Abstract Data Types Stack, Queue Amortized analysis

1. Abstract Data TypesStack, QueueAmortized analysis

2. Queue • Inject(x,Q) : Insert lastelement x into Q • Pop(Q) : Delete the first element in Q • Empty?(Q): Return yes if Q is empty • Front(Q): Return the first element in Q • Size(Q) • Make-queue()

3. Implementation with lists head size=3 5 1 12 tail inject(4,Q)

4. Implementation with lists head size=3 4 5 1 12 tail inject(4,Q)

5. Implementation with lists head size=3 4 5 1 12 tail inject(4,Q) Complete the details by yourself

6. Implementation with stacks S2 S1 13 5 4 17 21 size=5 inject(x,Q): push(x,S2); size ← size + 1 inject(2,Q)

7. Implementation with stacks S2 S1 13 5 4 17 21 2 size=5 inject(x,Q): push(x,S2); size ← size + 1 inject(2,Q)

8. Implementation with stacks S2 S1 13 5 4 17 21 2 size=6 inject(x,Q): push(x,S2); size ← size + 1 inject(2,Q)

9. Pop S2 S1 13 5 4 17 21 2 size=6 pop(Q): if empty?(Q) error if empty?(S1) then move(S2, S1) pop( S1); size ← size -1 pop(Q)

10. Pop S2 S1 5 4 17 21 2 size=6 pop(Q): if empty?(Q) error if empty?(S1) then move(S2, S1) pop( S1); size ← size -1 pop(Q)

11. Pop S2 S1 5 4 17 21 2 size=5 pop(Q): if empty?(Q) error if empty?(S1) then move(S2, S1) pop( S1); size ← size -1 pop(Q) pop(Q)

12. Pop S2 S1 2 5 4 17 21 size=5 pop(Q): if empty?(Q) error if empty?(S1) then move(S2, S1) pop( S1); size ← size -1 pop(Q) pop(Q)

13. Pop S2 S1 2 5 4 17 21 size=5 pop(Q): if empty?(Q) error if empty?(S1) then move(S2, S1) pop( S1); size ← size -1 pop(Q) pop(Q)

14. Pop S2 S1 2 17 5 4 21 size=5 pop(Q): if empty?(Q) error if empty?(S1) then move(S2, S1) pop( S1); size ← size -1 pop(Q) pop(Q)

15. Pop S2 S1 4 2 17 5 21 size=5 pop(Q): if empty?(Q) error if empty?(S1) then move(S2, S1) pop( S1); size ← size -1 pop(Q) pop(Q)

16. Pop S2 S1 4 2 5 17 21 size=5 pop(Q): if empty?(Q) error if empty?(S1) then move(S2, S1) pop( S1); size ← size -1 pop(Q) pop(Q)

17. Pop S2 S1 4 2 17 21 size=4 pop(Q): if empty?(Q) error if empty?(S1) then move(S2, S1) pop( S1); size ← size -1 pop(Q) pop(Q)

18. move(S2, S1) while not empty?(S2) do x ←pop(S2) push(x,S1)

19. Analysis • O(n) worst case time per operation

20. Amortized Analysis • How long it takes to perform m operations on the worst case ? • O(nm) • Is tha tight ?

21. Key Observation • An expensive operation cannot occur too often !

22. THM: If we start with an empty queue and perform m operations then it takes O(m) time

23. Proof Consider Recall that: Amortized(op) = actual(op) + ΔΦ This is O(1) if a move does not occur Say we move S2: Then the actual time is |S2| + O(1) ΔΦ = -|S2| So the amortized time is O(1)

24. Double ended queue (deque) • Push(x,D) : Insert x as the first in D • Pop(D) : Delete the first element of D • Inject(x,D): Insert x as the last in D • Eject(D): Delete the last element of D • Size(D) • Empty?(D) • Make-deque()

25. x x.next x.prev x.element Implementation with doubly linked lists head tail size=2 13 5

26. Empty list head tail size=0 We use two sentinels here to make the code simpler

27. Push head tail size=1 5 push(x,D): n = new node n.element ←x n.next ← head.next (head.next).prev ← n head.next ← n n.prev← head size ← size + 1

28. 4 head tail size=1 5 push(x,D): n = new node n.element ←x n.next ← head.next (head.next).prev ← n head.next ← n n.prev← head size ← size + 1 push(4,D)

29. 4 head tail size=1 5 push(x,D): n = new node n.element ←x n.next ← head.next (head.next).prev ← n head.next ← n n.prev← head size ← size + 1 push(4,D)

30. 4 head tail size=1 5 push(x,D): n = new node n.element ←x n.next ← head.next (head.next).prev ← n head.next ← n n.prev← head size ← size + 1 push(4,D)

31. 4 head tail size=2 5 push(x,D): n = new node n.element ←x n.next ← head.next (head.next).prev ← n head.next ← n n.prev← head size ← size + 1 push(4,D)

32. Implementation with stacks S2 S1 13 5 4 17 21 size=5 push(x,D): push(x,S1) push(2,D)

33. Implementation with stacks S2 S1 2 13 5 4 17 21 size=6 push(x,D): push(x,S1) push(2,D)

34. Pop S2 S1 2 13 5 4 17 21 size=6 pop(D): if empty?(D) error if empty?(S1) then split(S2, S1) pop( S1) pop(D)

35. Pop S2 S1 13 5 4 17 21 size=5 pop(D): if empty?(D) error if empty?(S1) then split(S2, S1) pop( S1) pop(D) pop(D)

36. Pop S2 S1 5 4 17 21 size=4 pop(D): if empty?(D) error if empty?(S1) then split(S2, S1) pop( S1) pop(D) pop(D)

37. Pop S2 S1 5 4 17 21 size=4 pop(D): if empty?(D) error if empty?(S1) then split(S2, S1) pop( S1) pop(D)

38. Pop S2 S1 5 4 17 21 size=4 pop(D): if empty?(D) error if empty?(S1) then split(S2, S1) pop( S1) pop(D)

39. Pop S2 5 4 S1 17 21 size=4 pop(D): if empty?(D) error if empty?(S1) then split(S2, S1) pop( S1) pop(D)

40. Pop S2 S1 4 17 21 size=4 pop(D): if empty?(D) error if empty?(S1) then split(S2, S1) pop( S1) pop(D)

41. Pop S2 S1 4 17 21 size=3 pop(D): if empty?(D) error if empty?(S1) then split(S2, S1) pop( S1) pop(D)

42. Split S2 S1 5 4 17 21 S3

43. Split S2 S1 5 4 17 S3 21

44. Split S2 S1 5 4 S3 17 21

45. Split S2 S1 4 5 S3 17 21

46. Split S2 S1 5 4 S3 17 21

47. Split S2 S1 17 5 4 S3 21

48. Split S2 S1 17 21 5 4 S3

49. split(S2, S1) S3←make-stack() d ←size(S2) while (i ≤⌊d/2⌋) do x ←pop(S2) push(x,S3) i ← i+1 while (i ≤⌈d/2⌉) do x ←pop(S2) push(x,S1) i ← i+1 while (i ≤⌊d/2⌋) do x ←pop(S3) push(x,S2) i ← i+1

50. Analysis • O(n) worst case time per operation