Download Threshold in N(n,p)

2/7/2007
Every monotone property has a threshold
This was a review of the previous session’s material.
∀ε > 0, ∃ integer m s.t. p(1-ε)≤mp(ε)
Suduko is a practical example of the existence of thresholds. Namely, the fact that in order to create an
interesting Suduko puzzle one has to erase enough numbers of the solution in order to keep the solution
unique and possible. There is a threshold on the number of integers that can be erased and the puzzle is
still solvable.
2/7/2007
Consider: Np=N(n,p) where N={1,2,…,n}
We are interested in the threshold for when there is an arithmetic progression of length k. Where the
arithmetic progression looks as:
a, a+b, a+2b, a+3b; all of which are elements of the progression set.
Let Xk be the number of arithmetic progressions of length k:
E(Xk)=n2pk
This is justified by the fact that the total number of expected arithmetic progressions is n2, and that
there are k integers each with the probability.
−2
If p= n k , then E(Xk)=n2n-2≈1.
−2
If p<< n k , then lim E(Xk)=0, meaning that almost surely N(n,p) does not have an arithmetic progression
n→∞
−2
of length k. If p>> n k , lim E(Xk)= ∞ .
n→∞
Apply Second Moment Method:
E
lim
n →∞
(( X ) ) ≤ 1
2
k
E2 ( Xk )
Let Ii be the indicator variable for the I-th arithmetic progression.
Xk=I1+I2+…+In^2
E
(( X ) ) =
2
k
E
2
(Xk )
?
n p 2k
4
If the indicators were statistically independent then we should know:
E ( Ii I j ) = E ( Ii )E ( I j )
Let’s say:
E
(( X ) ) = E ( I I ) = E ( I ) E ( I ) = En( IpI )
2
k
i
i
j
i
j
4
j
2k
If i=j, or if Ii and Ij overlap, then they are not independent.
What if Ii and Ij overlap in one element:
Number of such pairs: k 2 (kn ) = k 2 n 3
Expected number of such pairs: k 2 n 3 p 2 k −1
Overlap in 2 elements:
Number of such pairs: k 2 n 2
Expected number of such pairs: k 2 n 2 p 2 k −2
2/7/2007
At this point we take the limit for the second moment method:
n 2 p k + n 4 p 2 k + n 3 p 2 k −1 + n 2 p 2 k −2
1
1
1
lim
= lim 2 k + 1 +
+ 2 2 =1
4 2k
n →∞
n→∞
n p
n p
np n p
−2
If p>> n k , since k≥3 then p=n-c << 1.
Hence by second moment method, almost surely it has an arithmetic progression.
Introduction to 3-CNF:
Formula f is a Boolean formula constructed of clauses with 3 literals.
f=(x1+x2+x3)(x1+x2’+x4)…( )
If fixed number of variables, and there are few clauses, then there is almost surely satisfiability.