In this section we prove that Connected Dominating Set in Unit Disk Graphs is \({\mathsf {W[1]}}\)-hard parameterized by the solution size; our proof heavily relies on the proof of the \({\mathsf {W[1]}}\)-hardness of Dominating Set in Unit Disk Graphs by Marx [9].
Sketch of the Construction by Marx for Dominating Set in Unit Disk Graphs
Marx uses a reduction from Grid Tiling, see the book [13] (note that in [9] the Grid Tiling problem is not stated explicitly). In a Grid Tiling instance we are given an integer k, an integer n, and a collection \({\mathcal {S}}\) of \(k^2\) non-empty sets \(U_{a,b} \subseteq [n]\times [n]\) for \(1 \leqslant a,b \leqslant k\). The goal is to select an element \(u_{a,b}\in U_{a,b}\) for each \(1 \leqslant a,b \leqslant k\) such that
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If \(u_{a,b}=(x,y)\) and \(u_{a+1,b}=(x',y')\), then \(x=x'\).
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If \(u_{a,b}=(x,y)\) and \(u_{a,b+1}=(x',x')\), then \(y=y'\).
One can picture these sets in a \(k\times k\) matrix: in each cell (a, b), we need to select a representative from the set \(U_{a,b}\) so that the representatives selected from horizontally neighboring cells agree in the first coordinate, and representatives from vertically neighboring sets agree in the second coordinate (see Fig. 1).
Marx’s reduction places \(k^2\) gadgets, one for each \(U_{a,b}\). A gadget contains 16 blocks of disks, labeled \(X_1,Y_1,X_2,Y_2,\ldots ,X_8,Y_8\), that are arranged along the edges of a square—see Fig. 2a. Initially, each block \(X_{\ell }\) contains \(n^2\) disks, denoted by \(X_\ell (1),\ldots , X_\ell (n^2)\) and each block \(Y_{\ell }\) contains \(n^2+1\) disks denoted by \(Y_\ell (0),\ldots , Y_\ell (n^2)\). The argument j of \(X_\ell (j)\) can be thought of as a pair (x, y) with \(1\leqslant x,y\leqslant n\) for which \(f(x,y):=(x-1)n+y=j\). Let \(f^{-1}(j)=\left( \iota _1(j),\iota _2(j)\right) = \left( 1+\lfloor j/n\rfloor ,1+(j \mod n)\right) \). For the final construction, in each gadget at position (a, b), delete all disks \(X_\ell (j)\) for each \(\ell =1,\ldots ,8\) and \(\left( \iota _1(j),\iota _2(j)\right) \not \in U_{a,b}\). This deletion ensures that the gadgets represent the corresponding set \(U_{a,b}\). The construction is such that a minimum dominating set uses only disks in the X-blocks, and that for each gadget (a, b) the same disk \(X_{\ell }(j)\) is chosen for each \(1\leqslant \ell \leqslant 8\). This choice signifies a specific choice \(u_{a,b}=(x,y)\). To ensure that the choice for \(u_{a,b}\) in the same row and column agrees on their first and second coordinate, respectively, there are special connector blocks between neighboring gadgets. The connector blocks are denoted by A, B, C and D in Fig. 2a, and they each contain \(n+1\) disks—see Sect. 2.3 for further details.
Overview of Our Construction for Connected Dominating Set in Unit Disk Graphs
To extend the construction to Connected Dominating Set in Unit Disk Graphs, we have to make sure there is a minimum-size dominating set that is connected. This requires two things. First, we must add new disks inside the gadgets—that is, in the empty space surrounded by the X- and Y-blocks—to guarantee a connection between all chosen \(X_\ell (j)\) disks without interfering with the disks in the Y-blocks. Second, we need to connect all the different gadgets. This time, in addition to avoiding the Y-blocks, we also need to avoid interfering with the connector blocks.
The idea is as follows. Inside each gadget we add several pairs of disks, consisting of a parent disk and a leaf disk. The parent disks are placed such that, for any choice of one disk from each of the X-blocks, the parent disks together with the eight chosen disks from the X-blocks form a connected set. Moreover, the parent disks do not intersect any disk in a Y-block. See Fig. 2b for an illustration; the parent disks are blue in the figure. For each parent disk we add a leaf disk—the red disks in the figure—that only intersects its parent disk. The following is a key observation.
