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Stacks Example: Stack of plates in cafeteria. Stack => a list data structure in which elements are inserted in and removed from the same end, the top of.

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Presentation on theme: "Stacks Example: Stack of plates in cafeteria. Stack => a list data structure in which elements are inserted in and removed from the same end, the top of."— Presentation transcript:

1 Stacks Example: Stack of plates in cafeteria. Stack => a list data structure in which elements are inserted in and removed from the same end, the top of the stack. Use of Stacks => Computer system software such as compilers and operating system. A stack is called a last in, first out (LIFO) list. The last element stored is the first to be removed. Operations on a stack - push a new object on the top of the stack - pop the top object from the stack Push Pop A B A B C

2 Stacks Because of the push operation which adds elements to a stack, a stack is sometimes called a pushdown list. If a stack contains a single item and the stack is popped, the resulting stack contains no items and is called the empty stack. Push operation is applicable to any stack. Pop operation cannot be applied to the empty stack because such a stack has no element to pop. Therefore, before applying the pop operation to a stack, we must ensure that the stack is not empty. The result of an illegal attempt to pop or access an item from an empty stack is called underflow.

3 Implementing a stack using arrays An integer variable stack_top tells us which is the top element stack_top Push 7 Pop returns 7 Problem: Cannot hold more than 4 elements

4 Implementing a Stack using Link Lists A pointer topp points to the top of the stack. C+ 2 /C 2 + Stack of three characters Stack after insertion (push) of ‘/’ topp

5 List Representation of a Stack typedef char stack_element_t; typedef struct stack_node_s { stack_element_t element; struct stack_node_s *restp; } stack_node_t; typedef struct { stack_node_t *topp; } stack_t;

6 Push Function void push(stack_t *sp, // pointer to the stack stack_element_t c) // element to add { stack_node_t *newp; // pointer to new stack // creates and defines new node newp = (stack_node_t *) malloc(sizeof(stack_node_t)); newp->element = c; newp->restp = sp->topp; sp->topp = newp; //sets stack pointer to point to new node }

7 Implementation of the pop Operation The possibility of underflow must be considered in implementing the pop operation. The user may attempt to pop an element from an empty stack. Such an attempt is illegal and should be avoided. The function Pop should perform the following actions: - If the stack is empty, print a warning message and halt execution. - Remove the top element from the stack. - Return this element to the calling program. If we don’t check for empty stack and call pop with an empty stack, the value topp would be -1 and an attempt would be made to access the nonexistent element stack[-1].

8 Pop Function stack_element_t pop(stack_t *sp) { stack_node_t *to_freep; // pointer to node removed stack_element_t ans; // value at top of stack to_freep = sp->topp; // saves pointer to node // being deleted ans = to_freep->element; // retrieves value to return sp->topp = to_freep->restp; // deletes top node free(to_freep); // deallocates space return (ans); }

9 Stack Manipulation with Functions push & pop #include void push(stack_t *sp, stack_element_t c); stack_element_t pop(stack_t *sp); int main(void) { stack_t s = {NULL}; // stack is initially empty push(&s, ‘2’); // builds the stack push(&s, ‘+’); push(&s, ‘C’); push(&s, ‘/’); printf(“\nEmptying stack:\n”); // empties stack element while (s.topp != NULL) printf (“%c\n”, pop(&s)); return (0); }

10 Problem solving using Stacks Mathematical expression that includes several sets of nested parentheses: 7 - ((X * ((X + Y) / (J -3)) + Y) / ( )) We want to ensure that the parentheses are nested correctly so that - 1. There are an equal number of right and left parentheses Every right parenthesis is preceded by a matching left parenthesis. Check 1. ((A + B) violates condition 1 2. A + B( violates condition 1 3. )A + B(-C violates condition 2 4. (A + B)) - (C + D violates condition 2

11 Problem solving using Stacks We can solve this problem by using parenthesis count. Parenthesis count => the number of left parentheses minus the number of right parentheses that have been encountered from the left end up to a particular point. The expression will be valid if - The parenthesis count at the end of the expression is 0. This means that the number of left parentheses is the same as the number of right parentheses. - The parenthesis count at each point in the expression is nonnegative. This means that no right parenthesis is encountered for which a matching left parenthesis not previously been encountered.

