Here's an apparently elementary problem which has a rich history and has produced some really interesting combinatorial mathematics. Define FF(k) to be the largest integer t for which there exists an interval graph G and a linear order L on the vertices of G so that: (1) the maximum clique size of G is k, and (2) First Fit uses t colors on G when the vertices are processed in the order specified by L. In 1976, Woodall showed that FF(k) = O(k log k). In 1988, Kierstead showed that FF(k) <= 40k, i.e., FF(k) is linear in k. Kierstead and Qin improved this in 1995 to FF(k) <= 25.8k. From below, it follows from our work on on-line coloring of interval graphs that there exists an e > 0 so that FF(k) > (3 + e)k when k is large. In 1990, Chrobak and Slusarek proved that FF(k) > 4k - 10 when k > 6, and they later improved the lower bound to 4.4k.
In 2003, Pemmaraju, Raman and Varadarajan made a major breakthrough by showing that FF(k) <= 10k and commented that their upper bound might be improved but that the technique wouldn't yield a result better than 8k. Later in 2003, their predictions were confirmed, and their technique was refined by Brightwell, Kierstead and Trotter to obtain an upper bound of 8k. In 2004, Narayansamy and Babu found an even cleaner argument for this bound that actually yields the slightly stronger result: FF(k) <= 8k - 3.
Later in 2004, Kierstead and Trotter improved the lower bound by showing that for every e > 0, FF(k) > (5 - e)k when k is sufficiently large. Most recently, Kierstead and Trotter have shown that the tree-like structures used by various authors to obtain lower bounds cannot yield a better performance ratio than 5. So as k goes to infinity, the ratio FF(k)/k tends to a limit that is somewhere between 5 and 8, and I will bet that 5 is the right answer.
Now we want to tackle the log-concave results developed
by Stanley using the Alexandrov/Fenchel inequalities
for mixed volumes: For a fixed element x, the height
sequence of x is log-concave. To date, we have only some partial
success to report. Specifically, we have been able
to prove the result when x is a fixed point and
there are exactly two point incomparable with x. In this case, the height sequence of x contains only three
non-zero terms and we are able to show that the product of
the middle term is at least as large as the product of
the other two terms.
Hamiltonian Path and Cycle Problems
In 1973, I conjectured that every finite partially ordered set with
three or more points contains a pair whose removal decreases the
dimension by at most one. Papers by Hiraguchi, Bogart, Reuter
and me contain many examples of sufficient conditions which
guarantee the existence of such a pair. In making the
original conjecture, I made the stronger assertion that
the removal of any critical pair decreased
the dimension by at most 1. But Reuter constructed a 4-dimensional
counterexample on 10 points. Subsequently, Kierstead and I generalized
this counterexample and constructed an infinite sequence of
counterexamples for this stronger assertion for
all dimensions starting with 4. Still it is reasonable
to believe that (1) there is some critical pair whose removal
decreases the dimension by at most 1; and (2) for every x, there is some other point y so that the removal of x and y decreases the dimension by at most
1.
The "1/3 - 2/3" Conjecture
Every finite partially ordered set
which is not a chain contains an incomparable pair (x,y) so that 1/3 <= Prob(x < y) <= 2/3. Kahn and
Saks proved that there is always a pair for which 3/11 < Prob(x < y) < 8/11. Brightwell, Felsner and I improved these bounds by showing
that there exists a pair (x,y) with (5-5^{.5})/10 <= Prob(x < y) <= (5+5^{.5})/10.
In a certain sense, this result is best possible, since it
is tight for countably infinite posets. The BFT proof
makes use of the Alexandrov/Fenchel inequalities and
the machinery developed by Kahn and Saks. An interesting
correlation inequality called the "Cross Product Inequality"
is conjectured in BFT, and a special case is proven.
This case is enough for the inequality on balancing pairs. In a
recent popular lecture titled The Top Ten Theorems on Partially Ordered Sets,
I listed this result at the very top and credited it to the comprehensive list:
Ahlswede, Alexandrov, Brightwell, Felsner, Fenchel, Fishburn, Kahn, Saks,
Stanley and Trotter.
Geometric
Inclusion Orders
Felsner, Fishburn and I showed
that exists a finite 3-dimensional partially ordered
set which cannot be represented as the inclusion order of a family of
spheres in d-dimensional Euclidean space, regardless of how large d is. This proof solves a long standing problem first posed
by Fishburn and me (for circles) in 1984 at the Banff Conference
on Ordered Sets. Lately we have been looking at other geometric objects.
Although every 3-dimensional poset is an ellipse order, in fact we may
require all the major axes to be co-linear. On the other hand, we
conjecture that not all 4-dimensional posets are ellipse orders.
In a similar vein, every n-dimensional poset has an inclusion
representation using cubes (in general position) in euclidean 2n-space.
This is best possible for n = 1 and n = 2, but already for
n = 3, we don't know whether every 7 - dimensional poset has an inclusion
representation using cubes in 3-space.
Distances in the Euclidean Plane
Paul Erdos posed the problem of determining (or estimating)
the functions u(n) and d(n) so that a set of n points (a) has at
most u(n) pairs at unit distance, and (b) determines at least d(n) distinct distances.
In the 1980's, Spencer, Szemeredi and I proved that u(n) < c n^{4/3}, while Chung, Szemeredi and I showed d(n) > n^{4/5}/log^c n. In the last year, Szekely has used
crossing arguments to dramatically lower the constant in the
bound for u(n) and to remove the log term
from the denominator in the bound for d(n). Szekeley is also able to show that this many distances
occur from a single point. But we are still far away from
the answer conjectured by Erdos: n points determine at
most n^{1+c/log log n} unit distances and
at least n/log^.5 n distinct distances.
On-Line Graph Coloring
For many years, I have enjoyed a long standing
and very productive research collaboration with my
colleague Hal Kierstead. The next two
problems represent topics on which we have worked
together - and in fact he knows more about these
subjects than I do. The first problems is estimate the
function f(k,n), defined as the least number of colors
required to color on-line a k-chromatic graph. Alon showed f(3,n) is
at least log^2 n, and Saks, Lovasz and I showed f(3,n) is at
most n/log log n. In a major breakthrough in
the last year, Kierstead has shown
that f(3,n) = O(n ^2/3 / log^2/3 n). Kierstead's
argument gives a sublinear bound for f(k,n).
Still, the lower bounds are considerably weaker.
On-Line Chain Partitioning
Estimate the function (n),
defined as the least number of chains
required to partition on-line any partially ordered set of width n. Kierstead showed that
(n) <= (5^{n+1} - 1)/4, and Szemeredi showed
that f(n) >= n(n+1)/2. In particular settle, whether f(n) is polynomial
in n. Recently, Felsner settled the n = 2 case, by
showing that a poset of width 2 can be partitioned into 5
chains. This is already a complicated result, and no progress has been
made on the general problem for the last 15 years.
The Dimension of Cartesian Products
f P and Q are posets, then the dimension of the
cartesian product P x Q s at most dim P + dim Q. Baker showed that equality holds if P and Q have
(distinct) greatest and least elements. But what about
the lower bound? Trivially, dim P x Q is at least
the maximum of dim P and dim Q. In 1984, I showed that
for every n, m t least 3, there exists an n-dimensional
poset P and an m-dimensional poset Q so that dim P x Q = n + m - 2. These posets are taken from
the so called "standard examples." But the question is
whether these examples tell the whole story. It is
conceivable that there is an n-dimensional poset P for which P x P is also n-dimensional. Why not?
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