Observation 3
There is a minimum dominating set which contains all parent disks.
Proof
Since the leaf needs to be dominated, either it or its parent needs to be in any given dominating set D. Note that if the leaf is in D but its parent is not, then they can be exchanged; the resulting set is not larger than D and it is dominating. \(\square \)
This observation can be used to show that any canonical minimum dominating set in our construction is connected.
In Fig. 2b we used disks of different sizes. Unfortunately this is not allowed, which makes the construction significantly more tricky. To be able to place the pairs in a suitable way, we need to create more space inside the gadget. To this end we use a gadget consisting of 16 (instead of eight) X- and Y-blocks. This will also give us sufficient space to put parent–leaf pairs in between the gadgets, so the dominating sets from adjacent gadgets are connected through the parent disks. Thus the size of a minimum connected dominating set in the new construction is equal to the size of a minimum dominating set in the old construction plus the number of parent disks. Hence, we can decide if the Grid Tiling instance has a solution by checking the size of the minimum connected dominating set in our construction. Thus Connected Dominating Set in Unit Disk Graphs is \({\mathsf {W[1]}}\)-hard, and the reduction together with the the Exponential Time Hypothesis (ETH) yields the desired lower bound.
To review our construction, we need to delve into some of the details of the construction in [9].
Some Details of the Construction in [9]
In every block, the place of each disk center is defined with regard to the midpoint of the block, (x(z), y(z)). The center of each circle is of the form \((x(z)+\alpha \epsilon ,y(z)+\beta \epsilon )\) where \(x(z),y(z), \alpha \) and \(\beta \) are integers, and \(\epsilon >0\) a small constant. We say that the offset of the disk centered at \((x(z)+\alpha \epsilon , y(z)+\beta \epsilon )\) is \((\alpha ,\beta )\). Note that \(|\alpha |,|\beta |\leqslant n^2+1\), and \(\epsilon < n^{-3}\), so the disks in a block all intersect each other. The offsets of X and Y-blocks are defined as follows.
$$\begin{aligned} \begin{array}{l@{\quad }l} \hbox {offset}(X_1(j))=( j,-\iota _2(j)) &{} \hbox {offset}(Y_1(j))=(j+0.5,j+0.5) \\ \hbox {offset}(X_2(j))=( j, \iota _2(j)) &{} \hbox {offset}(Y_2(j))=(j+0.5,-n) \\ \hbox {offset}(X_3(j))=(-\iota _1(j),-j) &{} \hbox {offset}(Y_3(j))=(j+0.5,-j-0.5) \\ \hbox {offset}(X_4(j))=( \iota _1(j),-j) &{} \hbox {offset}(Y_4(j))=(-n,-j-0.5) \\ \hbox {offset}(X_5(j))=(-j, \iota _2(j)) &{} \hbox {offset}(Y_5(j))=(-j-0.5,-j-0.5) \\ \hbox {offset}(X_6(j))=(-j,-\iota _2(j)) &{} \hbox {offset}(Y_6(j))=(-j-0.5,n) \\ \hbox {offset}(X_7(j))=( \iota _1(j), j) &{} \hbox {offset}(Y_7(j))=(-j-0.5,j+0.5) \\ \hbox {offset}(X_8(j))=(-\iota _1(j), j) &{} \hbox {offset}(Y_8(j))=(n,j+0.5) \\ \end{array} \end{aligned}$$
We remark some important properties. First, two disks can intersect only if they are in the same or in neighboring blocks. Consequently, one needs at least eight disks to dominate a gadget. The second important property is that disk \(X_\ell (j)\) dominates exactly \(Y_{\ell }(j),\ldots ,Y_{\ell }(n^2)\) from the “previous” block \(Y_\ell \), and \(Y_{\ell +1}(0),\ldots ,Y_{\ell +1}(j-1)\) from the “next” block \(Y_{\ell +1}\). This property can be used to prove the following key lemma.
