The Stack And Recursion

Simple as it sounds
Fig. stack of items
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Definition
• A stack is an ordered collection of items into which new items
may be inserted and from which items may be deleted at one end,
called the top of the stack.
• Stack is a linear data structure where all the insertions and deletions
are done at end rather than in the middle.
• Stacks are also called Last in First Out (LIFO) lists.
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Intro
• We may come across situations, where insertion or deletion is
required only at one end, either at the beginning or end of the list.
• The suitable data structures to fulfil such requirements are stacks and
queues.
• For ex. a stack of plates, a stack of coins, a stack of books etc.
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Stack as an abstract data type
• A stack of elements of type T is a finite sequence of elements together
with the operations
1. CreateEmptyStack(S): create or make stack S be an empty stack
2. Push(S,x): Insert x at one end of the stack, called its top
3. Top(S): If stack S is not empty; then retrieve the element at its top
4. Pop(S): If stack S is not empty; then delete the element at its top
5. IsFull(S): Determine if S is full or not. Return true if S is full stack; return false
otherwise
6. IsEmpty(S): Determine if S is empty or not. Return true if S is an empty stack;
return false otherwise.
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Example of Push Pop
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Thing to consider
• Stack underflow happens when
we try to pop (remove) an item from
the stack, when nothing is actually
there to remove.
• Stack overflow happens when we
try to push one more item onto our
stack than it can actually hold.
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Stack As a Linked List
• The stack as linked list is represented as a single linked list.
• Each node in the list contains data and a pointer to the next node.
• The structure defined to represent stack is as follows:
struct node{
int data;
node *next;
};
Fig. Stack as a linked list
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Infix, Postfix and Prefix
• 2+3 <operand><operator><operand>
• A-b
• (P*2)
Operands are basic objects on which operation is performed
INFIX
<operand><operator><operand>
• (2+3) * 4 (2+3) * 4
• (P+Q)*(R+S) (P+Q) * (R+S)
Common expressions
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Infix
• What about 4+6*2and what about 4+6+2
• To clear ambiguity we remember school Mathematics
• Order of precedence
1. Parentheses (Brackets)
2. Order (Exponents)
3. Division
4. Multiplication
5. Addition
6. Subtraction
• So 4+6*2 | 2*6/2 -3 +7 | {(2*6)/2}-(3+7)
=4+12 | = 2*3-3+7 | = ???
=16 | = 6-3+7 | = ??
| = 3+7=10 | = ?
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Prefix (polish notation)
• In prefix notation the operator proceeds the two operands. i.e. the
operator is written before the operands.
<operator><operand><operand>
infix prefix
2+3 +23
p-q -pq
a+b*c +a*bc
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Postfix (Reverse Polish Notation)
• In postfix notation the operators are written after the operands so it is
called the postfix notation (post means after).
<operand><operand><operator>
infix prefix postfix
2+3 +23 23+
p-q -pq pq-
a+b*c +a*bc abc*+
Human-readable Good for machines
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Conversion of Infix to Postfix Expression
• While evaluating an infix expression, there is an evaluation order according to
which the operations are executed
• Brackets or Parentheses
• Exponentiation
• Multiplication or Division
• Addition or Subtraction
• The operators with same priority are evaluated from left to right.
• Eg:
• A+B+C means (A+B)+C
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Evaluation of Prefix and Postfix Expressions
• Suppose an expression
a*b+c*d-e
• We can write this as:
{(a*b)+(c*d)}-e
• As we want to change it into postfix expression
{(ab*)+(cd*)}-e
{(ab*)(cd*)+}-e
{(ab*)(cd*)+} e-
• Final form
ab*cd*+e-
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Evaluation of Postfix Expressions
• Suppose we have a postfix expression
ab*cd*+e-
• And we want to evaluate it so lets say
a=2, b=3,c=5,d=4,e=9
• We can solve this as,
2 3 * 5 4 * + 9 – <operand><operand><operator>
6 5 4 * + 9 –
6 20 + 9 –
26 9 –
17
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Evaluation of Postfix Expressions
• From above we get,
2 3 * 5 4 * + 9 –
Stack
EvaluatePostfix(exp)
{
create a stack S
for i=0 to length (exp) -1
{
if (exp[i] is operand)
PUSH(exp[i])
elseif (exp[i] is operator)
op2 <- pop()
op1 <- pop()
res– Perform(exp[i],op1,op2)
PUSH(res)
}
return top of STACK
}
2
3
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Evaluation of Prefix expressions
• An expression: 2*3 +5*4-9
• Can be written as:
{(2*3)+(5*4)}-9
{(*2 3)+(*5 4)}-9
{+(*2 3) (*5 4)}-9
-{+(*2 3) (*5 4)}9
• We can get Rid of Paranthesis
-+*2 3 *5 4 9
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Evaluation of Prefix expressions
• We have to scan it from right
-+*2 3 *5 4 9
Stack
9
4
5
Stack
9
20
6
Stack
9
26
Stack
17
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Algorithm to convert Infix to Postfix
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Examples:
1. A * B + C becomes A B * C +
NOTE:
• When the '+' is read, it has lower precedence than the '*', so the '*' must be printed first.
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current symbol Opstack Poststack
A A
* * A
B * AB
+ + AB* {pop and print the '*' before pushing the '+'}
C + AB*C
AB*C+

