Stacks Objective After this lecture you will be able to: Describe a stack Describe the representation of stack using linear array Describe the representation.

Stacks Objective After this lecture you will be able to: Describe a stack Describe the representation of stack using linear array Describe the representation of stack using linear linked list Implementation of various operations on stack Describe some applications of stacks

Stack Stack is one of the commonly used data structures. Stack is also called last in first out (LIFO) system Stack is a linear list in which insertion and deletion can take place only at one end called top. This structure operates in much the same way as stack of trays.

Stack of Trays

The following figure illustrate a stack, which can accommodate maximum of 10 elements figure shows stack after pushing elements 8,10,12,-5,6 6 -5 12 10 8 0 1 2 3 4 5 6 7 8 9 top

Stack after popping top two elements 12 10 8 0 1 2 3 4 5 6 7 8 9 top

stack after pushing elements 8,10,12,-5,6,9,55 55 9 6 -5 12 10 8 0 1 2 3 4 5 6 7 8 9 top

Operations on stacks Createstack(s)—to create s as an empty stack Push(s,i)--to push elementi onto stack s. Pop(s)—to access and remove the top element of the stack s Peek(s)—to access the top element of the stacks without removing it from the stack s. Isfull(s)—to check whether the stack s is full isempty—to check whether the stack s is empty

Representation of stack in memory Representation of stack using array: Suppose elements of the stack are integer type and stack can store maximum 10 elements.. #define MAX 10 typedef struct { int top; int elements[MAX]; }stack; stack s; Here we have defined our own data type named stack. First element top will be used to index top element Array elements hold the elements of the stack Last line declares variable s of type stack

stack In addition to the previous declaration, we will use the declaration typedef enum {false, true } Boolean; This statement defined new data type named Boolean which can take value false or true.

Representation of stack in memory 81012-56 0 1 2 3 4 5 6 7 8 9 4 top 81012 2 top 81012-56955 6 top 0 1 2 3 4 5 6 7 8 9

Creating an empty stack Before we can use a stack, it is to be initialized. As the index of array elements can take any value in the range 0 to MAX-1, the purpose of initializing the stack is served by assigning value -1 to the top of variable. This simple task can be accomplished by the following function. Void createstack( stack *ps) { ps=-1; }

Testing stack for underflow Boolean isempty(stack *ps) { if(ps->top==-1) return true; else return false; } or Boolean is empty(stack *ps) { return ((ps->top==-1)?true:false); }

Testing stack for overflow Boolean isfull(stack *ps) { if(ps->top==MAX-1) return true; else return false; } or Boolean is empty(stack *ps) { return ((ps->top==MAX-1)?true:false); }

Push Operation Before the push operation, if the stack is empty, then the value of the top will be -1and if the stack is not empty then the value of the top will be the index of the element currently on the top. Therefore we place the value onto the stack, the value of top is incremented so that it points to the new top of stack, where incoming element is placed. Void push(stack *ps, int value) { ps->top++; ps->elements[ps->top]=value; }

Pop Operation The element on the top of the stack is assigned to a local variable, which later on will be returned via the return statement. After assigning the top element to a local variable, the variable top is decremented so that it points to a new top Int pop(stack *ps) { int temp; temp=ps->elements[ps->top]; ps->top--; return temp; }

Accessing top element There may be instances where we want to access the top element of the stack without removing it from the stack. Int peek( stack *ps) { return(ps->elements[ps->top]); }

Representing a stack using a linked list A stack represented using a linked list is also known as linked stack. The array based representation of stack suffers from following limitations. Size of the stack must be known in advance We may come across situation when an attempt to push an element causes overflow. However stack is an abstract data structure can not be full. Hence, abstractly, it is always possible to push an element onto stack. Therefore stack as an array prohibits the growth of the stack beyond the finite number of elements.

Declaration of stack The linked list representation allows a stack to grow to a limit of the computer’s memory. Typedef struct nodetype { int info; struct nodetype *next; }stack; Stack *top; Here I have defined my own data type named stack, which is a self referential structure and whose first element info hold the element of the stack and the second element next hold the address of the element under it in the stack. The last line declares a pointer variable top of type stack.

Representation of stack in memory top 6-5128 X10 top 6-512 X top 55968 X-51210

Creating an empty stack Before we can use a stack, it is to be initialized. To initialize a stack, we will create an empty linked list. The empty linked list is created by setting pointer variable top to value NULL. Void createstack(stack **top) { *top=NULL; }

Testing stack for underflow Boolean isempty(stack *top) { if(top==NULL) return true; else return false; } or Boolean is empty(stack *top) { return ((top==NULL)?true:false); }

Testing stack for overflow Since stack represented using a linked list can grow to a limit of computers memory, there overflow condition never occurs. Hence this operation is not implemented for linked list.