Lemma 4
(Lemma 1 of [9]) Assume that a gadget is part of an instance such that none of the blocks \(Y_i\) are intersected by disks outside the gadget. If there is a dominating set \(\varDelta \) of the instance that contains exactly \(8k^2\) disks, then there is a canonical dominating set \(\varDelta '\) with \(|\varDelta '| = |\varDelta |\), such that for each gadget \({\mathcal {G}}\), there is an integer \(1 \leqslant j^G \leqslant n\) such that \(\varDelta '\) contains exactly the disks \(X_1(j^G),\ldots , X_8(j^G)\) from \({\mathcal {G}}\).
In the gadget \(G_{a,b}\), the value j defined in the above lemma represents the choice of \(s_{a,b}=\left( \iota _1(j),\iota _2(j)\right) \) in the grid tiling problem. Our deletion of certain disks in X-blocks ensures that \(\left( \iota _1(j),\iota _2(j)\right) \in U_{a,b}\). Finally, in order to get a feasible grid tiling, gadgets in the same row must agree on the first coordinate, and gadgets in the same column must agree on the second coordinate. These blocks have \(n+1\) disks each, with indices \(0,1,\ldots ,n\). We define the offsets in the connector gadgets the following way.
$$\begin{aligned} \begin{array}{l@{\quad }l} \hbox {offset}(A_j)=(-j-0.5,-n^2-1) &{} \hbox {offset}(B_j)=(j+0.5,n^2+1) \\ \hbox {offset}(C_j)=(n^2+1,-\iota _2(j)) &{} \hbox {offset}(D_j)=(-n^2-1,\iota _2(j)) \\ \end{array} \end{aligned}$$
Using this definition, it is easy to prove the following lemma.
Lemma 5
Let \(\varDelta \) be a canonical dominating set. For horizontally neighboring gadgets \({\mathcal {G}}\) and H representing \(j_G\) and \(j_H\), the disks of the connector block A are dominated if and only if \(\iota _1(j_G)\leqslant \iota _1(j_H)\); the disks of B are dominated if and only if \(\iota _1(j_G)\geqslant \iota _1(j_H)\). Similarly, for vertically neighboring blocks \(G'\) and \(H'\), the disks of block C are dominated if and only if \(\iota _2(j_{G'})\leqslant \iota _2(j_{H'})\); the disks of D are dominated if and only if \(\iota _2(j_{G'})\geqslant \iota _2(j_{H'})\).
With the above lemmas, the correctness of the reduction follows. A feasible grid tiling defines a dominating set of size \(8k^2\): in gadget \(G_{a,b}\), the dominating disks are \(X_\ell \left( f(u_{a,b})\right) , \;\ell =1,\ldots ,8\). On the other hand, if there is a dominating set of size \(8k^2\), then there is a canonical dominating set of the same size that defines a feasible grid tiling.
Gadgetry of Our Connected Dominating Set Construction
To extend the construction to Connected Dominating Set in Unit Disk Graphs, we want to make sure that minimum-size dominating set is connected. This requires two things. First, we must add new disks “inside” the gadgets—that is, in the empty space surrounded by the X and Y-blocks—such that a canonical minimum dominating set includes some new disks that connect the chosen \(X_\ell (j)\) disks without interfering with disks in the Y-blocks. Second, we need to connect all the different gadgets. This time in addition to avoiding the Y-blocks, we also need to avoid interfering with the connector blocks.
In order to have enough space, our gadgets contain 16 X-blocks and 16 Y-blocks instead of eight. The offsets of disks inside the blocks are not modified: we use the same building blocks. Figure 3 shows how we arrange these blocks, and depicts the connector block placement.
The analogue of Lemmas 4 and 5 are true here; we have a construction that could be used to prove the \({\mathsf {W[1]}}\)-hardness of Dominating Set in Unit Disk Graphs, with canonical sets of size \(16k^2\), that contain one disk from each X-block and \(X'\)-block. We extend this construction with parent–leaf pairs so that we have canonical dominating sets that span a connected subgraph.
