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Abstract Data Types Stack, Queue Amortized analysis

Explore the implementation and analysis of stack and queue ADTs using lists and stacks. Learn about push, pop, inject, eject operations and amortized analysis. Understand the efficiency and complexities involved in managing data structures.

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Abstract Data Types Stack, Queue Amortized analysis

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  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

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