Exercise
Question Explanation
The correct answer is: 6 7 8
Here are the sets of reachable states after reading in the first i characters (and following epsilon-transitions):
i substring set of reachable states ------------------------------------------------ 0 0 1 1 B 2 3 5 6 7 8 11 12 13 2 B A 2 3 4 5 6 7 8 9 10 11 12 13 3 B A A 2 3 4 5 6 7 8 9 10 11 12 13 4 B A A A 2 3 4 5 6 7 8 9 10 11 12 13 5 B A A A C 6 7 8 6 B A A A C C 6 7 8
Here are all of the edges in the epsilon-transition digraph: 0-> 1 0-> 3 2->10 3-> 4 3-> 9 5-> 6 6-> 5 6-> 7 8-> 9 9-> 3 9->10 10->11
Interview Questions
/*************************************************************************
* Compilation: javac NFA.java
* Execution: java NFA regexp text
* Dependencies: Stack.java Bag.java Digraph.java DirectedDFS.java
*
* % java NFA "(A*B|AC)D" AAAABD
* true
*
* % java NFA "(A*B|AC)D" AAAAC
* false
*
* % java NFA "(a|(bc)*d)*" abcbcd
* true
*
* % java NFA "(a|(bc)*d)*" abcbcbcdaaaabcbcdaaaddd
* true
*
* Remarks
* -----------
* - This version does not suport the + operator or multiway-or.
*
* - This version does not handle character classes,* metacharacters (either in the text or pattern),capturing
* capabilities,greedy vs. relucantant modifier,and
* other features in industrial-strength implementations such
* as java.util.regexp.
*
*************************************************************************/
public class NFA {
private Digraph G; // digraph of epsilon transitions
private String regexp; // regular expression
private int M; // number of characters in regular expression
// Create the NFA for the given RE
public NFA(String regexp) {
this.regexp = regexp;
M = regexp.length();
Stack<Integer> ops = new Stack<Integer>();
G = new Digraph(M+1);
for (int i = 0; i < M; i++) {
int lp = i;
if (regexp.charAt(i) == '(' || regexp.charAt(i) == '|')
ops.push(i);
else if (regexp.charAt(i) == ')') {
int or = ops.pop();
// 2-way or operator
if (regexp.charAt(or) == '|') {
lp = ops.pop();
G.addEdge(lp,or+1);
G.addEdge(or,i);
}
else if (regexp.charAt(or) == '(')
lp = or;
else assert false;
}
// closure operator (uses 1-character lookahead)
if (i < M-1 && regexp.charAt(i+1) == '*') {
G.addEdge(lp,i+1);
G.addEdge(i+1,lp);
}
if (regexp.charAt(i) == '(' || regexp.charAt(i) == '*' || regexp.charAt(i) == ')')
G.addEdge(i,i+1);
}
}
// Does the NFA recognize txt?
public boolean recognizes(String txt) {
DirectedDFS dfs = new DirectedDFS(G,0);
Bag<Integer> pc = new Bag<Integer>();
for (int v = 0; v < G.V(); v++)
if (dfs.marked(v)) pc.add(v);
// Compute possible NFA states for txt[i+1]
for (int i = 0; i < txt.length(); i++) {
Bag<Integer> match = new Bag<Integer>();
for (int v : pc) {
if (v == M) continue;
if ((regexp.charAt(v) == txt.charAt(i)) || regexp.charAt(v) == '.')
match.add(v+1);
}
dfs = new DirectedDFS(G,match);
pc = new Bag<Integer>();
for (int v = 0; v < G.V(); v++)
if (dfs.marked(v)) pc.add(v);
// optimization if no states reachable
if (pc.size() == 0) return false;
}
// check for accept state
for (int v : pc)
if (v == M) return true;
return false;
}
public static void main(String[] args) {
String regexp = "(" + args[0] + ")";
String txt = args[1];
if (txt.indexOf('|') >= 0) {
throw new IllegalArgumentException("| character in text is not supported");
}
NFA nfa = new NFA(regexp);
StdOut.println(nfa.recognizes(txt));
}
} (编辑:李大同)
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