We are going to add 72 extra disks to every gadget, and 4 “connector” disks between every pair of horizontally or vertically neighboring gadgets, resulting in canonical dominating sets of size \(16k^2 + 36k^2 + 4k(k-1)=56k^2 - 4k\) (Note that only the parent disks are included in the canonical set). In other words, the new construction has a connected dominating set of size \(56k^2 - 4k\) if and only if there is a feasible grid tiling.
An important property of the blocks that we use is that for a small enough value \(\epsilon \), the boundaries of the disks in a block all lie inside a small width annulus - for this reason, the blocks in our pictures are depicted with thick boundary disks. In order for a parent disk p to intersect every disk in a block it is sufficient if the boundary of p crosses this annulus.
Inside any of the blocks, all offsets are in the rectangle with bottom left \((-n^2-1,-n^2-1)\) and top right \((n^2+1,n^2+1)\). Consequently, every disk in the block with center offset \((\alpha ,\beta )\) where \(|\alpha |,|\beta |\in \{0,\ldots ,n^2+1\}\) intersects the square with bottom left \(\big ((-n^2-1)\epsilon , 1-(n^2+1)\epsilon \big )\) and top right \(\big ((n^2+1)\epsilon , 1+(n^2+1)\epsilon \big )\). There are three similar squares that also have this property, which we can get by rotating the square around the midpoint of the block by 90, 180 and 270 degrees. Consequently, a unit disk that contains such a square intersects all the disks in the given block. For an example with \(n=3\) and \(\epsilon = 0.02\) for the block \(X_2\), see Fig. 4.
Connecting Neighboring Gadgets
For a pair of horizontally neighboring gadgets, we add two pairs of disks that connect \(X'_3\) from the left gadget to \(X'_8\) in the right gadget. This arrangement is depicted in Fig. 5. The parent disk with center \(T_1\) intersects every disk in the block \(X'_3\) of the left gadget, and the other parent intersects every disk in the block \(X'_8\). The two leaf disks (red disks in the figure) only intersect their parent. Let the origin be the center of the block \(X'_3\) in the left gadget. The coordinates for the disk centers are:
$$\begin{aligned} T_1&= (1.3,0.4) \quad U_1 = (2,1.55) \\ T_2&= (2.7,-\,0.4) \quad U_2 = (2,-\,1.55) \end{aligned}$$
We use a rotated version of these four disks for vertical connections, where the parents connect \(X'_5\) from the upper gadget and \(X'_2\) from the lower gadget.
Disks Inside Gadgets
We begin by adding eight disk pairs to the center. The parents are arranged in the vertices and edge midpoints of a square, touching the neighbors. There are four parent disks whose leaf is placed towards the center of the gadget, leaving the portion of their boundary that faces the X- and Y-blocks available for connecting to further parent disks. See Fig. 6 for a picture: the corresponding leaf disks are displayed using a fill pattern of parallel lines.
Let \(\delta >0\) be a small constant to be specified later. From now on, we fix the origin in the center of the bottom left block, \(Y_7\). The disks that are placed in the middle of the gadget have their centers defined below; in each pair we specify the coordinates of a parent and its leaf.
$$\begin{aligned} \begin{matrix} (6,6),(6-\delta ,6) &{} \quad (8,6),(8,6+4\delta ) &{}\quad (10,6),(10,6-\delta ) &{} \quad (10,8),(10-4\delta ,8) \\ (10,10),(10,10+\delta ) &{}\quad (8,10),(8,10-4\delta ) &{}\quad (6,10),(6,10+\delta ) &{}\quad (6,8),(6+4\delta ,8)\\ \end{matrix} \end{aligned}$$
In order to connect the X-blocks, we add parent disks that together connect \(X_7, X_6, X'_6, X'_5\) to the center; rotated versions of these parent disks will allow us to connect all X-blocks to the center. For this purpose, we are going to use a zigzag pattern of disks. The first parent disk intersects all disks in \(X_6\) and \(X_7\) (i.e., it contains the small squares of \(X_6\) and \(X_7\) that are facing the inside of the gadget). The second parent is above the block \(Y_6\), but it is disjoint from it. The next with center \(p_3\) intersects all disks in \(X'_6\), and the disk around \(p_4\) is disjoint from the disks in \(Y'_6\). Finally, the disk around \(p_5\) intersects all disks in \(X'_5\). See Fig. 7 for an example. The leafs follow a more complicated pattern. In our zigzag pattern, two neighboring parents touch each other. We need the centers to have distance \(2\delta \) along the y-axis, so the distance along the x-axis is \(\sqrt{4-4\delta ^2}\). Let \(\xi =2-\sqrt{4-4\delta ^2}\). Note that
$$\begin{aligned} 2-\delta ^2-\delta ^4< \sqrt{4-4\delta ^2} < 2-\delta ^2, \end{aligned}$$
so \(\delta ^2< \xi <\delta ^2 + \delta ^4\). We add two more disk pairs to this pattern, and some modifications to the leafs. These seven disk pairs are depicted in Fig. 8. Each pair consists of a parent centered at \(p_i\) and a leaf centered at \(\ell _i\); their coordinates are defined as follows.
