CS301 Data Structures Lecture No. 07
CS301 – Data Structures Lecture No. 07
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Data Structures
Lecture No. 07
Reading Material
Data Structures and Algorithm Analysis in C++ Chapter. 3
3.3.3
Summary
1) Evaluating postfix expressions
2) An example
3) Infix to postfix Conversion
Evaluating postfix expressions
In the previous lecture, we talked about ‘infix and postfix expression’ and tried to understand how to write the postfix notation of mathematical expressions. A programmer can write the operators either after the operands i.e. postfix notation or before the operands i.e. prefix notation. Some of the examples are as under:
Infix / PostfixA + B / A B +
12 + 60 – 23 / 12 60 + 23 –
(A + B)*(C – D ) / A B + C D – *
A B * C – D + E/F / A B C*D – E F/+
The last expression seems a bit confusing but may prove simple by following the rules in letter and spirit. In the postfix form, parentheses are not used. Consider the infix expressions as ‘4+3*5’ and ‘(4+3)*5’. The parentheses are not needed in the first but are necessary in the second expression. The postfix forms are:
4+3*5 435*+
(4+3)*5 43+5*
In case of not using the parenthesis in the infix form, you have to see the precedence rule before evaluating the expression. In the above example, if we want to add first then we have to use the parenthesis. In the postfix form, we do not need to use parenthesis. The position of operators and operands in the expression makes it clear in which order we have to do the multiplication and addition.
Now we will see how the infix expression can be evaluated. Suppose we have a postfix expression. How can we evaluate it? Each operator in a postfix expression refers to the previous two operands. As the operators are binary (we are not talking about unary operators here), so two operands are needed for each operator. The nature of these operators is not affected in the postfix form i.e. the plus operator (+) will apply on two operands. Each time we read an operand, we will push it on the stack. We are going to evaluate the postfix expression with the help of stack. After reaching an operator, we pop the two operands from the top of the stack, apply the operator and push the result back on the stack. Now we will see an example to comprehend the working of stack for the evaluation of the postfix form. Here is the algorithm in pseudo code form. After reading this code, you will understand the algorithm.
Stack s; // declare a stackwhile( not end of input ) { // not end of postfix expression
e = get next element of input
if( e is an operand )
s.push( e );
else {
op2 = s.pop();
op1 = s.pop();
value = result of applying operator ‘e’ to op1 and op2;
s.push( value );
}
}
finalresult = s.pop();
We have declared a Stack‘s’. There is a ‘while loop’ along with ‘not end of input’ condition. Here the input is our postfix expression. You can get the expression from the keyboard and use the enter key to finish the expression. In the next statement, we get the next element and store it in ‘e’. This element can be operator or operand. The operand needs not to be single digit. It may be of two digits or even more like 60 or 234 etc. The complete number is stored in the ‘e’. Then we have an ‘if statement’ to check whether ‘e’ is an operand or not. If ‘e’ is an operand than we wrote s.push(e) i.e. we pushed the ‘e’ onto the stack. If ‘e’ is not the operand, it may be an operator. Therefore we will pop the two elements and apply that operator. We pop the stack and store the operand in ‘op2’. We pop the stack again and store the element in ‘op1’. Then the operator in ‘e’ is applied to ‘op1’ and ‘op2’ before storing the result in value. In the end, we push the ‘value’ on the stack. After exiting the loop, a programmer may have only one element in the stack. We pop this element which is the final result.
Consider the example of 4+3*2 having a postfix form of 432*+. Here 4, 3, and 2 are operands whereas + and * are operators. We will push the numbers 4, 3 and 2 on the stack before getting the operator *. Two operands will be popped from the stack and * is being applied on these. As stack is a LIFO structure, so we get 2 first and then 3 as a result of pop. So 2 is store in ‘op1’ and 3 in ‘op2’. Let’s have a look on the program again. On applying * on these, we will push the result (i.e. 6) on the stack. The ‘while loop’ will be executed again. In case of getting the next input as operand, we will push it on the stack otherwise we will pop the two operands and apply the operator on these. Here the next element is the operator +. So two operands will be popped from the stack i.e. 6 and 4. We will apply the operator plus on these and push the result (i.e. 10) on the stack. The input is finished. Now we will pop the stack to get the final result i.e. 10.
