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This is an excellent question but we know very little about such conditions. As Boris Bukh remarked the issue is about points in special positions, because for points in sufficiently general position, even the affine hulls of parts for every partition to r parts will have an empty intersection. However, configurations of points in special positions are of great interest in combinatorial geometry.

Let me start with an example. Suppose you have $2d+2$ points in $R^d$. This is one less than the number of points required to a Tverberg partition with three parts. Of course, one condition that guarantee a Tverberg 3-partition is that all the points (or all the points except one) belongs to a $(d-1)$ dimensional affine space. It is conjectured, more generally, that:

If the dimension of Radon points for a set of $2d+2$ points in $R^d$ is not $d$ then there is a Tevrberg partition to three parts.

The general conjecture in this direction (that I made in 1974) is: For a set $A$, denote by $T_r(A)$ those points in $R^d$ which belong to the convex hull of $r$ pairwise disjoint subsets of $latex X$. We call these points Tverberg points of order $r$.

Conjecture: For every $A \subset R^d$ , $$\sum_{r=1}^{|A|} {\rm dim} T_r(A) \ge 0.$$

(Note that $\dim \emptyset = -1$.)

Thus, if you have a set of points so that the dimension of (r-1)-Tverberg points is below what can be expected in the generic case, then there is a nonempty Tverberg partition to r parts. There are various strengthening and weakening of this conjecture. It was proved by Kadari (unpublished except his M Sc thesis in Hebrew from the early 90s) for the plane.

There are few more things that can be said:

1) While the computational complexity of finding a Tverberg 3-partition for $2d+3$ points is unknown, the computational complexity of finding a Tverberg 3 partition for less points is NP-hard. As observed by Shmuel Onn 3-colorability of cubic graphs reduces to finding such a Tverberg partition.

2) It will be interesting to come with a topological strengthening of the above conjecture.

3) The conjecture about $2d+2$ points in $R^d$ motivated and is related to the graph-theoretic conjecture (which turned out to be false) in this Overflow question.

4) The first section of my paper Combinatorics with a Geometric Flavor gives more information and connections.

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This is an excellent question but we know very little about such conditions. As Boris Bukh remarked the issue is about points in special positions, because for points in sufficiently general position, even the affine hulls of parts for every partition to r parts will have an empty intersection. However, configurations of points in special positions are of great interest in combinatorial geometry.

Let me start with an example. Suppose you have $2d+2$ points in $R^d$. This is one less than the number of points required to a Tverberg partition with three parts. Of course, one condition that guarantee a Tverberg 3-partition is that all the points (or all the points except one) belongs to a $(d-1)$ dimensional affine space. It is conjectures thatconjectured, more generally, if that:

If the dimension of Radon points for a set of $2d+2$ points in $R^d$ is not $d$ then there is a Tevrberg partition to three parts.

The general conjecture in this direction (that I made in 1974) is: For a set $A$, denote by $T_r(A)$ those points in $R^d$ which belong to the convex hull of $r$ pairwise disjoint subsets of $latex X$. We call these points Tverberg points of order $r$.

Conjecture: For every $A \subset R^d$ , $$\sum_{r=1}^{|A|} {\rm dim} T_r(A) \ge 0.$$

(Note that $\dim \emptyset = -1$.)

Thus, if you have a set of points so that the dimension of (r-1)-Tverberg points is below what can be expected in the generic case, then there is a nonempty Tverberg partition to r parts. There are various strengthening and weakening of this conjecture. It was proved by Kadari for the plane.

There are few more things that can be said:

1) While the computational complexity of finding a Tverberg 3-partition for $2d+3$ points is unknown, the computational complexity of finding a Tverberg 3 partition for less points is NP-hard. As observed by Shmuel Onn 3-colorability of cubic graphs reduces to finding such a Tverberg partition.

2) It will be interesting to come with a topological strengthening of the above conjecture.

3) The conjecture about $2d+2$ points in $R^d$ motivated and is related to the graph-theoretic conjecture in this Overflow question.

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Let me start with an example. Suppose you have $2d+2$ points in $R^d$. This is one less than the number of points required to a Tverberg partition with three parts. Of course, one condition that guarantee a Tverberg 3-partition is that all the points (or all the points except one) belongs to a $(d-1)$ dimensional affine space. It is conjectures that, more generally, if the dimension of Radon points is not $d$ then there is a Tevrberg partition to three parts.

The general conjecture in this direction (that I made in 1974) is: For a set $A$, denote by $T_r(A)$ those points in $R^d$ which belong to the convex hull of $r$ pairwise disjoint subsets of $latex X$. We call these points Tverberg points of order $r$.

Conjecture: For every $A \subset R^d$ , $$\sum_{r=1}^{|A|} {\rm dim} T_r(A) \ge 0.$$

(Note that $\dim \emptyset = -1$.)

Thus, if you have a set of points so that the dimension of (r-1)-Tverberg points is below what can be expected in the generic case, then there is a nonempty Tverberg partition to r parts. There are various stengthening strengthening and weakening of this conjecture. It was proved by Kadari for the plane.

There are few more things that can be said:

1) While the computational complexity of finding a Tverberg 3-partition for $2d+3$ points is unknown, the computational complexity of finding a Tverberg 3 partition for less points is NP-hard. As observed by Shmuel Onn 3-colorability of cubic graphs reduces to finding such a Tverberg partition.

2) It will be interesting to come with a topological strengthening of the above conjecture.

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