in the transition function. δN (q, a) = {δD (q, a)} for a ∈ Σ and δN (q, ) = ∅ for all q ∈ Q. Simulating an NFA on Your Computer NFA Acceptance Problem Given an NFA N and an input string w , does N accept w ? Simulating an NFA on Your Computer NFA Acceptance Problem Given an NFA N and an input string w , does N accept w ? How do we write a computer program to solve the NFA Acceptance problem? Two Views of Nondeterminsm Guessing View Parallel View At each step, the NFA “guesses” one of the choices available; the NFA will guess an “accepting sequence of choices” if such a one exists. At each step the machine “forks” threads corresponding to each of the possible next states. Two Views of Nondeterminsm Guessing View Parallel View At each step, the NFA “guesses” one of the choices available; the NFA will guess an “accepting sequence of choices” if such a one exists. Very useful in reasoning about NFAs and in designing NFAs. At each step the machine “forks” threads corresponding to each of the possible next states. Two Views of Nondeterminsm Guessing View Parallel View At each step, the NFA “guesses” one of the choices available; the NFA will guess an “accepting sequence of choices” if such a one exists. Very useful in reasoning about NFAs and in designing NFAs. At each step the machine “forks” threads corresponding to each of the possible next states. Very useful in simulating/running NFA on inputs. Algorithm for Simulating an NFA Algorithm Keep track of the current state of each of the active threads. Algorithm for Simulating an NFA Algorithm Keep track of the current state of each of the active threads. Example 0, 1 q0 0, 1 1 q1 Example NFA N Consider the input w = 111. The execution (listing only the states of currently active threads) is Algorithm for Simulating an NFA Algorithm Keep track of the current state of each of the active threads. Example 0, 1 q0 0, 1 1 q1 Example NFA N Consider the input w = 111. The execution (listing only the states of currently active threads) is hq0 i Algorithm for Simulating an NFA Algorithm Keep track of the current state of each of the active threads. Example 0, 1 q0 0, 1 1 q1 Example NFA N Consider the input w = 111. The execution (listing only the states of currently active threads) is 1 hq0 i −→ hq0 , q1 i Algorithm for Simulating an NFA Algorithm Keep track of the current state of each of the active threads. Example 0, 1 q0 1 0, 1 Consider the input w = 111. The execution (listing only the states of currently active threads) is q1 hq0 i −→ hq0 , q1 i −→ hq0 , q1 , q1 i 1 1 Example NFA N 1 −→ hq0 , q1 , q1 , q1 i Algorithm With optimizations Observations I Exponentially growing memory: more threads for longer inputs. Can we do better? Algorithm With optimizations Observations I I Exponentially growing memory: more threads for longer inputs. Can we do better? Exact order of threads is not important Algorithm With optimizations Observations I I Exponentially growing memory: more threads for longer inputs. Can we do better? Exact order of threads is not important I It is unimportant whether the 5th thread or the 1st thread is in state q. Algorithm With optimizations Observations I I Exponentially growing memory: more threads for longer inputs. Can we do better? Exact order of threads is not important I I It is unimportant whether the 5th thread or the 1st thread is in state q. If two threads are in the same state, then we can ignore one of the threads Algorithm With optimizations Observations I I Exponentially growing memory: more threads for longer inputs. Can we do better? Exact order of threads is not important I I It is unimportant whether the 5th thread or the 1st thread is in state q. If two threads are in the same state, then we can ignore one of the threads I Threads in the same state will “behave” identically; either one of the descendent threads of both will reach a final state, or none of the descendent threads of both will reach a final state Parsimonious Algorithm in