Stack (abstract data type)
{{#invoke:Hatnote|hatnote}}Template:Main other Template:More footnotes
In computer science, a stack is a particular kind of abstract data type or collection in which the principal (or only) operations on the collection are the addition of an entity to the collection, known as push and removal of an entity, known as pop.[1] The relation between the push and pop operations is such that the stack is a Last-In-First-Out (LIFO) data structure. In a LIFO data structure, the last element added to the structure must be the first one to be removed. This is equivalent to the requirement that, considered as a linear data structure, or more abstractly a sequential collection, the push and pop operations occur only at one end of the structure, referred to as the top of the stack. Often a peek or top operation is also implemented, returning the value of the top element without removing it.
A stack may be implemented to have a bounded capacity. If the stack is full and does not contain enough space to accept an entity to be pushed, the stack is then considered to be in an overflow state. The pop operation removes an item from the top of the stack. A pop either reveals previously concealed items or results in an empty stack, but, if the stack is empty, it goes into underflow state, which means no items are present in stack to be removed.
A stack is a restricted data structure, because only a small number of operations are performed on it. The nature of the pop and push operations also means that stack elements have a natural order. Elements are removed from the stack in the reverse order to the order of their addition. Therefore, the lower elements are those that have been on the stack the longest.[2]
History
The stack was first proposed in 1946, in the computer design of Alan M. Turing (who used the terms "bury" and "unbury") as a means of calling and returning from subroutines.Template:Clarify The Germans Klaus Samelson and Friedrich L. Bauer of Technical University Munich proposed the idea in 1955 and filed a patent in 1957.[3] The same concept was developed, independently, by the Australian Charles Leonard Hamblin in the first half of 1957.[4]
Abstract definition
A stack is a basic computer science data structure and can be defined in an abstract, implementation-free manner, or it can be generally defined as a linear list of items in which all additions and deletion are restricted to one end that is Top.
This is a VDM (Vienna Development Method) description of a stack:[5]
Function signatures:
init: -> Stack push: N x Stack -> Stack top: Stack -> (N U ERROR) pop: Stack -> Stack isempty: Stack -> Boolean
(where N indicates an element (natural numbers in this case), and U indicates set union)
Semantics:
top(init()) = ERROR top(push(i,s)) = i pop(init()) = init() pop(push(i, s)) = s isempty(init()) = true isempty(push(i, s)) = false
Inessential operations
In many implementations, a stack has more operations than "push" and "pop". An example is "top of stack", or "peek", which observes the top-most element without removing it from the stack.[6] Since this can be done with a "pop" and a "push" with the same data, it is not essential. An underflow condition can occur in the "stack top" operation if the stack is empty, the same as "pop". Often implementations have a function which just returns if the stack is empty.
Software stacks
Implementation
In most high level languages, a stack can be easily implemented either through an array or a linked list. What identifies the data structure as a stack in either case is not the implementation but the interface: the user is only allowed to pop or push items onto the array or linked list, with few other helper operations. The following will demonstrate both implementations, using C.
Array
The array implementation aims to create an array where the first element (usually at the zero-offset) is the bottom. That is, array[0] is the first element pushed onto the stack and the last element popped off. The program must keep track of the size, or the length of the stack. The stack itself can therefore be effectively implemented as a two-element structure in C:
typedef struct {
size_t size;
int items[STACKSIZE];
} STACK;
The push() operation is used both to initialize the stack, and to store values to it. It is responsible for inserting (copying) the value into the ps->items[] array and for incrementing the element counter (ps->size). In a responsible C implementation, it is also necessary to check whether the array is already full to prevent an overrun.
void push(STACK *ps, int x)
{
if (ps->size == STACKSIZE) {
fputs("Error: stack overflow\n", stderr);
abort();
} else
ps->items[ps->size++] = x;
}
The pop() operation is responsible for removing a value from the stack, and decrementing the value of ps->size. A responsible C implementation will also need to check that the array is not already empty.
int pop(STACK *ps)
{
if (ps->size == 0){
fputs("Error: stack underflow\n", stderr);
abort();
} else
return ps->items[--ps->size];
}
If we use a dynamic array, then we can implement a stack that can grow or shrink as much as needed. The size of the stack is simply the size of the dynamic array. A dynamic array is a very efficient implementation of a stack, since adding items to or removing items from the end of a dynamic array is amortized O(1) time.
Linked list
The linked-list implementation is equally simple and straightforward. In fact, a simple singly linked list is sufficient to implement a stack—it only requires that the head node or element can be removed, or popped, and a node can only be inserted by becoming the new head node.
Unlike the array implementation, our structure typedef corresponds not to the entire stack structure, but to a single node:
typedef struct stack {
int data;
struct stack *next;
} STACK;
Such a node is identical to a typical singly linked list node, at least to those that are implemented in C.
The push() operation both initializes an empty stack, and adds a new node to a non-empty one. It works by receiving a data value to push onto the stack, along with a target stack, creating a new node by allocating memory for it, and then inserting it into a linked list as the new head:
void push(STACK **head, int value)
{
STACK *node = malloc(sizeof(STACK)); /* create a new node */
if (node == NULL){
fputs("Error: no space available for node\n", stderr);
abort();
} else { /* initialize node */
node->data = value;
node->next = empty(*head) ? NULL : *head; /* insert new head if any */
*head = node;
}
}
A pop() operation removes the head from the linked list, and assigns the pointer to the head to the previous second node. It checks whether the list is empty before popping from it:
int pop(STACK **head)
{
if (empty(*head)) { /* stack is empty */
fputs("Error: stack underflow\n", stderr);
abort();
} else { //pop a node
STACK *top = *head;
int value = top->data;
*head = top->next;
free(top);
return value;
}
}
Stacks and programming languages
Some languages, like LISP and Python, do not call for stack implementations, since push and pop functions are available for any list. All Forth-like languages (such as Adobe PostScript) are also designed around language-defined stacks that are directly visible to and manipulated by the programmer. Examples from Common Lisp:
(setf list (list 'a 'b 'c))
;; ⇒ (A B C)
(pop list)
;; ⇒ A
list
;; ⇒ (B C)
(push 'new list)
;; ⇒ (NEW B C)
C++'s Standard Template Library provides a "stack" templated class which is restricted to only push/pop operations. Java's library contains a Template:Javadoc:SE class that is a specialization of Template:Javadoc:SE---this could be considered a design flaw, since the inherited get() method from Template:Javadoc:SE ignores the LIFO constraint of the Template:Javadoc:SE. PHP has an SplStack class.