Push operation To push a new element onto the stack, the element is inserted in the beginning of the linked list. void push(stack **top, int value) { stack *ptr; ptr=(stack*)malloc(sizeof(stack)); if(ptr==NULL) { printf(“\n unable to allocate memory for new node…”); printf(“\npress any key to exit..”); getch(); return; } ptr->info=value; ptr->next=*top; *top=ptr; }

Pop operation To pop an element from the stack, the element is removed from the beginning of the linked list. Int pop(stack **top) { int temp; stack *ptr; temp=(*top)->info; ptr=*top; *top=(*top)->next; free(ptr); return temp; }

Accessing top element Int peek(stack *top) { return(top->info) }

Dispose a stack Because the stack is implemented using linked lists, therefore it is programmers job to write the code to release the memory occupied by the stack. Void disposestack(stack **top) { stack *ptr; while(*top!=NULL) { ptr=*top; *top=(*top)->next; free(ptr); }

Applications of Stacks Stacks are used to pass parameters between functions. On a call to function, parameter and local variables are stored on stack. High level programming languages, such as Pascal c etc. that provide support for recursion use stack for book keeping. In each recursive call, there is need to save the current values of parameters, local variables and the return address. In addition to above stack are used to solve the various problems…. 1.Parenthesis checker 2.Mathematical notation translation 1.Polish (prefix) notation 2.Reverse polish (postfix) Notation 3.Quick sort algorithm

Parenthesis checker Parenthesis checker is a program that checks whether a mathematical expression is properly parenthesized. We will consider three sets of grouping symbols: –The standard parenthesis ”( )” –The braces “{ }” – the brackets “[ ]” For an input expression, it verifies that for each left parenthesis, braces or racket, there is a corresponding closing symbol and the symbols are appropriately nested.

Examples of valid inputs Valid Input ( ) ( { } [ ]) ( { [ ] [ ] } ) [ { { { } ( )} [ ] } [ ] ( ) { } ] invalid Input [(( ) ( { } [ ])) (( { [ ] [ ] } ) ([ { { { } ( )} [ ] } }[ ] ( ) { } ] Inside parenthesis there can be any valid arithmetic expression.

Parenthesis Checker Algo parenthesisChecker(exp) Begin Read: exp Create empty stack For each character c in exp if(current character is left symbol) then push the character onto stack else if(current character is right symbol) then if(stack is empty) then print: “Error No matching open symbol” exit else pop a symbol s from the stack if( s doesn’t correspond to c) then print: “Error incorrect nesting of symbol” exit endif Endfor If(stack is not empty) then print:”Error missing closing symbol” Else print: input expression is ok” End

Mathematical notation Translation Symbol usedOperation performed Precedence * (asterisk)MultiplicationHighest / (slash)DivisionHighest % (percentage)ModulusHighest + (plus)AdditionLowest - (hyphen)subtractionlowest

Infix notation In this notation, the operator symbol is placed between its two operands. To add A to B we can write as A+B or B+A To subtract D from C we write as C-D, but we can not write D-C as this operation is not commutative. In this notation we must distinguish between (A+B)/C and A+(B/C)

Polish (prefix) notation In this notation, named after the polish mathematician Jan Lukasiewiez, the operator symbol is placed before its two operands. To add A to B we write as +AB or +BA To subtract D from C we have to writ as –CD not as -DC

Infix to polish notation In order to translate an arithmetic expression in infix notation to polish notation, we do step by step using rackets ([]) to indicate the partial translations. Consider the following expression in infix notation: (A-B/C)*(A*K-L) The partial translation may look like: (A-[/BC])*([*AK]-L) [-A/BC]*[-*AKL] *-A/BC-*AKL The fundamental property of polish notation is that the order in which the operations to perform is completely determined by the position of the operators and operands in the expression. Accordingly one never needs parenthesis when writing expression in polish notation.

Reverse Polish (Postfix) Notation In this notation the operator symbol is placed after its two operands. To add A to B we can write as AB+ or BA+ To subtract D from C we have to write as CD- not as DC-.

