Balancing pairs and the cross product conjecture

@article{Brightwell1995BalancingPA,
  title={Balancing pairs and the cross product conjecture},
  author={Graham R. Brightwell and Stefan Felsner and William T. Trotter},
  journal={Order},
  year={1995},
  volume={12},
  pages={327-349}
}
In a finite partially ordered set, Prob (x>y) denotes the proportion of linear extensions in which elementx appears above elementy. In 1969, S. S. Kislitsyn conjectured that in every finite poset which is not a chain, there exists a pair (x,y) for which 1/3⩽Prob(x>y)⩽2/3. In 1984, J. Kahn and M. Saks showed that there exists a pair (x,y) with 3/11y)<8/11, but the full 1/3–2/3 conjecture remains open and has been listed among ORDER's featured unsolved problems for more than 10 years.In this… Expand
On the 1/3–2/3 Conjecture
TLDR
The1/3–2/3 Conjecture states that every finite partially ordered set which is not a chain has a 1/3-balanced pair, and is made progress on this conjecture by showing that it holds for certain families of posets. Expand
A combinatorial approach to height sequences in finite partially ordered sets
TLDR
This paper provides a purely combinatorial proof of two important special cases of Stanley's theorem by applying Daykin's inequality to an appropriately defined distributive lattice and proves a somewhat stronger result that may be possible to analyze the error terms when the log-concavity bound is not tight. Expand
Balanced pairs in partial orders
TLDR
The1/3–2/3 Conjecture states that, in every finite partial order P, not a chain, there is a 1/3-balanced pair. Expand
Sorting under partial information (without the ellipsoid algorithm)
TLDR
This work revisits the well-known problem of sorting under partial information, and develops efficient algorithms that approximate the entropy, or make sure it is computed only once, in a restricted class of graphs, permitting the use of a simpler algorithm. Expand
Exact exponential algorithms for two poset problems
  • L. Kozma
  • Computer Science, Mathematics
  • SWAT
  • 2020
Partially ordered sets (posets) are fundamental combinatorial objects with important applications in computer science. Perhaps the most natural algorithmic task, given a size-$n$ poset, is to computeExpand
Counting Linear Extensions of a Partial Order
A partially ordered set (P,<) is a set P together with an irreflexive, transitive relation. A linear extension of (P,<) is a relation (P,≺) such that (1) for all a, b ∈ P either a ≺ b or a = b or b ≺Expand
Sorting under partial information (without the ellipsoid algorithm)
TLDR
This work revisits the well-known problem of sorting under partial information and develops efficient algorithms for sorting underpartial information that approximate the entropy, or make sure it is computed only once in a restricted class of graphs, permitting the use of a simpler algorithm. Expand
Log-Concave Functions And Poset Probabilities
TLDR
These results are mainly based on the Brunn–Minkowski Theorem and a theorem of Keith Ball, which allow us to reduce to a 2-dimensional version of the problem. Expand
Efficient computation of rank probabilities in posets
As the title of this work indicates, the central theme in this work is the computation of rank probabilities of posets. Since the probability space consists of the set of all linear extensions of aExpand
On Entropy and Extensions of Posets
A vast body of literature in combinatorics and computer science aims at understanding the structural properties of a poset P implied by placing certain marginal constraints on the uniformExpand
...
1
2
3
...

References

SHOWING 1-10 OF 31 REFERENCES
Balancing Pairs in Partially Ordered Sets
J. Kahn and M. Saks proved that if P is a partially ordered set and is not a chain, then there exists a pair x; y 2 P so that the number of linear extensions of P with x less than y is at least 3=11Expand
Semiorders and the 1/3–2/3 conjecture
A well-known conjecture of Fredman is that, for every finite partially ordered set (X, <) which is not a chain, there is a pair of elements x, y such that P(x<y), the proportion of linear extensionsExpand
Linear extensions of infinite posets
TLDR
It is shown how to extend the idea of the probability P(x<y|R), namely the proportion of linear extensions of R, to a certain class of infinite posets, and shows that fundamental correlation inequalities for finite posets hold in the infinite case. Expand
A correlational inequality for linear extensions of a poset
AbstractSuppose 1, 2, and 3 are pairwise incomparable points in a poset onn≥3 points. LetN (ijk) be the number of linear extensions of the poset in whichi precedesj andj precedesk. Define λ byExpand
Balancing poset extensions
We show that any finite partially ordered setP (not a total order) contains a pair of elementsx andy such that the proportion of linear extensions ofP in whichx lies belowy is between 3/11 and 8/11.Expand
How Good is the Information Theory Bound in Sorting?
  • M. Fredman
  • Computer Science, Mathematics
  • Theor. Comput. Sci.
  • 1976
TLDR
If X and Y are n element sets of real numbers, then the n 2 element set X + Y can be sorted with O ( n 2 ) comparisons, improving upon the n2 log 2 n bound established by Harper et al. Expand
The 1/3-2/3 Conjecture for 5-Thin Posets
TLDR
The 1/3–2/3 conjecture is established in the case where every element of the poser is incomparable with at most five others, and the proof involves the use of a computer to eliminate a large number of cases. Expand
The FKG Inequality and Some Monotonicity Properties of Partial Orders
  • L. Shepp
  • Computer Science, Mathematics
  • SIAM J. Algebraic Discret. Methods
  • 1980
TLDR
A simple example is given to show that the more general inequality (*) where P is allowed to contain inequalities of the form $a_i < b_j $ is false, which is surprising because as Graham, Yao, and Yao proved, the general inequality(*) does hold if P totally orders both the a's and the b’s separately. Expand
An inequality for the weights of two families of sets, their unions and intersections
then ~(A) fi(B) < 7(A v B) cS(A A B) for all A, B ~ S, (2) where e(A) = ~(a~A) e(a) and A v B = {awb; aeA, b~B} and A A B = {ac~b; a~A, b~B}. Since every distributive lattice can be embedded in theExpand
The Information-Theoretic Bound is Good for Merging
  • N. Linial
  • Mathematics, Computer Science
  • SIAM J. Comput.
  • 1984
TLDR
There exists an algorithm which will take no more than C\log _2 N comparisons where C = (\log_2 ((\sqrt 5 + 1)/ 2))^{ -1} $ and the constant C is best possible. Expand
...
1
2
3
4
...