data structures and algorithms · characters in expression are valid validate() infix expression...

IIT Bombay

Data Structures and AlgorithmsProf. Ajit A. Diwan

Prof. Ganesh RamakrishnanProf. Deepak B. Phatak

Department of Computer Science and EngineeringIIT Bombay

Session: Infix to Postfix (Program)

1Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, IIT Bombay

IIT Bombay

Algorithm

Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, IIT Bombay 2

Input: Infix ExpressionOutput: Postfix Expression

1. Validate infix expression1.1 If not valid, display appropriate error message, else proceed for conversion

2. Traverse the infix expression (character by character)2.1 If character is operand, then append to the string ‘postfix’2.2 else, do the following:

2.2.1 If operator is ‘(‘, push it to the operator stack and continue with point 2.

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Algorithm


2.2.2 If stack is empty and operator is not ‘)’, then push the operator on the stack

2.2.3 If stack is not empty, then compare the current operator with the one which is present on top of the stack.(a) If it has higher precedence: push the operator on the stack(b) If it has lower precedence then,

i. Append the top element of the stack to string ‘postfix’ii. Pop the top operator from the stack iii. Continue i and ii till ‘(‘ is encountered or the stack

becomes empty.iv. Finally, push the current operator, but not ‘)’ on the

stack3. Append the remaining operators from the stack to the string ‘postfix’ by using

‘top’ followed by ‘pop’, till stack becomes empty.

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Program: Functions


Name Parameter Returns Purpose

checkParen-theses()

Infix expression (character array)

Called from validate(). Returns ‘true’ or ‘false’

To check whether openand close braces match

isValid() Current character Called from validate(). Returns ‘true’ or ‘false’

To check whether characters in expression are valid

validate() Infix expression(character array)

Returns ‘true’ or ‘false’ to the main() function

To check whether the expression is valid

isOperand() Current character Called from ‘infix2Postfix()’. Returns ‘true’ or ‘false’

To check whether the current character is an operand i.e. a to z or A to Z

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Program: Functions


Name Parameter Returns Purpose

priorityNumber() characteri.e. Operator

Called from‘checkPriority()’. Returns an integer.

To determine the priority of the operator

checkPriority() Current operator, and top operator on stack

Called from ‘infix2Postfix()’. Returns ‘true’ or ‘false’

To determine the priority of the operator

infix2Postfix() Infix expression,Postfix expression. (Both are character arrays)

Returns the postfix string to the main() function

Converts a valid infix expression to postfix expression

IIT Bombay

Program


bool checkParentheses(char infix[]) {int open=0, close=0, i=0;while (infix[i] != '\0') {

if(infix[i] == '(')open++;

if(infix[i] == ')') {close++;if (close>open)

return false;}i++;

}if(open==close) return true;else return false;

} // End of function

1. Traverse the entire infix expression

i. Count the number of ‘(‘

i. Count the number of ‘)’a. If count of ‘)’ is more than ‘(‘,

then return false

2. If the total count of ‘(‘ and ‘)’ matches, return true3. Else, return false

IIT Bombay

Program


bool isValid(char currentChar) {return ((currentChar>=65 && currentChar<=90) ||

(currentChar>=97 && currentChar<=122) ||(currentChar>=40 && currentChar<=43) || currentChar==45 ||currentChar==47 || currentChar==94);

} // end of fucntion

Check whether the character is either of the followingA to Z ora to z or(, ), *, + or- or/ or^Return true if the above condition is satisfied, else return false

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Program


bool validate(char infix[]) {int i=0;while (infix[i] != '\0') {

if (! isValid(infix[i]) )return false;

elsei++;

}if (checkParentheses(infix) == false)

return false;

return true;} // end of fucntion

1. Traverse the string i. If character is not an operand or

operator, then return falseii. Else, point to the next character

2. Examine whether ‘checkParentheses()’i. If ‘false’ was returned, return ‘false

3. Return ‘true’. This means that the expression passed all tests and it is valid.

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Program


bool checkPriority(char currentOperator, char topOperator) {if ( priorityNumber(currentOperator) < priorityNumber(topOperator))

return true;else

return false;} // End of function

int priorityNumber(char Operator) {if (Operator=='^') return 0;if (Operator=='*' || Operator=='/') return 1;if (Operator=='+' || Operator=='-') return 2;if (Operator=='(') return 3;