Hardware stacks
A common use of stacks at the architecture level is as a means of allocating and accessing memory.
Basic architecture of a stack
A typical stack is an area of computer memory with a fixed origin and a variable size. Initially the size of the stack is zero. A stack pointer, usually in the form of a hardware register, points to the most recently referenced location on the stack; when the stack has a size of zero, the stack pointer points to the origin of the stack.
The two operations applicable to all stacks are:
- a push operation, in which a data item is placed at the location pointed to by the stack pointer, and the address in the stack pointer is adjusted by the size of the data item;
- a pop or pull operation: a data item at the current location pointed to by the stack pointer is removed, and the stack pointer is adjusted by the size of the data item.
There are many variations on the basic principle of stack operations. Every stack has a fixed location in memory at which it begins. As data items are added to the stack, the stack pointer is displaced to indicate the current extent of the stack, which expands away from the origin.
Stack pointers may point to the origin of a stack or to a limited range of addresses either above or below the origin (depending on the direction in which the stack grows); however, the stack pointer cannot cross the origin of the stack. In other words, if the origin of the stack is at address 1000 and the stack grows downwards (towards addresses 999, 998, and so on), the stack pointer must never be incremented beyond 1000 (to 1001, 1002, etc.). If a pop operation on the stack causes the stack pointer to move past the origin of the stack, a stack underflow occurs. If a push operation causes the stack pointer to increment or decrement beyond the maximum extent of the stack, a stack overflow occurs.
Some environments that rely heavily on stacks may provide additional operations, for example:
- Duplicate: the top item is popped, and then pushed again (twice), so that an additional copy of the former top item is now on top, with the original below it.
- Peek: the topmost item is inspected (or returned), but the stack pointer is not changed, and the stack size does not change (meaning that the item remains on the stack). This is also called top operation in many articles.
- Swap or exchange: the two topmost items on the stack exchange places.
- Rotate (or Roll): the n topmost items are moved on the stack in a rotating fashion. For example, if n=3, items 1, 2, and 3 on the stack are moved to positions 2, 3, and 1 on the stack, respectively. Many variants of this operation are possible, with the most common being called left rotate and right rotate.
Stacks are often visualized growing from the bottom up (like real-world stacks). They may also be visualized growing from left to right, so that "topmost" becomes "rightmost", or even growing from top to bottom. The important feature is that the top of the stack is in a fixed position. The image to the right is an example of a top to bottom growth visualization: the top (28) is the stack 'bottom', since the stack 'top' is where items are pushed or popped from. Sometimes stacks are also visualized metaphorically, such as coin holders or Pez dispensers.
A right rotate will move the first element to the third position, the second to the first and the third to the second. Here are two equivalent visualizations of this process:
apple banana banana ===right rotate==> cucumber cucumber apple
cucumber apple banana ===left rotate==> cucumber apple banana
A stack is usually represented in computers by a block of memory cells, with the "bottom" at a fixed location, and the stack pointer holding the address of the current "top" cell in the stack. The top and bottom terminology are used irrespective of whether the stack actually grows towards lower memory addresses or towards higher memory addresses.
Pushing an item on to the stack adjusts the stack pointer by the size of the item (either decrementing or incrementing, depending on the direction in which the stack grows in memory), pointing it to the next cell, and copies the new top item to the stack area. Depending again on the exact implementation, at the end of a push operation, the stack pointer may point to the next unused location in the stack, or it may point to the topmost item in the stack. If the stack points to the current topmost item, the stack pointer will be updated before a new item is pushed onto the stack; if it points to the next available location in the stack, it will be updated after the new item is pushed onto the stack.
Popping the stack is simply the inverse of pushing. The topmost item in the stack is removed and the stack pointer is updated, in the opposite order of that used in the push operation.
Hardware support
Stack in main memory
Most CPUs have registers that can be used as stack pointers. Processor families like the x86, Z80, 6502, and many others have special instructions that implicitly use a dedicated (hardware) stack pointer to conserve opcode space. Some processors, like the PDP-11 and the 68000, also have special addressing modes for implementation of stacks, typically with a semi-dedicated stack pointer as well (such as A7 in the 68000). However, in most processors, several different registers may be used as additional stack pointers as needed (whether updated via addressing modes or via add/sub instructions).
Stack in registers or dedicated memory
The x87 floating point architecture is an example of a set of registers organised as a stack where direct access to individual registers (relative the current top) is also possible. As with stack-based machines in general, having the top-of-stack as an implicit argument allows for a small machine code footprint with a good usage of bus bandwidth and code caches, but it also prevents some types of optimizations possible on processors permitting random access to the register file for all (two or three) operands. A stack structure also makes superscalar implementations with register renaming (for speculative execution) somewhat more complex to implement, although it is still feasible, as exemplified by modern x87 implementations.
Sun SPARC, AMD Am29000, and Intel i960 are all examples of architectures using register windows within a register-stack as another strategy to avoid the use of slow main memory for function arguments and return values.
