1. Abstract Data Types Stack, Queue Amortized analysis

2. Queue Inject(x,Q) : Insert last element 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 12 1 5
tail
inject(4,Q)

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

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

6. Implementation with stacks13
size=5
inject(x,Q): push(x,S2); size ← size + 1
inject(2,Q)

7. Implementation with stacks13
size=5
inject(x,Q): push(x,S2); size ← size + 1
inject(2,Q)

8. Implementation with stacks13
size=6
inject(x,Q): push(x,S2); size ← size + 1
inject(2,Q)

9. Pop S1 S2 13 5 4 17 21 2 size=6 pop(Q): if empty?(Q) error
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
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
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 S1 S2 2 5 4 17 21 size=5 pop(Q): if empty?(Q) error
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 S1 S2 21 2 5 4 17 size=5 pop(Q): if empty?(Q) error
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 S1 S2 17 21 2 5 4 size=5 pop(Q): if empty?(Q) error
5 4
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 S1 S2 4 17 21 2 5 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 S1 S2 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)

17. Pop S1 S2 4 17 21 2 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. Implementation with doubly linked lists
head
tail
size=2 13 5
x.next
x.element
x.prev x

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

27. Push size=1 5 push(x,D): n = new node n.element ←x n.next ← head.next
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 size=1 5 push(x,D): n = new node n.element ←x n.next ← head.next
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 size=1 5 push(x,D): n = new node n.element ←x n.next ← head.next
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 size=1 5 push(x,D): n = new node n.element ←x n.next ← head.next
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 size=2 5 push(x,D): n = new node n.element ←x n.next ← head.next
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 stacks13
size=5
push(x,D): push(x,S1)
push(2,D)

33. Implementation with stacks
2 13
size=6
push(x,D): push(x,S1)
push(2,D)

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

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

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

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

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

42. Split S1 S2 S3

43. Split S1 S2 S3 21

44. Split S1 S2
5 4 S3 17 21

45. Split S1 S2 4 5 S3 17 21

46. Split S1 S2 5 4 S3 17 21

47. Split S1 17 S2 5 4 S3 21

48. Split S1 S2 5 4 17 21 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
x ← pop(S3) push(x,S2) i ← i+1

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

51. Thm: If we start with an empty deque and perform m operations then its takes O(m) time

52. A better bound Consider Recall that: Amortized(op) = actual(op) + ΔΦ
This is O(1) if no splitting occurs
Say we split S1:
Then the actual time is |S1| + O(1)
ΔΦ = -|S1|
So the amortized time is O(1)