Stacks

What is a Stack?

A stack (sometimes called a “push-down stack”) is an ordered collection of items where the addition of new items and the removal of existing items always takes place at the same end. This end is commonly referred to as the “top.” The end opposite the top is known as the “base.”

The base of the stack is significant since items stored in the stack that are closer to the base represent those that have been in the stack the longest. The most recently added item is the one that is in position to be removed first. This ordering principle is sometimes called LIFO, last-in first-out. It provides an ordering based on length of time in the collection. Newer items are near the top, while older items are near the base.

Many examples of stacks occur in everyday situations. Almost any cafeteria has a stack of trays or plates where you take the one at the top, uncovering a new tray or plate for the next customer in line. Imagine a stack of books on a desk (Figure 1). The only book whose cover is visible is the one on top. To access others in the stack, we need to remove the ones that are sitting on top of them. Figure 2 shows another stack. This one contains a number of primitive Python data objects.

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Figure 1: A Stack of Books

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Figure 2: A Stack of Primitive Python Objects

One of the most useful ideas related to stacks comes from the simple observation of items as they are added and then removed. Assume you start out with a clean desktop. Now place books one at a time on top of each other. You are constructing a stack. Consider what happens when you begin removing books. The order that they are removed is exactly the reverse of the order that they were placed. Stacks are fundamentally important, as they can be used to reverse the order of items. The order of insertion is the reverse of the order of removal. Figure 3 shows the Python data object stack as it was created and then again as items are removed. Note the order of the objects.

../_images/simplereversal.png

Figure 3: The Reversal Property of Stacks

Considering this reversal property, you can perhaps think of examples of stacks that occur as you use your computer. For example, every web browser has a Back button. As you navigate from web page to web page, those pages are placed on a stack (actually it is the URLs that are going on the stack). The current page that you are viewing is on the top and the first page you looked at is at the base. If you click on the Back button, you begin to move in reverse order through the pages.

The Stack Abstract Data Type

The stack abstract data type is defined by the following structure and operations. A stack is structured, as described above, as an ordered collection of items where items are added to and removed from the end called the “top.” Stacks are ordered LIFO. The stack operations are given below.

  • Stack() creates a new stack that is empty. It needs no parameters and returns an empty stack.
  • push(item) adds a new item to the top of the stack. It needs the item and returns nothing.
  • pop() removes the top item from the stack. It needs no parameters and returns the item. The stack is modified.
  • peek() returns the top item from the stack but does not remove it. It needs no parameters. The stack is not modified.
  • isEmpty() tests to see whether the stack is empty. It needs no parameters and returns a boolean value.
  • size() returns the number of items on the stack. It needs no parameters and returns an integer.

For example, if s is a stack that has been created and starts out empty, then Table 1 shows the results of a sequence of stack operations. Under stack contents, the top item is listed at the far right.

Table 1: Sample Stack Operations
Stack Operation Stack Contents Return Value
s.isEmpty() [] True
s.push(4) [4]  
s.push('dog') [4,'dog']  
s.peek() [4,'dog'] 'dog'
s.push(True) [4,'dog',True]  
s.size() [4,'dog',True] 3
s.isEmpty() [4,'dog',True] False
s.push(8.4) [4,'dog',True,8.4]  
s.pop() [4,'dog',True] 8.4
s.pop() [4,'dog'] True
s.size() [4,'dog'] 2

Implementing a Stack in Python

Now that we have clearly defined the stack as an abstract data type we will turn our attention to using Python to implement the stack. Recall that when we give an abstract data type a physical implementation we refer to the implementation as a data structure.

As we described in Chapter 1, in Python, as in any object-oriented programming language, the implementation of choice for an abstract data type such as a stack is the creation of a new class. The stack operations are implemented as methods. Further, to implement a stack, which is a collection of elements, it makes sense to utilize the power and simplicity of the primitive collections provided by Python. We will use a list.

