Data Structure - Stack

1. Data Structure - Stack Sang Yong Han http://ec.cse.cau.ac.kr/ Chung-Ang University Spring 2012

2. Stacks • Stack: what is it? • ADT • Applications • Implementation(s) • Tower of Hanoi • Maze (Maze Routing) • Mathematical Expression Chung-Ang University Spring 2012

3. What is a stack? • Stores a set of elements in a particular order • Stack principle: LAST IN FIRST OUT • = LIFO • It means: the last element inserted is the first one to be removed • Example • Which is the first element to pick up? Chung-Ang University Spring 2012

4. Last In First Out top B A C B A D C B A E D C B A D C B A top top top top A Chung-Ang University Spring 2012

5. Stack Applications • Real life • Pile of books • Plate trays • More applications related to computer science • Program execution stack • Evaluating expressions Chung-Ang University Spring 2012

6. Method Invocation And Return publicvoid a() { …; b(); …} publicvoid b() { …; c(); …} publicvoid c() { …; d(); …} publicvoid d() { …; e(); …} publicvoid e() { …; c(); …} return address in d() return address in c() return address in e() return address in d() return address in c() return address in b() return address in a() Chung-Ang University Spring 2012

7. objects: a finite ordered list with zero or more elements.methods: for all stack Stack, item  element, max_stack_size  positive integerStack createS(max_stack_size) ::= create an empty stack whose maximum size is max_stack_size Boolean isFull(stack, max_stack_size) ::= if (number of elements in stack == max_stack_size) return TRUEelse return FALSEStack push(stack, item) ::=if (IsFull(stack)) stack_fullelse insert item into top of stack and return Stack ADT Chung-Ang University Spring 2012

8. Boolean isEmpty(stack) ::= if(stack == CreateS(max_stack_size)) return TRUEelse return FALSEElement pop(stack) ::= if(IsEmpty(stack)) returnelse remove and return the item on the top of the stack. Stack ADT (cont’d) Chung-Ang University Spring 2012

9. Array-based Stack Implementation • Allocate an array of some size (pre-defined) • Maximum N elements in stack • Bottom stack element stored at element 0 • last index in the array is the top • Increment top when one element is pushed, decrement after pop Chung-Ang University Spring 2012

10. Stack createS(max_stack_size) ::= #define MAX_STACK_SIZE 100 /* maximum stack size */ typedef struct { int key; /* other fields */ } element; element stack[MAX_STACK_SIZE]; int top = -1;BooleanisEmpty(Stack) ::= top< 0;BooleanisFull(Stack) ::= top >= MAX_STACK_SIZE-1; Stack Implementation: CreateS, isEmpty, isFull Chung-Ang University Spring 2012

11. top a b c d e 0 1 2 3 4 void push(element item){ /* add an item to the global stack */ if (top >= MAX_STACK_SIZE-1) { stack_full( ); return; } stack[++top] = item;} Push Chung-Ang University Spring 2012

12. StackFull() void StackFull() { fprintf(stderr, “Stack is full, cannot add element.”); exit(EXIT_FAILURE); } Chung-Ang University Spring 2012

13. top a b c d e 0 1 2 3 4 element pop(){ /* return the top element from the stack */ if (top == -1) return stack_empty( ); /* returns and error key */ return stack[top--]; } Pop Chung-Ang University Spring 2012

14. StackFull()/Dynamic Array • Use a variable called capacity in place of MAX_STACK_SIZE • Initialize this variable to (say) 1 • When stack is full, double the capacity using REALLOC • This is called array doubling Chung-Ang University Spring 2012

15. StackFull()/Dynamic Array void StackFull() { REALLOC(stack, 2*capacity*sizeof(*stack); capacity *= 2; } Chung-Ang University Spring 2012

16. A LegendThe Towers of Hanoi • In the great temple of Brahma in Benares, on a brass plate under the dome that marks the center of the world, there are 64 disks of pure gold that the priests carry one at a time between these diamond needles according to Brahma's immutable law: No disk may be placed on a smaller disk. In the begging of the world all 64 disks formed the Tower of Brahma on one needle. Now, however, the process of transfer of the tower from one needle to another is in mid course. When the last disk is finally in place, once again forming the Tower of Brahma but on a different needle, then will come the end of the world and all will turn to dust. Chung-Ang University Spring 2012

17. The Towers of HanoiA Stack-based Application • GIVEN: three poles • a set of discs on the first pole, discs of different sizes, the smallest discs at the top • GOAL: move all the discs from the left pole to the right one. • CONDITIONS: only one disc may be moved at a time. • A disc can be placed either on an empty pole or on top of a larger disc. Chung-Ang University Spring 2012

18. Towers of Hanoi Chung-Ang University Spring 2012

19. Towers of Hanoi Chung-Ang University Spring 2012

20. Towers of Hanoi Chung-Ang University Spring 2012

21. Towers of Hanoi Chung-Ang University Spring 2012

22. Towers of Hanoi Chung-Ang University Spring 2012

23. Towers of Hanoi Chung-Ang University Spring 2012

24. Towers of Hanoi Chung-Ang University Spring 2012

25. Towers of Hanoi Chung-Ang University Spring 2012

26. 4 3 2 1 A B C Towers Of Hanoi/Brahma • 64 gold disks to be moved from tower A to tower C • each tower operates as a stack • cannot place big disk on top of a smaller one Chung-Ang University Spring 2012

27. 1 A B C Recursive Solution • n > 0 gold disks to be moved from A to C using B • move top n-1 disks from A to B using C Chung-Ang University Spring 2012

28. 1 B C Recursive Solution • move top disk from A to C A Chung-Ang University Spring 2012

29. 1 B C Recursive Solution • move top n-1 disks from B to C using A A Chung-Ang University Spring 2012

30. 1 B C Recursive Solution • moves(n) = 0 when n = 0 • moves(n) = 2*moves(n-1) + 1 = 2n-1 when n > 0 A Chung-Ang University Spring 2012

31. Towers of Hanoi – Recursive Solution void hanoi (int discs, Stack fromPole, Stack toPole, Stack aux) { Disc d; if( discs >= 1) { hanoi(discs-1, fromPole, aux, toPole); d = fromPole.pop(); toPole.push(d); hanoi(discs-1,aux, toPole, fromPole); } Chung-Ang University Spring 2012

32. Is the End of the World Approaching? • Problem complexity 2n • 64 gold discs • moves(64) = 1.8 * 1019 (approximately) • Given 1 move a second  600,000,000,000 years until the end of the world  Chung-Ang University Spring 2012