Block partitions: an extended view
I. B\'ar\'any, E. Cs\'oka, Gy. K\'arolyi, and G. T\'oth

TL;DR
This paper extends the concept of partitioning sequences into blocks with nearly equal sums to higher-dimensional settings, broadening the scope of the original one-dimensional problem.
Contribution
It generalizes the existing one-dimensional block partitioning problem to higher dimensions, providing new theoretical insights.
Findings
Partitioning sequences into nearly equal blocks is extended to multi-dimensional data.
Theoretical proof of existence of such partitions in higher dimensions.
Potential applications in data segmentation and multidimensional resource allocation.
Abstract
Given a sequence , a block of is a subsequence . The size of a block is the sum of its elements. It is proved in [1] that for each positive integer , there is a partition of into blocks with for every . In this paper, we consider a generalization of the problem in higher dimensions.
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Taxonomy
Topicsgraph theory and CDMA systems · Limits and Structures in Graph Theory · Analytic Number Theory Research
Block partitions: an extended view
I. Bárány, E. Csóka, Gy. Károlyi, and G. Tóth
Abstract.
Given a sequence , a block of is a subsequence . The size of a block is the sum of its elements. It is proved in [1] that for each positive integer , there is a partition of into blocks with for every . In this paper, we consider a generalization of the problem in higher dimensions.
1. Introduction
This paper is a follow-up to [1], which is about block partitions of sequences of real numbers. A block of is either a sequence where or the empty set. The size of a block is the sum of its elements. One of the main results of [1] says the following.
Theorem 1.1**.**
Given a sequence of real numbers with for all , and an integer , there is a partition of into blocks such that for all .
The bound given here is best possible as shown by the example when all the and does not divide .
Here, we rephrase or generalize the setting and the result in the following way. Define (where ) and set . A partition of into blocks is the same as choosing indices so that is represented by , and then for all . Here denotes the set . For the more general setting, we consider closed sets satisfying the following conditions:
- (i)
, . 2. (ii)
for every and for every .
A transversal of the system is simply a selection of elements for every . Given a transversal, we define for all . In this setting, corresponds to the size of the th block.
Theorem 1.2**.**
Under the above conditions, there is a transversal such that
[TABLE]
The bound is again best possible as shown by (essentially the same) example: , for and , with an integer not divisible by . Without the closedness of the sets , we would only have a transversal with for each , as shown by the following example: let , for odd , for even .
Theorem 1.2 could be easily deduced from Theorem 1.1, but we will present a new and shorter direct proof in the next section. Then we extend the new setting to higher dimensions.
Let be the unit ball of a norm in . Let be a sequence of closed sets in . It is called a grid-like sequence if (i) and for a fixed element , and (ii) each of intersect with all unit balls, or formally,
[TABLE]
Note that in the one dimensional case we required while the above condition would translate to . So there is a factor of 2 in the new setting.
Given a transversal , we define again for and set . The goal is to find a transversal such that
[TABLE]
is as small as possible. Let
[TABLE]
for grid-like sequences and transversals . Due to the closedness of each , this minimum always exists. It is easy to see the following two propositions.
Proposition 1.3**.**
**
Proof.
If is a grid-like sequence, then is also a grid-like sequence, and the -dimensional projection of any transversal of the latter sequence is a transversal of the former sequence with at most the same diameter. ∎
Proposition 1.4**.**
, or in other words, there is always a transversal with .
Proof.
Set and choose a point from for , so with , for all . Then , and so , clearly in . ∎
One could hope that the better bound , which is valid for , also holds in higher dimensions. But this is not the case, at least with Euclidean norm:
Theorem 1.5**.**
[TABLE]
In more detail, for every and , there exists a grid-like sequence such that for any transversal , .
We have a stronger bound , see Theorem 1.6 below, which may be sharp for or maybe even for all . Its proof is based on the same ideas as that of Theorem 1.5, but it is much longer and more complicated case analysis. Therefore, we prove Theorem 1.5 and give an informal description of the construction for Theorem 1.6, but omit the proof.
Theorem 1.6**.**
.
