1. Sorting
We have actually seen already two efficient ways to sort:

2. A kind of “insertion” sort
Insert the elements into a red-black tree one by one
Traverse the tree in in-order and collect the keys
Takes O(nlog(n)) time

3. Heapsort (Willians, Floyd, 1964)
Put the elements in an array
Make the array into a heap
Do a deletemin and put the deleted element at the last position of the array

4. Quicksort (Hoare 1961)

5. quicksort Input: an array A[p, r] Quicksort (A, p, r) if (p < r)
then q = Partition (A, p, r)
//q is the position of the pivot element
Quicksort (A, p, q-1)
Quicksort (A, q+1, r)

6. p r i j 2 8 7 1 3 5 6 4 2 8 7 1 3 5 6 4 i j 2 8 7 1 3 5 6 4 i j 2 8 7 1 3 5 6 4 i j 2 1 7 8 3 5 6 4 i j

7. 2 1 7 8 3 5 6 4 i j 2 1 3 8 7 5 6 4 i j 2 1 3 8 7 5 6 4 i j 2 1 3 8 7 5 6 4 i j 2 1 3 4 7 5 6 8 i j

8. 2 8 7 1 3 5 6 4 p r Partition(A, p, r) x ←A[r] i ← p-1
for j ← p to r-1
do if A[j] ≤ x
then i ← i+1
exchange A[i] ↔ A[j]
exchange A[i+1] ↔A[r]
return i+1

9. Analysis Running time is proportional to the number of comparisons
Each pair is compared at most once  O(n2)
In fact for each n there is an input of size n on which quicksort takes cn2  Ω(n2)

10. But
Assume that the split is even in each iteration

11. T(n) = 2T(n/2) + n
How do we solve linear recurrences like this ? (read Chapter 4)

12. Recurrence tree n
T(n/2)
T(n/2)

13. Recurrence tree n
n/2
n/2
T(n/4)
T(n/4)
T(n/4)
T(n/4)

14. Recurrence tree n n/2 n/2 logn T(n/4) T(n/4) T(n/4) T(n/4)
In every level we do bn comparisons
So the total number of comparisons is O(nlogn)

15. Observations We can’t guarantee good splits
But intuitively on random inputs we will get good splits

16. Randomized quicksort Use randomized-partition rather than partition
Randomized-partition (A, p, r)
i ← random(p,r)
exchange A[r] ↔ A[i]
return partition(A,p,r)

17. On the same input we will get a different running time in each run !
Look at the average for one particular input of all these running times

18. Expected # of comparisons
Let X be the expected # of comparisons
This is a random variable
Want to know E(X)

19. Expected # of comparisons
Let z1,z2,.....,zn the elements in sorted order
Let Xij = 1 if zi is compared to zj and 0 otherwise
So,

20. by linearity of expectation

21. by linearity of expectation

22. Consider zi,zi+1, ,zj ≡ Zij
Claim: zi and zj are compared  either zi or zj is the first chosen in Zij
Proof: 3 cases:
{zi, …, zj} Compared on this partition, and never again.
{zi, …, zj} the same
{zi, …, zk, …, zj} Not compared on this partition. Partition separates them, so no future partition uses both.

23. Pr{zi is compared to zj}
= Pr{zi or zj is first pivot chosen from Zij}
just explained
= Pr{zi is first pivot chosen from Zij} +
Pr{zj is first pivot chosen from Zij}
mutually exclusive
possibilities
= 1/(j-i+1) + 1/(j-i+1)
= 2/(j-i+1)

24. Simplify with a change of variable, k=j-i+1.
Simplify and overestimate, by adding terms.

25. Lower bound for sorting in the comparison model

26. A lower bound
Comparison model: We assume that the operation from which we deduce order among keys are comparisons
Then we prove that we need Ω(nlogn) comparisons on the worst case

27. Model the algorithm as a decision tree

28. Insertion sort 1:2 > < 2:3 2:3 < > > 1:2 1:2 > >
A[1] < A[2] < A[3]
A[2] < A[1] < A[3]
<
>
<
>
A[1] < A[3] < A[2]
A[3] < A[1] < A[2]
A[2] < A[3] < A[1]
A[3] < A[2] < A[1]

29. Quicksort 1:3 < > 2:3 2:3 < < > > 1:2 2:3 < >
A[1] < A[3] < A[2]
A[2] < A[3] < A[1]
<
>
<
>
A[1] < A[2] < A[3]
A[2] < A[1] < A[3]
A[3] < A[1] < A[2]
A[3] < A[2] < A[1]

30. Important Observations
Every algorithm can be represented as a (binary) tree like this
Assume that for every node v there is an input on which the algorithm reaches v
Then the # of leaves is n!

31. Important Observations
Each path corresponds to a run on some input
The worst case # of comparisons corresponds to the longest path

32. The lower bound Let d be the length of the longest path n! ≤
#leaves ≤ 2d
log2(n!) ≤d

33. Lower Bound for Sorting
Any sorting algorithm based on comparisons between elements requires (N log N) comparisons.

34. Beating the lower bound
We can beat the lower bound if we can deduce order relations between keys not by comparisons
Examples:
Count sort
Radix sort

35. Linear time sorting
Or assume something about the input: random, “almost sorted”

36. Sorting an almost sorted input
Suppose we know that the input is “almost” sorted
Let I be the number of “inversions” in the input: The number of pairs ai,aj such that i<j and ai>aj

37. Example
1, 4 , 5 , 8 , 3
I=3
8, 7 , 5 , 3 , 1
I=10

38. Think of “insertion sort” using a list
When we insert the next item ak, how deep it gets into the list?
As the number of inversions ai,ak for i < k lets call this Ik

39. Analysis
The running time is:

40. Thoughts When I=Ω(n2) the running time is Ω(n2)
But we would like it to be O(nlog(n)) for any input, and faster whan I is small

41. Finger red black trees

42. Finger tree
Take a regular search tree and reverse the direction of the pointers on the rightmost spine
We go up from the last leaf until we find the subtree containing the item and we descend into it

43. Finger trees Say we search for a position at distance d from the end
Then we go up to height O(log(d))
So search for the dth position takes O(log(d)) time
Insertions and deletions still take O(log n) worst case time but O(log(d)) amortized time

44. Back to sorting
Suppose we implement the insertion sort using a finger search tree
When we insert item k then d=O(Ik) and it take O(log(Ik)) time