Infix to reverse polish notation Consider the following expression in infix notation: (A-B/C)*(A/K-L) The partial translation may look like:] (A-[BC/])*([AK/]-L) [ABC/-]*[AK/L-] ABC/-AK/L-*

Evaluating Mathematical Expressions Generally we use infix notation, where one can not tell the order in which the operator should be applied by looking at the expression. The expression in postfix notation is very easy to evaluate, as the operands appear before the operator, there is no need of operator precedence or parentheses for operation. In order to evaluate a postfix expression it is scanned from left to right. As operands are encountered, they are pushed on a stack. When an operator encountered, pop top one or two operands depending on the operator, perform the operation and place the result back on the stack.

Infix to postfix procedure Consider the following infix expression q: (7-5)*(9/2) Character scannedstackExpression p (( 7(7 -(-7 5 7 5 )empty7 5 - ** (*(7 5 - 9*(7 5 – 9 /*(/7 5 – 9 2*(/7 5 – 9 2 )*7 5 – 9 2 / End of expressionEmpty7 5 – 9 2 / *

Evaluating expression in postfix notation Evaluate postfixnotation(p,result) Begin Create empty stack while(not end of exp p) do if(element is operand) then push element onto stack else pop two elements and let first one is a and the second one is b evalute b@a, let its value be c, where @ is operator push c onto stack endif endwhile pop stack and assign this value to parameter result end

Evaluating expression in postfix notation Consider the following postfix expression p: 7 5 – 9 2 / * Character ScannedStack 77 57 5 -2 92 9 22 9 2 /2 4.5 *9 End of expression9

Additional Notes Stacks structures are usually implemented using arrays or linked lists. For both implementations, the running time is O(n). We will be examining common Stack Applications.

Stack Applications Reversing Data: We can use stacks to reverse data. (example: files, strings) Very useful for finding palindromes. Consider the following pseudocode: 1)read (data) 2)loop (data not EOF and stack not full) 1) push (data) 2) read (data) 3)Loop (while stack notEmpty) 1) pop (data) 2) print (data)

Stack Applications Converting Decimal to Binary: Consider the following pseudocode 1)Read (number) 2)Loop (number > 0) 1) digit = number modulo 2 2) print (digit) 3) number = number / 2 // from Data Structures by Gilbert and Forouzan The problem with this code is that it will print the binary number backwards. (ex: 19 becomes 11001000 instead of 00010011. ) To remedy this problem, instead of printing the digit right away, we can push it onto the stack. Then after the number is done being converted, we pop the digit out of the stack and print it.

Stack Applications Postponement: Evaluating arithmetic expressions. Prefix: + a b Infix: a + b (what we use in grammar school) Postfix: a b + In high level languages, infix notation cannot be used to evaluate expressions. We must analyze the expression to determine the order in which we evaluate it. A common technique is to convert a infix notation into postfix notation, then evaluating it.

Infix to Postfix Conversion Rules: –Operands immediately go directly to output –Operators are pushed into the stack (including parenthesis) -Check to see if stack top operator is less than current operator -If the top operator is less than, push the current operator onto stack -If the top operator is greater than the current, pop top operator and push onto stack, push current operator onto stack -Priority 2: * / -Priority 1: + - -Priority 0: ( If we encounter a right parenthesis, pop from stack until we get matching left parenthesis. Do not output parenthesis.

Infix to Postfix Example A + B * C - D / E Infix Stack(bot->top)Postfix a) A + B * C - D / E b) + B * C - D / EA c) B * C - D / E+ A d) * C - D / E+A B e) C - D / E+ *A B f) - D / E+ *A B C g) D / E+ -A B C * h) / E+ -A B C * D i) E+ - /A B C * D j)+ - /A B C * D E k)A B C * D E / - +

Infix to Postfix Example #2 A * B - ( C + D ) + E Infix Stack(bot->top)Postfix a)A * B - ( C - D ) + Eemptyempty b) * B - ( C + D ) + EemptyA c) B - ( C + D ) + E*A d) - ( C + D ) + E*A B e) - ( C + D ) + EemptyA B * f) ( C + D ) + E-A B * g) C + D ) + E- (A B * h) + D ) + E- (A B * C i) D ) + E- ( +A B * C j) ) + E- ( +A B * C D k) + E-A B * C D + l) + EemptyA B * C D + - m) E+A B * C D + - n) +A B * C D + - E o) emptyA B * C D + - E +

Postfix Evaluation Operand: push Operator: pop 2 operands, do the math, pop result back onto stack 1 2 3 + * PostfixStack( bot -> top ) a)1 2 3 + * b) 2 3 + *1 c) 3 + *1 2 d) + *1 2 3 e) *1 5 // 5 from 2 + 3 f) 5// 5 from 1 * 5

Backtracking Stacks can be used to backtrack to achieve certain goals. Usually, we set up backtrack tokens to indicate a backtrack opportunity.

