Convert Infix to Prefix Expression

Data StructureAlgorithmsMathematical Problems

To solve expressions by the computer, we can either convert it in postfix form or to the prefix form. Here we will see how infix expressions are converted to prefix form.

At first infix expression is reversed. Note that for reversing the opening and closing parenthesis will also be reversed.

for an example: The expression: A + B * (C - D)

after reversing the expression will be: ) D – C ( * B + A

so we need to convert opening parenthesis to closing parenthesis and vice versa.

After reversing, the expression is converted to postfix form by using infix to postfix algorithm. After that again the postfix expression is reversed to get the prefix expression.

Input and Output

Input:
Infix Expression: x^y/(5*z)+2
Output:
Prefix Form Is: +/^xy*5z2

Algorithm

infixToPrefix(infix)

Input − Infix expression to convert into prefix form.

Output − The prefix expression.

Begin
   reverse the infix expression
   for each character ch of reversed infix expression, do
      if ch = opening parenthesis, then
         convert ch to closing parenthesis
      else if ch = closing parenthesis, then
         convert ch to opening parenthesis
   done

   postfix := find transformed infix expression to postfix expression
   prefix := reverse recently calculated postfix form
   return prefix
End

Example

#include<iostream>
#include<stack>
#include<locale> //for function isalnum()
#include<algorithm>
using namespace std;

int preced(char ch) {
   if(ch == '+' || ch == '-') {
      return 1;    //Precedence of + or - is 1
   }else if(ch == '*' || ch == '/') {
      return 2;    //Precedence of * or / is 2
   }else if(ch == '^') {
      return 3;    //Precedence of ^ is 3
   }else {
      return 0;
   }
}

string inToPost(string infix) {
   stack<char> stk;
   stk.push('#');    //add some extra character to avoid underflow
   string postfix = "";   //initially the postfix string is empty
   string::iterator it;

   for(it = infix.begin(); it!=infix.end(); it++) {
      if(isalnum(char(*it)))
         postfix += *it;    //add to postfix when character is letter or number
      else if(*it == '(')
         stk.push('(');
      else if(*it == '^')
         stk.push('^');
      else if(*it == ')') {
         while(stk.top() != '#' && stk.top() != '(') {
            postfix += stk.top();    //store and pop until ( has found
            stk.pop();
         }

         stk.pop();    //remove the '(' from stack
      }else {
         if(preced(*it) > preced(stk.top()))
            stk.push(*it);    //push if precedence is high
         else {
            while(stk.top() != '#' && preced(*it) <= preced(stk.top())) {
               postfix += stk.top();    //store and pop until higher precedence is found
               stk.pop();
            }
            stk.push(*it);
         }
      }
   }

   while(stk.top() != '#') {
      postfix += stk.top();    //store and pop until stack is not empty

      stk.pop();

   }
   return postfix;
}

string inToPre(string infix) {
   string prefix;
   reverse(infix.begin(), infix.end());    //reverse the infix expression
   string::iterator it;

   for(it = infix.begin(); it != infix.end(); it++) {    //reverse the parenthesis after reverse
      if(*it == '(')
         *it = ')';
      else if(*it == ')')
         *it = '(';
   }

   prefix = inToPost(infix);                 //convert new reversed infix to postfix form.
   reverse(prefix.begin(), prefix.end());    //again reverse the result to get final prefix form
   return prefix;
}

int main() {
   string infix = "x^y/(5*z)+2";
   cout << "Prefix Form Is: " << inToPre(infix) << endl;
}

Output

Prefix Form Is: +/^xy*5z2
raja
Published on 11-Jul-2018 13:04:05
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