Action Example 0, 1 q0 0, 1 1 q1 Example NFA N Consider the input w = 111. The execution (listing only the states of currently active threads) is Parsimonious Algorithm in Action Example 0, 1 q0 0, 1 1 q1 Example NFA N Consider the input w = 111. The execution (listing only the states of currently active threads) is {q0 } Parsimonious Algorithm in Action Example 0, 1 q0 0, 1 1 q1 Example NFA N Consider the input w = 111. The execution (listing only the states of currently active threads) is 1 {q0 } −→ {q0 , q1 } Parsimonious Algorithm in Action Example 0, 1 q0 0, 1 1 q1 Example NFA N Consider the input w = 111. The execution (listing only the states of currently active threads) is 1 1 {q0 } −→ {q0 , q1 } −→ {q0 , q1 } Parsimonious Algorithm in Action Example 0, 1 q0 0, 1 1 q1 Consider the input w = 111. The execution (listing only the states of currently active threads) is 1 1 Example NFA N 1 {q0 } −→ {q0 , q1 } −→ {q0 , q1 } −→ {q0 , q1 } Revisiting NFA Simulation Algorithm I Need to keep track of the states of the active threads I I Unordered: Without worrying about exactly which thread is in what state No Duplicates: Keeping only one copy if there are multiple threads in same state Revisiting NFA Simulation Algorithm I Need to keep track of the states of the active threads I I I Unordered: Without worrying about exactly which thread is in what state No Duplicates: Keeping only one copy if there are multiple threads in same state How much memory is needed? Revisiting NFA Simulation Algorithm I Need to keep track of the states of the active threads I I I Unordered: Without worrying about exactly which thread is in what state No Duplicates: Keeping only one copy if there are multiple threads in same state How much memory is needed? I I If Q is the set of states of the NFA N, then we need to keep a subset of Q! Can be done in |Q| bits of memory (i.e., 2|Q| states), which is finite!! Constructing an Equivalent DFA I The DFA runs the simulation algorithm Constructing an Equivalent DFA I The DFA runs the simulation algorithm I DFA remembers the current states of active threads without duplicates, i.e., maintains a subset of states of the NFA Constructing an Equivalent DFA I The DFA runs the simulation algorithm I DFA remembers the current states of active threads without duplicates, i.e., maintains a subset of states of the NFA I When a new symbol is read, it updates the states of the active threads Constructing an Equivalent DFA I The DFA runs the simulation algorithm I DFA remembers the current states of active threads without duplicates, i.e., maintains a subset of states of the NFA I When a new symbol is read, it updates the states of the active threads I Accepts whenever one of the threads is in a final state Example of Equivalent DFA 0, 1 q0 0, 1 1 q1 Example NFA N Example of Equivalent DFA 0, 1 q0 0, 1 1 q1 Example NFA N 0, 1 0, 1 {q1 } {} 0, 1 0 {q0 } 1 {q0 , q1 } DFA D equivalent to N Recall . . . Definition For an NFA M = (Q, Σ, δ, q0 , F ), string w , and state q1 ∈ Q, we ˆ 1 , w ) to denote states of all the active threads of say ∆(q computation on input w from q1 . Recall . . . Definition For an NFA M = (Q, Σ, δ, q0 , F ), string w , and state q1 ∈ Q, we ˆ 1 , w ) to denote states of all the active threads of say ∆(q computation on input w from q1 . Formally, w ˆ 1 , w ) = {q ∈ Q | q1 −→ ∆(q M q} Formal Construction Given NFA N = (Q, Σ, δ, q0 , F ), construct DFA det(N) = (Q 0 , Σ, δ 0 , q00 , F 0 ) as follows. I Q0 = I q00 = I F0 = Formal Construction Given NFA N = (Q, Σ, δ, q0 , F ), construct DFA det(N) = (Q 0 , Σ, δ 0 , q00 , F 0 ) as follows. I Q 0 = P(Q) I q00 = I F0 = Formal Construction Given NFA N = (Q, Σ, δ, q0 , F ), construct DFA det(N) = (Q 0 , Σ, δ 0 , q00 , F 0 ) as follows. I I I Q 0 = P(Q) ˆ 0 , ) q 0 = ∆(q 0 F0 = Formal Construction Given NFA N = (Q, Σ, δ, q0 , F ), construct DFA det(N) = (Q 0 , Σ, δ 0 , q00 , F 0 ) as follows. I I I Q 0 = P(Q) ˆ 0 , ) q 0 = ∆(q 0 F0 = {A ⊆ Q | A ∩ F 6= ∅} Formal