Recall that the list class in Python provides an ordered collection mechanism and a set of methods. For example, if we have the list [2,5,3,6,7,4], we need only to decide which end of the list will be considered the top of the stack and which will be the base. Once that decision is made, the operations can be implemented using the list methods such as append and pop.

The following stack implementation (ActiveCode 1) assumes that the end of the list will hold the top element of the stack. As the stack grows (as push operations occur), new items will be added on the end of the list. pop operations will manipulate that same end.


Implementing a Stack class using Python lists (stack_1ac)


Remember that nothing happens when we click the run button other than the definition of the class. We must create a Stack object and then use it. ActiveCode 2 shows the Stack class in action as we perform the sequence of operations from Table 1.


(stack_ex_1)


It is important to note that we could have chosen to implement the stack using a list where the top is at the beginning instead of at the end. In this case, the previous pop and append methods would no longer work and we would have to index position 0 (the first item in the list) explicitly using pop and insert. The implementation is shown in CodeLens 1.

Alternative Implementation of the Stack class (stack_cl_1)

This ability to change the physical implementation of an abstract data type while maintaining the logical characteristics is an example of abstraction at work. However, even though the stack will work either way, if we consider the performance of the two implementations, there is definitely a difference. Recall that the append and pop() operations were both O(1). This means that the first implementation will perform push and pop in constant time no matter how many items are on the stack. The performance of the second implementation suffers in that the insert(0) and pop(0) operations will both require O(n) for a stack of size n. Clearly, even though the implementations are logically equivalent, they would have very different timings when performing benchmark testing.

Self Check

stk-1: Given the following sequence of stack operations, what is the top item on the stack when the sequence is complete?

m = Stack()
m.push('x')
m.push('y')
m.pop()
m.push('z')
m.peek()





stk-2: Given the following sequence of stack operations, what is the top item on the stack when the sequence is complete?

m = Stack()
m.push('x')
m.push('y')
m.push('z')
while not m.isEmpty():
   m.pop()
   m.pop()





Write a function revstring(mystr) that uses a stack to reverse the characters in a string.


(stack_stringrev)


Simple Balanced Parentheses

We now turn our attention to using stacks to solve real computer science problems. You have no doubt written arithmetic expressions such as

\((5+6)*(7+8)/(4+3)\)

where parentheses are used to order the performance of operations. You may also have some experience programming in a language such as Lisp with constructs like

(defun square(n)
     (* n n))

This defines a function called square that will return the square of its argument n. Lisp is notorious for using lots and lots of parentheses.

In both of these examples, parentheses must appear in a balanced fashion. Balanced parentheses means that each opening symbol has a corresponding closing symbol and the pairs of parentheses are properly nested. Consider the following correctly balanced strings of parentheses:

(()()()())

(((())))

(()((())()))

Compare those with the following, which are not balanced:

((((((())

()))

(()()(()

The ability to differentiate between parentheses that are correctly balanced and those that are unbalanced is an important part of recognizing many programming language structures.

The challenge then is to write an algorithm that will read a string of parentheses from left to right and decide whether the symbols are balanced. To solve this problem we need to make an important observation. As you process symbols from left to right, the most recent opening parenthesis must match the next closing symbol (see Figure 4). Also, the first opening symbol processed may have to wait until the very last symbol for its match. Closing symbols match opening symbols in the reverse order of their appearance; they match from the inside out. This is a clue that stacks can be used to solve the problem.

../_images/simpleparcheck.png

Figure 4: Matching Parentheses

Once you agree that a stack is the appropriate data structure for keeping the parentheses, the statement of the algorithm is straightforward. Starting with an empty stack, process the parenthesis strings from left to right. If a symbol is an opening parenthesis, push it on the stack as a signal that a corresponding closing symbol needs to appear later. If, on the other hand, a symbol is a closing parenthesis, pop the stack. As long as it is possible to pop the stack to match every closing symbol, the parentheses remain balanced. If at any time there is no opening symbol on the stack to match a closing symbol, the string is not balanced properly. At the end of the string, when all symbols have been processed, the stack should be empty. The Python code to implement this algorithm is shown in ActiveCode 4.