An Example
In the earlier example, we have used the stack to solve the postfix expression. Let’s see another comprehensive example. The postfix expression is:
6 2 3 + - 3 8 2 / + * 2 3 +
We want to evaluate this long expression using stack. Let’s try to solve it on paper. We have five columns here i.e. input, op1, op2, value and stack. We will run our pseudo code program. In the start, we read the input as a result we get number 6. As 6 is operand, so it will be pushed on the stack. Then we have number 2 which will also be pushed on the stack. Now 2 is the most recent element. The next element is the number 3 that will also be pushed on the stack. Now, there are three elements on the stack i.e. 3, 2 and 6. The number 3 is the most recent. On popping, we will get the number 3 first of all. The next element is ‘+’, an operator. Now the else part of our pseudo code is executed. We will pop two operands from the stack and apply the operator (+) on these. The number 3 will be stored in variable op2 and number 2 in op1. The operator (+) will be applied on these i.e. 2+3 and the result is stored in value. Now we will push the value (i.e. 5) on the stack. Now we have two numbers on the stack i.e. 5 and 6. The number 5 is the most recent element. The next element is ‘-‘. As it is also an operator, so we will pop the two elements from the stack i.e. 5 and 6. Now we have 5 in op2 and 6 in op1. On applying the operator (-), we will get the result as 1 (6-5). We can’t say op2 - op1. The result (1) will be pushed on stack. Now on the stack, we have only one element i.e. 1. Next three elements are operands so we pushed 3, 8 and 2 on the stack. The most recent element is 2. The next input is an operator in the expression i.e. ‘/’, we will pop two elements from the stack. The number 2 will be stored in op2 while number 8 in op1. We apply the operator (/) on the op1 and op2 i.e. (op1/op2), the result is 4 (i.e. 8/2). We push the result on the stack. We have, now, three elements i.e. 4, 3, and 1 on the stack. The next element is operator plus (+). We will pop the two elements i.e. 4 and 3 and will apply the operator (+). The result (7) will be pushed on the stack. The next input element is operator multiply (*). We will pop the two elements i.e. 7 and 1 and the result (7*1 = 7) is pushed on the stack. You have noted that whenever we have an operator in the input expression, we have two or more elements on the stack. As the operators we are using are binary and we need two operands for them. It will never be the case that you want to pop two elements from the stack and there is only one or no element on the stack. If this happens than it means there is an error in the program and you have popped more values than required. The next input element is 2 that is pushed on the stack. We have, now, the operator ( ) in the input. So we will pop the two elements, op2 will hold 2 and op1 will have the number 7. The operator ( ) will be applied on the operands i.e. (7 2) and the result (49) is pushed on the stack. We have, now, the number 3 in the element being pushed on the stack. The last element is the operator plus (+). So we pop the two elements i.e. 49 and 2 and apply the operator on these. The result (49+3 = 52) is pushed on the stack. The input expression is finished, resulting in the final result i.e. 52.
This the tabular form of the evaluation of the postfix expression.
Input / op1 / op2 / value / stack6 / 6
2 / 2
6
3 / 3
2
6
+ / 2 / 3 / 5 / 5
6
- / 6 / 5 / 1 / 1
3 / 6 / 5 / 1 / 3
1
8 / 6 / 5 / 1 / 8
3
1
2 / 6 / 5 / 1 / 2
8
3
1
/ / 8 / 2 / 4 / 4
3
1
+ / 3 / 4 / 7 / 7
1
* / 1 / 7 / 7 / 7
2 / 1 / 7 / 7 / 2
7
/ 7 / 2 / 49 / 49
3 / 7 / 2 / 49 / 3
49
+ / 49 / 3 / 52 / 52
With the help of stack we can easily solve a very big postfix expression. Suppose you want to make a calculator that is a part of some application e.g. some spreadsheet program. This calculator will be used to evaluate expressions. You may want to calculate the value of a cell after evaluating different cells. Evaluation of the infix form programmatically is difficult but it can be done. We will see another data structure which being used to solve the expressions in infix form. Currently, we have to evaluate the values in different cells and put this value in another cell. How can we do that? We will make the postfix form of the expression associated with that cell. Then we can apply the above algorithm to solve the postfix expression and the final result will be placed at that cell. This is one of the usages of the stack.
