The k-metric dimension

Abstract

For an undirected graph $G=(V,E)$, a vertex $\tau \in V$ separates vertices u and v (where $u,v\in V$, $u\ne v$) if their distances to $\tau$ are not equal. Given an integer parameter $k \ge 1$, a set of vertices $L\subseteq V$ is a feasible solution, if for every pair of distinct vertices, u, v, there are at least k distinct vertices $\tau _{1},\tau _{2},\ldots ,\tau _{k}\in L$, each separating u and v. Such a feasible solution is called a landmark set, and the k-metric dimension of a graph is the minimal cardinality of a landmark set for the parameter k. The case $k=1$ is a classic problem, where in its weighted version, each vertex v has a non-negative cost, and the goal is to find a landmark set with minimal total cost. We generalize the problem for $k \ge 2$, introducing two models, and we seek for solutions to both the weighted version and the unweighted version of this more general problem. In the model of all-pairs (AP), k separations are needed for every pair of distinct vertices of V, while in the non-landmarks model (NL), such separations are required only for pairs of distinct vertices in $V \setminus L$. We study the weighted and unweighted versions for both models (AP and NL), for path graphs, complete graphs, complete bipartite graphs, and complete wheel graphs, for all values of $k \ge 2$. We present algorithms for these cases, thus demonstrating the difference between the two new models, and the differences between the cases $k=1$ and $k \ge 2$.

Appendix

1.1 The case $n=5$ for complete wheels in AP and NL

Theorem 39

Consider the case $n=5$.

• For AP, $md_{2}^{AP}(W_5)=4$, $wmd_{2}^{AP}(W_5)=w(C)$, and $md_{3}^{AP}(W_5)=md_{4}^{AP}(W_5)=\infty$.

• For NL, $md_{2}^{NL}(W_5)=3$, and

  $$\begin{aligned} wmd_{2}^{NL}(W_5)=\min \{w(C),w(V)-\max _{1 \le i \le 4} (w(c_{i})+w(c_{i+1}))\} \ . \end{aligned}$$

  Additionally, $md_{3}^{NL}(W_5)=4$, and

  $$\begin{aligned} wmd_{3}^{NL}(W_5)=\min \{w(C), W(V)-\max _{1 \le i \le 4} w(c_{i}) \} \ . \end{aligned}$$

Proof

First, consider AP and $k=2$. A pair of cycle vertices $c_{i}$ and $c_{i+2}$ cannot be separated by any vertex except for $c_{i}$ and $c_{i+2}$, as they have distances of 1 to all other vertices. Thus, all the cycle vertices (which can be split into two such pairs) must belong to any landmark set $L \in LS_{2}^{AP}(W_5)$. We claim that $C \in LS_{2}^{AP}(W_5)$, and it is a unique minimal landmark set. By Property 1, and the above explanation, it is sufficient to show two separations between $c_{i}$ (for $1\le i\le 4$) and h. One separation results from $c_{i}$ being a member of the landmark set. Additionally, $d(h,c_{i+2})=1$ while $d(c_{i},c_{i+2})=2$, so $c_{i+2}$ separates $c_{i}$ and h. Note that V is a landmark set as well, but it is not minimal.

Next, consider NL and $k=2$, and let $L \in LS_{2}^{NL}(W_5)$. In this case, for $i=1,2$, at least one of $c_{i}$ and $c_{i+2}$ must belong to L (otherwise they cannot be separated, as explained above). If there is at least one index j such that $c_{j}\notin L$ and, in addition, $h\notin L$, then only $c_{j+2}$ (if it is in L) separates $c_{j}$ and h (as their distances to both $c_{j-1}$ and $c_{j+1}$ is 1), and $L \notin LS_{2}^{NL}(W_5)$. Thus, if $h\notin L$, then $L=C$, where C is a trivial landmark set for NL. On the other hand, a set of the form $\{c_{i},c_{i+1},h\} \in LS_{2}^{NL}(W_5)$, as $c_{i-1}$ and $c_{i+2}$ are separated by both $c_{i}$ and $c_{i+1}$. Therefore, a minimal landmark set (with respect to set inclusion) consists of either of C, or h and two adjacent cycle vertices.