Solving the Balanced Parentheses Problem (parcheck1)


This function, parChecker, assumes that a Stack class is available and returns a boolean result as to whether the string of parentheses is balanced. Note that the boolean variable balanced is initialized to True as there is no reason to assume otherwise at the start. If the current symbol is (, then it is pushed on the stack (lines 9–10). Note also in line 15 that pop simply removes a symbol from the stack. The returned value is not used since we know it must be an opening symbol seen earlier. At the end (lines 19–22), as long as the expression is balanced and the stack has been completely cleaned off, the string represents a correctly balanced sequence of parentheses.

Balanced Symbols (A General Case)

The balanced parentheses problem shown above is a specific case of a more general situation that arises in many programming languages. The general problem of balancing and nesting different kinds of opening and closing symbols occurs frequently. For example, in Python square brackets, [ and ], are used for lists; curly braces, { and }, are used for dictionaries; and parentheses, ( and ), are used for tuples and arithmetic expressions. It is possible to mix symbols as long as each maintains its own open and close relationship. Strings of symbols such as

{ { ( [ ] [ ] ) } ( ) }

[ [ { { ( ( ) ) } } ] ]

[ ] [ ] [ ] ( ) { }

are properly balanced in that not only does each opening symbol have a corresponding closing symbol, but the types of symbols match as well.

Compare those with the following strings that are not balanced:

( [ ) ]

( ( ( ) ] ) )

[ { ( ) ]

The simple parentheses checker from the previous section can easily be extended to handle these new types of symbols. Recall that each opening symbol is simply pushed on the stack to wait for the matching closing symbol to appear later in the sequence. When a closing symbol does appear, the only difference is that we must check to be sure that it correctly matches the type of the opening symbol on top of the stack. If the two symbols do not match, the string is not balanced. Once again, if the entire string is processed and nothing is left on the stack, the string is correctly balanced.

The Python program to implement this is shown in ActiveCode 5. The only change appears in line 16 where we call a helper function, matches, to assist with symbol-matching. Each symbol that is removed from the stack must be checked to see that it matches the current closing symbol. If a mismatch occurs, the boolean variable balanced is set to False.


Solving the General Balanced Symbol Problem (parcheck2)


These two examples show that stacks are very important data structures for the processing of language constructs in computer science. Almost any notation you can think of has some type of nested symbol that must be matched in a balanced order. There are a number of other important uses for stacks in computer science. We will continue to explore them in the next sections.

Converting Decimal Numbers to Binary Numbers

In your study of computer science, you have probably been exposed in one way or another to the idea of a binary number. Binary representation is important in computer science since all values stored within a computer exist as a string of binary digits, a string of 0s and 1s. Without the ability to convert back and forth between common representations and binary numbers, we would need to interact with computers in very awkward ways.

Integer values are common data items. They are used in computer programs and computation all the time. We learn about them in math class and of course represent them using the decimal number system, or base 10. The decimal number \(233_{10}\) and its corresponding binary equivalent \(11101001_{2}\) are interpreted respectively as

\(2\times10^{2} + 3\times10^{1} + 3\times10^{0}\)

and

\(1\times2^{7} + 1\times2^{6} + 1\times2^{5} + 0\times2^{4} + 1\times2^{3} + 0\times2^{2} + 0\times2^{1} + 1\times2^{0}\)

But how can we easily convert integer values into binary numbers? The answer is an algorithm called “Divide by 2” that uses a stack to keep track of the digits for the binary result.

The Divide by 2 algorithm assumes that we start with an integer greater than 0. A simple iteration then continually divides the decimal number by 2 and keeps track of the remainder. The first division by 2 gives information as to whether the value is even or odd. An even value will have a remainder of 0. It will have the digit 0 in the ones place. An odd value will have a remainder of 1 and will have the digit 1 in the ones place. We think about building our binary number as a sequence of digits; the first remainder we compute will actually be the last digit in the sequence. As shown in Figure 5, we again see the reversal property that signals that a stack is likely to be the appropriate data structure for solving the problem.

