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

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

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

Quicksort (Hoare 1961)

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)

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

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

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

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 Ω(n2) time

But Assume that the split is even in each iteration

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

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

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

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)

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

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)

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

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

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,

by linearity of expectation

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.

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)

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

Lower bound for sorting in the comparison model

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

Model the algorithm as a decision tree

Insertion sort x y z 1:2 < > x y z 2:3 2:3 y x z < > >

Quicksort x y z 1:3 < > 2:3 2:3 < < > > 2:3 x y z 1:2 x z y y z x z x y > < > < x y z y x z z x y z y x

Important observations Every algorithm can be represented as a (binary) tree like this For every node v there is an input on which the algorithm reaches v The # of leaves is n!

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

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

Lower bound for sorting Any sorting algorithm based on comparisons between elements requires (n log n) comparisons.

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

Count sort Assume that keys are integers between 0 and k A 2 3 5 3 5 2

Count sort Allocate a temporary array of size k: cell x counts the # of keys =x A 2 3 5 3 5 2 5 C

Count sort A 2 3 5 3 5 2 5 C 1

Count sort A 2 3 5 3 5 2 5 C 1 1

Count sort A 2 3 5 3 5 2 5 C 1 1 1

Count sort A 2 3 5 3 5 2 5 C 2 2 2 3

Count sort Compute prefix sums of C: cell x holds the # of keys ≤ x (rather than =x) A 2 3 5 3 5 2 5 C 2 2 2 3

Count sort Compute prefix sums of C: cell x holds the # of keys ≤ x (rather than =x) A 2 3 5 3 5 2 5 C 2 2 4 6 6 9

Count sort Move items to output array A 2 3 5 3 5 2 5 C 2 2 4 6 6 9 B 5 3 5 2 5 C 2 2 4 6 6 9 B / / / / / / / / /

Count sort 2 3 5 A 4 6 9 C / B

Count sort 2 3 5 A 4 6 8 C / B

Count sort 2 3 5 A 6 8 C / B

Count sort 2 3 5 A 1 6 8 C / B

Count sort 2 3 5 A 1 6 7 C / B

Count sort 2 3 5 A 1 6 7 C / B

Count sort 2 3 5 A 4 6 C B

Count sort Complexity: O(n+k) The sort is stable Note that count sort does not perform any comparison

Radix sort Say we have numbers with d digits each between 0 and k 2 8 7 1 4 5 9 1 6 5 7 2 1 3 1 2 4 7 2 3 5 5 5 7 2 2 8 3 9 4 4 8 4 4 3 5 3 6

Radix sort Use a stable sort to sort by the least significant digit (e.g. count sort) 2 8 7 1 4 5 9 1 6 5 7 2 1 3 1 2 4 7 2 3 5 5 5 7 2 2 8 3 9 4 4 8 4 4 3 5 3 6

Radix sort 2 8 7 1 2 8 7 1 4 5 9 1 4 5 9 1 6 5 7 2 1 3 1 1 3 1 6 5 7 2 2 4 7 2 2 4 7 2 3 5 5 5 7 2 2 7 2 2 8 3 9 4 8 3 9 4 4 8 4 4 4 8 4 4 3 5 5 5 3 5 3 6 3 5 3 6

Radix sort 2 8 7 1 2 8 7 1 4 5 9 1 4 5 9 1 6 5 7 2 1 3 1 1 3 1 6 5 7 2 2 4 7 2 2 4 7 2 3 5 5 5 7 2 2 7 2 2 8 3 9 4 8 3 9 4 4 8 4 4 4 8 4 4 3 5 5 5 3 5 3 6 3 5 3 6

Radix sort 2 8 7 1 2 8 7 1 1 3 1 4 5 9 1 4 5 9 1 7 2 2 6 5 7 2 1 3 1 3 5 3 6 1 3 1 6 5 7 2 4 8 4 4 2 4 7 2 2 4 7 2 3 5 5 5 3 5 5 5 7 2 2 2 8 7 1 7 2 2 8 3 9 4 6 5 7 2 8 3 9 4 4 8 4 4 2 4 7 2 4 8 4 4 3 5 5 5 4 5 9 1 3 5 3 6 3 5 3 6 8 3 9 4

