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

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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
  node [shape=doublecircle,fixedsize=true,width=0.2,fontsize=10]; 8
  node [shape=circle,fixedsize=true,width=0.2,fontsize=10];
  start -> 0
  0 -> 1; 0 -> 8
  1 -> 2; 1 -> 4
  2 -> 3;
  3 -> 6
  4 -> 5 [label="a",fontsize=10]
  5 -> 6
  6 -> 7 [label="b",fontsize=10]
  7 -> 1; 7 -> 8
  fontsize=10
  //label="NFA for ((ε|a)b)*"
}
</graph>
=== DFA ===
Determination table for the above NFA:
{| cellspacing="2"
! style="padding-left: 20px; padding-right: 20px; background: wheat;" | I<sub>n</sub>
! style="padding-left: 20px; padding-right: 20px; background: wheat;" | α∈Σ
! style="padding-left: 20px; padding-right: 20px; background: wheat;" | move(I<sub>n</sub>, α)
! style="padding-left: 20px; padding-right: 20px; background: wheat;" | ε-closure(move(I<sub>n</sub>, α))
! style="padding-left: 20px; padding-right: 20px; background: wheat;" | I<sub>n+1</sub> = ε-closure(move(I<sub>n</sub>, α))
|-
! style="font-weight: normal; align: center; background: #ffffcc;" | -
! style="font-weight: normal; align: center; background: #ffffcc;" | -
! style="font-weight: normal; align: center; background: #ffffcc;" | 0
! style="font-weight: normal; align: left;  background: #ffffcc;" | 0, 1, 2, 3, 4, 6, '''8'''
! style="font-weight: normal; align: center; background: #ffffcc;" | '''0'''
|-
! style="font-weight: normal; align: center; background: #e6e6e6;" | 0
! style="font-weight: normal; align: center; background: #e6e6e6;" | a
! style="font-weight: normal; align: center; background: #e6e6e6;" | 5
! style="font-weight: normal; align: left;  background: #e6e6e6;" | 5, 6
! style="font-weight: normal; align: center; background: #e6e6e6;" | 1
|-
! style="font-weight: normal; align: center; background: #e6e6e6;" | 0
! style="font-weight: normal; align: center; background: #e6e6e6;" | b
! style="font-weight: normal; align: center; background: #e6e6e6;" | 7
! style="font-weight: normal; align: left;  background: #e6e6e6;" | 1, 2, 3, 4, 6, 7, '''8'''
! style="font-weight: normal; align: center; background: #e6e6e6;" | '''2'''
|-
! style="font-weight: normal; align: center; background: #ffffcc;" | 1
! style="font-weight: normal; align: center; background: #ffffcc;" | a
! style="font-weight: normal; align: center; background: #ffffcc;" | -
! style="font-weight: normal; align: left;  background: #ffffcc;" | -
! style="font-weight: normal; align: center; background: #ffffcc;" | -
|-
! style="font-weight: normal; align: center; background: #ffffcc;" | 1
! style="font-weight: normal; align: center; background: #ffffcc;" | b
! style="font-weight: normal; align: center; background: #ffffcc;" | 7
! style="font-weight: normal; align: left;  background: #ffffcc;" | 1, 2, 3, 4, 6, 7, '''8'''
! style="font-weight: normal; align: center; background: #ffffcc;" | '''2'''
|-
! style="font-weight: normal; align: center; background: #e6e6e6;" | 2
! style="font-weight: normal; align: center; background: #e6e6e6;" | a
! style="font-weight: normal; align: center; background: #e6e6e6;" | 5
! style="font-weight: normal; align: left;  background: #e6e6e6;" | 5, 6
! style="font-weight: normal; align: center; background: #e6e6e6;" | 1
|-
! style="font-weight: normal; align: center; background: #e6e6e6;" | 2
! style="font-weight: normal; align: center; background: #e6e6e6;" | b
! style="font-weight: normal; align: center; background: #e6e6e6;" | 7
! style="font-weight: normal; align: left;  background: #e6e6e6;" | 1, 2, 3, 4, 6, 7, '''8'''
! style="font-weight: normal; align: center; background: #e6e6e6;" | '''2'''
|}
{| 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
  node [shape=doublecircle,fixedsize=true,width=0.2,fontsize=10]; 0 2
  node [shape=circle,fixedsize=true,width=0.2,fontsize=10];
  start -> 0
  0 -> 1 [label="a"]
  0 -> 2 [label="b"]
  1 -> 2  [label="b"]
  2 -> 1 [label="a"]
  2 -> 2 [label="b"]
  fontsize=10
  //label="DFA for ((ε|a)b)*"
}
</graph>
Given the minimization tree to the right, the final minimal DFA is:
<graph>
digraph dfamin {
    { node [shape=circle style=invis] start }
  rankdir=LR; ratio=0.5
  node [shape=doublecircle,fixedsize=true,width=0.3,fontsize=10]; 02
  node [shape=circle,fixedsize=true,width=0.2,fontsize=10]; 1
  start -> 02
  02 -> 1 [label="a"]
  02 -> 02 [label="b"]
  1 -> 02 [label="b"]
  fontsize=10
  //label="DFA for (a|b)*"
}
</graph>
! 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.
<graph>
digraph mintree {
  node [shape=none,fixedsize=true,width=0.2,fontsize=10]
  "{0, 1, 2}" -> "{1}" [label="NF",fontsize=10]
  "{0, 1, 2}" -> "{0, 2}" [label="  F",fontsize=10]
  "{0, 2}" -> "{0,2} " [label="  a,b",fontsize=10]
  fontsize=10
  //label="Minimization tree"
}
</graph>
|}