There are also a number of small microprocessors that implements a stack directly in hardware and some microcontrollers have a fixed-depth stack that is not directly accessible. Examples are the PIC microcontrollers, the Computer Cowboys MuP21, the Harris RTX line, and the Novix NC4016. Many stack-based microprocessors were used to implement the programming language Forth at the microcode level. Stacks were also used as a basis of a number of mainframes and mini computers. Such machines were called stack machines, the most famous being the Burroughs B5000.
Applications
Stacks have numerous applications. We see stacks in everyday life, from the books in our library, to the blank sheets of paper in our printer tray. All of them follow the Last In First Out (LIFO) logic, that is when we add a book to a pile of books, we add it to the top of the pile, whereas when we remove a book from the pile, we generally remove it from the top of the pile.
Given below are a few applications of stacks in the world of computers:
Converting a decimal number into a binary number
The logic for transforming a decimal number into a binary number is as follows:
- Read a number
- Iteration (while number is greater than zero)
- Find out the remainder after dividing the number by 2
- Print the remainder
- End the iteration
However, there is a problem with this logic. Suppose the number, whose binary form we want to find, is 23. Using this logic, we get the result as 11101, instead of getting 10111.
To solve this problem, we use a stack.[7] We make use of the LIFO property of the stack. Initially we push the binary digit formed into the stack, instead of printing it directly. After the entire number has been converted into the binary form, we pop one digit at a time from the stack and print it. Therefore we get the decimal number converted into its proper binary form.
Algorithm:
function outputInBinary(Integer n)
Stack s = new Stack
while n > 0 do
Integer bit = n modulo 2
s.push(bit)
if s is full then
return error
end if
n = floor(n / 2)
end while
while s is not empty do
output(s.pop())
end while
end function
Towers of Hanoi
One of the most interesting applications of stacks can be found in solving a puzzle called Tower of Hanoi. According to an old Brahmin story, the existence of the universe is calculated in terms of the time taken by a number of monks, who are working all the time, to move 64 disks from one pole to another. But there are some rules about how this should be done, which are:
- move only one disk at a time.
- for temporary storage, a third pole may be used.
- a disk of larger diameter may not be placed on a disk of smaller diameter.[8]
For algorithm of this puzzle see Tower of Hanoi.
Assume that A is the first tower, B is the second tower, and C is the third tower.
{{#invoke:Multiple image|render}}
{{#invoke:Multiple image|render}}
Output: (when there are 3 disks)
Let 1 be the smallest disk, 2 be the disk of medium size and 3 be the largest disk.
| Move disk | From peg | To peg |
|---|---|---|
| 1 | A | C |
| 2 | A | B |
| 1 | C | B |
| 3 | A | C |
| 1 | B | A |
| 2 | B | C |
| 1 | A | C |
The C++ code for this solution can be implemented in two ways:
First implementation (using stacks implicitly by recursion)
#include <stdio.h>
void TowersofHanoi(int n, int a, int b, int c)
{
if(n > 0)
{
TowersofHanoi(n-1, a, c, b); //recursion
printf("> Move top disk from tower %d to tower %d.\n", a, b);
TowersofHanoi(n-1, c, b, a); //recursion
}
}
Second implementation (using stacks explicitly)
// Global variable , tower [1:3] are three towers
arrayStack<int> tower[4];
void TowerofHanoi(int n)
{
// Preprocessor for moveAndShow.
for (int d = n; d > 0; d--) //initialize
tower[1].push(d); //add disk d to tower 1
moveAndShow(n, 1, 2, 3); /*move n disks from tower 1 to tower 3 using
tower 2 as intermediate tower*/
}
void moveAndShow(int n, int a, int b, int c)
{
// Move the top n disks from tower a to tower b showing states.
// Use tower c for intermediate storage.
if(n > 0)
{
moveAndShow(n-1, a, c, b); //recursion
int d = tower[a].top(); //move a disc from top of tower x to top of
tower[a].pop(); //tower y
tower[c].push(d);
showState(); //show state of 3 towers
moveAndShow(n-1, b, a, c); //recursion
}
}
However complexity for above written implementations is . So it's obvious that problem can only be solved for small values of n (generally ).
In case of the monks, the number of turns taken to transfer 64 disks, by following the above rules, will be 18,446,744,073,709,551,615; which will surely take a lot of time![8][9]
Expression evaluation and syntax parsing
Calculators employing reverse Polish notation use a stack structure to hold values. Expressions can be represented in prefix, postfix or infix notations and conversion from one form to another may be accomplished using a stack. Many compilers use a stack for parsing the syntax of expressions, program blocks etc. before translating into low level code. Most programming languages are context-free languages, allowing them to be parsed with stack based machines.