../_images/dectobin.png

Figure 5: Decimal-to-Binary Conversion

The Python code in ActiveCode 6 implements the Divide by 2 algorithm. The function divideBy2 takes an argument that is a decimal number and repeatedly divides it by 2. Line 7 uses the built-in modulo operator, %, to extract the remainder and line 8 then pushes it on the stack. After the division process reaches 0, a binary string is constructed in lines 11-13. Line 11 creates an empty string. The binary digits are popped from the stack one at a time and appended to the right-hand end of the string. The binary string is then returned.


Converting from Decimal to Binary (divby2)


The algorithm for binary conversion can easily be extended to perform the conversion for any base. In computer science it is common to use a number of different encodings. The most common of these are binary, octal (base 8), and hexadecimal (base 16).

The decimal number \(233\) and its corresponding octal and hexadecimal equivalents \(351_{8}\) and \(E9_{16}\) are interpreted as

\(3\times8^{2} + 5\times8^{1} + 1\times8^{0}\)

and

\(14\times16^{1} + 9\times16^{0}\)

The function divideBy2 can be modified to accept not only a decimal value but also a base for the intended conversion. The “Divide by 2” idea is simply replaced with a more general “Divide by base.” A new function called baseConverter, shown in ActiveCode 7, takes a decimal number and any base between 2 and 16 as parameters. The remainders are still pushed onto the stack until the value being converted becomes 0. The same left-to-right string construction technique can be used with one slight change. Base 2 through base 10 numbers need a maximum of 10 digits, so the typical digit characters 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 work fine. The problem comes when we go beyond base 10. We can no longer simply use the remainders, as they are themselves represented as two-digit decimal numbers. Instead we need to create a set of digits that can be used to represent those remainders beyond 9.


Converting from Decimal to any Base (baseconvert)


A solution to this problem is to extend the digit set to include some alphabet characters. For example, hexadecimal uses the ten decimal digits along with the first six alphabet characters for the 16 digits. To implement this, a digit string is created (line 4 in Listing 6) that stores the digits in their corresponding positions. 0 is at position 0, 1 is at position 1, A is at position 10, B is at position 11, and so on. When a remainder is removed from the stack, it can be used to index into the digit string and the correct resulting digit can be appended to the answer. For example, if the remainder 13 is removed from the stack, the digit D is appended to the resulting string.

Self Check

stk-3: What is value of 25 expressed as an octal number

stk-4: What is value of 256 expressed as a hexidecimal number

stk-5: What is value of 26 expressed in base 26

Infix, Prefix and Postfix Expressions

When you write an arithmetic expression such as B * C, the form of the expression provides you with information so that you can interpret it correctly. In this case we know that the variable B is being multiplied by the variable C since the multiplication operator * appears between them in the expression. This type of notation is referred to as infix since the operator is in between the two operands that it is working on.

Consider another infix example, A + B * C. The operators + and * still appear between the operands, but there is a problem. Which operands do they work on? Does the + work on A and B or does the * take B and C? The expression seems ambiguous.

In fact, you have been reading and writing these types of expressions for a long time and they do not cause you any problem. The reason for this is that you know something about the operators + and *. Each operator has a precedence level. Operators of higher precedence are used before operators of lower precedence. The only thing that can change that order is the presence of parentheses. The precedence order for arithmetic operators places multiplication and division above addition and subtraction. If two operators of equal precedence appear, then a left-to-right ordering or associativity is used.

Let’s interpret the troublesome expression A + B * C using operator precedence. B and C are multiplied first, and A is then added to that result. (A + B) * C would force the addition of A and B to be done first before the multiplication. In expression A + B + C, by precedence (via associativity), the leftmost + would be done first.