[[category:Teaching]]
[[category:Teaching]]
[[category:Compilers]]
[[category:Compilers]]
[[en:Theoretical Aspects of Lexical Analysis]]
[[en:Theoretical Aspects of Lexical Analysis]]

Revision as of 00:31, 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.

<graph> digraph nfa {

    { node [shape=circle style=invis] start }
 rankdir=LR; ratio=0.5
 node [shape=doublecircle,fixedsize=true,width=0.2,fontsize=10]; 8
 node [shape=circle,fixedsize=true,width=0.2,fontsize=10];
 start -> 0
 0 -> 1; 0 -> 8
 1 -> 2; 1 -> 4
 2 -> 3;
 3 -> 6
 4 -> 5 [label="a",fontsize=10]
 5 -> 6
 6 -> 7 [label="b",fontsize=10]
 7 -> 1; 7 -> 8
 fontsize=10
 //label="NFA for ((ε|a)b)*"

} </graph>

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:

<graph> digraph dfa {

    { node [shape=circle style=invis] start }
 rankdir=LR; ratio=0.5
 node [shape=doublecircle,fixedsize=true,width=0.2,fontsize=10]; 0 2
 node [shape=circle,fixedsize=true,width=0.2,fontsize=10];
 start -> 0
 0 -> 1 [label="a"]
 0 -> 2 [label="b"]
 1 -> 2  [label="b"]
 2 -> 1 [label="a"]
 2 -> 2 [label="b"]
 fontsize=10
 //label="DFA for ((ε|a)b)*"

} </graph>

Given the minimization tree to the right, the final minimal DFA is: <graph> digraph dfamin {

    { node [shape=circle style=invis] start }
 rankdir=LR; ratio=0.5
 node [shape=doublecircle,fixedsize=true,width=0.3,fontsize=10]; 02
 node [shape=circle,fixedsize=true,width=0.2,fontsize=10]; 1
 start -> 02
 02 -> 1 [label="a"]
 02 -> 02 [label="b"]
 1 -> 02 [label="b"]
 fontsize=10
 //label="DFA for (a|b)*"

} </graph>

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

<graph> digraph mintree {

 node [shape=none,fixedsize=true,width=0.2,fontsize=10]
 "{0, 1, 2}" -> "{1}" [label="NF",fontsize=10]
 "{0, 1, 2}" -> "{0, 2}" [label="  F",fontsize=10]
 "{0, 2}" -> "{0,2} " [label="  a,b",fontsize=10]
 fontsize=10
 //label="Minimization tree"

} </graph>