The Probabilistic Method in Computer Science

The Probabilistic Method in Computer Science

Combinatorics in Computer Science

Combinatorics in Computer Science

Paul Erdós (1913-1996)

Mathematical Achievements• ~1500 papers

• ~500 co-authors

• Invented new branches of mathematics

• Solved a LARGE number of beautiful problems

• Posed a HUGE number of beautiful problems

Some Facts

• Devoted his entire life to Math:• never had a woman• never had an house• never had an office• never had possessions (except for

a small suitcase with silk shirts and underwears necessary for his skin)

• never had much money (once he won 50000$ and he kept just 720$ for himself)

Some Other Facts• Took Amphetamines to be

able to sleep 4 hours per day (clearly spending the remaining 20 hours doing math)• Waking you up at 4 o’clock: “Is your brain open?”

• “Another roof, another proof”• Prizes for problems, for a total of

15000$: What if all problems are solved at once?

Number Theory• Branch of math studying integer

numbers and, in particular, prime numbers• First Erdós success: at 17 years, a proof “from the Book” that there is always a prime number between n and 2n (Bertrand’s postulate)

• Prime Number Theorem (simple proof by Erdós and Selberg): The number of primes up to x approaches x/log(x) as x->∞.

Number Theory

• Exercise: Prove that prime numbers are infinite

• Exercise: Design a method for generating all prime numbers

Number Theory

• Question 1: Does a closed formula exists generating (all and) only prime numbers?

Number Theory

• Question 2: Does an infinite number of friendly numbers exist?• Two numbers are friendly if each

one is the sum of the divisors of the other one.

• 284=1+2+4+5+10+11+20+22+44+55+110

• 220=1+2+4+71+142• Divisori(284)={1,2,4,71,142}• Divisori(220)={1,2,4,5,10,11,20,22,

44, 55,110}

Number Theory

• Question 3: Does an odd perfect number exists?• A perfect number is such that the

sum of its divisors is equal to itself• 6=1+2+3• 28=1+2+4+7+14• ….

Erdós Vocabulary• Supreme Fascist -> God• Epsilon ->

Children• Bosses -> Women• Slaves -> Men• Poison -> Alcohol• Preaching -> Lecturing• Dying -> Stop doing Math• Leaving -> Dying

Happy Ending Problem• Esther Klein:

• Is it true that, for all n, there exists an integer g(n) such that any set of g(n) points in general position contains a convex n-gon?

Happy Ending Problem• g(4)=5

Happy Ending Problem

• Erdós and Szekeres:

)(12 2 ngn ( 42 n2n )

• Szekeres and Klein got married

Ramsey Theory

• About unavoidable occurrences of patterns in a large instance of the problem• In a party, every two persons either know each other or they don’t

• Is it true that in a party with a sufficiently large number r(n) of persons, there are always n persons pairwise knowing each other or n persons pairwise not knowing each other?

Ramsey Theory

• Exercise: r(3)=?

• Exercise: translate the problem into a graph problem

hint: (if two persons know each other, then the graph has an edge…)

• Exercise: translate the problem into a graph coloring problem hint: (use complete graphs…)

Ramsey Theory

• Actual bounds for r(n)

n

cn

nnrne

log2

1

222

( 42 n2n )

Ramsey Theory

• Erdós LB:• There are ways to bicolor ( m2 )2 mK

( m2 )2( mn ) ( n2 ) 1

nK

nK

( m2 )2 ( m2 )2( mn ) ( n2 ) 1

• There are at most colorings with a monochromatic

• Therefore, a coloring with no monochromatic exists if

Ramsey Theory

• Exercise: construct a sequence of the first 100 integers containing no subsequence of 11 decreasing numbers and no subsequence of 11 increasing numbers• Exercise: construct a sequence of the first 101 integers containing no subsequence of 11 decreasing numbers and no subsequence of 11 increasing numbers

Ramsey Theory

• Every sequence of n2 +1 distinct numbers contains a decreasing sequence of n+1 numbers or an increasing sequence of n+1 numbers

Ramsey Theory• Proof: • Consider any sequence S of n2 +1

distinct numbers, and associate to the i-th number of S a pair (ai,bi) such that ai and bi are the sizes of the longest increasing and decreasing subsequences of S till the i-th number.