Although all this may be obvious to you, remember that computers need to know exactly what operators to perform and in what order. One way to write an expression that guarantees there will be no confusion with respect to the order of operations is to create what is called a fully parenthesized expression. This type of expression uses one pair of parentheses for each operator. The parentheses dictate the order of operations; there is no ambiguity. There is also no need to remember any precedence rules.

The expression A + B * C + D can be rewritten as ((A + (B * C)) + D) to show that the multiplication happens first, followed by the leftmost addition. A + B + C + D can be written as (((A + B) + C) + D) since the addition operations associate from left to right.

There are two other very important expression formats that may not seem obvious to you at first. Consider the infix expression A + B. What would happen if we moved the operator before the two operands? The resulting expression would be + A B. Likewise, we could move the operator to the end. We would get A B +. These look a bit strange.

These changes to the position of the operator with respect to the operands create two new expression formats, prefix and postfix. Prefix expression notation requires that all operators precede the two operands that they work on. Postfix, on the other hand, requires that its operators come after the corresponding operands. A few more examples should help to make this a bit clearer (see Table 2).

A + B * C would be written as + A * B C in prefix. The multiplication operator comes immediately before the operands B and C, denoting that * has precedence over +. The addition operator then appears before the A and the result of the multiplication.

In postfix, the expression would be A B C * +. Again, the order of operations is preserved since the * appears immediately after the B and the C, denoting that * has precedence, with + coming after. Although the operators moved and now appear either before or after their respective operands, the order of the operands stayed exactly the same relative to one another.

Table 2: Examples of Infix, Prefix, and Postfix
Infix Expression Prefix Expression Postfix Expression
A + B + A B A B +
A + B * C + A * B C A B C * +

Now consider the infix expression (A + B) * C. Recall that in this case, infix requires the parentheses to force the performance of the addition before the multiplication. However, when A + B was written in prefix, the addition operator was simply moved before the operands, + A B. The result of this operation becomes the first operand for the multiplication. The multiplication operator is moved in front of the entire expression, giving us * + A B C. Likewise, in postfix A B + forces the addition to happen first. The multiplication can be done to that result and the remaining operand C. The proper postfix expression is then A B + C *.

Consider these three expressions again (see Table 3). Something very important has happened. Where did the parentheses go? Why don’t we need them in prefix and postfix? The answer is that the operators are no longer ambiguous with respect to the operands that they work on. Only infix notation requires the additional symbols. The order of operations within prefix and postfix expressions is completely determined by the position of the operator and nothing else. In many ways, this makes infix the least desirable notation to use.

Table 3: An Expression with Parentheses
Infix Expression Prefix Expression Postfix Expression
(A + B) * C * + A B C A B + C *

Table 4 shows some additional examples of infix expressions and the equivalent prefix and postfix expressions. Be sure that you understand how they are equivalent in terms of the order of the operations being performed.

Table 4: Additional Examples of Infix, Prefix, and Postfix
Infix Expression Prefix Expression Postfix Expression
A + B * C + D + + A * B C D A B C * + D +
(A + B) * (C + D) * + A B + C D A B + C D + *
A * B + C * D + * A B * C D A B * C D * +
A + B + C + D + + + A B C D A B + C + D +

Conversion of Infix Expressions to Prefix and Postfix

So far, we have used ad hoc methods to convert between infix expressions and the equivalent prefix and postfix expression notations. As you might expect, there are algorithmic ways to perform the conversion that allow any expression of any complexity to be correctly transformed.

The first technique that we will consider uses the notion of a fully parenthesized expression that was discussed earlier. Recall that A + B * C can be written as (A + (B * C)) to show explicitly that the multiplication has precedence over the addition. On closer observation, however, you can see that each parenthesis pair also denotes the beginning and the end of an operand pair with the corresponding operator in the middle.

Look at the right parenthesis in the subexpression (B * C) above. If we were to move the multiplication symbol to that position and remove the matching left parenthesis, giving us B C *, we would in effect have converted the subexpression to postfix notation. If the addition operator were also moved to its corresponding right parenthesis position and the matching left parenthesis were removed, the complete postfix expression would result (see Figure 6).