Finally, we deal with $k = 3$. For AP, due to the above, there cannot be more than two separations between $c_{i}$ and $c_{i+2}$, so $md_{3}^{AP}(W_5)=\infty$. Since $md_{4}^{AP}(W_n) \ge md_{3}^{AP}(W_5)$, we get $md_{4}^{AP}(W_5)=\infty$.

Consider the case of NL and $k=3$. We already showed that any landmark set (for NL and $k=2$) with at most three vertices must contain h and two cycle vertices. For such a set, the other two cycle vertices will only have two separations. Thus, for NL and $k=3$ only trivial solutions exist. $\square$

1.2 Minimal landmark sets for complete wheels with $6 \le n \le 8$ in AP

1.2.1 $k=2$

We discuss minimal landmark sets (with respect to set inclusion) for AP, $n=6,7,8$ and $k=2$. By Corollary 19, for $n=8$, such sets do not contain h, and the sets of cycle vertices that satisfy Condition 1 have at least four (cycle) vertices. Any subset of at least five cycle vertices is a landmark set (for AP and $k=2$) , but only those with two consecutive non-landmarks are minimal, with respect to set inclusion (sets of the form $c_{j},c_{j+1},\ldots ,c_{j+4}$ for $1\le j\le 7$), since if there is no pair of consecutive non-landmarks, there must be three consecutive landmarks $c_{i-1},c_{i},c_{i+1}$, and $c_{i}$ is redundant (i.e. it is a landmark set without it). Any subset of four cycle vertices for which there is no pair of consecutive non-landmarks satisfies Condition 1, and it has the form $c_{i},c_{i+1},c_{i+3},c_{i+5}$ for some $1\le i \le 7$.

For $n=6,7$, a set satisfying Condition 1 has at least three cycle vertices, and we already considered such sets. The next corollary specifies all minimal landmark sets for $n=6,7$, where unlike the case $n\ge 8$, there exist such sets that contain h.

Corollary 40

Let $L\subseteq C$, such that $|L|\ge 4$. Then, $L \in LS_{2}^{AP}(W_n)$, for $n=6,7$. Furthermore, assume that $X \subseteq C$ satisfies Condition 1, and $|X|=3$. Then, for $n=6,7$, $Y=X\cup \{h\} \in LS_{2}^{AP}(W_n)$, while $X \notin LS_{2}^{AP}(W_n)$.

Proof

For $n=6,7$, any set with four cycle vertices (particularly L) satisfies Condition 1; for $n=6$, there is at most one cycle vertex not in L, and for $n=7$, there are at most two cycle vertices not in L, and Condition 1 is satisfied no matter whether they are adjacent or not. By Proposition 17, there are two separations for any pair of the form $h,c_{i}$. By Lemma 18, any pair of cycle vertices also has at least two separations. Hence, $L \in LS_{2}^{AP}(W_n)$ for $n=6,7$.

The set X has two separations for any pair of cycle vertices, by Lemma 18, and in Y, a pair $h,c_{j}$ is separated by h and $X\setminus \{c_{j-1},c_{j+1}\}$, which must contain at least one vertex, so $Y \in LS_{2}^{AP}(W_n)$ for $n=6,7$. We show that $X \notin LS_{2}^{AP}(W_n)$, for $n=6,7$.

For $n=7$, the only form of a set X that satisfies Condition 1 is $c_{i},c_{i+2},c_{i+4}$, since in the case of two consecutive vertices not in X, all remaining cycle vertices must be in X. For this set, h and $c_{i+1}$ are only separated by $c_{i+4}$.