$$\begin{aligned} p_1&= (2- \delta ,2-\delta ) \;&\; \ell _1&= (2-\delta ,3-\delta ) \\ p_2&= (4-\delta -\xi ,2+\delta ) \;&\; \ell _2&= (4-\delta -\xi ,2+2\delta ) \\ p_3&= (6-\delta -2\xi ,2-\delta ) \;&\; \ell _3&= (6-\delta -2\xi ,2+5\delta )\\ p_4&= (8-\delta -3\xi ,2+\delta ) \;&\; \ell _4&= (8-\delta -3\xi ,2+2\delta )\\ p_5&= (10-\delta -4\xi ,2-\delta ) \;&\; \ell _5&= (11,2-\delta )\\ p_6&= (10-\delta -4\xi ,4-\delta ) \;&\; \ell _6&= (11,4)\\ p_7&= (8,4+3\delta ) \;&\; \ell _7&= (7,4+3\delta ) \end{aligned}$$
By analyzing the coordinates carefully, it can be verified that only the intended intersections arise among these seven disk pairs and the central eight disk pairs. Also not that the seven disk pairs are disjoint from the segment \((12+\delta /2,1)(12+\delta /2,5)\) and also from the segment on egets by rotating this by 90 degrees around (8, 8), namely \((1,4-\delta /2)(5,4-\delta /2)\).
Our final gadget can be attained by rotating the above seven disk pairs around the center (8, 8) by 90, 180 and 270 degrees: see Fig. 9. We added the spanned edges of a canonical dominating set to this picture.
We can now finish the proof of Theorem 1.
A feasible grid tiling defines \(k^2\) values \(u_{a,b}\) for \((a,b)\in [k]\times [k]\). We can use this to define \(16k^2\) disks in our blocks. Recall that f is the function assigning pairs of integers to a single index: for \(1\leqslant x,y \leqslant n\), \(f(x,y)=(x-1)n + y\). In the gadget with index (a, b), we include the disks \(X_\ell \left( f(u_{a,b})\right) \) and \(X'_\ell \left( f(u_{a,b})\right) \) for all \(\;\ell =1,\ldots ,8\). We add all parent disks of the construction, this results in a connected dominating set of size \(56k^2-4k\). In the other direction, if there is a connected dominating set of size \(56k^2-4k\), then there is a canonical dominating set of the same size, whose disks inside X-blocks and \(X'\)-blocks define a feasible grid tiling. Thus, it is sufficient to prove that the intersection patterns are as described.
It can be verified using the coordinates that our final leaf disks only intersect their parent disk, and also that the parent disks form a connected subgraph both inside gadgets and at every connection. We need to show that the parents inside the gadget connect all the X-blocks of the gadget, and that the horizontal and vertical connectors intersect the two X-blocks that they need to connect. In all of these cases, it is sufficient to show that the parent disk contains one of the four squares that we associated with each block. For connector disks, it is easy to see that the center of one of the four squares is covered by the interior of the corresponding parent disk (i.e., the square around (1, 0) is contained in the interior of \({\mathrm {disk}}(T_1)\)). By choosing a small enough value for \(\epsilon \), the square is contained in the parent disk.
