语言(111+11111)*的NFA转DFA分步转换请求

Got it, let's walk through converting the NFA for (111+11111)* to a DFA step by step using the subset construction method—this is the standard approach for NFA-to-DFA conversion, so I'll break it down clearly.

First, let's align on the NFA structure for this regex (it matches strings made of any number of 3-1 sequences or 5-1 sequences, including the empty string):

• States: Q = {0, 1, 2, 3, 4, 5, 6, 7, 8}
• Start state: 0 (also an accept state, since the empty string is allowed via *)
• Accept states: F = {0}
• Transitions:
  • ε-transitions: δ(0, ε) = {1, 4} (start can branch into either the 3-1 or 5-1 path)
  • 1-transitions:
    • δ(1,1)={2}, δ(2,1)={3}, δ(3,1)={0} (completes a 3-1 sequence and loops back to start)
    • δ(4,1)={5}, δ(5,1)={6}, δ(6,1)={7}, δ(7,1)={8}, δ(8,1)={0} (completes a 5-1 sequence and loops back to start)
  • All other transitions are empty sets.

We'll use the subset construction method: each DFA state represents a set of NFA states reachable via ε-transitions (called an ε-closure).

Step 1: Initialize the DFA start state #

The DFA start state is the ε-closure of the NFA's start state (0):

• ε-closure(0) = {0} ∪ δ(0, ε) = {0, 1, 4}
• Name this DFA state S0 (accept state, since it includes NFA accept state 0)

Step 2: Compute transitions for S0 on input 1 #

First, collect all NFA states reachable from S0 via input 1:

• δ(0,1)=∅, δ(1,1)={2}, δ(4,1)={5} → union is {2,5}
• Compute ε-closure({2,5}) = {2,5} (no ε-transitions from these states)
• Name this state S1 (not an accept state, no NFA accept state included)

Step 3: Compute transitions for S1 on input 1 #

• δ(2,1)={3}, δ(5,1)={6} → union is {3,6}
• ε-closure({3,6}) = {3,6}
• Name this state S2 (not an accept state)

Step 4: Compute transitions for S2 on input 1 #

• δ(3,1)={0}, δ(6,1)={7} → union is {0,7}
• ε-closure({0,7}) = {0,1,4,7} (add ε-transitions from 0: {1,4})
• Name this state S3 (accept state, includes 0)

Step 5: Compute transitions for S3 on input 1 #

• δ(0,1)=∅, δ(1,1)={2}, δ(4,1)={5}, δ(7,1)={8} → union is {2,5,8}
• ε-closure({2,5,8}) = {2,5,8}
• Name this state S4 (not an accept state)

Step 6: Compute transitions for S4 on input 1 #

• δ(2,1)={3}, δ(5,1)={6}, δ(8,1)={0} → union is {3,6,0}
• ε-closure({3,6,0}) = {0,1,3,4,6} (add ε-transitions from 0: {1,4})
• Name this state S5 (accept state)

Step 7: Compute transitions for S5 on input 1 #

• δ(0,1)=∅, δ(1,1)={2}, δ(3,1)={0}, δ(4,1)={5}, δ(6,1)={7} → union is {2,0,5,7}
• ε-closure({2,0,5,7}) = {0,1,2,4,5,7} (add ε-transitions from 0: {1,4})
• Name this state S6 (accept state)

Step 8: Compute transitions for S6 on input 1 #

• δ(0,1)=∅, δ(1,1)={2}, δ(2,1)={3}, δ(4,1)={5}, δ(5,1)={6}, δ(7,1)={8} → union is {2,3,5,6,8}
• ε-closure({2,3,5,6,8}) = {2,3,5,6,8}
• Name this state S7 (not an accept state)

Step 9: Compute transitions for S7 on input 1 #

• δ(2,1)={3}, δ(3,1)={0}, δ(5,1)={6}, δ(6,1)={7}, δ(8,1)={0} → union is {3,0,6,7}
• ε-closure({3,0,6,7}) = {0,1,3,4,6,7} (add ε-transitions from 0: {1,4})
• Name this state S8 (accept state)

Step 10: Compute transitions for S8 on input 1 #

• δ(0,1)=∅, δ(1,1)={2}, δ(3,1)={0}, δ(4,1)={5}, δ(6,1)={7}, δ(7,1)={8} → union is {2,0,5,7,8}
• ε-closure({2,0,5,7,8}) = {0,1,2,4,5,7,8} (add ε-transitions from 0: {1,4})
• Name this state S9 (accept state)

Step 11: Compute transitions for S9 on input 1 #

• δ(0,1)=∅, δ(1,1)={2}, δ(2,1)={3}, δ(4,1)={5}, δ(5,1)={6}, δ(7,1)={8}, δ(8,1)={0} → union is {2,3,5,6,8,0}
• ε-closure({2,3,5,6,8,0}) = {0,1,2,3,4,5,6,8} (add ε-transitions from 0: {1,4})
• Name this state S10 (accept state)

Step 12: Compute transitions for S10 on input 1 #

• δ(0,1)=∅, δ(1,1)={2}, δ(2,1)={3}, δ(3,1)={0}, δ(4,1)={5}, δ(5,1)={6}, δ(6,1)={7}, δ(8,1)={0} → union is {2,3,0,5,6,7}
• ε-closure({2,3,0,5,6,7}) = {0,1,2,3,4,5,6,7} (add ε-transitions from 0: {1,4})
• Name this state S11 (accept state)

Step 13: Check for new states #

Compute transitions for S11 on input 1:

• δ(0,1)=∅, δ(1,1)={2}, δ(2,1)={3}, δ(3,1)={0}, δ(4,1)={5}, δ(5,1)={6}, δ(6,1)={7}, δ(7,1)={8} → union is {2,3,0,5,6,8}
• ε-closure of this set is {0,1,2,3,4,5,6,8} → which is S10 (already exists)

No new states are generated, so we're done with subset construction.

Final DFA Summary #

DFA States & Acceptance #

DFA Transition Function (Input: 1) #

• S0 → S1
• S1 → S2
• S2 → S3
• S3 → S4
• S4 → S5
• S5 → S6
• S6 → S7
• S7 → S8
• S8 → S9
• S9 → S10
• S10 → S11
• S11 → S10

You can verify this DFA works for test cases:

• Empty string: stays in S0 (accept)
• "111": S0→S1→S2→S3 (accept)
• "11111": S0→S1→S2→S3→S4→S5 (accept)
• "111111" (3+3): S0→S1→S2→S3→S1→S2→S3 (accept)

内容的提问来源于stack exchange，提问作者Atul Kumar