Evaluation of an infix expression that is fully parenthesized
Input: (((2 * 5) - (1 * 2)) / (11 - 9))
Output: 4
Analysis: Five types of input characters
- Opening bracket
- Numbers
- Operators
- Closing bracket
- New line character
Data structure requirement: A character stack
Algorithm
1. Read one input character
2. Actions at end of each input
Opening brackets (2.1) Go to step (1)
Number (2.2) Push into stack and then Go to step (1)
Operator (2.3) Push into stack and then Go to step (1)
Closing brackets (2.4) Pop from stack
(2.4.1) Pop is used three times
The first popped element is assigned to op2
The second popped element is assigned to op
The third popped element is assigned to op1
Evaluate op1 op op2
Convert the result into character and
push into the stack
Go to step (1)
New line character (2.5) Pop from stack and print the answer
STOP
Result: The evaluation of the fully parenthesized infix expression is printed as follows:
Input String: (((2 * 5) - (1 * 2)) / (11 - 9))
| Input Symbol | Stack (from bottom to top) | Operation |
|---|---|---|
| ( | ||
| ( | ||
| ( | ||
| 2 | 2 | |
| * | 2 * | |
| 5 | 2 * 5 | |
| ) | 10 | 2 * 5 = 10 and push |
| - | 10 - | |
| ( | 10 - | |
| 1 | 10 - 1 | |
| * | 10 - 1 * | |
| 2 | 10 - 1 * 2 | |
| ) | 10 - 2 | 1 * 2 = 2 & Push |
| ) | 8 | 10 - 2 = 8 & Push |
| / | 8 / | |
| ( | 8 / | |
| 11 | 8 / 11 | |
| - | 8 / 11 - | |
| 9 | 8 / 11 - 9 | |
| ) | 8 / 2 | 11 - 9 = 2 & Push |
| ) | 4 | 8 / 2 = 4 & Push |
| New line | Empty | Pop & Print |
Evaluation of infix expression which is not fully parenthesized
Input: (2 * 5 - 1 * 2) / (11 - 9)
Output: 4
Analysis There are five types of input characters which are:
- Opening parentheses
- Numbers
- Operators
- Closing parentheses
- New line character (\n)
We do not know what to do if an operator is read as an input character. By implementing the priority rule for operators, we have a solution to this problem.
The Priority rule: we should perform a comparative priority check if an operator is read, and then push it. If the stack top contains an operator of priority higher than or equal to the priority of the input operator, then we pop it and print it. We keep on performing the priority check until the top of stack either contains an operator of lower priority or if it does not contain an operator.
Data Structure Requirement for this problem: a character stack and an integer stack
Algorithm:
1. Read an input character
2. Actions that will be performed at the end of each input
Opening parentheses (2.1) Push it into character stack and then Go to step (1)
Number (2.2) Push into integer stack, Go to step (1)
Operator (2.3) Do the comparative priority check
(2.3.1) if the character stack's top contains an operator with equal
or higher priority, then pop it into op
Pop a number from integer stack into op2
Pop another number from integer stack into op1
Calculate op1 op op2 and push the result into the integer
stack
Closing parentheses (2.4) Pop from the character stack
(2.4.1) if it is an opening parentheses, then discard it and Go to
step (1)
(2.4.2) To op, assign the popped element
Pop a number from integer stack and assign it op2
Pop another number from integer stack and assign it
to op1
Calculate op1 op op2 and push the result into the integer
stack
Convert into character and push into stack
Go to the step (2.4)
New line character (2.5) Print the result after popping from the stack
STOP
Result: The evaluation of an infix expression that is not fully parenthesized is printed as follows:
Input String: (2 * 5 - 1 * 2) / (11 - 9)
| Input Symbol | Character Stack (from bottom to top) | Integer Stack (from bottom to top) | Operation performed |
|---|---|---|---|
| ( | ( | ||
| 2 | ( | 2 | |
| * | ( * | Push as * has higher priority | |
| 5 | ( * | 2 5 | |
| - | ( * | Since '-' has less priority, we do 2 * 5 = 10 | |
| ( - | 10 | We push 10 and then push '-' | |
| 1 | ( - | 10 1 | |
| * | ( - * | 10 1 | Push * as it has higher priority |
| 2 | ( - * | 10 1 2 | |
| ) | ( - | 10 2 | Perform 1 * 2 = 2 and push it |
| ( | 8 | Pop - and 10 - 2 = 8 and push, Pop ( | |
| / | / | 8 | |
| ( | / ( | 8 | |
| 11 | / ( | 8 11 | |
| - | / ( - | 8 11 | |
| 9 | / ( - | 8 11 9 | |
| ) | / | 8 2 | Perform 11 - 9 = 2 and push it |
| New line | 4 | Perform 8 / 2 = 4 and push it | |
| 4 | Print the output, which is 4 |
Evaluation of prefix expression
Input: / - * 2 5 * 1 2 - 11 9
Output: 4
Analysis There are three types of input characters
- Numbers
- Operators
- New line character (\n)
Data structure requirement: a character stack and an integer stack
Algorithm:
1. Read one character input at a time and keep pushing it into the character stack until the new
line character is reached
2. Perform pop from the character stack. If the stack is empty, go to step (3)
Number (2.1) Push in to the integer stack and then go to step (1)
Operator (2.2) Assign the operator to op
Pop a number from integer stack and assign it to op1
Pop another number from integer stack
and assign it to op2
Calculate op1 op op2 and push the output into the integer
stack. Go to step (2)
3. Pop the result from the integer stack and display the result
Result: the evaluation of prefix expression is printed as follows:
Input String: / - * 2 5 * 1 2 - 11 9
| Input Symbol | Character Stack (from bottom to top) | Integer Stack (from bottom to top) | Operation performed |
|---|---|---|---|
| / | / | ||
| - | /- | ||
| * | / - * | ||
| 2 | / - * 2 | ||
| 5 | / - * 2 5 | ||
| * | / - * 2 5 * | ||
| 1 | / - * 2 5 * 1 | ||
| 2 | / - * 2 5 * 1 2 | ||
| - | / - * 2 5 * 1 2 - | ||
| 11 | / - * 2 5 * 1 2 - 11 | ||
| 9 | / - * 2 5 * 1 2 - 11 9 | ||
| \n | / - * 2 5 * 1 2 - 11 | 9 | |
| / - * 2 5 * 1 2 - | 9 11 | ||
| / - * 2 5 * 1 2 | 2 | 11 - 9 = 2 | |
| / - * 2 5 * 1 | 2 2 | ||
| / - * 2 5 * | 2 2 1 | ||
| / - * 2 5 | 2 2 | 1 * 2 = 2 | |
| / - * 2 | 2 2 5 | ||
| / - * | 2 2 5 2 | ||
| / - | 2 2 10 | 5 * 2 = 10 | |
| / | 2 8 | 10 - 2 = 8 | |
| Stack is empty | 4 | 8 / 2 = 4 | |
| Stack is empty | Print 4 |
Evaluation of postfix expression
The calculation: 1 + 2 * 4 + 3 can be written down like this in postfix notation with the advantage of no precedence rules and parentheses needed:
1 2 4 * + 3 +
The expression is evaluated from the left to right using a stack:
- when encountering an operand: push it
- when encountering an operator: pop two operands, evaluate the result and push it.
