CSL758Instructors: Naveen Garg

Kavitha Telikepalli

Scribe: Manish SinghVaibhav Rastogi

February 7 & 11, 2008.

The Quick Sort ProblemTo sort a given set of numbers

In traditional quick sort algorithm, we pick a particular index element as pivot for splitting. Worst Case: O(n2

) Average Case: O(n log n)

A good pivot can be selected using median finding algorithm but the total complexity will again be O(n2

).So, what if we pick a random element uniformly

as pivot and do the partition. We will show that this takes expected O(n log n) time.

Randomized Quick Sort algorithmAssuming all elements are distinctWe pick a random element x as the pivot and

partition the input set S into two sets L and R such:L = numbers less than xR = numbers greater than x

Recursively sort L and R.Return LxR

Analysis of Randomized Quick SortThe running time of this algorithm is variable.Running time = # of comparisons in this algorithm.Expected Running Time, Let S be the sorted sequence of the n input

numbers.S =

Let Xij = 1 if Si and Sj are compared in the algo = 0 otherwiseRunning time = # of comparisons =

S1 S2 Si SnSn-1Sj

n

i ijijX

1

[ ] *Pr[ ]iiE X x X x

Analysis cont…The expected running time = = Now, = 1 * Pr[Si and Sj are compared in

our alg.] + 0 * Pr[Si and Sj are not compared]Suppose we have a set of numbers:

2, 7, 15, 18, 19, 23, 35In this 18 and 19 will always be compared.2 and 35 will be compared only if compared at

root.

][1

n

i ijijXE

n

i ijijXE

1

][][ ijXE

Analysis cont…Pr[Si and Sj are compared in our algo] =

Pr[the first element chosen as pivot in set {si,si+1,….,sj} is either si or sj].

To elements get compared only if they have ancestor relationship in the tree.

Pr[Picking Si or Sj] =

S1 S2 Si SnSn-1Sj

3rd pivot

1st pivot

2nd pivot

4th pivot 5th pivot

j i j i j i

Analysis cont…Thus the expected runtime:

This algorithm will always give the right answer though the running time may be different. This is an example of Las Vegas algorithms.

[ ] ( log )n n n n

iji j i i j i i j

E X O n nj i j

Since, ... lnn

n

The global min cut problemInput: A connected

undirected multigraph G(V, E).

Output: A minimum cardinality subset of edges whose removal makes G disconnected.

a b

c d

x

First IdeaRecall the max flow problem

An s-t min cut is the min cut which makes s and t disconnected.

Run a max flow algorithm on G for all (s, t) pairs and return the smallest of the output cuts.

Time:Since G is undirected we could fix s and

iterate t over all other vertices.Time:

(time for max flow) ( ) ( )n n

O mn O mn

( )(time for max flow) ( )n O mn

Reduce G to G1 with one vertex less such that1. 2.

Contract a random edge. Each edge is contracted with probability

Contract a random edge

Another idea

mincut of mincut of G G

with high probability, mincut of mincut of .G G

G G1

.n

d c

a b

d c

(a,b)

Observation 1Any cut of G1 is also a cut of G, with a and b

on the same side.If mincut of G1 < mincut of G

This mincut of G1 should be the mincut of G, by the above observation.

Thus, mincut of G1 ≥ mincut of G.

Observation 2Fix our favorite

global mincut C, with

Let e be the edge contracted.

If e is not in C thenC is also a cut in G1.Using Observation

1, mincut of G1 = mincut of G.

.C k

e

A V A

( , )C A V A

Observation 2 cont…

which is quite a high probability.

deg( ) , since is the global mincut value

deg

Pr[One of the edges is chosen as ]

mincut of ]

Pr[ is not one of the

( )

Thus, Pr[mi

ncut of

ed es ,g ]

u V

kE

u k k

E u nk

kE

k e

k kE

e

nG

n

G

The algorithmRepeat

Choose an edge of G uniformly at random and contract it. Let the resulting graph be called G.

Until there are only two vertices in G.Output the set of edges between the two

vertices as our candidate mincut.

Pr[ is ] Pr[ ]Pr[ ]Pr[ | ]Pr |Pr

]| ][

[n

n n

F C EE

E

E E EE E E E E

EE

Bounding the probability of correctnessLet C be our favorite mincut and F be the cut

output by the algorithm.Pr[F is C]=Pr[no edge of C is contracted in

any iteration].Let Ei be the event that no edge of C got

contracted in the ith iteration.

Bounding the probability of correctnessWe know that,Because C remains the mincut we can also

bound the conditional following conditional probabilities as we bounded Thus,

Pr[ ] .nn n

E

Pr( ).E

|

Pr[

[

|

Pr ]

]i i

nEn n

n iE E En i n

E

i

Bounding the probability of correctness is the product of these conditional

probabilities. So we have,

This probability bound is too low. It can be improved by repeating the algorithm N times and returning the least cut.

Pr[ is ]F C

· ·

( )Pr[ is ] n n n i

n n n i n nC

nF

(Failure Probability ) for .N nNn e

Running timeChoosing a random edge.

Choose a vertex u with probability Now choose one of u’s with probability

Time: O(n).

deg( ).um

.deg( )u

Pr[edge ( , ) is select deg( ) deg( ) .deg( ) deg( )

ed] u vu m

um

vm v

Running timeContracting edges: Use adjacency matrix.

Add the rows and columns of u and v.O(n) time.

u

u

v

v

u+v

u+v

Running timeThe basic algorithm has iterations.Each iteration in the basic algorithm involves

selecting and contracting a random edge. O(n) time.

The basic algorithm is repeated times.

Therefore running time =

n

n

( ) ( ).n n O n