# Higher-dimensional DHJ numbers

For any n, k let cn,k denote the cardinality of the largest subset of [k]n that does not contain a combinatorial line. When k=3, the quantity cn,k = cn is studied for instance in this page. The density Hales-Jewett theorem asserts that for any fixed k, $\lim_{n \to \infty} c_n / k^n = 0$.

A computer search has found the following cn values for different values of dimension n and edgelength k. Several of these values reach the upper bound of (k − 1)kn − 1.

 n\k 2 3 4 5 6 7 2 2 6 12 20 30 42 3 3 18 48 100 180 294 4 6 52 183 500 1051-1079 2058 5 10 150 712-732 2500 6325-6480 14406 6 20 450

We trivially have

cn,1 = 0 for n > 0 (and c0,0 = 1)

and Sperner's theorem tells us that

$c_{n,2} = \binom{n}{\lfloor n/2\rfloor}$.

Now we look at the opposite regime, in which n is small and k is large. We easily have

c0,k = 1

and

c1,k = k − 1;

together with the trivial bound

$c_{n+1,k} \leq k c_{n,k}$

this implies that

$c_{n,k} \leq (k-1) k^{n-1}$

for any $n \geq 1$. Let us call a pair (n,k) with n > 0 saturated if cn,k = (k − 1)kn − 1, thus there exists a line-free set with exactly one point omitted from every row and column.

Question: Which pairs (n,k) are saturated?

From the above discussion we see that (1,k) is saturated for all k >= 1, and (n,1) is (rather trivially) saturated for all n. Sperner's theorem tells us that (n,2) is saturated only for n= 1, 2. Note that if (n,k) is unsaturated then (n',k) will be unsaturated for all n' > n.

## (3,k) is saturated when k is at least 3

Let S be a latin square of side k on the symbols 1…k, with colour i in position (i,i) ( This is not possible for k=2 )

Let axis one in S correspond to coordinate 1 in [k]^3, axis two to coordinate 2 and interpret the colour in position (i,j) as the third coordinate. Delete the points so defined.

The line with three wild cards has now been removed. A line with two wildcards will be missing the point corresponding to the diagonal in S. A line with a single wildcard will be missing a point corresponding to an off diagonal point in S.

Something similar should work in higher dimensions if one can find latin cubes etc with the right diagonal properties.

## (n,k) is saturated when all prime divisors of k are at least n

First consider the case when k is prime and at least n: Delete those points whose coordinates add up to a multiple of k. Every combinatorial line has one point deleted, except for the major diagonal of n=k, which has all points deleted.

Now consider for instance the case (n,k) = (4,35). Select one value modulo 35 and eliminate it. Combinatorial lines with one, two, three or four moving coordinates will realize all values modulo 35 as one, two, three, or four are units modulo 35, thus (4,35) is saturated.

The same argument tells us that (n,k) is saturated when all prime divisors of k are at least n.

On the other hand, computer data shows that (4,4) and (4,6) are not saturated.

## (4,k)

Divide the points into five types: xyzw, xxyz, xxyy, xxxy and xxxx. Let p(xyzw) be the number of points removed whose coordinates are all different, and so on.

There are seven types of line: *xyz, *xxy, *xxx, **xy, **xx, ***x, ****. Enough points must be removed to remove all lines. That leads to the following inequalities

• $4p(xyzw)+2p(xxyz) \ge 4k(k-1)(k-2)$
• $2p(xxyz)+4p(xxyy)+3p(xxxy) \ge 12k(k-1)$
• $p(xxxy)+4p(xxxx) \ge 4k$
• $p(xxyz)+3p(xxxy) \ge 6k(k-1)$
• $2p(xxyy)+6p(xxxx) \ge 6k$
• $p(xxxx) \ge 1$

If (4,k) is saturated, then for some h between 0 and k-1 inclusive the k^3 missing points fall into the following types

• (k-1)(k-2)(k-3) - 6h of type xyzw
• 6(k-1)(k-2) + 12h of type xxyz
• 3(k-1) - 3h of type xxyy
• 4(k-1) - 4h of type xxxy
• 1 + h of type xxxx

As a result, the only combinatorial line with more than one deleted point is the major diagonal.

The saturated solutions described above for (n,k) have h=0, so the only xxxx point deleted is kkkk. When k is coprime to 6, the following saturated solution has all xxxx points deleted: the set of points (x,y,z,w) for which x+y+z+(k-3)w is a multiple of k.

## (5,k)

Any saturated coordinate-line-free solution of the five-dimensional k-cube, with k4 missing points, must have the following missing points of each type.

• p(vwxyz) = 24a + (k-1)(k-2)(k-3)(k-4)
• p(wwxyz) = -60a + 10(k-1)(k-2)(k-3)
• p(xxyyz) = 30a + 15(k-1)(k-2)
• p(xxxyz) = 20a + 10(k-1)(k-2)
• p(xxxyy) = -10a + 10(k-1)
• p(xxxxy) = -5a + 5(k-1)
• p(xxxxx) = a+1
• (a=0..k-1)

Again, the only combinatorial line with more than one missing point is the major diagonal. The number of missing points on the main diagonal (xxxxx) can be 1 or k, but then the number of each other type is fixed.

## General lower bounds

There are kn − 1 disjoint lines *abcd..m, so the density of removed points must be at least 1/k, and retained points at most (k-1)/k.

If k is prime and k ≥ n, then one can remove all combinatorial lines by deleting all points whose coordinates sum to a multiple of k. So the density of deleted points in the optimal configuration is 1/k when k is prime.

The next two bounds rely on prime numbers close to k. The following paper shows there is a prime between xx0.525 and x.

Baker, R. C.(1-BYU); Harman, G.(4-LNDHB); Pintz, J.(H-AOS)
The difference between consecutive primes. II.
Proc. London Math. Soc. (3) 83 (2001), no. 3, 532–562.


If n ≤ p ≤ k then one can get a density of (p-1)/p by deleting points whose coordinates sum to a multiple of p. The lower bound (p-1)/p approaches (k-1)/k as $k\rightarrow \infty$.

Let p be the smallest prime greater than or equal to both k and n. One can remove all combinatorial lines by deleting all points whose coordinates sum to 0 ≤ x ≤ p-k (mod p), So the density of deleted points is at most (p-k+1)/p. This approaches zero as $k\rightarrow\infty$.

I think these results can be used to get lower bounds on lines free sets for large n for all values of k. For any k and any n we can find a prime prelatively close to k^n then we remove the first k+1 values mod p then we pick a value then we remove the must k+1 so we only have k+2, 2k+3, etch the idea is to prevent any two values on a line because two points on a combinatorial line increase by at most k. This has density 1/k so we have a line free density of 1/(k+1).

I think the above bound could possibly be improved. First by getting most of the set concentrated around the point with equal numbers of ones twos and threes or the point with values closes to equality the standard deviation should be something like the square root of n. Then we could apply the near prime with sets c(k^.5 + 1) and get a density of roughly c/k^.5 which I think will be better than the Behrend-Elkin construction as e^-x will eventually be less than 1/x as x increases without limit and the square root of k will increase without limit.