12 Parenthesis Count at various points of Expressions Expressions: Parenthesis count at the end Validity ( ( A + B ) 1 invalid A + B ( 1 invalid ) A + B ( - ( 1 invalid ( ( A + B ) ) 0 at the end and valid nonnegative at each point

13 Extension of this problem In the earlier problem, expressions contain only parentheses as scope delimiter. However, an expression can contain three different types of scope delimiters: parentheses(), brackets [], and braces {}. Examples: (A + B]){ or {(A - B) - (C - [D + E])} Therefore, we need to count not only how many scopes are opened but also their types. Using stacks we can easily solve this problem. Left parenthesis => opening a scope Right parenthesis => closing a scope

14 Solution of this problem using Stack Whenever a scope opener is encountered, it is pushed onto the stack. Whenever a scope ender is encountered, the stack is examined. If the stack is empty, the scope ender does not have a matching opener and the string is invalid. If the stack is nonempty, we pop the stack and check whether the popped item corresponds to the scope ender. If a match occurs, we continue. If does not, the string is invalid. When the end of the string is reached, the stack must be empty. Otherwise one or more scopes have been opened which have not been closed, and the string is invalid.

15 Parenthesis stack at Various stages of Processing Expression: {(A - B) - (C - [D + E])} { { ( { { ( { ( { {….{( ….{(A - B)... {(A - B) - (.. {(A - B) -(C- [.. { ( [ {(A -B) - (C - [D+E]... {(A-B) - (C - [D+E]).. {(A- B) - (C -[D+E])}

16 Why we have used Stacks ? The last scope (parenthesis) to be opened must be first to be closed. This is simulated by a stack in which the last element arriving is the first to leave. Each item on the stack represents a scope that has been opened but that has not been closed. Pushing an item onto the stack corresponds to opening a scope and popping an item from the stack corresponds to closing a scope, leaving one less scope open. Notice that at any point, we examine only the element at the top of the stack. In general, a stack is used in any situations that calls for a last-in, first-out discipline or that displays a nesting pattern.

17 Infix, Postfix, and Prefix The sum of A and B can be represented in three different ways. Infix notation: In infix notation, the operator is between the two operands. Humans generally use this notation. E.g. A + B Prefix notation: In prefix notation, the operator precedes the two operands. Computers prefer this notation. E.g + A B Postfix notation: In postfix notation, the operator follows the two operands. E.g A B +

18 Use of postfix Notation To evaluate a complex infix expression, a compiler would first convert the expression to postfix notation. Then evaluate the postfix version of the expression. Each of these algorithms requires only a single left-to- right pass of the expression. Each algorithm uses a stack in support of its operation. In each the stack is used for a different purpose.

19 Rules of Conversions of different notations The rules to convert an infix to postfix: - The operation with highest precedence are converted first. - After a portion of the expression has been converted to postfix it is to be treated as a single operand. The order of precedence (highest to lowest) - Exponentiation - Multiplication/ Division - Addition/ Subtraction When unparenthesized operators of the same precedence are scanned, the order is from left to right except in the case of exponentiation, where the order from right to left.

20 Example Expression: A + B * C (standard infix notation) It requires knowledge of which of the two operations, + or *, is to be performed first. We know that multiplication has higher precedence over addition. So, A + B * C is interpreted as A + ( B * C ). Suppose we want to rewrite A + B * C in postfix. The steps are: A + ( B * C ) parentheses for emphasis A + ( B C * ) convert the multiplication A ( B C * ) + convert the addition A B C * + postfix form

21 Another Example Expression: (A + B) * C (infix notation) In this example, addition will be converted before multiplication because parentheses have higher precedence than multiplication. Suppose we want to rewrite ( A + B ) * C in postfix. The steps are: ( A + B ) * C infix form (A B + ) * C convert the addition (A B + ) C * convert the multiplication A B + C * postfix form

22 More Examples Rewrite the infix ( A + B ) * ( C - D ) in postfix. The steps are: ( A + B ) * ( C - D ) infix form ( A B + ) * ( C D -) convert the parenthesized operations ( A B + C D -) * convert the multiplication. A B + C D - * postfix form