Like the following way (the Stack is displayed after Operation has taken place):
| Input | Operation | Stack (after op) |
|---|---|---|
| 1 | Push operand | 1 |
| 2 | Push operand | 2, 1 |
| 4 | Push operand | 4, 2, 1 |
| * | Multiply | 8, 1 |
| + | Add | 9 |
| 3 | Push operand | 3, 9 |
| + | Add | 12 |
The final result, 12, lies on the top of the stack at the end of the calculation.
Example in C
#include<stdio.h>
int main()
{
int a[100], i;
printf("To pop enter -1\n");
for(i = 0;;)
{
printf("Push ");
scanf("%d", &a[i]);
if(a[i] == -1)
{
if(i == 0)
{
printf("Underflow\n");
}
else
{
printf("pop = %d\n", a[--i]);
}
}
else
{
i++;
}
}
}
Evaluation of postfix expression (Pascal)
This is an implementation in Pascal, using marked sequential file as data archives.
{
programmer : clx321
file : stack.pas
unit : Pstack.tpu
}
program TestStack;
{this program uses ADT of Stack, I will assume that the unit of ADT of Stack has already existed}
uses
PStack; {ADT of STACK}
{dictionary}
const
mark = '.';
var
data : stack;
f : text;
cc : char;
ccInt, cc1, cc2 : integer;
{functions}
IsOperand (cc : char) : boolean; {JUST Prototype}
{return TRUE if cc is operand}
ChrToInt (cc : char) : integer; {JUST Prototype}
{change char to integer}
Operator (cc1, cc2 : integer) : integer; {JUST Prototype}
{operate two operands}
{algorithms}
begin
assign (f, cc);
reset (f);
read (f, cc); {first elmt}
if (cc = mark) then
begin
writeln ('empty archives !');
end
else
begin
repeat
if (IsOperand (cc)) then
begin
ccInt := ChrToInt (cc);
push (ccInt, data);
end
else
begin
pop (cc1, data);
pop (cc2, data);
push (data, Operator (cc2, cc1));
end;
read (f, cc); {next elmt}
until (cc = mark);
end;
close (f);
end
}
Conversion of an Infix expression that is fully parenthesized into a Postfix expression
Input: (((8 + 1) - (7 - 4)) / (11 - 9))
Output: 8 1 + 7 4 - - 11 9 - /
Analysis: There are five types of input characters which are:
* Opening parentheses
* Numbers
* Operators
* Closing parentheses
* New line character (\n)
Requirement: A character stack
Algorithm:
1. Read an character input
2. Actions to be performed at end of each input
Opening parentheses (2.1) Push into stack and then Go to step (1)
Number (2.2) Print and then Go to step (1)
Operator (2.3) Push into stack and then Go to step (1)
Closing parentheses (2.4) Pop it from the stack
(2.4.1) If it is an operator, print it, Go to step (2.4)
(2.4.2) If the popped element is an opening parentheses,
discard it and go to step (1)
New line character (2.5) STOP
Therefore, the final output after conversion of an infix expression to a postfix expression is as follows:
| Input | Operation | Stack (after op) | Output on monitor |
|---|---|---|---|
| ( | (2.1) Push operand into stack | ( | |
| ( | (2.1) Push operand into stack | ( ( | |
| ( | (2.1) Push operand into stack | ( ( ( | |
| 8 | (2.2) Print it | 8 | |
| + | (2.3) Push operator into stack | ( ( ( + | 8 |
| 1 | (2.2) Print it | 8 1 | |
| ) | (2.4) Pop from the stack: Since popped element is '+' print it | ( ( ( | 8 1 + |
| (2.4) Pop from the stack: Since popped element is '(' we ignore it and read next character | ( ( | 8 1 + | |
| - | (2.3) Push operator into stack | ( ( - | |
| ( | (2.1) Push operand into stack | ( ( - ( | |
| 7 | (2.2) Print it | 8 1 + 7 | |
| - | (2.3) Push the operator in the stack | ( ( - ( - | |
| 4 | (2.2) Print it | 8 1 + 7 4 | |
| ) | (2.4) Pop from the stack: Since popped element is '-' print it | ( ( - ( | 8 1 + 7 4 - |
| (2.4) Pop from the stack: Since popped element is '(' we ignore it and read next character | ( ( - | ||
| ) | (2.4) Pop from the stack: Since popped element is '-' print it | ( ( | 8 1 + 7 4 - - |
| (2.4) Pop from the stack: Since popped element is '(' we ignore it and read next character | ( | ||
| / | (2.3) Push the operand into the stack | ( / | |
| ( | (2.1) Push into the stack | ( / ( | |
| 11 | (2.2) Print it | 8 1 + 7 4 - - 11 | |
| - | (2.3) Push the operand into the stack | ( / ( - | |
| 9 | (2.2) Print it | 8 1 + 7 4 - - 11 9 | |
| ) | (2.4) Pop from the stack: Since popped element is '-' print it | ( / ( | 8 1 + 7 4 - - 11 9 - |
| (2.4) Pop from the stack: Since popped element is '(' we ignore it and read next character | ( / | ||
| ) | (2.4) Pop from the stack: Since popped element is '/' print it | ( | 8 1 + 7 4 - - 11 9 - / |
| (2.4) Pop from the stack: Since popped element is '(' we ignore it and read next character | Stack is empty | ||
| New line character | (2.5) STOP |
Rearranging railroad cars
Problem Description
This is one useful application of stacks. Consider that a freight train has n railroad cars, each to be left at different station. They're numbered 1 through n and freight train visits these stations in order n through 1. Obviously, the railroad cars are labeled by their destination. To facilitate removal of the cars from the train, we must rearrange them in ascending order of their number (i.e. 1 through n). When cars are in this order, they can be detached at each station. We rearrange cars at a shunting yard that has input track, output track and k holding tracks between input & output tracks (i.e. holding track).