• Then (ai,bi)≠(ak,bk)

• Hence, an > n or bn > n

Ramsey Theory• The largest value of k such that r(k) is

known is…

• Erdós, aliens, and Ramsey numbers…

Erdós skills• Never done laundry• Never payed bills• Never cooked (“I can make very good

cold cereals and I could probably boil an egg but I never tried”)

• Never drived• János Pach: “I entered the kitchen and

saw pools of blood-like red liquid trailing to the refrigerator, where there was a tomato juice carton with a large hole on its side. Erdós must have been thirsty…”

Random Graph Theory

• Exercise: design a method to construct a random graph with n vertices

• Exercise: design a method to construct a random graph with n vertices and m edges

Random Graph Theory• Start with G having n vertices and no

edge. At each step choose a random edge among those not belonging to G. Stop when G has m edges.

• For a given p, 0 ≤ p ≤ 1, each potential edge is chosen with probability p, independent of the other edges. The obtained graph is Gn,p .

Random Graph Theory• Random Graphs are useful for:

• Analyzing deterministic algorithms• Designing random solutions

Random Graph Theory• Which is the right value of p?

• p=0?• p=1?• p=1/2?• p=1/2n?• p=1/n?

Random Graph Theory• As G acquires more and more edges,

various properties and substructures emerge. The problem of interest is to study the (sudden) appearance of graph properties as p varies.

• A random graph has property A, if the probability that Gn,p has property A approaches to 1, as n approaches infinity.

Random Graph Theory• Erdós-Rényi phases• Phase 1:• Gn,p is the disjoint union of trees on k

vertices• Trees on k vertices appear when

n

op1

1

1

k

k

n

p

Random Graph Theory• Phase 2:• Gn,p contains cycles of any given size

with probability tending to a positive limit

• Almost all vertices are in trees or in connected components with a single cycle

• The largest component is a tree with Θ(log n) vertices

10, cn

cp

Random Graph Theory• Phase 3: the double jump • The behavior of Gn,p when is

dramatically different from when • Most of the vertices are into a giant

component which has Θ(n) vertices• All other vertices are in trees or in

connected components with a single cycle. Each small component has O(log n) vertices.

nnp

1

np

1

np

1

Random Graph Theory• Phase 4:• All components other than the giant one

are very small and are trees

1, cn

cp

• Phase 5:• There is one connected component

1,log

cn

ncp

• Phase 6:• There is one connected component and

the degrees of all vertices are asymptotically equal.

nwn

nnwp ,

log

Erdós Number• Erdós had so many co-authors

that they started classifying mathematicians by their Erdós number:• Erdós has Erdós number 0• Its co-authors have Erdós number 1• The co-authors of its co-authors

have Erdós number 2• …• Erdós number is either ≤7 or ∞

The Probabilistic Method• Trying to prove that a structure with

certain desired properties exists, one defines an appropriate space of structures and then shows that the desired property holds in this space with positive probability.

The Probabilistic Method

• Consider a random 2-coloring of .• For any set R of k vertices, let AR be the

event that R is monochromatic.• Then,• Since there are choices for R, the

probability that one of events AR occurs is at most

22)(

k

kr

nK

12)Pr( RAk( 2)

( )nk

12 ( )( )nkk2

The Probabilistic Method• If , then 12 ( )( )nk

k2

22

k

n

!2!2

!

2

2

!!

!

k

kkkn

n

2

1

2

2

!

11kkk

knnn

2

2

2

2

22

!

11k

k

k

knnn 1

2!

2

2

12

2

k

kk

n

k

22)(

k

kr• Hence,

The Probabilistic Method• Tournament T: orientation of the

edges of Kn. Vertices are players. An edge (u,v) means that a player u beats a player v.

• T has property Sk if for every set K of k players there is a player not in K that beats all players in K.

• Is it true that for every k there is a tournament T with property Sk ?

The Probabilistic Method• Erdós solution:• If ,then there is a

tournament on n players which has property Sk

• Take a random tournament on Kn=(V,E).

• For any fixed set K of k players, let AK be the event that no player in V-K beats them all.

121 knk( )nk

The Probabilistic Method• Then, , as for every

player v in V-K, the probability that v does not beat all players in K is and all the n-k events corresponding to the different choices of v are independent. It follows that:

• Therefore, with positive probability there is a tournament on n players with property Sk

kn

kkA

2

11)Pr(

k2

11

)Pr( ||, kkKVK A )Pr(||, kkKVK A 12

11

kn

k( )nk

Finally I’m becoming stupider no more