Difference between revisions of "Theoretical Aspects of Lexical Analysis/Exercise 3"

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< Theoretical Aspects of Lexical Analysis
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== Solution ==
 
== Solution ==
 +
 +
=== NFA ===
 +
 +
The following is the result of applying Thompson's algorithm.
 +
 +
<graph>
 +
digraph nfa {
 +
    { node [shape=circle style=invis] start }
 +
  rankdir=LR; ratio=0.5
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  node [shape=doublecircle,fixedsize=true,width=0.2,fontsize=10]; 8
 +
  node [shape=circle,fixedsize=true,width=0.2,fontsize=10];
 +
  start -> 0
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  0 -> 1; 0 -> 8
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  1 -> 2; 1 -> 4
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  2 -> 3;
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  3 -> 6
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  4 -> 5 [label="a",fontsize=10]
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  5 -> 6
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  6 -> 7 [label="b",fontsize=10]
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  7 -> 1; 7 -> 8
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  fontsize=10
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  //label="NFA for ((ε|a)b)*"
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}
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</graph>
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 +
=== DFA ===
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 +
Determination table for the above NFA:
 +
 +
{| cellspacing="2"
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! style="padding-left: 20px; padding-right: 20px; background: wheat;" | I<sub>n</sub>
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! style="padding-left: 20px; padding-right: 20px; background: wheat;" | α∈Σ
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! style="padding-left: 20px; padding-right: 20px; background: wheat;" | move(I<sub>n</sub>, α)
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! style="padding-left: 20px; padding-right: 20px; background: wheat;" | ε-closure(move(I<sub>n</sub>, α))
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! style="padding-left: 20px; padding-right: 20px; background: wheat;" | I<sub>n+1</sub> = ε-closure(move(I<sub>n</sub>, α))
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|-
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! style="font-weight: normal; align: center; background: #ffffcc;" | -
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! style="font-weight: normal; align: center; background: #ffffcc;" | -
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! style="font-weight: normal; align: center; background: #ffffcc;" | 0
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! style="font-weight: normal; align: left;  background: #ffffcc;" | 0, 1, 2, 3, 4, 6, '''8'''
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! style="font-weight: normal; align: center; background: #ffffcc;" | '''0'''
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|-
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! style="font-weight: normal; align: center; background: #e6e6e6;" | 0
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! style="font-weight: normal; align: center; background: #e6e6e6;" | a
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! style="font-weight: normal; align: center; background: #e6e6e6;" | 5
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! style="font-weight: normal; align: left;  background: #e6e6e6;" | 5, 6
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! style="font-weight: normal; align: center; background: #e6e6e6;" | 1
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|-
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! style="font-weight: normal; align: center; background: #e6e6e6;" | 0
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! style="font-weight: normal; align: center; background: #e6e6e6;" | b
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! style="font-weight: normal; align: center; background: #e6e6e6;" | 7
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! style="font-weight: normal; align: left;  background: #e6e6e6;" | 1, 2, 3, 4, 6, 7, '''8'''
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! style="font-weight: normal; align: center; background: #e6e6e6;" | '''2'''
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|-
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! style="font-weight: normal; align: center; background: #ffffcc;" | 1
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! style="font-weight: normal; align: center; background: #ffffcc;" | a
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! style="font-weight: normal; align: center; background: #ffffcc;" | -
 +
! style="font-weight: normal; align: left;  background: #ffffcc;" | -
 +
! style="font-weight: normal; align: center; background: #ffffcc;" | -
 +
|-
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! style="font-weight: normal; align: center; background: #ffffcc;" | 1
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! style="font-weight: normal; align: center; background: #ffffcc;" | b
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! style="font-weight: normal; align: center; background: #ffffcc;" | 7
 +
! style="font-weight: normal; align: left;  background: #ffffcc;" | 1, 2, 3, 4, 6, 7, '''8'''
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! style="font-weight: normal; align: center; background: #ffffcc;" | '''2'''
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|-
