IMPLEMENTATION OF STACKS PART 1

INTRODUCTION ABOUT STACKSThe stack is a very common data structure used in programs which has

lot of potential.

Stacks hold objects, usually all of the same type.

It follows the concept of LIFO – last in first out.

Stack is a linear list of items in which all additions and deletion are

restricted to one end.

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 PhotoScript) are also designed

around language-defined stacks that are directly visible to and

manipulated by the programmer.

STACKS VS ARRAYS

An array is a contiguous block of memory.

A stack is a first-in-last-out data structure with access only to the top

of the data.

Since many languages does not provide facility for stack, it is backed

by either arrays or linked list.

The values can be added and deleted on any side from an array.

But in stack, insertion and deletion is possible on only one side of the

stack. The other side is sealed.

Eg: a[10] –array a[10] - stack

The bottom of a stack is a sealed end. Stack may have a capacity

which is a limitation on the number of elements in a stack. The operations

on stack are

• Push: Places an object on the top of the stack.

• Pop: Removes an object from the top of the stack.

• IsEmpty: Reports whether the stack is empty or not.

• IsFull: Reports whether the stack exceeds limit or not.

(i)stack (ii)push(s,a) (iii)push(s,d)-stack overflow (iv)pop(s)

(v)pop(s)-stack underflow

a abc

STACK OPERATIONS

CA B D EADEB

STACK OPERATIONS

STACK IMPLEMENTATION Stack data structure is not inherently provided by many programming

languages.

Stack is implemented using arrays or linked lists.

Let S be a stack, n be the capacity, x be the element to be pushed, then

push and pop will be given as

Push(S,x) and Pop(S)

Here we use “top” which keeps track of the top element in the stack.

When top = = 0 , and pop() operation gives stack underflow as result.

When top = = n, and push() operation gives stack overflow as result.

The pop() operation just gives an illusion of deletion, but the elements

are retained. Only the top is decremented.

S[1:6]

Top=1

Push(S,a)

a

Top=6

Push(S,a)Push(S,b)Push(S,c) push(S,d)Push(S,e) push(S,f)

a b c d e f

Top=6 = nStack overflow

Push(S,g)

a b c d e f

Top=5

Pop(S) a b c d e

Top=1

Pop(S)Pop(S)Pop(S)Pop(S) pop(S)

a

Top=0

Pop(S)

Top=0 stack underflow

Pop(S)

STACK APPLICATIONS

Recursion handling

Evaluation of expression

Conversion of infix to postfix expression

Computation of postfix expression

Parenthesis handling

Backtracking

Conversion of decimal to other number system

Maze tracer

Undo operations

RECURSION HANDLING Without stack, recursion is difficult

Compiler automatically uses stack data structure while handling

recursion.

All computer needs to remember for each active function call, values

of arguments & local variables and the location of the next statement

to be executed when control goes back.

Essentially what is happening when we call that method is that our

current execution point is pushed onto the call stack and the runtime

starts executing the code for the internal method call. When that

method finally returns, we pop our place from the stack and continue

executing.

Eg: int f(int n){ int k,r; if(n==0) return 0; k=n*n; r=f(n-1);Return k+r; }

3

-

-

n

k

r

3

9

.

3 2

9 4

. .

3 2 1

9 4 1

. . .

n

k

r

3 2 1

9 4 1

. . 0

Ans: 14

3 2

9 4

. 1

3

9

5

PARENTHESIS CHECKINGProcedure check()

Declare a character stack S.

Now traverse the expression.

    a) If the current character is a starting bracket then push it to

stack.

    b) If the current character is a closing bracket then pop from

stack and if the popped character is the matching starting bracket

then fine else parenthesis are not balanced.

After complete traversal, if there is some starting bracket left in stack

then “not balanced”

End procedure

Eg: [a+(b*c)+{(d-e)}]

[

[ (

[

[ {

[ { (

[ {

[

Push [

Push (

) and ( matches, Pop (

Push {

Push (

matches, pop (

Matches, pop {

Matches, pop [

Thus, parenthesis match here

CONVERSION OF INFIX TO POSTFIX Expressions can be represented in prefix, postfix or infix notations.

Conversion from one form of the expression to another form may be

accomplished using a stack.

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 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.

Eg: (A*B)+C

Element Stack Prefix Action

$

(

A

*

B

)

+

C

$

A

AB

AB*

AB*C

AB*C+

Work stack

Push (

Print A

Push(*)

Print B

Pop *,( print *

Push +

Print C

Pop +

$

$ ($ ($ ( *$ ( *

$ $ + $ +

EVALUATION OF POSTFIX EXPRESSION Postfix expression is easily evaluated by the compiler by using stack.