Examples:
2. A * (B + C) becomes A B C + *
NOTE:
• Since expressions in parentheses must be done first, everything on the stack is saved and the left
parenthesis is pushed to provide a marker.
• When the next operator is read, the stack is treated as though it were empty and the new operator
(here the '+' sign) is pushed on.
• Then when the right parenthesis is read, the stack is popped until the corresponding left parenthesis is
found.
• Since postfix expressions have no parentheses, the parentheses are not printed.
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current symbol Opstack Poststack
A A
* * A
( *( A
B *( AB
+ *(+ AB
C *(+ ABC
) * ABC+
ABC+*

Classwork
• A - B + C
• A * B ^ C + D
• A * (B + C * D) + E
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References
• For Code Check Github
• For Assignment Check Github
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Recursion
• Recursion is a process by which a function calls itself repeatedly, until
some specified condition has been satisfied
• The process is used for repetitive computations in which each action is
stated in terms of a previous result.
• To solve a problem recursively, two conditions must be satisfied.
• First, the problem must be written in a recursive form
• Second the problem statement must include a stopping condition
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Factorial of an integer number using recursive function
void main(){
int n;
long int facto;
scanf(“%d”,&n);
facto=factorial(n);
printf(“%d!=%ld”,n,facto);
}
long int factorial(int n){
if(n==0){
return 1;
}else{
return n*factorial(n-1);
}
Factorial(5)=
5*Factorial(4)=
5*(4*Factorial(3))=
5*(4*(3*Factorial(2)))=
5*(4*(3*(2*Factorial(1))))=
5*(4*(3*(2*(1*Factorial(0)))))=
5*(4*(3*(2*(1*1))))= 5*(4*(3*(2*1)))=
5*(4*(3*2))= 5*(4*6)= 5*24=
120
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Recursion Pros and Cons
• Pros
• The code may be much easier to write.
• To solve some problems which are naturally recursive such as tower of Hanoi.
• Cons
• Recursive functions are generally slower than non-recursive functions.
• May require a lot of memory to hold intermediate results on the system stack.
• It is difficult to think recursively so one must be very careful when writing
recursive functions.
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Tower of Hanoi Problem
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Tower of Hanoi Problem
• The mission is to move all the disks to some another tower without
violating the sequence of arrangement.
• The below mentioned are few rules which are to be followed for tower
of hanoi −
• Only one disk can be moved among the towers at any given time.
• Only the "top" disk can be removed.
• No large disk can sit over a small disk.
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A recursive algorithm for Tower of Hanoi can be driven as follows −
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The Stack And Recursion

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The Stack And Recursion