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! style="font-weight: normal; align: center; background: #e6e6e6;" | 2
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! style="font-weight: normal; align: center; background: #e6e6e6;" | a
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! style="font-weight: normal; align: center; background: #e6e6e6;" | 5
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! style="font-weight: normal; align: left;  background: #e6e6e6;" | 5, 6
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! style="font-weight: normal; align: center; background: #e6e6e6;" | 1
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|-
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! style="font-weight: normal; align: center; background: #e6e6e6;" | 2
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! style="font-weight: normal; align: center; background: #e6e6e6;" | b
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! style="font-weight: normal; align: center; background: #e6e6e6;" | 7
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! style="font-weight: normal; align: left;  background: #e6e6e6;" | 1, 2, 3, 4, 6, 7, '''8'''
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! style="font-weight: normal; align: center; background: #e6e6e6;" | '''2'''
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|}
 +
 +
{| width="100%"
 +
! style="text-align: left; font-weight:normal; vertical-align: top; width: 50%;" |Graphically, the DFA is represented as follows:
 +
 +
<graph>
 +
digraph dfa {
 +
    { node [shape=circle style=invis] start }
 +
  rankdir=LR; ratio=0.5
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  node [shape=doublecircle,fixedsize=true,width=0.2,fontsize=10]; 0 2
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  node [shape=circle,fixedsize=true,width=0.2,fontsize=10];
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  start -> 0
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  0 -> 1 [label="a"]
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  0 -> 2 [label="b"]
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  1 -> 2  [label="b"]
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  2 -> 1 [label="a"]
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  2 -> 2 [label="b"]
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  fontsize=10
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  //label="DFA for ((ε|a)b)*"
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}
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</graph>
 +
 +
Given the minimization tree to the right, the final minimal DFA is:
 +
<graph>
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digraph dfamin {
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    { node [shape=circle style=invis] start }
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  rankdir=LR; ratio=0.5
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  node [shape=doublecircle,fixedsize=true,width=0.3,fontsize=10]; 02
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  node [shape=circle,fixedsize=true,width=0.2,fontsize=10]; 1
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  start -> 02
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  02 -> 1 [label="a"]
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  02 -> 02 [label="b"]
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  1 -> 02 [label="b"]
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  fontsize=10
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  //label="DFA for (a|b)*"
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}
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</graph>
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! style="text-align: left; font-weight:normal; vertical-align: top; width: 50%;" | The minimization tree is as follows. As can be seen, the states are indistinguishable.
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 +
<graph>
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digraph mintree {
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  node [shape=none,fixedsize=true,width=0.2,fontsize=10]
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  "{0, 1, 2}" -> "{1}" [label="NF",fontsize=10]
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  "{0, 1, 2}" -> "{0, 2}" [label="  F",fontsize=10]
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  "{0, 2}" -> "{0,2} " [label="  a,b",fontsize=10]
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  fontsize=10
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  //label="Minimization tree"
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}
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</graph>
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|}
  
 
[[category:Teaching]]
 
[[category:Teaching]]
 
[[category:Compilers]]
 
[[category:Compilers]]
 
[[en:Theoretical Aspects of Lexical Analysis]]
 
[[en:Theoretical Aspects of Lexical Analysis]]

Revision as of 02:31, 22 March 2009

Use Thompson's algorithm to build the NFA for the following regular expression. Build the corresponding DFA and minimize it.

  • ((ε|a)b)*

Solution

NFA

The following is the result of applying Thompson's algorithm.


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, 3, 4, 6, 8 0
0 a 5 5, 6 1
0 b 7 1, 2, 3, 4, 6, 7, 8 2
1 a - - -
1 b 7 1, 2, 3, 4, 6, 7, 8 2
2 a 5 5, 6 1
2 b 7 1, 2, 3, 4, 6, 7, 8 2
Graphically, the DFA is represented as follows:


Given the minimization tree to the right, the final minimal DFA is:


The minimization tree is as follows. As can be seen, the states are indistinguishable.


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