References Our class textbook Data Structures: A Pseudocode Apporoach with C. Gilberg, Richard F., Forouzan, Behrouz A.. PWS Publishing Company: 1998

user compiler Postfix: no parentheses, no precedence

#define MAX_STACK_SIZE 100 /* maximum stack size */ #define MAX_EXPR_SIZE 100 /* max size of expression */ typedef enum{lparan, rparen, plus, minus, times, divide, mod, eos, operand} precedence; int stack[MAX_STACK_SIZE]; /* global stack */ char expr[MAX_EXPR_SIZE]; /* input string */ Assumptions: operators: +, -, *, /, % operands: single digit integer Infix to Postfix

int eval(void) { /* evaluate a postfix expression, expr, maintained as a global variable, ‘\0’ is the the end of the expression. The stack and top of the stack are global variables. get_token is used to return the token type and the character symbol. Operands are assumed to be single character digits */ precedence token; char symbol; int op1, op2; int n = 0; /* counter for the expression string */ int top = -1; token = get_token(&symbol, &n); while (token != eos) { if (token == operand) push(&top, symbol-’0’); /* stack insert */ Evaluation of Postfix Expressions

else { /* remove two operands, perform operation, and return result to the stack */ op2 = pop(&top); /* stack delete */ op1 = pop(&top); switch(token) { case plus: push(&top, op1+op2); break; case minus: push(&top, op1-op2); break; case times: push(&top, op1*op2); break; case divide: push(&top, op1/op2); break; case mod: push(&top, op1%op2); } } token = get_token (&symbol, &n); } return pop(&top); /* return result */ }

precedence get_token(char *symbol, int *n) { /* get the next token, symbol is the character representation, which is returned, the token is represented by its enumerated value, which is returned in the function name */ *symbol =expr[(*n)++]; switch (*symbol) { case ‘(‘ : return lparen; case ’)’ : return rparen; case ‘+’: return plus; case ‘-’ : return minus;

case ‘/’ : return divide; case ‘*’ : return times; case ‘%’ : return mod; case ‘\0‘ : return eos; default : return operand; /* no error checking, default is operand */ } }

Infix to Postfix Conversion (Intuitive Algorithm) (1)Fully parenthesized expression a / b - c + d * e - a * c --> ((((a / b) - c) + (d * e)) – (a * c)) (2)All operators replace their corresponding right parentheses. ((((a / b) - c) + (d * e)) – (a * c)) (3) Delete all parentheses. ab/c-de*+ac*- two passes / - * + * -

The orders of operands in infix and postfix are the same. a + b * c, * > +

(1)Operators are taken out of the stack as long as their in-stack precedence is higher than or equal to the incoming precedence of the new operator. (2)( has low in-stack precedence, and high incoming precedence. ()+-*/%eos isp0 1912 121313130 icp201912121313130 Rules

precedence stack[MAX_STACK_SIZE]; /* isp and icp arrays -- index is value of precedence lparen, rparen, plus, minus, times, divide, mod, eos */ static int isp [ ] = {0, 19, 12, 12, 13, 13, 13, 0}; static int icp [ ] = {20, 19, 12, 12, 13, 13, 13, 0}; isp: in-stack precedence icp: incoming precedence

void postfix(void) { /* output the postfix of the expression. The expression string, the stack, and top are global */ char symbol; precedence token; int n = 0; int top = 0; /* place eos on stack */ stack[0] = eos; for (token = get _token(&symbol, &n); token != eos; token = get_token(&symbol, &n)) { if (token == operand) printf (“%c”, symbol); else if (token == rparen ){ Infix to Postfix

/*unstack tokens until left parenthesis */ while (stack[top] != lparen) print_token(delete(&top)); pop(&top); /*discard the left parenthesis */ } else{ /* remove and print symbols whose isp is greater than or equal to the current token’s icp */ while(isp[stack[top]] >= icp[token] ) print_token(delete(&top)); push(&top, token); } } while ((token = pop(&top)) != eos) print_token(token); print(“\n”); } Infix to Postfix (cont’d)

Stacks Objective After this lecture you will be able to: Describe a stack Describe the representation of stack using linear array Describe the representation.

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Stacks Objective After this lecture you will be able to: Describe a stack Describe the representation of stack using linear array Describe the representation.

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