Solution Strategy
To rearrange cars, we examine the cars on the input from front to back. If the car being examined is next one in the output arrangement, we move it directly to output track. If not, we move it to the holding track & leave it there until it's time to place it to the output track. The holding tracks operate in a LIFO manner as the cars enter & leave these tracks from top. When rearranging cars only following moves are permitted:
- A car may be moved from front (i.e. right end) of the input track to the top of one of the holding tracks or to the left end of the output track.
- A car may be moved from the top of holding track to left end of the output track.
The figure shows a shunting yard with k = 3, holding tracks H1, H2 & H3, also n = 9. The n cars of freight train begin in the input track & are to end up in the output track in order 1 through n from right to left. The cars initially are in the order 5,8,1,7,4,2,9,6,3 from back to front. Later cars are rearranged in desired order.
A Three Track Example
- Consider the input arrangement from figure, here we note that the car 3 is at the front, so it can't be output yet, as it to be preceded by cars 1 & 2. So car 3 is detached & moved to holding track H1.
- The next car 6 can't be output & it is moved to holding track H2. Because we have to output car 3 before car 6 & this will not possible if we move car 6 to holding track H1.
- Now it's obvious that we move car 9 to H3.
The requirement of rearrangement of cars on any holding track is that the cars should be preferred to arrange in ascending order from top to bottom.
- So car 2 is now moved to holding track H1 so that it satisfies the previous statement. If we move car 2 to H2 or H3, then we've no place to move cars 4,5,7,8.The least restrictions on future car placement arise when the new car λ is moved to the holding track that has a car at its top with smallest label Ψ such that λ < Ψ. We may call it an assignment rule to decide whether a particular car belongs to a specific holding track.
- When car 4 is considered, there are three places to move the car H1,H2,H3. The top of these tracks are 2,6,9.So using above mentioned Assignment rule, we move car 4 to H2.
- The car 7 is moved to H3.
- The next car 1 has the least label, so it's moved to output track.
- Now it's time for car 2 & 3 to output which are from H1(in short all the cars from H1 are appended to car 1 on output track).
The car 4 is moved to output track. No other cars can be moved to output track at this time.
- The next car 8 is moved to holding track H1.
- Car 5 is output from input track. Car 6 is moved to output track from H2, so is the 7 from H3,8 from H1 & 9 from H3.
Backtracking
Another important application of stacks is backtracking. Consider a simple example of finding the correct path in a maze. There are a series of points, from the starting point to the destination. We start from one point. To reach the final destination, there are several paths. Suppose we choose a random path. After following a certain path, we realise that the path we have chosen is wrong. So we need to find a way by which we can return to the beginning of that path. This can be done with the use of stacks. With the help of stacks, we remember the point where we have reached. This is done by pushing that point into the stack. In case we end up on the wrong path, we can pop the last point from the stack and thus return to the last point and continue our quest to find the right path. This is called backtracking.
Quicksort
Sorting means arranging the list of elements in a particular order. In case of numbers, it could be in ascending order, or in the case of letters, alphabetic order.
Quicksort is an algorithm of the divide and conquer type. In this method, to sort a set of numbers, we reduce it to two smaller sets, and then sort these smaller sets.
This can be explained with the help of the following example:
Suppose A is a list of the following numbers:
In the reduction step, we find the final position of one of the numbers. In this case, let us assume that we have to find the final position of 48, which is the first number in the list.
To accomplish this, we adopt the following method. Begin with the last number, and move from right to left. Compare each number with 48. If the number is smaller than 48, we stop at that number and swap it with 48.
In our case, the number is 24. Hence, we swap 24 and 48.
The numbers 96 and 72 to the right of 48, are greater than 48. Now beginning with 24, scan the numbers in the opposite direction, that is from left to right. Compare every number with 48 until you find a number that is greater than 48.
In this case, it is 60. Therefore we swap 48 and 60.
Note that the numbers 12, 24 and 36 to the left of 48 are all smaller than 48. Now, start scanning numbers from 60, in the right to left direction. As soon as you find lesser number, swap it with 48.
In this case, it is 44. Swap it with 48. The final result is:
Now, beginning with 44, scan the list from left to right, until you find a number greater than 48.
Such a number is 84. Swap it with 48. The final result is:
Now, beginning with 84, traverse the list from right to left, until you reach a number lesser than 48. We do not find such a number before reaching 48. This means that all the numbers in the list have been scanned and compared with 48. Also, we notice that all numbers less than 48 are to the left of it, and all numbers greater than 48, are to its right.
The final partitions look as follows:
Therefore, 48 has been placed in its proper position and now our task is reduced to sorting the two partitions. This above step of creating partitions can be repeated with every partition containing 2 or more elements. As we can process only a single partition at a time, we should be able to keep track of the other partitions, for future processing.