Construction Given NFA N = (Q, Σ, δ, q0 , F ), construct DFA det(N) = (Q 0 , Σ, δ 0 , q00 , F 0 ) as follows. I I I Q 0 = P(Q) ˆ 0 , ) q 0 = ∆(q 0 F0 = {A ⊆ Q | A ∩ F 6= ∅} I δ 0 ({q1 , q2 , . . . qk }, a) = Formal Construction Given NFA N = (Q, Σ, δ, q0 , F ), construct DFA det(N) = (Q 0 , Σ, δ 0 , q00 , F 0 ) as follows. I I I Q 0 = P(Q) ˆ 0 , ) q 0 = ∆(q 0 F0 = {A ⊆ Q | A ∩ F 6= ∅} ˆ 1 , a) ∪ ∆(q ˆ 2 , a) ∪ · · · ∪ ∆(q ˆ k , a) = ∆(q I δ 0 ({q1 , q2 , . . . qk }, a) Formal Construction Given NFA N = (Q, Σ, δ, q0 , F ), construct DFA det(N) = (Q 0 , Σ, δ 0 , q00 , F 0 ) as follows. I I I Q 0 = P(Q) ˆ 0 , ) q 0 = ∆(q 0 F0 = {A ⊆ Q | A ∩ F 6= ∅} ˆ 1 , a) ∪ ∆(q ˆ 2 , a) ∪ · · · ∪ ∆(q ˆ k , a) or = ∆(q more concisely, [ ˆ δ 0 (A, a) = ∆(q, a) I δ 0 ({q1 , q2 , . . . qk }, a) q∈A Correctness Lemma For any NFA N, the DFA det(N) is equivalent to it, i.e., L(N) = L(det(N)). Correctness Lemma For any NFA N, the DFA det(N) is equivalent to it, i.e., L(N) = L(det(N)). Proof Idea Need to show Correctness Lemma For any NFA N, the DFA det(N) is equivalent to it, i.e., L(N) = L(det(N)). Proof Idea Need to show ∀w ∈ Σ∗ . det(N) accepts w iff N accepts w Correctness Lemma For any NFA N, the DFA det(N) is equivalent to it, i.e., L(N) = L(det(N)). Proof Idea Need to show ∀w ∈ Σ∗ . det(N) accepts w iff N accepts w ˆ 0 , w ) ∩ F 6= ∅ ∀w ∈ Σ∗ .δ̂(q00 , w ) ∈ F 0 iff ∆(q Correctness Lemma For any NFA N, the DFA det(N) is equivalent to it, i.e., L(N) = L(det(N)). Proof Idea Need to show ∀w ∈ Σ∗ . det(N) accepts w iff N accepts w ˆ 0 , w ) ∩ F 6= ∅ ∀w ∈ Σ∗ .δ̂(q00 , w ) ∈ F 0 iff ∆(q ∗ 0 ˆ 0 , w ) ∩ F 6= ∅ ∀w ∈ Σ . for A = δ̂(q0 , w ), A ∩ F 6= ∅ iff ∆(q Correctness Lemma For any NFA N, the DFA det(N) is equivalent to it, i.e., L(N) = L(det(N)). Proof Idea Need to show ∀w ∈ Σ∗ . det(N) accepts w iff N accepts w ˆ 0 , w ) ∩ F 6= ∅ ∀w ∈ Σ∗ .δ̂(q00 , w ) ∈ F 0 iff ∆(q ∗ 0 ˆ 0 , w ) ∩ F 6= ∅ ∀w ∈ Σ . for A = δ̂(q0 , w ), A ∩ F 6= ∅ iff ∆(q We will instead prove the stronger claim ∀w ∈ Σ∗ . δ̂(q00 , w ) = A iff ˆ 0 , w ) = A. ∆(q Correctness Proof Lemma ˆ 0 , w ) = A. ∀w ∈ Σ∗ . δ̂(q00 , w ) = A iff ∆(q Correctness Proof Lemma ˆ 0 , w ) = A. ∀w ∈ Σ∗ . δ̂(q00 , w ) = A iff ∆(q Proof. By induction on |w | Correctness Proof Lemma ˆ 0 , w ) = A. ∀w ∈ Σ∗ . δ̂(q00 , w ) = A iff ∆(q Proof. By induction on |w | I Base Case |w | = 0: Then w = . Now Correctness Proof Lemma ˆ 0 , w ) = A. ∀w ∈ Σ∗ . δ̂(q00 , w ) = A iff ∆(q Proof. By induction on |w | I Base Case |w | = 0: Then w = . Now δ̂(q00 , ) = q00 defn. of δ̂ Correctness Proof Lemma ˆ 0 , w ) = A. ∀w ∈ Σ∗ . δ̂(q00 , w ) = A iff ∆(q Proof. By induction on |w | I Base Case |w | = 0: Then w = . Now δ̂(q00 , ) = q00 ˆ 0 , ) = ∆(q defn. of δ̂ defn. of q00 Correctness Proof Lemma ˆ 0 , w ) = A. ∀w ∈ Σ∗ . δ̂(q00 , w ) = A iff ∆(q Proof. By induction on |w | I Base Case |w | = 0: Then w = . Now δ̂(q00 , ) = q00 ˆ 0 , ) = ∆(q I defn. of δ̂ defn. of q00 Induction Hypothesis: Assume inductively that the statement holds ∀w . |w | = n ··→ Correctness Proof Induction Step Proof (contd). I Induction Step: If |w | = n + 1 then w = ua with |u| = n and a ∈ Σ. δ̂(q00 , ua) = δ(δ̂(q00 , u), a) defn. of δ̂ Correctness Proof Induction Step Proof (contd). I Induction Step: If |w | = n + 1 then w = ua with |u| = n and a ∈ Σ. δ̂(q00 , ua) = δ(δ̂(q00 , u), a) ˆ 0 , u), a) = δ(∆(q defn. of δ̂ ind. hyp. Correctness Proof Induction Step Proof (contd). I Induction Step: If |w | = n + 1 then w = ua with |u| = n and a ∈ Σ. δ̂(q00 , ua) = δ(δ̂(q00 , u), a) ˆ 0 , u), a) = δ(∆(q [ ˆ = ∆(q, a) defn. of δ̂ ind. hyp. defn. of δ ˆ 0 ,u) q∈∆(q Correctness Proof Induction Step Proof (contd). I Induction Step: If |w | = n + 1 then w = ua with |u| = n and a ∈ Σ. δ̂(q00 , ua) = δ(δ̂(q00 , u), a) ˆ 0 , u), a) = δ(∆(q [ ˆ = ∆(q, a) defn. of δ̂ ind. hyp. defn. of δ ˆ 0 ,u) q∈∆(q ˆ 0 , ua) = ∆(q ˆ prop. about ∆ Another Example 0, 1 q1 {q3 } 0 q0 q3 1 q2 Example NFA N 0, 1 0, 1 0 {q0 , q1 } {} 1 {q2 , q3 } DFA D0 for N (only relevant states)