On b-Matchings and b-Edge Dominating Sets: A 2-Approximation Algorithm for the 4-Edge Dominating Set Problem

Abstract

We consider a multiple domination version of the edge dominating set problem, called b-EDS. Here, an edge set $D\subseteq E$ of minimum cardinality is sought in a given graph $G=(V,E)$ with a demand vector $b\in \mathbb {Z}^E$ such that each edge $e\in E$ is dominated by $b_e$ edges of D. When a solution D is not allowed to be a multi-set, it is called simple b-EDS. We present a 2-approximation algorithm for simple b-EDS for the case of $\max _{e\in E}b_e\le 4$. The 2-approximation on general graphs has been known when $\max _{e\in E}b_e\le 3$, and in the most general cases 8/3 is the best approximation ratio known attainable in polynomial time [2]. Our algorithms are designed based on the most natural LP relaxation of b-EDS and maximal b-matchings (or its generalization).

This work is supported in part by JSPS KAKENHI under Grant Numbers 20K11676 and 17K00013.

Appendices

A Proof of Lemma 1

Proof

We consider only $y(\delta (v))$ for $e=\{u,v\}\not \in M$ (as the same arguments apply to $y(\delta (u))$). Moreover, if no edge in $\delta _M(v)$ is upgraded, $y(\delta (v))\le 1/2$ (recall Observation 2), and we thus consider the case when at least one in $\delta _M(v)$ is upgraded.

Let $\delta _M(v)=\{e_1,e_2,\dots ,e_s\}$ for $s=d_M(v)$ in what follows, sorted in the ascending order of $b_e$ (i.e., $i< j$ implies that $b_{e_i}\le b_{e_j}$).

• Case $d_M(v)=0$. Clearly $y(\delta (v))=0$ since $y_e=0, \forall e\in \delta (v)$.

• Case $d_M(v)=1$. Since $y_{e}\le 1/2$ for any $e\in M$, $y(\delta (v))\le 1/2$.

• Case $d_M(v)=4$. For each $e\in \delta _M(v)$, $b_e=4$, and none in $\delta _M(v)$ can be upgraded as it is fully dominated by M.

• Case $d_M(v)=3$. For each $e\in \delta _M(v)$, $b_e\in \{3,4\}$.

  • Subcase $|\delta _M(v)\cap E_3|=3$. Then, none in $\delta _M(v)$ can be upgraded as it is fully dominated by M.

  • Subcase $|\delta _M(v)\cap E_3|=2$. The only edge that can be possibly upgraded is $e_3$ with $b_{e_3}=4$. Such an upgrade on $e_3$ results in $y_{e_3}=1/2$ while the initial values assigned on $e_1, e_2$ being nullified at the same time, and hence, $y(\delta (v))=1/2$.

  • Subcase $|\delta _M(v)\cap E_3|=1$. The edges in $\delta _M(v)$ that can be possibly upgraded are only $e_2 \text { and }e_3$ since $e_1$ with $b_{e_1}=3$ is fully dominated by M. Suppose $e_2$ is upgraded before $e_3$ is. This means that $|\delta _M(e_2)|=3$, and when $e_2$ is upgraded, an edge to be added to D must be chosen form $\delta (v)\setminus F$ if $e_3$ is not yet fully dominated, by addition of such an edge, $e_3$ becomes fully dominated. Therefore, $e_3$ already is or becomes fully dominated when $e_2$ is upgraded, and hence, it never happens to upgrade $e_3$ itself. The upgrading stage also shifts the values of y such that, while resetting $y_{e_1}$ to 0, $y_{e_2}$ is increased from initial 1/8 by 1/8 for $e_1$ and another 1/8 for an edge newly added to D, thus $y_{e_2}$ totaling to 3/8. Meanwhile, $y_{e_3}$ remains at the initial value of 1/8, and hence, $y(\delta (v))=\sum _{i=1}^3y_{e_i}=0+3/8+1/8=1/2$.

  • Subcase $|\delta _M(v)\cap E_3|=0$. So, $\delta _M(v)\subseteq E_4$, and if $e_i\in \delta _M(v)$ is to be upgraded, it must be the case that $|\delta _M(e_i)|=3$, which means that $\delta _M(e_i)=\delta _M(v)$. If only one of $e_1,e_2,e_3$ is to be upgraded, then $y(\delta (v))=\sum _{i=1}^3y_{e_i}=2/8+1/8+1/8=1/2$ after the upgrade. If two of them are possibly upgradable, then such an upgrade would hinder any subsequent upgrade for the following reason; the first upgrade on e would cause an addition of one edge to D, and that edge must come from $\delta (v)\setminus F$ because $|\delta _M(e)|=3$ and $\delta (v)$ contains other edges not yet fully dominated. So the first upgrade makes all edges in $\delta _M(v)$ fully dominated. Therefore, at most one out of three is possibly upgraded again in this case, and we have $y(\delta (v))=\sum _{i=1}^3y_{e_i}=2/8+1/8+1/8=1/2$.