23 More Examples Rewrite the infix A$B*C-D+E/F/(G+H) in postfix. The steps are: A$B*C-D+E/F/(G+H) infix form A$B*C-D+E/F/(GH+) convert the parenthesized operation (AB$)*C-D+E/F/(GH+) convert the exponentiation. (AB$C*)-D+E/F/(GH+) convert the multiplication. (AB$C*)-D+(EF/)/(GH+) convert the division (left most). (AB$C*)-D+(EF/GH+/) convert the division (AB$C*D-)+(EF/GH+/) convert the subtraction. (AB$C*D-EF/GH+/)+ convert the addition. AB$C*D-EF/GH+/+ postfix form

24 Conversion from Infix to Prefix The precedence rules for converting an expression from infix to prefix and from infix to postfix are identical. The only difference is that the operator is placed before the operands rather than after them. Example: Rewrite the infix A + B - C in prefix. The steps are: A + B - C infix form (+ A B ) - C convert the addition - (+A B C ) convert the subtraction - + A B C prefix form

25 Another Example Rewrite the infix ( A + B ) * ( C - D ) in prefix. The steps are: ( A + B ) * ( C - D ) infix form ( +A B ) * ( - C D ) convert the parenthesized operations * ( + A B - C D ) convert the multiplication. * + A B - C D prefix form Note that the prefix form of a complex expression is not the mirror image of the postfix form.

26 Advantage of using Postfix and prefix Postfix and prefix forms of expressions allow us an unambiguous form of the original expressions without the use of cumbersome parentheses. Consider two expressions: A + ( B * C ) and (A + B) * C (infix forms) The parentheses in the first expression are superfluous. However, the parentheses in the second expression are necessary to avoid confusion with the first. The corresponding postfix expressions are: ABC*+ AB+C* The order of operators in the postfix expressions determines the actual order of operations in evaluating it. No parentheses

27 Algorithm to Evaluate a Postfix Expression The algorithm is as follows: 1Append the NULL character (‘\0’) to the end of the postfix expression. When the NULL character is encountered, no further processing is necessary. 2While ‘\0’ has not been encountered, read the expression from left to right. If the current character is a digit, push its integer value on the stack Otherwise, if the current character is an operator, pop the two elements of the stack into variables x and y. Calculate y operator x. Push the result of the calculation on the stack. 3When the NULL character is encountered in the expression, pop the top value of the stack. This is the result of the postfix expression.

28 Example Postfix Expression: / + * 2 $ 3 + Steps for evaluation are as follows: / + * 2 $ / + * 2 $ / + * 2 $ * 2 $ * 2 $ $

29 Example: / + * 2 $ 3 + Symb opnd1 opnd2 value opndstk , 2 3 6, 2, , , , 3, , 3, 8, 2 / , 3, , 7 * , 2 $ ,

30 Another example Postfix Expression: * 8 4 / - Steps for evaluation are as follows: * 8 4 / * 8 4 / /

31 Algorithm for Creating a Postfix Expression 1Push a left parentheses ‘(‘ on the stack. 2Append a right parenthesis’)’ to the end of infix. 3While the stack is not empty, read infix from left to right and do the following: If the current character in infix is a digit, copy it to the next element of postfix. If the current character in infix is a left parenthesis, push it on the stack. If the current character in infix is an operator, Pop operators (if there are any) at the top of the stack while they have equal or higher precedence than the current operator, and insert the popped operators in postfix. Push the current character in infix on the stack. If the current character in infix is a right parenthesis Pop operators from the top of the stack and insert them in postfix until a left parenthesis is at the top of the stack. Pop (and discard) the left parenthesis from the stack

32 Explanation Expression: A + B * C (standard infix notation) symb postfix string opstk A A + A + B AB + * AB + * C ABC + * ABC * +

33 Infix to Prefix 1) A + B = +AB 2) A + B - C = (+AB) - C = - + ABC 3) A + B * C = A + (*BC) = + A * BC 4) A * (B + C) = A * (+ BC) = * A+ BC 5) A * B + C = (* AB) + C = + * ABC 6) (A + B) * (C - D) = (+ AB) * (- CD) = * + AB - CD

