Exploiting a hypergraph model for finding Golomb rulers

Abstract

Golomb rulers are special rulers where for any two marks it holds that the distance between them is unique. They find applications in radio frequency selection, radio astronomy, data encryption, communication networks, and bioinformatics. An important subproblem in constructing “compact” Golomb rulers is Golomb Subruler  (GSR), which asks whether it is possible to make a given ruler Golomb by removing at most $k$ marks. We initiate a study of GSR from a parameterized complexity perspective. In particular, we consider a natural hypergraph characterization of rulers and investigate the construction and structure of the corresponding hypergraphs. We exploit their properties to derive polynomial-time data reduction rules that reduce a given instance of GSR to an equivalent one with ${{\mathrm{O}}}(k^3)$ marks. Finally, we complement a recent computational complexity study of GSR by providing a simplified reduction that shows NP-hardness even when all integers are bounded by a polynomial in the input length.

Appendix: Further forbidden subgraphs

Proposition 3

(Forbidden subgraph “large hand”) The graph shown in Fig. 2c is a forbidden subgraph in a conflict hypergraph.

Proof

In a 4-conflict there are two pairs of vertices that measure the same distance. We choose one unordered pair from $\{a, b, c\}$ and, thus, define the distance that caused the conflict. Then, for the fourth mark, there are only two possible positions left. Multiplying this with the number of possible unordered pairs, one gets six as an upper bound for such edges intersecting in three marks.

In order to prove an upper bound of three, we show that in the previous argument every edge is actually counted twice. Assume $a < b$ has been chosen as pair. Then a fourth mark $d$ can assume only two values, given by

$$\begin{aligned} d = c - (b - a) \quad \text {or} \quad d = c + (b - a)\text {.} \end{aligned}$$

We can rewrite these conditions as

$$\begin{aligned} d = a + (c - b) \quad \text {or} \quad d = b + (c - a)\text {.} \end{aligned}$$

Now observe that the conditions correspond also to the case that the chosen pair is $b < c$ or $a < c$, respectively. That is, every possible location for $d$ is counted twice. This means that there are at most three 4-conflicts that intersect in three marks. $\square$

Proposition 4

(Forbidden subgraph “rotor”) The graph shown in Fig. 2d is a forbidden subgraph in a conflict hypergraph.

Proof

First, by definition of the rotor graph, $a, c, d$ are distinct. We fix a total ordering of the three marks in $\{a, c, d\}$ and then try to position $b$ in that ordering. We find that all possible locations lead to equality of two of the marks $a, c, d$, a contradiction. Because of the symmetry of the graph we can look at one specific ordering without loss of generality. Hence, let $a < c < d$. Now assume that $b < c$. Because of the conflict $\{b, c, d\}$, mark $c$ is half-way between $b$ and $d$. The conflict $\{a, b, d\}$ implies that either $a = c$ (a contradiction) or $a < b$. But in the latter case, because $a, b$ are in one conflict with $c$ and in one with $d$, we have $c = d$ which again is a contradiction. The case $b > c$ is symmetric. $\square$

Proposition 5

(Forbidden induced subgraph “scissors”) The graph shown in Fig. 2e is a forbidden induced subgraph in a conflict hypergraph.

Proof

We show that, in the configuration shown in Fig. 2e, an edge comprising $d_1, d_2$ and one mark $m \in \{a, b, c\}$ must also be present.

We again use the fact that 4-conflicts are due to two pairs of them having the same distances. Choose two pairs from $\{a, b, c\}$ corresponding to the two conflicts and hence defining a distance each conflict arises from. If the chosen pairs comprise the same marks, then the proposition holds: If $a, b$ is the pair measuring the same distance in both conflicts, then

$$\begin{aligned} |a - b| = |c - d_1| \quad \text {and} \quad |a - b| = |c - d_2|\text {,} \end{aligned}$$

and, hence, $\{c, d_1, d_2\}$ is a conflict. The cases that $a, c$, or $b, c$ are chosen in both conflicts are similar. If the two chosen pairs are not equal, then the pairs must share one mark. Without loss of generality, let the pairs be $a < b$ and $b < c$. The equations

$$\begin{aligned} d = c \pm (b - a) \quad \text {and}\quad e = a \pm (c - b) \end{aligned}$$

hold for appropriate choices of $+$ or $-$ instead of $\pm$. Note that the sign before $(b - a)$ cannot be negative at the same time with the sign of $(c - b)$ being positive. Otherwise this would imply that $d = e$. In any other case, the two terms on the right-hand side of the equations differ only in the sign of exactly two variables. This means that there exists an $m \in \{a, b, c\}$ such that the following equation holds:

$$\begin{aligned} |e - m| = |m - d|\text {.} \end{aligned}$$

Thus there is an additional conflict $\{d, e, m\}$. $\square$