This is done by using two stacks called LOWERBOUND and UPPERBOUND, to temporarily store these partitions. The addresses of the first and last elements of the partitions are pushed into the LOWERBOUND and UPPERBOUND stacks respectively. Now, the above reduction step is applied to the partitions only after its boundary values are popped from the stack.
We can understand this from the following example:
Take the above list A with 12 elements. The algorithm starts by pushing the boundary values of A, that is 1 and 12 into the LOWERBOUND and UPPERBOUND stacks respectively. Therefore the stacks look as follows:
LOWERBOUND: 1 UPPERBOUND: 12
To perform the reduction step, the values of the stack top are popped from the stack. Therefore, both the stacks become empty.
LOWERBOUND: {empty} UPPERBOUND: {empty}
Now, the reduction step causes 48 to be fixed to the 5th position and creates two partitions, one from position 1 to 4 and the other from position 6 to 12. Hence, the values 1 and 6 are pushed into the LOWERBOUND stack and 4 and 12 are pushed into the UPPERBOUND stack.
LOWERBOUND: 1, 6 UPPERBOUND: 4, 12
For applying the reduction step again, the values at the stack top are popped. Therefore, the values 6 and 12 are popped. Therefore the stacks look like:
LOWERBOUND: 1 UPPERBOUND: 4
The reduction step is now applied to the second partition, that is from the 6th to 12th element.
After the reduction step, 98 is fixed in the 11th position. So, the second partition has only one element. Therefore, we push the upper and lower boundary values of the first partition onto the stack. So, the stacks are as follows:
LOWERBOUND: 1, 6 UPPERBOUND: 4, 10
The processing proceeds in the following way and ends when the stacks do not contain any upper and lower bounds of the partition to be processed, and the list gets sorted.
The Stock Span Problem
In the stock span problem, we will solve a financial problem with the help of stacks.
Suppose, for a stock, we have a series of n daily price quotes, the span of the stock's price on a given day is defined as the maximum number of consecutive days just before the given day, for which the price of the stock on the current day is less than or equal to its price on the given day.
Let, Price(i) = price of the stock on day "i".
Then, Span(i) = Max{k : k>=0 and Price(j)<=Price(i) for j=i-k, .., i}
Thus, if Price(i-1)>Price(i), then Span(i)=0.
An algorithm which has Quadratic Time Complexity
Input: An array P with n elements
Output: An array S of n elements such that S[i] is the largest integer k such that k <= i + 1 and P[j] <= P[i] for j = i - k + 1,.....,i
Algorithm:
1. Initialize an array P which contains the daily prices of the stocks
2. Initialize an array S which will store the span of the stock
3. for i = 0 to i = n - 1
3.1 Initialize k to zero
3.2 Done with a false condition
3.3 repeat
3.3.1 if ( P[i - k] <= P[i] ) then
Increment k by 1
3.3.2 else
Done with true condition
3.4 Till (k > i) or done with processing
Assign value of k to S[i] to get the span of the stock
4. Return array S
Now, analyzing this algorithm for running time, we observe:
- We have initialized the array S at the beginning, hence takes O(n) time
- The repeat loop is nested within the for loop. The for loop, whose counter is i is executed n times. The statements which are not in the repeat loop, but in the for loop are executed n times. Therefore these statements and the increment and condition testing of i take O(n) time.
- In repetition of i for the outer for loop, the body of the inner repeat loop is executed maximum i + 1 times. In the worst case, element S[i] is greater than all the previous elements. So, testing for the if condition, the statement after that, as well as testing the until condition, will be performed i + 1 times during iteration i for the outer for loop. Hence, the total time taken by the inner loop is O(n(n + 1)/2), which is O()
- We returned the array S at the end. This is a constant time operation, hence takes O(1) time
The running time of all these steps is calculated by adding the time taken by all these four steps. The first two terms are O() while the last term is O(). Therefore the total running time of the algorithm is O().
An algorithm which has Linear Time Complexity
In order to calculate the span more efficiently, we see that the span on a particular day can be easily calculated if we know the closest day before i, such that the price of the stocks on that day was higher than the price of the stocks on the present day. If there exists such a day, we can represent it by h(i) and initialize h(i) to be -1. This is basically the same algorithm as the one used for efficient construction of Cartesian tree.
Therefore the span of a particular day is given by the formula, s = i - h(i).
To implement this logic, we use a stack as an abstract data type to store the days i, h(i), h(h(i)) and so on. When we go from day i-1 to i, we pop the days when the price of the stock was less than or equal to p(i) and then push the value of day i back into the stack.
Here, we assume that the stack is implemented by operations that take O(1) that is constant time. The algorithm is as follows:
Input: An array P with n elements and an empty stack N
Output: An array S of n elements such that P[i] is the largest integer k such that k <= i + 1 and P[j] <= P[i] for j = i - k + 1,.....,i
Algorithm:
1. Initialize an array P which contains the daily prices of the stocks
2. Initialize an array S which will store the span of the stock
3. for i = 0 to i = n - 1
3.1 Initialize k to zero
3.2 Done with a false condition
3.3 while not (Stack N is empty or done with processing)
3.3.1 if ( P[i] >= P[N.top())] then
Pop a value from stack N
3.3.2 else
Done with true condition
3.4 if Stack N is empty
3.4.1 Initialize h to -1
3.5 else
3.5.1 Initialize stack top to h
3.6 Put the value of i - h in S[i]
3.7 Push the value of i in N
4. Return array S
Now, analyzing this algorithm for running time, we observe:
- We have initialized the array S at the beginning and returned it at the end. This is a constant time operation, hence takes O(n) time
- The while loop is nested within the for loop. The for loop, whose counter is i is executed n times. The statements which are not in the repeat loop, but in the for loop are executed n times. Therefore these statements and the incrementing and condition testing of i take O(n) time.