Apart from the trivial bound given in Proposition 1.4, we cannot prove any upper bound,111Update: Endre Csóka recently claimed an unpublished upper bound using topology. not even in the case of the maximum norm . Note that the existence of a transversal with “ in the maximum norm” would imply the bound (in the Euclidean norm). For some related problems and results we refer to [2].
Remark. We may assume without any loss that . Indeed, with the previous meaning of , set . Then the system with satisfies condition (1) and it is easy to check that for the transversals and one has the same and then the diameters of and coincide. Note that .
2. Proof of Theorem 1.2
The idea is to find an such that the transversal we look for satisfies the condition
[TABLE]
With this strategy the condition is automatically guaranteed. The question is whether there is an , which admits a transversal satisfying condition (3) for every .
We analyze what happens when this strategy is followed. A partial transversal is just a selection of for , . We call it -good if it satisfies condition (3) for every . As a first step, is to be chosen from the interval , and is nonempty because of condition (ii). Thus, is an -good partial transversal for any .
It is easy to see that -good transversals exist for every and . To construct such partial transversals we define recursively as follows. Given for some and a fixed , we let
[TABLE]
A routine induction, based on condition (ii), shows that is a closed interval of length at least one for every , and so intersects if . The intervals of course depend on , and we write to express this dependence. The definition of implies that
[TABLE]
Note that both and are increasing functions of satisfying . Also, implying via an easy induction that and then tends to infinity as . A similar argument shows that as . It is also clear that if for some , then there exists an -good partial transversal .
To complete the proof of Theorem 1.2 we only have to show that there is an such that is non-empty, that is, . Actually we prove more:
Lemma 2.1**.**
For every the intervals () cover every point of : .
Proof. First we claim that is a left continuous function for every . This is evident for , so we assume that is left continuous and proceed to prove that is also left continuous. In view of (4) it amounts to checking that
[TABLE]
is a left continuous function of . Let be arbitrary; we have to show that , implies . It is clear that if for some , then is constant on the interval and for .
Otherwise there is a sequence such that
[TABLE]
Here by the left continuity of . Accordingly,
[TABLE]
because is closed. It follows that . The left continuity of at is thus established.
Similarly, is a right continuous function for every . To complete the proof of the lemma, consider any point and define
[TABLE]
Here neither nor is empty since and . The continuity and monotonicity properties of the functions and imply that both and are closed sets. Further, , as implies and implies , and then , which is impossible.
Now if two closed sets cover the connected space , then they have a point in common. So there is an , and then , so . Thus every is contained in some . ∎
Remark. The proof of Theorem 1.1, which is an algorithm of complexity can be modified to give another (algorithmic) proof of Theorem 1.2. But the above proof can be turned into an approximation algorithm the following way. By the remark at the end of the introduction we may assume that . Note that for no interval contains a positive number, and similarly, for no interval contains a negative one. Using binary search, after iteration, one finds an that is within distance of the solution.
3. Proof of Theorem 1.5
We begin with an informal description of the construction. We fix a large enough integer . The grid-like sequence consists of sets , where . Recall that in this case . Each other is the union of sets (corresponding to North, South, East, and West) plus four corners , see the figures below. The sets are symmetric about the and axes. The sets make up the first part of the construction. For , is the refection of about the line . In particular, .
The main characters in our construction are a square and segments of changing length that are either horizontal or vertical. The Minkowski sum is a hexagon shown on Figures 1 and 2. Its vertical and horizontal sides (or vertices) are drawn with heavy lines, its oblique sides with thin segments. The horizontal and vertical sides (or vertices) are extended to the sets .
The main step of the proof is to show that an optimal tranversal has no point in the corners and that it does not visit the same region ( or or or ) twice. More precisely, if and for some , then for all , and the same for the components of type .
Formally we define the various components of for as
[TABLE]
and
[TABLE]
with appropriately chosen parameters depending on . We call and the vertical and horizontal parts, and the set
[TABLE]
the union of the corners. Set first , where
[TABLE]
Here the values 4.5 and 1.5 are chosen so that , and later all other satisfy condition (ii). Next, for we define , where
[TABLE]
with . Note that is made up of four halflines and the four corners. For we let be the reflected copy of about the line . It is easy to check that is a grid-like sequence for .