For $n=6$, X can be of one of two forms. If X consists of $c_{i},c_{i+1},c_{i+3}$, then $c_{i+2}$ and h are only separated by $c_{i}$. If X consists of $c_{i},c_{i+1},c_{i+2}$, then $c_{i+1},h$ are only separated by $c_{i+1}$. Thus, $X \notin LS_{2}^{AP}(W_n)$ for $n=6,7$. $\square$

1.2.2 $k=3$

A minimal landmark set (with respect to set inclusion) for AP, $n=7,8$ and $k=3$ consists of $n-2$ cycle vertices, and any subset of $n-2$ vertices is a landmark set. For $n=6$, a set that satisfies Condition 2 has at least four cycle vertices. We now specify the structure of minimal landmark sets (with respect to set inclusion) for $n=6$, and show these sets are exactly all subsets of five vertices.

Corollary 41

For $n=6$, $k=3$ and AP, a set of four cycle vertices is not a landmark set. Any subset of V with five vertices is a landmark set.

Proof

A set of four (consecutive) cycle vertices $c_{i-2},c_{i-1},c_{i+1},c_{i+2}$ only makes two separations for the following three pairs: $h,c_{i-2}$, $h,c_{i}$, $h,c_{i+2}$. Any set of five vertices gives another separation to each one of the pairs, as both h and $c_{i}$ separate each one of the three pairs. $\square$

1.2.3 $k \ge 4$

Theorem 42

For AP, and $n\ge 6$, $md_{k}^{AP}(W_n)=\infty$ for $k\ge 5$. Additionally, $md_{4}^{AP}(W_n)=n-1$ if $n\ge 7$ (in this case $wmd_{4}^{AP}(W_n)=w(C)$), and $md_{4}^{AP}(W_n)=6$ for $n=6$ (in this case $wmd_{4}^{AP}(W_n)=w(V)$).

Proof

Since a pair of the form $c_{i},c_{i+1}$ can be separated by exactly four vertices, a landmark set for $k=4$ must contain all cycle vertices, and there is no valid landmark set for $k\ge 5$. Any cycle vertex except for $c_{i+1}$ and $c_{i-1}$ separates h and $c_{i}$, thus C is a landmark set for $k=4$ if $n\ge 7$. For $n=6$, there are only three separations by cycle vertices between h and $c_{i}$, and therefore V is the only landmark set (as h adds a separation). $\square$

1.3 Minimal landmark sets for complete wheels and $6 \le n \le 8$ in NL

1.3.1 $k=2$

The minimal landmark sets (with respect to set inclusion) for $n=8$ and $k=2$ are not the same as for AP, since the condition is slightly weaker. All subsets of four cycle vertices where there is no pair of consecutive non-landmarks (that were defined for AP) are still minimal landmark sets. Minimal landmark sets with a pair of consecutive non-landmarks are possible too, and have the form $c_{i},c_{i+1},c_{i+3},c_{i+4}$ (for some $1\le i\le 7$).

For $n=6,7$, we consider sets with at least three cycle vertices.

Corollary 43

Let $L\subseteq C$, such that $|L|\ge 4$. Then, $L \in LS_{2}^{NL}(W_n)$, for $n=6,7$. Furthermore, assume that $X \subseteq C$ satisfies Condition 3, and $|X|=3$. Then, $Y=X\cup \{h\} \in LS_{2}^{NL}(W_n)$, for $n=6,7$, while $X \in LS_{2}^{NL}(W_n)$ only if $n=6$, and X consists of three consecutive cycle vertices.

Proof

We showed in Corollary 40 that any subset of four vertices is a landmark set for AP, $n=6,7$ and $k=2$, hence (as stated in Sect. 1) it is also a landmark set for NL, $n=6,7$ and $k=2$.

For $n=7$, the proof that any landmark set (for NL and $k=2$) has at least four vertices is done in the same way as for AP.

For $n=6$, the proof that a triplet of cycle vertices that are not consecutive is not a landmark set (for NL and $k=2$) is done in the same way as for AP (for the proof of these two properties for AP see Corollary 40).

If X consists of $c_{i},c_{i+1},c_{i+2}$, then $c_{i-1},h$ are separated by $c_{i+1}$ and $c_{i+2}$, while $c_{i+3},h$ are separated by $c_{i}$ and $c_{i+1}$, and $c_{i-1},c_{i+3}$ are separated by $c_{i}$ and $c_{i+2}$. $\square$

We find that a minimal landmark set (with respect to set inclusion) for $n=7$ is the same as for AP. For $n=6$, a minimal landmark set consists of either three consecutive cycle vertices, or h together with three cycle vertices that are not consecutive.

