The polynomial Hirsch conjecture: Difference between revisions

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One approach to this problem is purely combinatorial.  It is known that this conjecture follows from
One approach to this problem is purely combinatorial.  It is known that this conjecture follows from


: '''Combinatorial polynomial Hirsch conjecture''': Consider t disjoint families of subsets <math>F_1,\ldots,F_t</math> of <math>\{1,\ldots,n\}</math>.  Suppose that
: '''Combinatorial polynomial Hirsch conjecture''': Consider t non-empty families of subsets <math>F_1,\ldots,F_t</math> of <math>\{1,\ldots,n\}</math> that are disjoint (i.e. no set S can belong to two of the families <math>F_i, F_j</math>).  Suppose that
:: For every <math>i < j < k</math>, and every <math>S \in F_i</math> and <math>T \in F_k</math>, there exists <math>R \in F_j</math> such that <math>S \cap T \subset R</math>.   
:: For every <math>i < j < k</math>, and every <math>S \in F_i</math> and <math>T \in F_k</math>, there exists <math>R \in F_j</math> such that <math>S \cap T \subset R</math>.  (*)
: Let f(n) be the largest value of t for which this is possible.
: Let f(n) be the largest value of t for which this is possible.
: Conjecture: f(n) is of polynomial size in n.
: Conjecture: f(n) is of polynomial size in n.


== Threads ==
== Threads ==
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What are the best upper and lower bounds on f(n) for small n?
What are the best upper and lower bounds on f(n) for small n?
Trivially, f(n) is non-decreasing in n, f(0)=1 (take <math>\{ \emptyset\}</math>), f(1)=2 (take e.g. <math>\{\emptyset\}, \{\{1\}\}</math>), and f(2)=4 (take e.g. <math> \{ \emptyset\}, \{ \{1 \}\}, \{ \{1,2\}\}, \{2\}</math>).
: <bf>Theorem 1</bf> For any <math>n > 1</math>, <math>f(n) \leq f(n-1) + 2 f(\lfloor n/2\rfloor)<math>.
'''Proof''' Consider t families <math>F_1,\ldots,F_t \subset \{1,\ldots,n\}</math> obeying (*).  Consider the largest s so that the union of all sets in <math>F_1 \cup \ldots \cup F_s</math> is at most n/2.  Clearly, <math>0 \leq s \leq f(\lfloor n/2\rfloor)</math>.
Consider the largest r so that the union <math>\bigcup (F_{n-r+1} \cup \ldots \cup F_n)</math> of all sets in <math>F_{n-r+1} \cup \ldots \cup F_n</math> is at most n/2.  Clearly, <math>0 \leq r \leq f(\lfloor n/2\rfloor)</math>.
If <math>t \leq s+r</math> then we are done, so suppose that <math>t > s+r</math>.  By construction, the sets <math>\bigcup(F_1 \cup \ldots \cup F_{s+1})</math> and <math>\bigcup(F_{n-r} \cup \ldots \cup F_n)</math> both have cardinality more than <math>n/2</math> and thus have a common element, say m.  By (*), each of the <math>t-r-s</math> families <math>F_{s+1},\ldots,F_{n-r}</math> must thus contain a set with this element m.  If we then throw away all the sets that don't contain m, and then delete m, we obtain t-r-s non-empty families of an n-1-element set that obey (*), hence <math>t-r-s \leq f(n-1)</math>, and the claim follows.  QED
Note: the same argument gives <math>f(n) \leq f(n-1) + f(a) + f(b)</math> for any positive integers a, b with <math>a+b+1 \geq n</math>.  In particular we have the slight refinement
: <math>f(n) \leq f(n-1) + f(\lfloor n/2\rfloor) + f(\lfloor (n-1)/2\rfloor).<math>





Revision as of 09:08, 30 September 2010

The polynomial Hirsch conjecture (or polynomial diameter conjecture) states the following:

Polynomial Diameter Conjecture: Let G be the graph of a d-polytope with n facets. Then the diameter of G is bounded above by a polynomial of d and n.