Infix to postfix Conversion
We have seen how to evaluate the postfix expressions while using the stack. How can we convert the infix expression into postfix form? Consider the example of a spreadsheet. We have to evaluate expressions. The users of this spreadsheet will employ the infix form of expressions. Consider the infix expressions ‘A+B*C’ and ‘(A+B)*C’. The postfix versions are ‘ABC*+’ and ‘AB+C*’ respectively. The order of operands in postfix is the same as that in the infix. In both the infix expressions, we have the order of operands as A, B and then C. In the postfix expressions too, the order is the same i.e. A, B, followed by C. The order of operands is not changed in postfix form. However, the order of operators may be changed. In the first expression ‘A+B*C’, the postfix expression is ‘ABC*+’. In the postfix form multiplication comes before the plus operator. In scanning from left to right, the operand ‘A’ can be inserted into postfix expression. First rule of algorithm is that if we find the operand in the infix form, put it in the postfix form. The rules for operators are different. The ‘+’ cannot be inserted in the postfix expression until its second operand has been scanned and inserted. Keep the expression A+B*C in your mind. What is the second operand of the plus? The first operand is A and the second operand is the result of B*C. The ‘+’ has to wait until the ‘*’ has not been performed. You do the same thing while using the calculator. First you will multiply the B*C and then add A into the result. The ‘+’ has to be stored away until its proper position is found. When ‘B’ is seen, it is immediately inserted into the postfix expression. As ‘B’ is the operand, we will send the operand to the postfix form. Can the ‘+’ be inserted now? In case of ‘A+B*C’, we cannot insert ‘+’ because ‘*’ has precedence. To perform multiplication, we need the second operand. The first operand of multiplication is ‘B’ while the second one is ‘C’. So at first, we will perform the multiplication before adding result to ‘A’.
In case of ‘(A+B)*C’, the closing parenthesis indicates that ‘+’ must be performed first. After sending the A and B to postfix perform, we can perform the addition due to the presence of the parenthesis. Then C will be sent to the postfix expression. It will be followed by the multiplication of the C and the result of A + B. The postfix form of this expression is AB+C*. Sometimes, we have two operators and need to decide which to apply first like in this case ‘+’ and ‘*’. In this case, we have to see which operator has higher precedence. Assume that we have a function ‘prcd(op1,op2)’ where op1 and op2 are two operators. The function ‘prcd(op1,op2)’ will return TRUE if op1 has precedence over op2, FASLE otherwise. Suppose we call this function with the arguments ‘*’ and ‘+’ i.e. prcd(*, +), it will return true. It will also return true in case both op1 and op2 are ‘+’ e.g. if we have A+B+C, then it does not matter which + we perform first. The call prcd(+ , *) will return false as the precedence of * is higher than the + operator. The ‘+’ has to wait until * is performed.
Now we will try to form an algorithm to convert infix form into postfix form. For this purpose, a pseudo code will be written. We will also write the loops and if conditions. The pseudo code is independent of languages. We will be using a stack in this algorithm. Here, the infix expression is in the form of a string. The algorithm is as follows:
Stack s;
while( not end of input ) {
c = next input character;
if( c is an operand )
add c to postfix string;
else {
while( !s.empty() & prcd(s.top(),c) ){
op = s.pop();
add op to the postfix string;
}
s.push( c );
}
while( !s.empty() ) {
op = s.pop();
add op to postfix string;
}
First we will declare a stack ‘s’. The ‘while loop’ will continue till the end of input. We read the input character and store it in the ‘c’. Here the input character does not mean one character, but an operand or an operator. Then we have a conditional if statement. If ‘c’ is an operand, then we will have to add it to postfix string. Whenever we get an operand in the infix form, it will be added to the postfix form. The order of operands does not change in the conversion.