../_images/moveright.png

Figure 6: Moving Operators to the Right for Postfix Notation

If we do the same thing but instead of moving the symbol to the position of the right parenthesis, we move it to the left, we get prefix notation (see Figure 7). The position of the parenthesis pair is actually a clue to the final position of the enclosed operator.

../_images/moveleft.png

Figure 7: Moving Operators to the Left for Prefix Notation

So in order to convert an expression, no matter how complex, to either prefix or postfix notation, fully parenthesize the expression using the order of operations. Then move the enclosed operator to the position of either the left or the right parenthesis depending on whether you want prefix or postfix notation.

Here is a more complex expression: (A + B) * C - (D - E) * (F + G). Figure 8 shows the conversion to postfix and prefix notations.

../_images/complexmove.png

Figure 8: Converting a Complex Expression to Prefix and Postfix Notations

General Infix-to-Postfix Conversion

We need to develop an algorithm to convert any infix expression to a postfix expression. To do this we will look closer at the conversion process.

Consider once again the expression A + B * C. As shown above, A B C * + is the postfix equivalent. We have already noted that the operands A, B, and C stay in their relative positions. It is only the operators that change position. Let’s look again at the operators in the infix expression. The first operator that appears from left to right is +. However, in the postfix expression, + is at the end since the next operator, *, has precedence over addition. The order of the operators in the original expression is reversed in the resulting postfix expression.

As we process the expression, the operators have to be saved somewhere since their corresponding right operands are not seen yet. Also, the order of these saved operators may need to be reversed due to their precedence. This is the case with the addition and the multiplication in this example. Since the addition operator comes before the multiplication operator and has lower precedence, it needs to appear after the multiplication operator is used. Because of this reversal of order, it makes sense to consider using a stack to keep the operators until they are needed.

What about (A + B) * C? Recall that A B + C * is the postfix equivalent. Again, processing this infix expression from left to right, we see + first. In this case, when we see *, + has already been placed in the result expression because it has precedence over * by virtue of the parentheses. We can now start to see how the conversion algorithm will work. When we see a left parenthesis, we will save it to denote that another operator of high precedence will be coming. That operator will need to wait until the corresponding right parenthesis appears to denote its position (recall the fully parenthesized technique). When that right parenthesis does appear, the operator can be popped from the stack.

As we scan the infix expression from left to right, we will use a stack to keep the operators. This will provide the reversal that we noted in the first example. The top of the stack will always be the most recently saved operator. Whenever we read a new operator, we will need to consider how that operator compares in precedence with the operators, if any, already on the stack.

Assume the infix expression is a string of tokens delimited by spaces. The operator tokens are *, /, +, and -, along with the left and right parentheses, ( and ). The operand tokens are the single-character identifiers A, B, C, and so on. The following steps will produce a string of tokens in postfix order.

  1. Create an empty stack called opstack for keeping operators. Create an empty list for output.
  2. Convert the input infix string to a list by using the string method split.
  3. Scan the token list from left to right.
    • If the token is an operand, append it to the end of the output list.
    • If the token is a left parenthesis, push it on the opstack.
    • If the token is a right parenthesis, pop the opstack until the corresponding left parenthesis is removed. Append each operator to the end of the output list.
    • If the token is an operator, *, /, +, or -, push it on the opstack. However, first remove any operators already on the opstack that have higher or equal precedence and append them to the output list.
  4. When the input expression has been completely processed, check the opstack. Any operators still on the stack can be removed and appended to the end of the output list.

Figure 9 shows the conversion algorithm working on the expression A * B + C * D. Note that the first * operator is removed upon seeing the + operator. Also, + stays on the stack when the second * occurs, since multiplication has precedence over addition. At the end of the infix expression the stack is popped twice, removing both operators and placing + as the last operator in the postfix expression.