For the inner connections of gadgets, it is sufficient to show that the inner squares of \(X_7, X_6, X'_6\) and \(X'_5\) are contained in \({\mathrm {disk}}(p_1),{\mathrm {disk}}(p_1),{\mathrm {disk}}(p_3)\) and \({\mathrm {disk}}(p_5)\) respectively: the other sides have the same containments since the rotation around (8, 8) by 90, 180 and 270 degrees are automorphisms on the small squares.
Concentrating on \({\mathrm {disk}}(p_1)\) now, notice that \(p_1=(2-\delta ,2-\delta )\) contains the right hand side square of \(X_7\) if and only if it contains the top square of \(X_6\) by the symmetry on the line \(x=y\). Furthermore, observe that \(p_1\) and the center of the top square of \(X_6\) are closer to each other than \(p_3\) and the center of the top square of \(X'_6\), which are in turn closer than \(p_5\) and the center of the top square of \(X'_5\). (This follows from the fact that the differences in y-coordinates are the same but are increasing in x-coordinates.) Therefore, if we can show that \({\mathrm {disk}}(p_5)\) contains the top square of \(X'_5\), than all the other desired intersection must also be present. The farthest corner of this square from \(p_5\) is \(\big (10+(n^2+1)\epsilon ,1-(n^2+1)\epsilon \big )\). Let \(\epsilon <\frac{1}{2n^3}\) and \(\delta <1\). The distance squared from \(p_5\) has to be at most 1:
$$\begin{aligned}&\left( 10-\delta -4\xi - (10+(n^2+1)\epsilon ) \right) ^2+\left( 2-\delta - (1-(n^2+1)\epsilon ) \right) ^2\\&\quad < \left( \delta +4\delta ^2+ \delta ^4+ \frac{1}{n} \right) ^2 +\left( 1-\delta + \frac{1}{n} \right) ^2\\&\quad = 1 - 2\delta + \frac{4}{n} + O\left( \frac{\delta }{n}\right) + O(\delta ^2) \end{aligned}$$
Let \(\delta = \frac{1}{\sqrt{n}}\). For n large enough,
$$\begin{aligned} 1 - 2\delta + \frac{4}{n} + O\left( \frac{\delta }{n}\right) + O(\delta ^2)\;=\;1 - \frac{2}{\sqrt{n}} + \frac{4}{n} + O\left( \frac{1}{n\sqrt{n}}\right) + O\left( \frac{1}{n}\right) < 1. \end{aligned}$$
Since \(\delta>1/n>(n^2+1)\epsilon \), we can also observe that \(p_2\) and \(p_4\) are disjoint from the y-blocks, since their projection on the y-axis is disjoint from the projection of the top squares of these blocks.
Note that the coordinates of each point can be represented with \(O(\log n)\) bits, since a precision of \(c/n^4\) is sufficient for the construction. Since Grid Tiling has no \(n^{o(k)}\) algorithm under ETH and the above is a parameterized reduction leading to a parameter \(16k^2\), there can be no \(n^{o(\sqrt{k})}\) algorithm for Connected Dominating Set in Unit Disk Graphs unless ETH fails. Finally, the containment in \({\mathsf {W[1]}}\) is a simple consequence of the proof for Dominating Set in Unit Disk Graphs in [10]. The key point of that proof is that a dominating set can be verified with a tail-deterministic machine; in our case, we only need to add a connectivity check on the solution set to the end of the Dominating Set verifier program. This concludes the proof for Connected Dominating Set. To prove \({\mathsf {W[1]}}\)-hardness and the same lower bound for the broadcast problem, we can let one of the blue parent disks be the source disk: in this way, the minimum broadcast sets equal the minimum connected dominating sets. Finally, the containment in \({\mathsf {W[1]}}\) needs an extra check that the source s is in the set. This concludes the proof of Theorem 1. \(\square \)
Since the whole construction fits in a strip of width \(w=O(k)\), we also get the following corollary.
Corollary 6
The broadcast problem in strips is W[1] -hard parameterized by the strip width w. Moreover, there is no \(f(w)n^{o(w)}\) algorithm for it, unless the Exponential Time Hypothesis fails.