Radix sort 2 8 7 1 2 8 7 1 1 3 1 4 5 9 1 4 5 9 1 7 2 2 6 5 7 2 1 3 1 3 5 3 6 1 3 1 6 5 7 2 4 8 4 4 2 4 7 2 2 4 7 2 3 5 5 5 3 5 5 5 7 2 2 2 8 7 1 7 2 2 8 3 9 4 6 5 7 2 8 3 9 4 4 8 4 4 2 4 7 2 4 8 4 4 3 5 5 5 4 5 9 1 3 5 3 6 3 5 3 6 8 3 9 4

Radix sort 2 8 7 1 2 8 7 1 1 3 1 7 2 2 4 5 9 1 4 5 9 1 7 2 2 1 3 1 6 5 7 2 1 3 1 3 5 3 6 8 3 9 4 1 3 1 6 5 7 2 4 8 4 4 2 4 7 2 2 4 7 2 2 4 7 2 3 5 5 5 3 5 3 6 3 5 5 5 7 2 2 2 8 7 1 3 5 5 5 7 2 2 8 3 9 4 6 5 7 2 6 5 7 2 8 3 9 4 4 8 4 4 2 4 7 2 4 5 9 1 4 8 4 4 3 5 5 5 4 5 9 1 4 8 4 4 3 5 3 6 3 5 3 6 8 3 9 4 2 8 7 1

Radix sort 2 8 7 1 2 8 7 1 1 3 1 7 2 2 4 5 9 1 4 5 9 1 7 2 2 1 3 1 6 5 7 2 1 3 1 3 5 3 6 8 3 9 4 1 3 1 6 5 7 2 4 8 4 4 2 4 7 2 2 4 7 2 2 4 7 2 3 5 5 5 3 5 3 6 3 5 5 5 7 2 2 2 8 7 1 3 5 5 5 7 2 2 8 3 9 4 6 5 7 2 6 5 7 2 8 3 9 4 4 8 4 4 2 4 7 2 4 5 9 1 4 8 4 4 3 5 5 5 4 5 9 1 4 8 4 4 3 5 3 6 3 5 3 6 8 3 9 4 2 8 7 1

Radix sort 2 8 7 1 2 8 7 1 1 3 1 7 2 2 1 3 1 4 5 9 1 4 5 9 1 7 2 2 1 3 1 2 4 7 2 6 5 7 2 1 3 1 3 5 3 6 8 3 9 4 2 8 7 1 1 3 1 6 5 7 2 4 8 4 4 2 4 7 2 3 5 3 6 2 4 7 2 2 4 7 2 3 5 5 5 3 5 3 6 3 5 5 5 3 5 5 5 7 2 2 2 8 7 1 3 5 5 5 4 5 9 1 7 2 2 8 3 9 4 6 5 7 2 6 5 7 2 4 8 4 4 8 3 9 4 4 8 4 4 2 4 7 2 4 5 9 1 6 5 7 2 4 8 4 4 3 5 5 5 4 5 9 1 4 8 4 4 7 2 2 3 5 3 6 3 5 3 6 8 3 9 4 2 8 7 1 8 3 9 4

Radix sort Complexity O(d(n+k)) if we use count sort and have d digits each between 0 and k

Assume something about the input Random, “almost sorted” For such inputs we want to sort faster

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

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

Insertion sort Think of “insertion sort” How long it takes to insert ak ? As the number of inversions ai,ak for i < k lets call this Ik

Analysis The running time is:

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 when I is small

Finger red black trees

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

Finger trees Say we search for a position at distance d from the end Then we go up to height O(1+log(d)) So search for the dth position takes O(1+log(d)) time Insertions and deletions still take O(log n) worst case time but O(1+log(d)) amortized time

Back to sorting Suppose we implement the insertion sort using a finger search tree When we insert item k then d=O(Ik+1) and it takes O(1+log(Ik+1)) time Total time is bounded by O(n+n log ((I+n)/n))