我试图实现基于维基百科伪代码的CYK算法。当我测试字符串" a b"对于语法输入:
S-> A B
A->一种
B-&GT,B
它让我虚伪,我认为应该是真的。我有一个名为AllGrammar的arraylist,其中包含所有规则。对于上面的示例,它将包含:
[0]:S-> AB
[1]:A-> a一[2]:B-> b
对于示例S- > hello和输入字符串hello它给了我应有的真实。更复杂的测试(更多制作)给我错误:S
public static boolean cyk(String entrada) {
int n = entrada.length();
int r = AllGrammar.size();
//Vector<String> startingsymbols = getSymbols(AllGrammar);
String[] ent = entrada.split("\\s");
n = ent.length;
System.out.println("length of entry" + n);
//let P[n,n,r] be an array of booleans. Initialize all elements of P to false.
boolean P[][][] = initialize3DVector(n, r);
//n-> number of words of string entrada,
//r-> number of nonterminal symbols
//This grammar contains the subset Rs which is the set of start symbols
for (int i = 1; i < n; i++) {
for(int j = 0; j < r; j++) {
String[] rule = (String[]) AllGrammar.get(j);
if (rule.length == 2) {
if (rule[1].equals(ent[i])) {
System.out.println("entrou");
System.out.println(rule[1]);
P[i][1][j + 1] = true;
}
}
}
}
for(int i = 2; i < n; i++) {
System.out.println("FIRST:" + i);
for(int j = 1; j < n - i + 1; j++) {
System.out.println("SECOND:" + j);
for(int k = 1; k < i - 1; k++) {
System.out.println("THIRD:" + k);
for(int g = 0; g < r; g++) {
String[] rule = (String[]) AllGrammar.get(g);
if (rule.length > 2) {
int A = returnPos(rule[0]);
int B = returnPos(rule[1]);
int C = returnPos(rule[2]);
System.out.println("A" + A);
System.out.println("B" + B);
System.out.println("C" + C);
if (A!=-1 && B!=-1 && C!=-1) {
if (P[j][k][B] && P[j + k][i - k][C]) {
System.out.println("entrou2");
P[j][i][A] = true;
}
}
}
}
}
}
}
for(int x = 0; x < r; x++) {
if(P[1][n][x]) return true;
}
return false;
}
答案 0 :(得分:0)
与CYK算法相比:
似乎问题在索引的使用中可能是相当基本的。如果您不熟悉该语言,可能需要复习一下。