• Case $d_M(v)=2$. Let $\delta _M(v)=\{e_1,e_2\}$ and we know that $2\le b_{e_1}\le b_{e_2}\le 4$.

  • Subcase $b_{e_1}=2$: So, $e_1$ is fully dominated by M (cannot be upgraded), and $e_2$ must be upgraded, implying that $b_{e_2}\ge 3$. Such an upgrade would reset $y_{e_1}$ to 0 while $y_{e_2}\le 1/2$, and hence, $y(\delta (v))\le 1/2$.

  • Subcase $b_{e_1}=3$: We first consider the case when $e_1$ is upgraded, which implies that $\delta _M(e_1)=\delta _M(v)$. So, this upgrade chooses an edge from $\delta (v)\setminus F$ and adds it to D. If $e_2\in E_3$, $e_2$ becomes fully dominated as a result of upgrading $e_1$, and hence, $y_{e_1}=2/6\text { and }y_{e_2}=1/6$. Suppose now that $e_2\in E_4$, and after the upgrade on $e_1$, either $e_2$ can be fully dominated or not. In the former case, $y(\delta (v))$ ends up with $y_{e_1}=2/6\text { and }y_{e_2}=1/8$. In the latter case, $e_2$ is also upgraded and it causes resetting 2/6 of $y_{e_1}$ to 0 while increasing $y_{e_2}$ to no larger than 1/2, and hence, $y(\delta (v))$ becomes $\le 1/2$.

    Let us consider next the case when $e_1$ is not upgraded, implying that upgraded is $e_2$. In this case again, $y_{e_1}=1/6\text { and }y_{e_2}=2/6$ if $e_2\in E_3$, or $y_{e_1}=0\text { and }y_{e_2}\le 1/2$ if $e_2\in E_4$, and hence, $y(\delta (v))\le 1/2$ in either case.

  • Subcase $b_{e_1}=4$: In this case $\delta _M(v)\subseteq E_4$. If only one of $e_1,e_2$ is upgraded, say $e_2$, then $y_{e_1}=1/8$ and $y_{e_2}$ becomes no larger than 3/8, and hence, $y(\delta (v))\le 1/2$. It remains to consider the case both $e_1,e_2$ are upgraded, and assume w.l.o.g. that $|\delta _M(e_1)|\le |\delta _M(e_2)|$ and that $e_1$ is upgraded before $e_2$ is.

    • Subsubcase $|\delta _M(e_1)|=3$: So, $|\delta _M(e_2)|=3$, and $y_{e_1}$ becomes 2/8 while $y_{e_2}$ does at most 2/8, and hence, $y(\delta (v))\le 1/2$.

    • Subsubcase $|\delta _M(e_1)|=2$ (or $\delta _M(e_1)=\delta _M(v)$): When $e_1$ is upgraded, two edges are added to D. The algorithm would choose the 1st of those two edges from $\delta (v)\setminus D$ since $e_2$, not yet fully dominated edge, is incident with v. Since $e_2$ is supposed to be still subsequently upgraded, it must be the case that $\delta _M(e_2)=\delta _M(v)$, and the algorithm would try to choose the 2nd of those two again from $\delta (v)\setminus D$, if it exists, and if it does, $e_2$ becomes fully dominated, unnecessitating an upgrade on $e_2$. Therefore, it is possible for both $e_1\text { and }e_2$ to be upgraded only when $\delta _M(e_1)=\delta _M(e_2)=\delta _M(v)$ and $d(v)=3=d_M(v)+1$, but this is exactly the case when e is an edge of Type 1.

   $\square$

B Proof of Lemma 2

Proof

First of all if no edge in $\delta (e)$ is upgraded, $y(\delta (e))=y(\delta (u))+y(\delta (v))-y_e\le 1$ because of Observation 2.

Because of Observation 3.2, clearly $y(\delta (e))\le 1$ if $|\delta _M(e)|\le 2$. So we consider only cases when $|\delta _M(e)|\ge 3$.

• Case $|\delta _M(e)|=3$. If none in $\delta _M(e)$ is upgraded, it can be verified that

  $$ y(\delta (e))\le 2\cdot \frac{1}{4}+\frac{1}{6} < 1 . $$

  So, we may assume that at least one in $\delta _M(e)$ is upgraded.

  • Subcase $d_M(u)=3, d_M(v)=1$ (or vice versa). For any $e'\in \delta _M(e)$ we have $b_{e'}\ge 3$. This means that $y_{e'}\le 2/8$ even after those edges are upgraded, and hence, $y(\delta (e))\le 6/8$.