- Now, observe the inner while loop during i repetitions of the for loop. The statement done with a true condition is done at most once, since it causes an exit from the loop. Let us say that t(i) is the number of times statement Pop a value from stack N is executed. So it becomes clear that while not (Stack N is empty or done with processing) is tested maximum t(i) + 1 times.
- Adding the running time of all the operations in the while loop, we get:
- An element once popped from the stack N is never pushed back into it. Therefore,
So, the running time of all the statements in the while loop is O()
The running time of all the steps in the algorithm is calculated by adding the time taken by all these steps. The run time of each step is O(). Hence the running time complexity of this algorithm is O().
Runtime memory management
{{#invoke:main|main}}
A number of programming languages are stack-oriented, meaning they define most basic operations (adding two numbers, printing a character) as taking their arguments from the stack, and placing any return values back on the stack. For example, PostScript has a return stack and an operand stack, and also has a graphics state stack and a dictionary stack.
Forth uses two stacks, one for argument passing and one for subroutine return addresses. The use of a return stack is extremely commonplace, but the somewhat unusual use of an argument stack for a human-readable programming language is the reason Forth is referred to as a stack-based language.
Many virtual machines are also stack-oriented, including the p-code machine and the Java Virtual Machine.
Almost all calling conventions – computer runtime memory environments – use a special stack (the "call stack") to hold information about procedure/function calling and nesting in order to switch to the context of the called function and restore to the caller function when the calling finishes. The functions follow a runtime protocol between caller and callee to save arguments and return value on the stack. Stacks are an important way of supporting nested or recursive function calls. This type of stack is used implicitly by the compiler to support CALL and RETURN statements (or their equivalents) and is not manipulated directly by the programmer.
Some programming languages use the stack to store data that is local to a procedure. Space for local data items is allocated from the stack when the procedure is entered, and is deallocated when the procedure exits. The C programming language is typically implemented in this way. Using the same stack for both data and procedure calls has important security implications (see below) of which a programmer must be aware in order to avoid introducing serious security bugs into a program.
Security
Some computing environments use stacks in ways that may make them vulnerable to security breaches and attacks. Programmers working in such environments must take special care to avoid the pitfalls of these implementations.
For example, some programming languages use a common stack to store both data local to a called procedure and the linking information that allows the procedure to return to its caller. This means that the program moves data into and out of the same stack that contains critical return addresses for the procedure calls. If data is moved to the wrong location on the stack, or an oversized data item is moved to a stack location that is not large enough to contain it, return information for procedure calls may be corrupted, causing the program to fail.
Malicious parties may attempt a stack smashing attack that takes advantage of this type of implementation by providing oversized data input to a program that does not check the length of input. Such a program may copy the data in its entirety to a location on the stack, and in so doing it may change the return addresses for procedures that have called it. An attacker can experiment to find a specific type of data that can be provided to such a program such that the return address of the current procedure is reset to point to an area within the stack itself (and within the data provided by the attacker), which in turn contains instructions that carry out unauthorized operations.
This type of attack is a variation on the buffer overflow attack and is an extremely frequent source of security breaches in software, mainly because some of the most popular compilers use a shared stack for both data and procedure calls, and do not verify the length of data items. Frequently programmers do not write code to verify the size of data items, either, and when an oversized or undersized data item is copied to the stack, a security breach may occur.
Programming tasks
There are many programming tasks which require application of a stack. The following tasks can be solved and evaluated online:
- SPOJ tasks: Transform the Expression
- Codility training tasks: Brackets, Fish, Stone-wall
See also
- List of data structures
- Queue
- Double-ended queue
- Call stack
- Stack-based memory allocation
- Stack machine
- Stack overflow
- Stack-oriented programming language
References
- ↑ Template:Cite web
- ↑ http://www.cprogramming.com/tutorial/computersciencetheory/stack.html cprogramming.com
- ↑ {{#invoke:Citation/CS1|citation |CitationClass=journal }}
- ↑ C. L. Hamblin, "An Addressless Coding Scheme based on Mathematical Notation", N.S.W University of Technology, May 1957 (typescript)
- ↑ Jones: "Systematic Software Development Using VDM"
- ↑ Horowitz, Ellis: "Fundamentals of Data Structures in Pascal", page 67. Computer Science Press, 1984
- ↑ {{#invoke:citation/CS1|citation |CitationClass=book }}
- ↑ 8.0 8.1 {{#invoke:citation/CS1|citation |CitationClass=book }}
- ↑ 9.0 9.1 9.2 Data structures, Algorithms and Applications in C++ by Sartaj Sahni
- ↑ 10.0 10.1 10.2 {{#invoke:citation/CS1|citation |CitationClass=book }}
- ↑ {{#invoke:citation/CS1|citation |CitationClass=book }}
- ↑ {{#invoke:citation/CS1|citation |CitationClass=book }}
Further reading
- Donald Knuth. The Art of Computer Programming, Volume 1: Fundamental Algorithms, Third Edition.Addison-Wesley, 1997. ISBN 0-201-89683-4. Section 2.2.1: Stacks, Queues, and Deques, pp. 238–243.
- Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. Introduction to Algorithms, Second Edition. MIT Press and McGraw-Hill, 2001. ISBN 0-262-03293-7. Section 10.1: Stacks and queues, pp. 200–204.
External links
Template:Sister Template:Sister
- Stacks and its Applications
- Stack Machines - the new wave
- Bounding stack depth
- Libsafe - Protecting Critical Elements of Stacks
- VBScript implementation of stack, queue, deque, and Red-Black Tree
- Stack Size Analysis for Interrupt-driven Programs (322 KB)
- Paul E. Black, Bounded stack at the NIST Dictionary of Algorithms and Data Structures.