## C program to check whether a mathematical statement is solvable or not

ALGORITHM
Start.
Declare two character arrays str[],token[] and initialize integer variables a=0,b=0,c,d.
Input the string from the user.
Repeat steps 5 to 12 till str[a] =‟\0‟.
If str[a] =='(‘ or str[a] =='{‘ then token[b] =‟4‟, b++.
If str[a] ==’)’ or str[a] ==’}‟ then token[b] =‟5‟, b++.
Check if isdigit(str[a]) then repeat steps 8 till isdigit(str[a])
a++.
a–, token[b] =‟6‟, b++.
If str[a]==’+‟ then token[b]=’2′,b++.
If(str[a]==’*’) then token[b]=‟3‟,b++.
a++.
token[b]=’\0′;
then print the token generated for the string .
b=0.
Repeat step 22 to 31 till token[b]!=’\0′
c=0.
Repeat step 24 to 30 till (token[b]==’6′ and token[b+1]==’2′ and token[b+2]==’6′) or
(token[b]==’6′ and token[b+1]==’3’and token[b+2]==’6′) or (token[b]==’4′ and
token[b+1]==’6′ and token[b+2]==’5′) or (token[c]!=’\0′).
token[c]=’6′;
c++;
Repeat step 27 to 28 till token[c]!=’\0′.
token[c]=token[c+2].
c++.
token[c-2]=‟\0‟.
print token.
b++.
Compare token with 6 and store the result in d.
If d=0 then print that the string is in the grammar.
Else print that the string is not in the grammar.
Stop.

				
PROGRAM:
#include<stdio.h>
#include<conio.h>
#include<ctype.h>
#include<string.h>
void main()
{
int a=0,b=0,c;
char str[20],tok[11];
clrscr();
printf("Input the expression = ");
gets(str);
while(str[a]!='\0')
{
if((str[a]=='(')||(str[a]=='{'))
{
tok[b]='4';
b++;
}
if((str[a]==')')||(str[a]=='}'))
{
tok[b]='5';
b++;
}
if(isdigit(str[a]))
{
while(isdigit(str[a]))
{
a++;
}
a--;
tok[b]='6';
b++;
}
if(str[a]=='+')
{
tok[b]='2';
b++;
}
if(str[a]=='*')
{
tok[b]='3';
b++;
}
a++;
}
tok[b]='\0';
puts(tok);
b=0;
while(tok[b]!='\0')
{
if(((tok[b]=='6')&&(tok[b+1]=='2')&&(tok[b+2]=='6'))||((tok[b]=='6')&&(tok[b+1]=='3')&&(tok[b+2]=='6'))||((tok[b]=='4')&&(tok[b+1]=='6')&&(tok[b+2]=='5'))/*||((tok[b]!=6)&&(tok[b+1]!='\0'))*/)
{
tok[b]='6';
c=b+1;
while(tok[c]!='\0')
{
tok[c]=tok[c+2];
c++;
}
tok[c]='\0';
puts(tok);
b=0;
}
else
{
b++;
puts(tok);
}
}
int d;
d=strcmp(tok,"6");
if(d==0)
{
printf("It is in the grammar.");
}
else
{
printf("It is not in the grammar.");
}
getch();
}



INPUT/OUTPUT
Input the expression = (23+)
4625
4625
4625
4625
4625
It is not in the grammar.
Input the expression = (2+(3+4)+5)
46246265265
46246265265
46246265265
46246265265
46246265265
462465265
462465265
462465265
462465265
4626265
4626265
46265
46265
465
6
6
It is in the grammar