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Please show me how it works in detail.
Description: Implement the CKY Algorithm as a recognizer to determine if a string is in a language in the languageplatform of your choice.
Program Requirements:
Input a grammar G in CNF. You can use any method to input the grammar. Your program will be tested with grammars of size: 5 – 10 variables, 10 – 20 productions, 2 – 6 terminals.
Input a string w.
Output : True if w ÎL(G), otherwise false.
* Extend your program to a parser that also constructs a parse tree for w.
Expert Answer
Algorithm:
Input: start category S and input string
n := length(string)
for j := 1 to n
chart(j − 1, j) := {X | X → wordj ∈ P}
for i := j − 2 down to 0
chart(i, j) := {}
for k := i + 1 to j − 1
for every A → BC ∈ P
if B ∈ chart(i, k) and C ∈ hart(k, j) then
chart(i, j) := chart(i, j) ∪ {A}
Output: if S ∈ chart(0, n) then accept else reject
Program:
#include<iostream>
#include<cstring>
#include<algorithm>
#include<string>
#include<cassert>
#include<iomanip>
using namespace std;
#define MAX 100
#define for(i,a,b) for(i=a;i<b; i++)
string gram[MAX][MAX]; //to store entered grammar
string dpr[MAX];
int p,np; //np-> number of productions
inline string concat( string a, string b) //concatenates unique non-terminals
{
int i;
string r=a;
for(i,0,b.length())
if(r.find(b[i]) > r.length())
r+=b[i];
return (r);
}
inline void break_gram(string a) //seperates right hand side of entered grammar
{
int i;
p=0;
while(a.length())
{
i=a.find(“|”);
if(i>a.length())
{
dpr[p++] = a;
a=””;
}
else
{
dpr[p++] = a.substr(0,i);
a=a.substr(i+1,a.length());
}
}
}
inline int lchomsky(string a) //checks if LHS of entered grammar is in CNF
{
if(a.length()==1 && a[0]>=’A’ && a[0]<=’Z’)
return 1;
return 0;
}
inline int rchomsky(string a) //checks if RHS of grammar is in CNF
{
if (a.length() == 1 && a[0]>=’a’ && a[0] <=’z’)
return 1;
if (a.length()==2 && a[0]>=’A’ && a[0]<=’Z’ && a[1]>=’A’ && a[1]<=’Z’ )
return 1;
return 0;
}
inline string search_prod(string p) //returns a concatenated string of variables which can produce string p
{
int j,k;
string r=””;
for(j,0,np)
{
k=1;
while(gram[j][k] != “”)
{
if(gram[j][k] == p)
{
r=concat(r,gram[j][0]);
}
k++;
}
}
return r;
}
inline string gen_comb(string a, string b) //creates every combination of variables from a and b . For eg: BA * AB = {BA, BB, AA, BB}
{
int i,j;
string pri=a,re=””;
for(i,0,a.length())
for(j,0,b.length())
{
pri=””;
pri=pri+a[i]+b[j];
re=re+search_prod(pri); //searches if the generated productions can be created or not
}
return re;
}
int main()
{
int i,pt,j,l,k;
string a,str,r,pr,start;
cout<<“nEnter the start Variable “;
cin >> start;
cout<<“nNumber of productions “;
cin >> np;
for(i,0,np)
{
cin >> a;
pt=a.find(“->”);
gram[i][0] = a.substr(0,pt);
if (lchomsky(gram[i][0]) == 0)
{
cout<<“nGrammar not in Chomsky Form”;
abort();
}
a = a.substr(pt+2, a.length());
break_gram(a);
for(j,0,p)
{
gram[i][j+1]=dpr[j];
if (rchomsky(dpr[j]) == 0)
{
cout<<“nGrammar not in Chomsky Form”;
abort();
}
}
}
string matrix[MAX][MAX],st;
cout<<“nEnter string to be checked : “;
cin >> str;
for(i,0,str.length()) //Assigns values to principal diagonal of matrix
{
r=””;
st = “”;
st+=str[i];
for(j,0,np)
{
k=1;
while(gram[j][k] != “”)
{
if(gram[j][k] == st)
{
r=concat(r,gram[j][0]);
}
k++;
}
}
matrix[i][i]=r;
}
int ii,kk;
for(k,1,str.length()) //Assigns values to upper half of the matrix
{
for(j,k,str.length())
{
r=””;
for(l,j-k,j)
{
pr = gen_comb(matrix[j-k][l],matrix[l+1][j]);
r = concat(r,pr);
}
matrix[j-k][j] = r;
}
}
for(i,0,str.length()) //Prints the matrix
{
k=0;
l=str.length()-i-1;
for(j,l,str.length())
{
cout<<setw(5)<<matrix[k++][j]<<” “;
}
cout<<endl;
}
int f=0;
for(i,0,start.length())
if(matrix[0][str.length()-1].find(start[i]) <= matrix[0][str.length()-1].length()) //Checks if last element of first row contains a Start variable
{
cout<<“String can be generatedn”;
return 0;
}
cout<<“String cannot be generatedn”;
return 0;
}
