Difference between revisions of "Theoretical Aspects of Lexical Analysis/Exercise 8"

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(New page: __NOTOC__ Compute the non-deterministic finite automaton (NFA) by using Thompson's algorithm. Compute the minimal deterministic finite automaton (DFA).<br/>The alphabet is Σ = { a, b }. I...)
 
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The input string ''aababb'' is, after 10 steps, split into three tokens: '''T1''' (corresponding to lexeme ''aa''), '''T2''' (''b''), '''T1''' (''a''), and '''T3''' (''bb'').
 
The input string ''aababb'' is, after 10 steps, split into three tokens: '''T1''' (corresponding to lexeme ''aa''), '''T2''' (''b''), '''T1''' (''a''), and '''T3''' (''bb'').
  
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Revision as of 16:39, 6 April 2015

Compute the non-deterministic finite automaton (NFA) by using Thompson's algorithm. Compute the minimal deterministic finite automaton (DFA).
The alphabet is Σ = { a, b }. Indicate the number of processing steps for the given input string.

  • G = { a*, a*|b, a|b* }, input string = aababb

NFA

The following is the result of applying Thompson's algorithm. State 4 recognizes the first expression (token T1); state 12 recognizes token T2; and state 20 recognizes token T3.


DFA

Determination table for the above NFA:

In α∈Σ move(In, α) ε-closure(move(In, α)) In+1 = ε-closure(move(In, α))
- - 0 0, 1, 2, 4, 5, 6, 7, 9, 10, 12, 13, 14, 16, 17, 19, 20 0 (T1)
0 a 3, 8, 15 2, 3, 4, 7, 8, 9, 12, 15, 20 1 (T1)
0 b 11, 18 11, 12, 17, 18, 19, 20 2 (T2)
1 a 3, 8 2, 3, 4, 7, 8, 9, 12 3 (T1)
1 b - - -
2 a - - -
2 b 18 17, 18, 19, 20 4 (T3)
3 a 3, 8 2, 3, 4, 7, 8, 9, 12 3 (T1)
3 b - - -
4 a - - -
4 b 18 17, 18, 19, 20 4 (T3)

Graphically, the DFA is represented as follows:


The minimization tree is as follows. Note that before considering transition behavior, states are split according to the token they recognize.


The tree expansion for non-splitting sets has been omitted for simplicity ("a" transitions for super-state {0, 1, 3}, and "a" and "b" transitions for super-state {1,3}).

Given the minimization tree, the final minimal DFA is as follows. Note that states 2 and 4 cannot be the same since they recognize different tokens.


Input Analysis

In Input In+1 / Token
0 aababb$ 13
13 ababb$ 13
13 babb$ T1 (aa)
0 babb$ 2
2 abb$ T2 (b)
0 abb$ 13
13 bb$ T1 (a)
0 bb$ 2
2 b$ 4
4 $ T3 (bb)

The input string aababb is, after 10 steps, split into three tokens: T1 (corresponding to lexeme aa), T2 (b), T1 (a), and T3 (bb).