1 Introduction
In a game Battleship, two players first secretly place a family of ships (often rectangles) on a domain (an integer grid), and then they aim to locate the opponent’s ships by querying individual grid cells. Inspired by the game, we consider the problem of finding a sparse shooting pattern: a subset of the grid cells that is guaranteed to hit at least one cell of each ship, no matter how the ships are translated within the domain. See Figure 1 for an example and Section 1.3 for a formal problem statement.
This problem is surprisingly intricate. In this note, we make two simplifying assumptions.
Infinite domains.
First, in order to avoid boundary effects, we assume the domain is an infinite grid . Since any shooting pattern on an infinite domain is also infinite, we measure its quality using the (asymptotic) density, refer to Figure 2. Note that the problem is subtle; for instance, for two shaped triominoes as ships, the lowest possible density of a shooting pattern depends on the relative orientation of the ships (see Figure 3).
Let and be families of two Lshaped triominoes from Fig. 3. Then and .
Proof.
The shooting patterns shown in Fig. 3 imply and . For the matching lower bound is trivial, since , where consists of a single Lshaped triomino. It remains to prove . Split the plane into infinite vertical slabs of width 2. We argue for each slab separately. Fix a row. If neither cell is shot, then in the row just above it, both cells must be shot. Hence the overall density is at least . ∎
One dimension.
Second, for the remainder of this note we focus on the case , which is far from trivial if we consider not only connected ships, but arbitrary finite subsets of integers (see Section 1.3). In Appendix C, we describe how our results extend to higher dimensions. Thus, our problem is: Given a family of ships in (or in ), find the minimum density of a shooting pattern that hits each translate of each ship .
1.1 Related work
We say that a family of ships in is of type if it consists of ships with sizes , respectively. Previous work has studied families consisting of a single (not necessarily connected) ship, that is, families of type for some . It is easy to see that for any family with . In 2008, Schmidt and Tuller conjectured a formula for for any family [9], but as of now its validity is still open.
Given this difficulty, other works studied the toughest instances of a given type. Formally, given a type , let be the smallest density that suffices to hit any family of type . Already in 1967, Newman [8] showed that (one toughest instance is the ship in Fig. 4(a)) and that as . Recently, it was shown that [3]. Also, given a ship , the density can be found using a “sliding window” algorithm [4]. The algorithm can be used to establish lower bounds such as .
To our knowledge, the problem for multiple ships has not been studied; however, we point out a similar work in the continuous setting for rectangles in two dimensions [6].
1.2 Our contribution
We propose to study the problem for multiple ships or, equivalently, for a single ship of uncertain shape. (Note that by considering suitable families, we can model mirrored or reflected ships on top of translated ships, cf. Fig. 3.) Apart from Lemma 1 above, we present results in 1D (see Appendix C for extensions to 2D). First, we note that the sliding window algorithm of [4] can be adapted to families of multiple ships in a straightforward way, see Fig. 4. We implement the algorithm and use it to obtain lower bounds such as (due to e.g. ). As our main contribution, we present three results for families of ships of small size (ships). First, for any family of two 2ships, we find an explicit formula for , see Section 2. Second, we determine , that is, we identify the toughest instances for families consisting of any number of 2ships, see Section 2. Third, we determine the toughest instances for families consisting of any 3ship together with its reflection, see Section 2. Finally, in Appendix B we present general bounds for the density of the toughest instances of ships of size each.
1.3 Preliminaries
A ship of size (a ship) is a tuple with . A span of a ship is . See Fig. 4 for an illustration. A finite family of ships has a span . A shooting pattern is a sequence . We say that a shooting pattern hits a ship (or that is a shooting pattern for ) if . The density of a shooting pattern is defined as The density of a ship and of a family of ships is then defined as
Further, we define and . That is, is the required density for the simplest instances, whereas is the required density for the toughest instances, among families that consist of ships of size each. Regarding , it is straightforward to prove (even when no two ships in the family are translates of each other). Regarding , in Appendix B we show when . We also list two auxiliary results (with proofs in Appendix A). [Sliding window algorithm] Given a family with span , the density can be computed in time polynomial in .