Claim 3.1**.**
There is a transversal with .
Proof. Consider the transversal where
[TABLE]
for . Then is an equilateral triangle whose vertices are
[TABLE]
and . ∎
Fix an optimal transversal , such a transversal exists by compactness. Write . As , the above claim implies .
We say that jumps if is in one type of component in but is in another type in . Then is called the corresponting jump. For instance jumps if but is in or in . The important property is that is large when is a jump. Therefore the structure of the sequence of jumps is rather restricted.
Assume by symmetry that . Then and . Similarly, we may assume that . Then and .
Fact 1. For every , and . Proof: Otherwise the component of or the -component of is larger than and then , which contradicts Claim 3.1 and the optimality of . ∎
This implies that the jumps and are forbidden, and so are the jumps and , see Figure 3. In particular, and is never visited by , because all other components could jump from into or from which could jump into are too far away. Moreover, as indicated on Figure 3, there cannot be or jumps either. Indeed, suppose that there is a jump from some to . To get back to the point there must be another jump to for some . But then and , yielding , a contradiction. A similar argument applies to an jump.
We say that a pair of jumps is opposite if either (a) one is or and the other is or , or (b) one is or and the other is or .
Fact 2. There cannot be an opposite pair of jumps if is large enough. Proof: If and are opposite jumps, then and . Thus if is small enough, a contradiction.∎
Lemma 3.2**.**
There is a single jump and it goes from to for some .
Proof. There must be a jump since and . Assume that the first jump is , that is for but . As we have seen, is either in or in , and too far away.
Suppose first that , we will see that it leads to a contradiction. Note that is fairly large: and . Also, there must be a further jump, say meaning that for but . So is in or in since is too far away if is large enough. Now which cannot be reached from without creating an opposite pair: along the way there is an jump or a one. A contradiction.
We can conclude that the first jump goes from to .
If there is a further jump, then the next jump, say, must go from to or or , as is too far away. Here is excluded as then are opposite. If , then . Assume finally that . Then, again, cannot be reached from without using an opposite pair: along the way there is an jump or an one. ∎
Let be the single jump along the way. Note that and . This leads to a simple minimisation problem. Given points , with , , and , find the minimum diameter of the triangle formed by these points. As tends to 0, the unique solution to this problem converges to the equilateral triangle specified in the proof of Claim 3.1. Thus, for . This completes the proof of Theorem 1.5 for odd values of . For even values of some straightforward modifications are needed, which are left to the reader.
4. Sketch of the proof of Theorem 1.6
In this construction is a large integer again, and the grid-like sequence of sets is , , , where . Further , where come from the previous construction to guarantee that and with the previous notation .
The main characters are a rhombus (instead of the square ) and the same segments . The length of the shorter diagonal of the rhombus is , its sides are of length , and its centre is the origin. But this time rotated copies of are needed, so let stand for the rotated copy of ; rotated by angle in clockwise direction around its centre.
The first sets come from the Minkowski sum of and the previous horizontal : This sum is a hexagon again and the components are just extended from the horizontal and vertical sides of this hexagon the same way as in Theorem 1.5. The corners are at the same distance from the horizontal and vertical components as before. See Figure 4. The last sets come from the Minkowski sum of and the corresponding vertical segments analogously.
Note that are halflines, and they remain halflines in all with . The sets come from gradually rotated copies of . Then the rhombus that defines is gradually deformed to a square (of side legth ) in , which is further deformed to the rhombus in , see Figure 5. Then rotates back to in .
The proof that this construction gives is based on ideas similar to those used in Theorem 1.5: First one shows that an optimal transversal has no point in the corners, and second, that does not visit the same type component twice. We omit the details.
Acknowledgements. IB was supported by ERC Advanced Research Grant no 267165 (DISCONV) and by National Research, Development and Innovation Office NKFIH Grants K 111827 and K 116769. ECs was supported by ERC grants 306493 and 648017 and by Marie Curie Fellowship, grant No. 750857. GyK was partially supported by bilateral research grant TÉT 12 MX–1–2013–0006. GT was supported by the National Research, Develpoment and Innovation Office NKFIH Grant K-111827.
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