The Rectangle Covering number of Random Boolean Matrices
Dirk Oliver Theis
University of Tartu Estonia
Mozhgan Pourmoradnasseri
University of Tartu Estonia
Estonian Computer Science Theory Day October 4, 2015
The Rectangle Covering number of Random Boolean Matrices Mozhgan - - PowerPoint PPT Presentation
The Rectangle Covering number of Random Boolean Matrices Mozhgan Pourmoradnasseri University of Tartu Estonia Dirk Oliver Theis University of Tartu Estonia Estonian Computer Science Theory Day October 4, 2015 What this talk is about
Estonian Computer Science Theory Day October 4, 2015
◮ Rectangle Covering Number, rc(M), of a 01-matrix M
◮ Rectangle Covering Number, rc(M), of a 01-matrix M ◮
(entries Bernoulli w/ parameter p)
⊠
⊠(M)
◮ Rectangle Covering Number, rc(M), of a 01-matrix M ◮
(entries Bernoulli w/ parameter p)
⊠
⊠(M)
◮ Rectangle Covering Number, rc(M), of a 01-matrix M ◮
(entries Bernoulli w/ parameter p)
⊠
⊠(M)
◮ Bound rc(Mn,p) = χ(Gn,p
⊠ )
◮ Rectangle Covering Number, rc(M), of a 01-matrix M ◮
(entries Bernoulli w/ parameter p)
⊠
⊠(M)
◮ Bound rc(Mn,p) = χ(Gn,p
⊠ )
◮ Bound other parameters related to χ,
⊠(M)
⊠
⊠(M)
⊠
⊠(M)
⊠(M)
⊠(M)
⊠(M)
⊠(M): 1-entries
⊠(M)
⊠(M)
⊠(M): 1-entries
⊠(M):
⊠(M)
⊠(M)
⊠(M): 1-entries
⊠(M):
⊠(M)
⊠(M)
⊠(M): 1-entries
⊠(M):
⊠(M)
⊠(M)
⊠(M): 1-entries
⊠(M):
⊠(M)
⊠(M)
⊠(M): 1-entries
⊠(M):
⊠(M)
⊠(M)
⊠(M): 1-entries
⊠(M):
⊠(M):
⊠(M):
⊠(M):
⊠(M))
⊠(M)
⊠(M))
⊠(M)
◮ ∀G:
⊠(1 − Adj(G))
1 . . . 1 . . . ... . . . 1 . . . 1 all-1 matrix of appropriate dimensions.
◮ This talk is not concerned with the Log-Rank conjecture.
⊠
⊠
⊠(M)
⊠(M))
⊠
⊠(M)
⊠(M))
⊠
⊠(M)
⊠(M))
⊠
H
M= Adj(H)
rectangle which contains a 0 1 A B D A 1 C E B C 1 F D E F 1 At least one of the As has to be 0. Same for B,C,D,E,F,. . .
G
⊠(M)
⊠(M))
⊠
H
M= Adj(H)
rectangle which contains a 0 1 A B D A 1 C E B C 1 F D E F 1 At least one of the As has to be 0. Same for B,C,D,E,F,. . .
G
⊠(M)
⊠(M))
⊠(M)
⊠
k,ℓ =
⊠
⊠
k,ℓ =
⊠
⊠
◮ ω, α, χ known when p = Θ(1) (or p = 1/2) ◮ Not when p = o(1) or p = 1 − o(1).
⊠
⊠
◮ Binomial random variable Bin(n2, p) ◮ nice concentration near pn2
(We couldn’t see anything interesting happening outside of that range.)
⊠
⊠
◮ Mean: E M = (1 + o(1))(1 − p2)
2
◮ No very good concentration. . . ◮ . . . particularly when p = 1 − o(1)
⊠
⊠
◮ Mean: E M = (1 + o(1))(1 − p2)
2
◮ No very good concentration. . . ◮ . . . particularly when p = 1 − o(1)
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 1 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
⊠ )
⊠ )
⊠ )
⊠ )
⊠ )
◮ For p = Ω(1), √n is large compared to E ω = O(ln n) ◮ For p = 1 − o(1), √n will be large compared to
⊠(M)
⊠
⊠ ):
(with ν := size of largest matching in the bipartite graph with adj. mtx. M)
⊠ ):
(with ν := size of largest matching in the bipartite graph with adj. mtx. M)
⊠ ):
(with ν := size of largest matching in the bipartite graph with adj. mtx. M)
⊠ ):
(with ν := size of largest matching in the bipartite graph with adj. mtx. M)
(recall δ := (1 − p)2 =density of Gn,p
⊠
)
⊠ ):
(with ν := size of largest matching in the bipartite graph with adj. mtx. M)
(recall δ := (1 − p)2 =density of Gn,p
⊠
)
◮ The constant in (b) is probably not optimal. ◮ In (c), how does ω decrease?
⊠(M)
⊠
⊠
(or vice versa).
⊠
(or vice versa).
◮ With p = 1 − λ/n, λ = o(n), λ ≥ C:
⊠
⊠(M)
⊠
◮ n/2 ≤ χ ≤ n. ◮ More careful look: χ = (1−o(1)) n = N
(Recall N := |V (Gn,p
⊠
)| = n
◮ n/2 ≤ χ ≤ n. ◮ More careful look: χ = (1−o(1)) n = N
(Recall N := |V (Gn,p
⊠
)| = n
◮ n/2 ≤ χ ≤ n. ◮ More careful look: χ = (1−o(1)) n = N
(Recall N := |V (Gn,p
⊠
)| = n
⊠ ) = O
⊠ )
folklore fact
◮
Take probability distribution µ on the independent sets
◮
v∈V
P
R∼µ(v ∈ R)
◮
µ
1 c(µ) =: χ∗
folklore fact
◮
Take probability distribution µ on the independent sets
◮
v∈V
P
R∼µ(v ∈ R)
◮
µ
1 c(µ) =: χ∗
⊠ ) = O
⊠ )
⊠ ) = O
⊠ )
⊠ ) = o
⊠ )
⊠ )= O
⊠ )
⊠ ) = (1−o(1)) eλ = χ∗(Gn,p ⊠ )
(1−o(1)) eλ ≤ χ(Gn,p
⊠ ) = O(ln λ · ln n).
⊠ ) = o
⊠ )
⊠ ) = o
⊠ )
⊠ ) = Θ(u(n)), or even = (1−o(1)) u(n)?
⊠ ) = o
⊠ )
⊠ ) = Θ(u(n)), or even = (1−o(1)) u(n)?
⊠ ) for Ω(1/√n) = p = o(1)!
⊠ ) = o
⊠ )
⊠ ) = Θ(u(n)), or even = (1−o(1)) u(n)?
⊠ ) for Ω(1/√n) = p = o(1)!
⊠ ) = o
⊠ )
⊠ ) = Θ(u(n)), or even = (1−o(1)) u(n)?
⊠ ) for Ω(1/√n) = p = o(1)!
⊠ ) = o
⊠ )
⊠ ) = Θ(u(n)), or even = (1−o(1)) u(n)?
⊠ ) for Ω(1/√n) = p = o(1)!