One approach to this problem is purely combinatorial. It is known that this conjecture follows from

Combinatorial polynomial Hirsch conjecture: Consider t non-empty families of subsets [math]\displaystyle{ F_1,\ldots,F_t }[/math] of [math]\displaystyle{ \{1,\ldots,n\} }[/math] that are disjoint (i.e. no set S can belong to two of the families [math]\displaystyle{ F_i, F_j }[/math]). Suppose that
For every [math]\displaystyle{ i \lt j \lt k }[/math], and every [math]\displaystyle{ S \in F_i }[/math] and [math]\displaystyle{ T \in F_k }[/math], there exists [math]\displaystyle{ R \in F_j }[/math] such that [math]\displaystyle{ S \cap T \subset R }[/math]. (*)
Let f(n) be the largest value of t for which this is possible.
Conjecture: f(n) is of polynomial size in n.

Threads

Possible strategies

Partial results and remarks

In the setting of the combinatorial Hirsch conjecture, one can obtain the bound [math]\displaystyle{ f(n) \leq n^{\log n + 1} }[/math] by an elementary argument, given here.

In [EHRR] it is noted that f(n) is at least quadratic in n.

What are the best upper and lower bounds on f(n) for small n?

Trivially, f(n) is non-decreasing in n, f(0)=1 (take [math]\displaystyle{ \{ \emptyset\} }[/math]), f(1)=2 (take e.g. [math]\displaystyle{ \{\emptyset\}, \{\{1\}\} }[/math]), and f(2)=4 (take e.g. [math]\displaystyle{ \{ \emptyset\}, \{ \{1 \}\}, \{ \{1,2\}\}, \{2\} }[/math]).

<bf>Theorem 1</bf> For any [math]\displaystyle{ n \gt 1 }[/math], [math]\displaystyle{ f(n) \leq f(n-1) + 2 f(\lfloor n/2\rfloor)\lt math\gt . '''Proof''' Consider t families \lt math\gt F_1,\ldots,F_t \subset \{1,\ldots,n\} }[/math] obeying (*). Consider the largest s so that the union of all sets in [math]\displaystyle{ F_1 \cup \ldots \cup F_s }[/math] is at most n/2. Clearly, [math]\displaystyle{ 0 \leq s \leq f(\lfloor n/2\rfloor) }[/math].

Consider the largest r so that the union [math]\displaystyle{ \bigcup (F_{n-r+1} \cup \ldots \cup F_n) }[/math] of all sets in [math]\displaystyle{ F_{n-r+1} \cup \ldots \cup F_n }[/math] is at most n/2. Clearly, [math]\displaystyle{ 0 \leq r \leq f(\lfloor n/2\rfloor) }[/math].

If [math]\displaystyle{ t \leq s+r }[/math] then we are done, so suppose that [math]\displaystyle{ t \gt s+r }[/math]. By construction, the sets [math]\displaystyle{ \bigcup(F_1 \cup \ldots \cup F_{s+1}) }[/math] and [math]\displaystyle{ \bigcup(F_{n-r} \cup \ldots \cup F_n) }[/math] both have cardinality more than [math]\displaystyle{ n/2 }[/math] and thus have a common element, say m. By (*), each of the [math]\displaystyle{ t-r-s }[/math] families [math]\displaystyle{ F_{s+1},\ldots,F_{n-r} }[/math] must thus contain a set with this element m. If we then throw away all the sets that don't contain m, and then delete m, we obtain t-r-s non-empty families of an n-1-element set that obey (*), hence [math]\displaystyle{ t-r-s \leq f(n-1) }[/math], and the claim follows. QED

Note: the same argument gives [math]\displaystyle{ f(n) \leq f(n-1) + f(a) + f(b) }[/math] for any positive integers a, b with [math]\displaystyle{ a+b+1 \geq n }[/math]. In particular we have the slight refinement

<math>f(n) \leq f(n-1) + f(\lfloor n/2\rfloor) + f(\lfloor (n-1)/2\rfloor).<math>


Background and terminology

(Maybe some history of the Hirsch conjecture here?)

The disproof of the Hirsch conjecture

The Hirsch conjecture: The graph of a d-polytope with n facets has diameter at most n-d.

This conjecture was recently disproven by Francisco Santos [S].

Bibliography

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