../_images/intopost.png

Figure 9: Converting A * B + C * D to Postfix Notation

In order to code the algorithm in Python, we will use a dictionary called prec to hold the precedence values for the operators. This dictionary will map each operator to an integer that can be compared against the precedence levels of other operators (we have arbitrarily used the integers 3, 2, and 1). The left parenthesis will receive the lowest value possible. This way any operator that is compared against it will have higher precedence and will be placed on top of it. Line 15 defines the operands to be any upper-case character or digit. The complete conversion function is shown in ActiveCode 8.


Converting Infix Expressions to Postfix Expressions (intopost)



A few more examples of execution in the Python shell are shown below.

>>> infixtopostfix("( A + B ) * ( C + D )")
'A B + C D + *'
>>> infixtopostfix("( A + B ) * C")
'A B + C *'
>>> infixtopostfix("A + B * C")
'A B C * +'
>>>

Postfix Evaluation

As a final stack example, we will consider the evaluation of an expression that is already in postfix notation. In this case, a stack is again the data structure of choice. However, as you scan the postfix expression, it is the operands that must wait, not the operators as in the conversion algorithm above. Another way to think about the solution is that whenever an operator is seen on the input, the two most recent operands will be used in the evaluation.

To see this in more detail, consider the postfix expression 4 5 6 * +. As you scan the expression from left to right, you first encounter the operands 4 and 5. At this point, you are still unsure what to do with them until you see the next symbol. Placing each on the stack ensures that they are available if an operator comes next.

In this case, the next symbol is another operand. So, as before, push it and check the next symbol. Now we see an operator, *. This means that the two most recent operands need to be used in a multiplication operation. By popping the stack twice, we can get the proper operands and then perform the multiplication (in this case getting the result 30).

We can now handle this result by placing it back on the stack so that it can be used as an operand for the later operators in the expression. When the final operator is processed, there will be only one value left on the stack. Pop and return it as the result of the expression. Figure 10 shows the stack contents as this entire example expression is being processed.

../_images/evalpostfix1.png

Figure 10: Stack Contents During Evaluation

Figure 11 shows a slightly more complex example, 7 8 + 3 2 + /. There are two things to note in this example. First, the stack size grows, shrinks, and then grows again as the subexpressions are evaluated. Second, the division operation needs to be handled carefully. Recall that the operands in the postfix expression are in their original order since postfix changes only the placement of operators. When the operands for the division are popped from the stack, they are reversed. Since division is not a commutative operator, in other words \(15/5\) is not the same as \(5/15\), we must be sure that the order of the operands is not switched.

../_images/evalpostfix2.png

Figure 11: A More Complex Example of Evaluation

Assume the postfix expression is a string of tokens delimited by spaces. The operators are *, /, +, and - and the operands are assumed to be single-digit integer values. The output will be an integer result.

  1. Create an empty stack called operandStack.
  2. Convert the string to a list by using the string method split.
  3. Scan the token list from left to right.
    • If the token is an operand, convert it from a string to an integer and push the value onto the operandStack.
    • If the token is an operator, *, /, +, or -, it will need two operands. Pop the operandStack twice. The first pop is the second operand and the second pop is the first operand. Perform the arithmetic operation. Push the result back on the operandStack.
  4. When the input expression has been completely processed, the result is on the stack. Pop the operandStack and return the value.

The complete function for the evaluation of postfix expressions is shown in ActiveCode 9. To assist with the arithmetic, a helper function doMath is defined that will take two operands and an operator and then perform the proper arithmetic operation.


Postfix Evaluation (postfixeval)


It is important to note that in both the postfix conversion and the postfix evaluation programs we assumed that there were no errors in the input expression. Using these programs as a starting point, you can easily see how error detection and reporting can be included. We leave this as an exercise at the end of the chapter.

Self Check

stk-6: Without using the activecode infixToPostfix function, convert the following expression to postfix 10 + 3 * 5 / (16 - 4)

stk-7: 17 10 + 3 * 9 / ==

stk-8: Modify the infixToPostfix function so that it can convert the following expression: 5 * 3 ^ (4 - 2) Paste the answer here: