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On the number of matroids

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### On the number of matroids

Nikhil Bansal (TU Eindhoven)

Rudi Pendavingh (TU Eindhoven)

Jorn van der Pol (TU Eindhoven)

Matroids

Matroid(U,C): U = Universe [n], C: collection of independent sets

- Subset closed: I independent, then also .
- Exchange: |I|>|I’| then some s.t also independent.
How does a typical matroid look ?

Can we generate themrandomly?

How many matroids m(n) on n elements?

Clearly, m(n)

Knuth’74: m(n)

(explicit construction of a class called sparse paving matroids)

i.e. log log m(n) [n – 3/2 log n – O(1), n]

Narrowing the gap

Why bother about this tiny 3/2 log n gap?

log log scale a bit deceptive.

x vs.

Conjecture: Most matroids are sparse paving (various versions)

Knuth’s bound perhaps close to optimal

Often counting ->

Sampling and generating matroids.

m(n)

Better bound

Known results

Knuth 74

Easy Upper bound:

Pf: Any rank r matroid is specified by its bases (max. indep. sets)

Number of rank r matroidsm(n,r) .

So,

m(n) (n+1) (can just focus on rank r=n/2)

Piff 73: ]

Our Result

Thm: log log m(n) Knuth’s lower bound + 1+o(1) .

(Piff had ½ log n gap)

Knuth 74: m

We show: m

Need only the most basic matroid facts.

Main Tool: Bounding number of stable sets in a graph.

Outline

- Knuth’s lower bound construction
- Counting stable sets
- The final upper bound

Knuth’s Lower bound

Matroid of rank r can be specified by r-sets that are non-bases.

Johnson Graph: J(n,r) V: r-subsets of [n] |V|=.

Edge (u,v): if

Fact: If non-bases form a stable set in J(n,r), then get a matroid.

These are called sparse paving matroids.

(various nice properties)

Knuth’s bound

Sparse Paving Matroids : precisely the stable sets of J(n,r).

For graph G: = size of max stable set.

i(G) = # stable sets. Note: .

J(n,r) is a regular graph of degree d = for .

So,

Knuth:

Proof: Color vertex by j if

Gives proper n-coloring.

Rest of the talk

Goal: Show log log m(n) Knuth’s lower bound + 1+o(1)

(Necessary) first step: Show this for sparse paving matroids

log log s(n) Knuth’s lower bound + 1+o(1)

( same as bounding i(G) for J(n,r))

The ideas developed there will be useful for bounding m(n).

Bounding s(n)

Claim: Max stable set in J(n,n/2) (2/n) N N = # of vertices

Fact: If is smallest eigenvalue of adj. matrix of a d-regular graph.

Then, (proof later)

Johnson graphs: for J(n/n/2) So,

Naïve bound: i(G) + + … +

≈

Recall, knuth Lower’s bound:

Niavely: i(J(n,n/2)) (note: base of exponent)

Better Bound

Refined bound: i(J(n,n/2))

Morally: All independent sets are subsets of few large independent sets.

Examples: n-Hypercube vertices

Naïve bound on i(G) =

Right answer: [Saphozhenko’83] (1+o(1)

Entropy Method [Kahn’01]: Any d-regular bipartite graph

Tight: Disjoint copies of (n/2d copies)

Holds even for general graphs [Zhao’10]

Our Result

Thm: In any d-regular graph G with min eigenvalue

i(G)

Idea: Encode an independent set using few bits of information.

Eg: Bipartite graphs: Our bound:

Our approach closest to Alon, Balogh, Morris, Samotij[arxiv’12]

(their bound not useful for our purposes)

This encoding idea is later used to encode matroids.

A useful lemma

G: d-regular with min eigenvalue . For any vertex subset A.

2

Proof:

Split

Corollary:

Corollary: If |A| + N, then G[A] has a vertex of degree

(For random set A of size , expected degree

A

Encoding a stable set

Associate to an independent set I of G the pair of vertices (S,A), s.t.

- |A|
- A is completely determined by S.
I can be uniquely specified by (S, ).

Key Point: A is completely determined by S.

Number of possibilities for I = (gives the result)

S

I

A

G

Encoding a stable set

Input: Independent set I.

Initialize: A = V and S =

While |A| >

Let v = largest degree vertex in G[A] (ties in lex order)

If v in I, add to S and set

Else discard v and set A = A\{v}

.

Encoding a stable set

Input: Independent set I.

Initialize: A = V and S =

While |A| >

Let v = largest degree vertex in G[A] (ties in lex order)

If v in I, add to S and set

Else discard v and set A = A\{v}

.

Encoding a stable set

Input: Independent set I.

Initialize: A = V and S =

While |A| >

Let v = largest degree vertex in G[A] (ties in lex order)

If v in I, add to S and set

Else discard v and set A = A\{v}

.

Encoding a stable set

Input: Independent set I.

Initialize: A = V and S =

While |A| >

Let v = largest degree vertex in G[A] (ties in lex order)

If v in I, add to S and set

Else discard v and set A = A\{v}

Observe: S completely determines A.

Encoding a stable set

Claim:

Pf: Alg in phase j if ]

A vertex in phase jhas at least j neighbors in G[A].

Must pick vertices in S. Sum over j=d,…1.

Phases:

d

…

d-1

1

2

Encoding Matroids

Matroid can be specified by listing r-sets that are non-bases.

Want a more compact representation (fewer bits).

Idea: r-set Y is dependent iff some X s.t. || > rk(X)

(i.e. X acts as a witness that Y contains a dependency).

E.g. X=Y trivially works. But not very efficient.

Want small witness set { (X,rk(X)) } that works for all Y.

Key Lemma: For a dependent r-set X, we can associate sets that are witnesses for all non-bases Y in .

If rank(X) < r-1. Witness = (X,rk(X))

Key Lemma: For a dependent r-set X, we can associate sets that are witnesses for all non-bases Y in .

If rank(X) = r-1. Then X has a unique circuit C.

Witness = (Cl(X), C)

Cl(X) : closure all z s.t. rank(X U z) = rank(X)

Proof:

(as Y=X-x+y)

Case 1: If rk(X+y) = r-1, then

= |Y|> rk(Cl(X))

Case 2: rk(X-x) r-2 < |X-x|

So X-x contains a circuit C’. But C’=C by uniqueness.

So (as Y=X-x+y). Hence || = |C| > rk(C)

Finish up

Given a matroid, let K = set of non-bases.

Apply stable set procedure to K obtain (S, A) with

- S and A small as before, S determined by A.
Encoding: {witness of } for each

List

1) Witness for all non-bases in

2) List of remaining non-bases.

Questions

Would be nice to reduce the gap to o(1).

The reason for +1+o(1) gap

Do not understand the size of max. stable set in J(n,n/2)

N/n (explicit construction) vs. 2N/n (eigenvalue methods)

Studied a lot in coding. Simulations suggest closer to N/n

Perhaps a new method for certifying that

would also bound m(n).

Narrowing the gap

Why bother about this tiny 3/2 log n gap?

log log scale a bit deceptive.

x vs.

Conjectures (various quantitative versions):

Most matroids are sparse paving. s(n): sparse paving matroids

- m_n/s_n \rightarrow 1
- log logm_n = log logs_n + o(1)
- log logm_n = log logs_n + O(log log n)
Perhaps counting -> Sampling and generating matroids.

Encoding Matroids

If rank(X) = r-1. Then X has a unique circuit C.

Witness = (Cl(X), C)

Cl(X) : closure all z s.t. rank(X U z) = rank(X)

Proof this witness works:

(as Y=X-x+y)

Case 1: rk(X+y) = r-1, but then then

Case 2: rk(X-x) r-2 < |X-x|

So X-x contains a circuit C’. But then C=C’ by uniqueness.

So Y=X-x+y contains C.

Finish up

Given a matroid, let K = set of non-bases.

Apply stable set procedure to K obtain (S, A) with

- S and A small as before, S determined by A.
For a non-basis X, we have a witness for non-bases in the neighborhood of X.

Encoding: (X, witness of X) for each

List

Knuth’s Lower bound

Matroid of rank r can be specified by r-sets that are non-bases.

Johnson Graph: J(n,r) V: r-subsets of [n] |V|=.

Edge (u,v): if

Claim: If non-bases form a stable set in J(n,r), then get a matroid.

These are called sparse paving matroids.

Proof: Base Exchange: M is matroidiff for every two bases B,B’ and e B\B’ there exists f in B’ such that B-e+f is base.

B

not a base

and not a base

…

…

B

B’

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