The procedure for the evaluation of postfix expression is given below:

procedure eval(E)

x=getnextchar(E);

case x:

x is an operand: push x into stack S.

x is an operator: pop elements, perform operation and

push result into stack.

x is null: pop stack and print result

end case

End procedure

Eg: AB*C+ A=2, B=3, C=5

Element Stack Action

A

B

*

C

+

$

Push A

Push B

Pop A and B, A*B, push 6

Push C

Pop C and 6, C+6, push 11

Pop

Result:11

A

A B

6

6 C

11

BACKTRACKINGBacktracking is a simple, elegant, recursive technique which can be put to a

variety of uses.

You start at the root of a tree, the tree probably has some good and bad

leaves. You want to get to a good leaf. At each node, you choose one of its

children to move to, and you keep this up in a stack until you get to a leaf.

Suppose you get to a bad leaf. You can backtrack to continue the search for

a good leaf by revoking your most recent choice, and trying out the next

option in that set of options.

If you run out of options, revoke the choice that got you here, and try

another choice at that node.

If you end up at the root with no options left, there are no good leaves to be

found.

•Starting at Root, your options are A and B. You choose A. •At A, your options are C and D. You choose C. •C is bad. Go back to A. •At A, you have already tried C, and it failed. Try D. •D is bad. Go back to A. •At A, you have no options left to try. Go back to Root. •At Root, you have already tried A. Try B. •At B, your options are E and F. Try E. •E is good.

Eg:

CONVERSION OF DECIMAL TO OTHER NUMBER SYSTEM

The given decimal number is divided by the base (2 or 8 or 16)

repeatedly and the corresponding remainders are pushed into stack.

At last the elements are popped out of the stack to give the result.

Eg. 84 – 1010100 in binary

124 in octal

54 in hexadecimal

84 / 2 = 42 042 / 2 = 21 021 / 2 = 10 110 / 2 = 5 05 / 2 = 2

12 / 2 = 1

01

1

0

1

0

1

0

0push

pop 1010100

84 / 8 = 10 410 / 8 = 1 2

1

1

2

4

124

push

pop

84 / 16 = 5 45 5

454push pop

MAZE TRACER

All our mazes will be two-dimensional arrays of n rows and n columns. Each row, column cell is either open, or blocked by an internal wall. From any open cell, you may move left, right, up, or down to an adjacent empty cell. To solve a maze, you must find a path of open cells from a given start cell to a specified end cell. By default, you should assume that the start cell is in position (0,0).

Sample:S O O O O OH H H H O HH O O O O HH O H H H HH O H O O OH O O O H E

The path will be: [(0,0),(0,1),(0,2),(0,3),(0,4), (1,4),(2,4),(2,3), (2,2),(2,1),(3,1), (4,1),(5,1),(5,2),(5,3),(4,3),(4,4), (4,5),(5,5)]

S O O O O OH H H H O HH O O O O HH O H H H HH O H O O OH O O O H E

0 0Push start (0,0)

0 0 Push

0 1

0 2

0 3

0 4

0 5

0 0 Pop

0 1

0 2

0 3

0 4

0 0 Push

0 1

0 2

0 3

0 4

1 4

0 0 Push

0 1

0 2

0 3

0 4

1 42 4

Push0 0

0 1

0 2

0 3

0 4

1 42 42 32 2

2 1

0 0

0 1

0 2

0 3

0 4

1 42 42 32 2

2 1

3 1

4 1

Push

0 0

0 1

0 2

0 3

0 4

1 42 42 32 2

2 1

3 1

4 1

Push

5 2

5 1

5 3

4 3

4 5 4 4

0 0

0 1

0 2

0 3

0 4

1 42 42 32 2

2 1

3 1

4 1

Push

5 2

5 1

5 3

4 3

4 54 4

5 5 End

start

UNDO OPERATION

An undo operation is a method for reverting a change to an object,

along with the arguments needed to revert the change.

Undo operations are typically collected in undo groups, which

represent whole revertible actions, and are stored on a stack.

To undo a single operation, it must still be packaged in a group.

Undo groups are stored on a stack, with the oldest groups at the

bottom and the newest at the top.

The undo stack is unlimited by default, but you can restrict it.

When the stack exceeds the maximum, the oldest undo groups are

dropped from the bottom.

Initially, both stacks are empty. Recording undo operations adds to the

undo stack, but the redo stack remains empty until undo is performed.

Performing undo causes the reverting operations in the latest group to

be applied to their objects.

Consecutive undos add to the redo stack. Subsequent redo operations

pull the operations off the redo stack, apply them to the objects, and

push them back onto the undo stack.

The redo stack’s contents last as long as undo and redo are performed

successively. However, because applying a new change to an object

invalidates the previous changes, as soon as a new undo operation is

registered, any existing redo stack is cleared.

UNDO OPERATION

Thank you…!