} // End of function

Return value 0, 1, 2, or 3 depending on the operator

IIT Bombay

Program


bool isOperand(char currentChar) {return ( (currentChar>=65 && currentChar<=90) ||

(currentChar>=97 && currentChar<=122) );} // End of function

Check whether the character is either of the followingA to Z ora to z orReturn true if the above condition is satisfied, else return false

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Program


void infix2Postfix(char infix[], char answer[]) {int i=0, j=0; char currentChar; char postfix[50]="";stack<char> operatorStack; //Operator stackwhile(infix[i]!='\0') { //Traverse the infix expression

currentChar = infix[i];

i++; //Point to next character of infix expression} //End of while

strcpy(answer,postfix); } //End of function

Code for 2.1

Code for 2.2

Code for 3

Note: 2.1, 2.2, and 3 are point number given in the ‘Algorithm’

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Program


// Code for 2.1if ( isOperand(currentChar) ) {

postfix[j]=currentChar;j++;

}

Note: 2.1, is point number given in the ‘Algorithm’

2.1 If character is operand, then append to the string ‘postfix’

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Program


// Code for 2.2else { //It is operator

}

Code for 2.2.1

Code for 2.2.2

Code for 2.2.3

if (currentChar=='(') {operatorStack.push(currentChar);i++;continue;

}

Note: 2.2, 2.2.1, 2.2.2, and 2.2.3 are point number given in the ‘Algorithm’

2.2 else, do the following (It is an operator): 2.2.1 If operator is ‘(‘, push it to the operator stack and continue with point 2.

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Program



}

Code for 2.2.1

Code for 2.2.2

Code for 2.2.3

if(operatorStack.empty() && currentChar!=')') {operatorStack.push(currentChar);

}

Note: 2.2, 2.2.1, 2.2.2, and 2.2.3 are point number given in the ‘Algorithm’

2.2.2 If stack is empty and operator is not ‘)’, then push the operator on the stack

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Program


2.2.3 If stack is not empty, then compare the current operator with the one which is present on top of the stack.(a) If it has higher precedence: push the operator on the stack(b) If it has less precedence then,

i. Append the top element of the stack to string ‘postfix’

ii. Pop the top operator from the stack iii. Continue i and ii till ‘(‘ is encountered or the stack

becomes empty.iv. Finally, push the current operator, but not ‘)’ on

the stack

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Program



}

Code for 2.2.1

Code for 2.2.2

Code for 2.2.3

else {if (checkPriority(currentChar,operatorStack.top()))

operatorStack.push(currentChar);else {

while (!operatorStack.empty()) {if (operatorStack.top()=='(') {

operatorStack.pop(); break;

}postfix[j]=operatorStack.top();j++;operatorStack.pop();

} //End of whileif(currentChar!=')')

operatorStack.push(currentChar);} //End of else

} //End of else

2.2.3.a

2.2.3.b

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Program


// Code for 3//Transfer remaining items in stackwhile(!operatorStack.empty()) {

postfix[j]=operatorStack.top();operatorStack.pop();j++;

}

Note: 3, is point number given in the ‘Algorithm’

3. Append the remaining operators from the stack to the string ‘postfix’ by using ‘top’ followed by ‘pop’, till stack becomes empty.

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Program


int main() {char infix[50], postfix[50];cout << "Enter an infix expression without space.

Use parentheses to indicate priority. Only +, -, *, /, ^ symbols are allowed\n";

cin >> infix;

if (validate(infix) == false) {cout << "The expression contains a number, or an unwanted symbol,

or the round braces do not match\n";return -1;

}cout << "Infix expression : " << infix << endl;infix2Postfix(infix,postfix);cout << "Postfix expression : " << postfix << endl;

return 0; }

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References


• https://en.wikipedia.org/wiki/Stack_(abstract_data_type)

• http://csis.pace.edu/~wolf/CS122/infix-postfix.htm

https://en.wikipedia.org/wiki/Polish_notation

http://csis.pace.edu/~wolf/CS122/infix-postfix.htm

IIT Bombay

Thank you


data structures and algorithms · characters in expression are valid validate() infix expression...

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