Finally, for an integer and a ship let , and likewise for a family let . Let be a positive integer and any family. Then .
2 Families of 2ships and 3ships
Here we study families that consist of ships of small size each. Note that when , we obviously have (for all ). Also, for a single 2ship it is straightforward to show that . Our first nontrivial result is an explicit formula for when consists of two 2ships.
[Formula for two 2ships] Let , , coprime and .
Proof.
Let . By Lemma 4, it suffices to determine . Clearly, . When both and are odd, a shooting pattern defined by “ if and only if is even” provides a matching construction. From now on, assume that precisely one of , is odd (that is, is odd).
Split into blocks of consecutive integers. Let be one such block and let be a shooting pattern for on (instead of on ). We will argue that needs to be hit at least times (that is, ). Consider a graph with nodes and directed edges . This graph records the constraints on the shooting pattern: For every edge , we must have . Since , each node in has indegree 1 and outdegree 1. Moreover, since and are coprime, the graph is connected. Hence it is a directed cycle on an odd number of nodes. Its minimum vertex cover has size , thus .
To prove that this bound is tight, consider any vertex cover of of the minimum size . Then the periodic shooting pattern defined by “ if and only if ” hits on : Indeed, consider any translate of the ship . Suppose for some . If then and both belong to the same block, thus , since is a vertex cover. On the other hand, if then by the periodicity of the shooting pattern we have . Since , both and belong to the same block, so we conclude as before. For translates of the ship we argue analogously. ∎
As a corollary, we have , as witnessed by families for any .
Next, we study the toughest instances in two other cases, namely for any number of ships, and for a 3ship together with its reflection. [Toughest families of 2ships] For any we have .
Proof.
For the claim is trivial. For it follows from Section 2. Assume .
First, note that for a family we have : Indeed, split into blocks of consecutive integers. Then any shooting pattern for may miss at most 1 number from each block. On the other hand, the pattern defined by “ if and only if is a multiple of ” hits .
To prove the upper bound, consider any positive integers , and the corresponding family . We construct a shooting pattern for with density at most . We proceed in steps. Initially, we set for all with . Then, we process integers in increasing order. Whenever is not yet set, we set and for each . (Note that some of might have already been set to 1, due to some .) By construction, hits all translates of within the interval . Moreover, since for every set to there are at most values newly set to 1, in the limit we obtain . Similarly, we process integers in decreasing order and get . Together with the finite initial segment this gives .∎
Given a ship , let be its reflection.
[Toughest 3ship with its reflection] Let be a 3ship and let . Then , with equality if and only if (or their reflections) for some .
Proof.
First, we argue that for a single 3ship , the toughest instance has a density of ; that is, . This fact was first proven by Newman [8]. Here, we present a geometric proof, which we then extend to the case of two symmetric 3ships.
Consider a 3ship for positive integers and with and . We arrange the integers into a 2dimensional grid by the bijection . Refer to Figure 5(a). Most translations of now correspond to an Ltriomino with the same orientation; therefore, we can hit all of them with a shooting pattern with density using the same solution as in Figure 2. However, this misses exactly the translations by an amount that is congruent to ; those translations correspond to a triomino that “wraps around” (Figure 5(d)). To hit these it is sufficient to increase the density of the first column to . This gives , which is strictly less than for . For the remaining cases with , we find an optimal solution by Fig. 4. The toughest instances, yielding , turn out to be as claimed.
We note that Theorems 2, 2 and 2 can be generalized to higher dimensions. Notes on these extensions can be found in Appendix C.
3 Conclusions
We introduced the problem of locating a battleship of an uncertain shape. Given the difficulty of the problem in general, we focused on the simplest possible setting, namely ships of size 2 or 3 in 1D (see Appendix C for extensions to 2D). We also implemented an algorithm for computing in 1D and used it to compute lower bounds on the minimum density required for the toughest families of ships of size each, see Fig. 6. Many open problems arise, e.g.:

Which values in Fig. 6 are tight? For instance, is it true that ?

What are the asymptotics of for fixed small ?

In 2D, is there an algorithm for computing ?
References
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Appendix A Proof of Fig. 4 and Lemma 4
Proof of Fig. 4.
This is a direct extension of [4, Remark 5.4].
Briefly, the algorithm finds the minimum mean cycle in an auxiliary graph whose nodes are 01sequences of length that are valid shooting patterns for , and where a directed edge connects sequences such that the last entries of coincide with the first entries of .
Proof of Lemma 4. Any shooting pattern for can be scaled by a factor of and used times (once for each remainder mod ) to give a shooting pattern for with the same density. (Formally, we set for any and .) And vice versa, starting with a shooting pattern for , for each , consider the pattern defined by . Then each hits and the sparsest of them has density at most .
Appendix B Proof of bounds for and
Here we state and prove our general bounds on the required densities and for the simplest and toughest instances among families consisting of ships of size each.
Theorem.
Let and be integers. Then and
Proof.
We start with the proof for . First, we prove that for any single ship . Let be the span of , take large and consider a block of length . Then contains translates of . Since any shot in hits at most translates of in , there must be at least shots in . Thus the density of shots within is at least .
On the other hand, for any the family
can be hit by a shooting pattern that has density .
Regarding , we show the lower bound first. The idea is to consider a block of a suitable length and define as the set of all ships of size which fit into this block and contain its first cell. Then every tuple within is either a ship of or a translate of a ship of . Hence, in order to hit we have to hit at least times, which implies .
Now we calculate a suitable . Note that we must make sure that and . Using it suffices to have
Setting and using a standard bound for , we get
thus
For the upper bound, the expression is inherited from Section 2. The second expression comes from calculations that N.Alon [1] did to compute an upper bound on minimal transversals of hypergraphs.
Namely, we take a period and we mark one cell of the ship with an anchor. Obviously, we can define a bijection between the translates of a ship modulo and by identifying each translate with point where the anchor of is positioned. Now, we construct a digraph with vertices and add a directed edge with color from point to point if a mark at point hits the translate of with anchor at point .
Note that a feasible shooting pattern corresponds to a directed dominating set in each monochromatic subgraph induced by of . Further, all vertices in have exactly incoming and outgoing edges of the same color. Now, we choose a random set from and for each , we add the set that consist of elements where an anchor of a translate of that is not hit by is located. Completely analogous calculations as in the proof of Theorem 1.2.2 in [2] then yield the desired upper bound of . ∎
Note that the proof also works for higher dimensions.
Appendix C Notes on generalizations to higher dimensions
In this section, we show how to generalize Theorems 2, 2, and 2 to two dimensions. Interestingly, it turns out that all three results are actually easier in ; intuitively this is because for small ships, either the ships are all in a onedimensional subspace, or they can be decomposed into smaller components for which the reasoning is simpler. We believe that for twodimensional ships of size , or for at least three twodimensional ships of size , the situation in 2D does become more complex.
First, we generalize Theorem 2.
[Formula for two 2ships in 2D] Let , for .

If and are linearly independent, then .

If and for some and , coprime and odd, then .

Otherwise, .
Proof.
A 2ship in is of the form . So, we can think of it as the vector . Now suppose we have two two ships, represented by two vectors and .
If and are linearly independent, then the optimal density is 1/2. To see this, consider the subgrid of cells of the form for integers and . Any shooting pattern needs to hit half of the cells of this grid (either the cells where is odd or where is even). Since we can tile the plane with independent copies of this subgrid, we have . The other inequality is trivial.
If and are not independent, then we are in the 1D case: let be the vector such that and for coprime integers and , then we can tile the plane with copies of the space generated by and apply the existing 1D result (Section 2) to each copy.
∎
Second, we generalize Theorem 2. [Toughest families of 2ships in 2D] Let for . Then .
Proof.
First, we divide the plane into the 4 quadrants and mark the cells along the axes by ’s such that the patterns in the quadrants are independent of each other. These areas have measure 0, so the density depends only on the density of the patterns within the individual quadrants.
Now we create a shooting pattern for the first quadrant with the same greedy algorithm as in the case. We anchor the ships at their leftmost cell (if there a two leftmost cells at the lowest leftmost cell) which will be our reference point such that they are contained in a box. Then we move the anchored cells along the plane as follows: Start at the leftmost lowest unmarked cell and go up cells. Now, go from to . Next to , then to and to and so on. If the anchored parts are at , we mark all cells where unanchored parts of the ships are located. By our moving pattern, we avoid marking any checked ’s. Since we add at most ’s at a , the density of the pattern is at most . Similarly, we can create patterns for all other quadrants.
∎
Third, we generalize Theorem 2. [Toughest 3ship with its reflection in 2D] Given , let be a 3ship in 2D, let its reflection, and set .

If and are linearly independent, then .

Otherwise .
Proof.
If and are linearly independent then, similarly to Appendix C, consider the lattice generated by and . Within the lattice , the ship is an Lshaped triomino, and is its rotation by , thus, as in Lemma 1, the subset is a shooting pattern for . Since is a disjoint union of copies of , using this shooting pattern for each copy of we get the desired .
The second claim follows immediately from Section 2. ∎