  • Subcase $d_M(u)=d_M(v)=2$. Let $\delta _M(u)=\{e,e_1\}$ and $\delta _M(v)=\{e,e_2\}$. Then, $b_{e_1},b_{e_2}\ge 2$ and $b_{e}\ge 3$. This means that $y_{e_1}, y_{e_2}\le 3/8$ and $y_e\le 2/8$ even after those edges are upgraded, and hence, $y(\delta (e))\le 1$.

• Case $|\delta _M(e)|=4$. If none in $\delta _M(e)$ is upgraded, it can be verified that

  $$ y(\delta (e))\le \frac{1}{4}+ \frac{1}{8}+2\cdot \frac{1}{6} < 1 . $$

  So, we may assume that at least one in $\delta _M(e)$ is upgraded.

  • Subcase $d_M(u)=4, d_M(v)=1$ (or vice versa). In this case, $\delta _M(e)\subseteq E_4$, and none can be upgraded.

  • Subcase $d_M(u)=3, d_M(v)=2$ (or vice versa). Let $\delta _M(u)=\{e,e_1,e_2\}$ and $\delta _M(v)=\{e,e_3\}$. If $e\in E_4$, then it cannot be upgraded while $y_{e_1}, y_{e_2}\le 2/8$ and $y_{e_3}\le 3/8$. Hence,

    $$ y(\delta (e))\le \frac{1+2+2+3}{8}=1 $$

    if $e\in E_4$.

    Consider next the case when $e\in E_3$, which cannot be upgraded. If $e_3\in E_3\cup E_2$, then $y_{e_3}\le 2/6$, and by the same reasoning as above,

    $$ y(\delta (e))\le \frac{2+2}{8}+\frac{1+2}{6}=1 . $$

    So, let us assume that $e_3\in E_4$. If $e_3$ is not upgraded, easily $y(\delta (e))\le 1$, and so suppose it is upgraded. But then, such an operation would absorb 1/6 of $y_e$ while $y_{e_3}\le 4/8$. Thus again,

    $$ y(\delta (e))\le \frac{2+2}{8}+0+\frac{4}{8}=1 . $$

• Case $|\delta _M(e)|=6$. So, $d_M(u)=4\text { and }d_M(v)=3$ (or vice versa). Then, $\delta _M(u)\subseteq E_4$, no edge in it can be upgraded, and $y(\delta _M(u))=4/8$. Meanwhile, $y_{e'}$ cannot become larger than 2/8 for each edge $e'\in \delta _M(v)\setminus \{e\}$. Therefore, $y(\delta (e))\le (4+2+2)/8=1$.

• Case $|\delta _M(e)|=7$. So, $d_M(u)=d_M(v)=4$, $\delta _M(e)\subseteq E_4$, and none in $\delta _M(e)$ can be upgraded.

• Case $|\delta _M(e)|=5$. If $d_M(u)=4\text { and }d_M(v)=2$ (or conversely, $d_M(u)=2\text { and }d_M(v)=5$), since every edge in $\delta _M(u)$ belongs to $E_4$ and is not upgradable,

  $$ y(\delta (e))\le \frac{4}{8}+\frac{3}{8}=\frac{7}{8} . $$

  It remains to consider the case when $d_M(u)=d_M(v)=3$. Observe first that $\delta _M(e)\subseteq E_3\cup E_4$, and e as well as any edge in $E_3$ is not upgradable. So, only such an edge $e'\in \delta _M(e)\cap E_4$ with $|\delta _M(e')|=3$ is upgradable. Let $\delta _M(u)=\{e,e_1,e_2\}$ and suppose $e_1$ is such an edge and the first edge upgraded within $\delta _M(e)$. Observe here that, if $e_2\in E_3$ or if $d(u)>d_M(u)$, $e_2$ cannot be upgraded after $e_1$ is, and when $e_1$ is upgraded, $y(\delta (u))$ becomes 4/8 whether $e, e_2$ are in $E_3$ or in $E_4$. So, in this case, $y(\delta (e))\le y(\delta (u))+2/8+2/8= 1$.

  Similarly, if any in $\delta _M(v)\setminus \{e\}$ is not upgraded, $y(\delta (e))\le 1$. Thus, the only remaining case is when all the edges in $\delta _M(e)$ except for e are upgraded, and this is possible if and only if $d(u)=d(v)=3$, $e'\in E_4$, and $|\delta _M(e')|=3$ for all edges $e'\in \delta _M(e)\setminus \{e\}$, and this is exactly when e is an edge of Type 2.

Thus, we may conclude that $y(\delta (e))>1$ for $e\in M$ only if e is an edge of Type 2.

   $\square$