Randomized Algorithms

* Why randomized algorithms?

* Two interesting algorithms

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WHY RANDOMIZED ALGORITHMS?

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Randomness is a useful resource to take advantage of in designing algorithms.

Ways randomness helps:

1. Use simple randomized algorithm to achieve same worst-case performance

as complicated deterministic algorithm. Or, sometimes have simple

deterministic heuristic with good performance "most of the time"

that can transform into an algorithm with good worst-case

performance by adding randomness.

E.g., median-finding, many standard algorithmic problems.

2. Sometimes can do things we don't know how to do deterministically.

E.g., primality testing (deterministic alg requires unproven

assumptions), linear-time min-spanning tree.

3. Sometimes can do things provably impossible deterministically

(protocol or oracle problems)

4. Sometimes can do things that just seem impossible...

This course will talk about:

* Techniques for using this resource in algorithms.

* Techniques for analyzing randomized algorithms.

* Some neat randomized algorithms.

Book is structured in this way. First half discusses important

techniques and 2nd half is a sampling of different areas where

randomized algorithms are useful. What I envision for class

presentations is that in 2nd half, there are a lot more subjects than

we can really go through. So, I'll pick some that I like, and you will

each have a chance to learn one in depth and talk about it.

Today: Some neat randomized algs. Categories 1 and (3,4).

SPLITTING A 3-COLORABLE GRAPH INTO TWO PIECES SO NEITHER HAS A TRIANGLE

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Definition of problem: You have a 3-colorable graph. Want to split

into 3 independent sets. NP-hard. Easier problem: split into two

pieces so that neither has a triangle. (If could do this for

diamonds or pentagons, then this would improve current approx ratio)

A solution: split arbitrarily. Find some bad triangle. Then pick one

of its 3 nodes AT RANDOM, and move it to the other side. Continue

until there are no more bad triangles.

* How to see as random walk. In this walk, the expected time to hit

boundary 0 or n is O(n^2) [*]. This means that expect to end in at most

that much time.

Proof of [*]:

First, let's consider 50/50 random walk --- it's easy to see proving

for this case is enough. Why? Now, in a random walk with 2m steps,

prob that make A steps to right is 2^{-2m} * (2m choose A). What is

the maximum value of this? It's maximized at A=m. Using Stirling's

formula: m! is approximately (m/e)^m * sqrt{2 pi m}, so for A=m we

have about 1/sqrt{pi m}. So, prob that after 2m steps you are between

0 and n is at most n/sqrt{pi m}, which is < 1/2 for m = 2n^2. So, with

probability at least 1/2, we end after 4n^2 steps. With prob > 1 -

1/2^k, we end after 4kn^2 steps. So, expectation is O(n^2).

More exact calculation: Let E_x be the expected time to reach 0 or n

given that you start at x. So, E_0 = E_n = 0. What is E_x in terms

of its neighbors? E_x = 1 + 1/3(E_x + E_{x+1} + E_{x-1}). Rewrite as:

E_{x+1} - E_x = E_x - E_{x-1} - 3. So ``second derivative'' of E

w.r.t. x is -3. Suggests to solve E_x = - 3/2 x^2 + cx + c' at

boundary conditions. Get: 3/2 x(n-x). Note: this is using

linearity of expectation, though of indep events.

FINDING THE AVERAGE GRADE, WITHOUT GIVING AWAY YOUR OWN SCORE

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* The problem: A group of N friends wants to determine their average

score on a test, but nobody wants to reveal their own score.

* Even if two people collaborate, they shouldn't be able to find your

score.

* More generally, if K of the N "friends" are dishonest, then they

should still learn no more than the sum of the remaining N-K honest ones.

* Possible protocols?

Well see one that works next time...

[Notes: Coloring algorithm is from: Colin McDiarmid, ``A random recouloring

method for graphs and hypergraphs.'' Combinatorics, Probability and

Computing, 2:363-365, 1993.]