34 Infix to Prefix 7) (( A + B ) * C - ( D - E )) $ (F + G) = ((+ AB) * C - (- DE)) $ (+ FG) = ((* + ABC) - (- DE)) $ (+ FG) = (- * + ABC - DE) $ (+ FG) = $ - * + ABC - DE + FG 8) A - B / ( C * D $ E) = A - B / (C * ($ DE)) = A - B / (* C $ DE) = A - (/ B * C $ DE) = - A / B * C $ DE 9) A + B * C + D - E * F = A + ( * BC) + D - (* EF) = (+ A * BC) + D - (* EF) = (+ + A * BCD) - (* EF) = A * BCD * EF 10) (A + B) * (C + D - E) * F = (+ AB) * (( + CD) - E) * F = (+ AB) * (- + CDE) * F = (* + AB - + CDE) * F = * * + AB - + CDEF 11) 7) A $ B * C - D + E / F / (G + H) = A $ B * C - D + E / F / (+GH) = ($AB) * C - D + E / F / (+GH) = (*$ABC) - D + E / F / (+GH) = (*$ABC) - D + (/EF) / (+GH) = (*$ABC) - D + (//EF + GH) = (- * $ ABCD) + (//EF + GH) = + - * $ ABCD//EF + GH

35 Infix to Postfix 1) A + B = AB + 2) A + B - C = (AB +) - C = AB + C - 3) A + B * C = A + ( BC *) = ABC * + 4) A * (B + C) = A * (BC +) = ABC + * 5) A * B + C = (AB *) + C = AB * C + 6) (A + B) * (C - D) = (AB +) * (CD-) = AB + CD - *

36 Infix to Postfix 7) A $ B * C - D + E / F / (G + H) = A $ B * C - D + E / F / (GH +) = (AB $) * C - D + E / F / (GH +) = (AB $ C *) - D + E / F / (GH +) = (AB $ C *) - D + (EF /) / (GH +) = (AB $ C *) - D + (EF / GH + /) = (AB $ C * D -) + (EF / GH + /) = AB $ C * D - EF / GH + / + 8) ((A + B) * C - (D - E)) $ (F + G) = ((AB +) * C - (DE -)) $ (FG +) = ((AB + C *) - (DE -)) $ (FG +) = (AB + C * DE - - ) $ (FG + ) = AB + C * DE - - FG + $ 9) A - B / (C * D $ E) = A - B / (C * (DE $)) = A - B / (CDE $ *) = A - (BCDE $ * /) = ABCDE $ * / - 10) (A + B) * (C + D - E) * F = (AB +) * ((CD +) - E) * F = (AB +) * (CD + E -) * F = (AB + CD + E - *) * F = AB + CD + E - * F * 11) A + B * C + D - E * F = A + (BC *) + D - (EF *) = (ABC * +) + D - (EF *) = (ABC * + D + ) - (EF*) = ABC * + D + EF * -

37 Prefix to Infix 1) + AB = A + B 2) - + ABC = - (A + B)C = A + B - C 3) + A * BC = +A(B * C) = A + (B * C) 4) * A + BC = * A(B + C) = A * (B + C) 5) + * ABC = + (A * B) C = (A * B) + C 6) * + AB - CD = * (A + B)(C - D) = (A + B) * (C - D)

38 Prefix to Infix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

39 Postfix to Infix 1) AB + = A + B 2) ABC * + = A (B * C) + = A + ( B * C) 3) ABC + * = A (B + C) * = A * (B + C) 4) AB * C + = ( A * B) C + = (A * B) + C

40 Postfix to Infix 5) ABC * + D + EF * - = A (B * C) + D + (E * F) - = (A + B * C)D + (E * F) - = (A + B * C + D)(E * F) - = (A + B * C + D) - (E * F) 6) AB + CD + E - * F * = (A + B) CD + E - * F * = (A + B)(C + D)E - * F * = (A + B)(C + D - E) * F * = (A + B) * (C + D - E) F * = (A + B) * (C + D - E) * F 7) ABC + - D * EF + $ = A(B + C) - D * (E + F) $ = ( A - (B + C)) D * (E + F) $ = ((A - (B + C)) * D) (E + F) $ = ((A - (B +C)) * D) $ (E +F)


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