1.3.2 $k=3,4$

Lemma 44

For NL, $n = 6$ and $k \in \{3,4\}$, only trivial landmark sets exist ($md_{k}^{NL}(W_n)=5$ for $k=3$ or $k=4$).

For NL, $n=7$ and $k=3$, $L\subseteq V$ is a landmark set if and only if it satisfies Condition 4 and $|L|\ge 5$ (thus $md_{3}^{NL}(W_n)=5$). For NL, $n=7$ and $k=4$, $L\subseteq V$ is a landmark set if and only if it either satisfies Condition 4 and in addition $h\in L$, or $L=C$ (thus $md_{4}^{NL}(W_n)=5$).

For NL, $n=8$ and $k=3$, $L\subseteq C$ is a landmark set if and only if it satisfies Condition 4. In addition, $md_{3}^{NL}(W_n)=5$.

For NL, $n=8$ and $k=4$, $L\subseteq V$ is a landmark set if and only if it either satisfies Condition 4 and in addition $h\in L$, or it has at least six cycle vertices (thus $md_{4}^{NL}(W_n)=6$).

Proof

For $n=6$, a set L that satisfies Condition 4 has at least four cycle vertices. If $c_{i},h \notin L$, then as $c_{i-1},c_{i+1}$ do not separate $c_{i}$ and h, there are two separations between them. Thus, L must contain another vertex, hence $|L|\ge 5$, and it is a trivial landmark set for NL.

For $n=7$, a set L that satisfies Condition 4 has at least four cycle vertices. If it has exactly four cycle vertices, then it has the form $c_{i-2},c_{i-1},c_{i+1},c_{i+2}$. If $h \notin L$, then similarly to the case $n=6$, there are two separations between $c_{i}$ and h. On the other hand if $h \in L$, this is a landmark set for $k=3,4$. Assume now that $h \notin L$, and thus L must contain at least five cycle vertices. If $c_{i} \notin L$, then there are three separations between h and $c_{i}$, so L is a landmark set for $k=3$ but not for $k=4$, and the only landmark set for $k=4$ not containing h is C.

For $n=8$ and $k=3$, a set that satisfies Condition 4 has at least five cycle vertices (using Proposition 33), and there are three separations for any pair of the form $h,c_{i}$. By Lemma 34, any pair of non-landmark cycle vertices also has three separations. The set $L=\{c_{1},c_{2},c_{4},c_{5},c_{7}\}$ is a landmark set with five vertices, thus $md_{3}^{NL}(W_n)=5$.

For $n=8$ and $k=4$, a set L that satisfies Condition 4 has at least five cycle vertices. If $h \in L$, then it is a landmark set. Otherwise, by the condition, if $c_{i} \notin L$, then $c_{i-1},c_{i+1} \in L$, and to obtain four separations between $c_{i}$ and h, L must contain $C\setminus \{c_{i}\}$, which results in a landmark set of this form. $\square$

1.3.3 $k\ge 5$

Theorem 45

Let $k\ge 5$. For NL and $n\ge 7$, $md_{k}^{NL}(W_n)=n-2$ if $n\ge k+4$, and $md_{k}^{NL}(W_n)=n-1$ if $n\le k+3$.

Proof

Since a pair of cycle vertices $c_{i},c_{j}$ that are not landmarks can be separated by at most four vertices, any landmark set must contain all cycle vertices except for at most one. Any set consisting of $C\setminus \{c_{i}\}$ for some $1\le i\le n-1$ is a landmark set (for NL and $k \ge 5$) if and only if h and $c_{i}$ can be separated by k cycle vertices. The number of cycle vertices that can separate them is $n-4$ (all cycle vertices except for $c_{i-1},c_{i},c_{i+1}$). Thus, if $k\ge n-3$, only trivial landmark sets exist, and otherwise the minimal landmark sets consist of all cycle vertices expect for one. $\square$