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1 Available online at ScienceDirect Procedia Computer Science 74 (05 ) 47 5 International Conference on Graph Theory and Information Security Fractional Metric Dimension of Tree and Unicyclic Graph Daniel A. Krismanto, Suhadi Wido Saputro Combinatorial Mathematics Research Group, Faculty of Mathematics and Natural Sciences, Institut Teknologi Bandung, Jalan Ganesa 0, Bandung 403, Indonesia Abstract A vertex v in a simple connected graph G resolves two vertices x and y in G if the distance from x to v is not equal to distance from y to v. The vertex set R{x, y} is defined as the set of vertices in G which resolve x and y. A function f : V(G) [0, ] is called a resolving function of G if f (R{x, y}) for any two distinct vertices x and y in G. The minimal value of f (V(G)) for all resolving functions f of G is called the fractional metric dimension of G. In this paper, we determine the fractional metric dimension of G where G is a tree or G is a unicyclic graph. c 05 The Authors. Published by Elsevier by Elsevier B.V. B.V. This is an open access article under the CC BY-NC-ND license Peer-review ( under responsibility of the Organizing Committee of ICGTIS 05. Peer-review under responsibility of the Organizing Committee of ICGTIS 05 Keywords: Fractional metric dimension, resolving function, tree graph, unicyclic graph. 00 MSC: 05C, 05C78. Introduction Throughout this paper, all graphs G are finite, connected, and simple. We denote by V the vertex set of G and by E the edge set of G. The distance between two vertices u, v V(G), denoted by d(u, v), is the length of a shortest path from u to v in G. Foru, v V(G), we define R{u, v} = {z V(G) d(u, z) d(v, z)}. A vertex set W V(G) is called a resolving set of G if W R{u, v} for any two distinct vertices u, v V(G). The minimum cardinality of all resolving sets of G is called the metric dimension of G. The metric dimension concept has been applied to many various areas including coin weighing problem [8], robot navigation [7], and strategies for the mastermind game [3]. Determining the metric dimension of a graph was formulated as an integer programming problem by Chartrand et al. []. Furthermore, Currie and Oellermann [4] defined fractional metric dimension as the optimal solution of the linear relaxation of the integer programming problem. Let f be a function assigning each vertex v of G a real number f (u) [0, ]. For W V(G), denote f (W) = v W f (v). The function f is called a resolving function of G if f (R{u, v}) for any two distinct vertices u and v in G. The minimum value of f (V(G)) for any resolving function f of G is called the fractional metric dimension of G, denoted by dim f (G). address: The Authors. Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND license ( Peer-review under responsibility of the Organizing Committee of ICGTIS 05 doi:0.06/j.procs

2 48 Daniel A. Krismanto and Suhadi Wido Saputro / Procedia Computer Science 74 ( 05 ) 47 5 The fractional metric dimension problem was initiated by Arumugam and Mathew [] in 0. They provided asufficient condition for a connected graph G whose fractional metric dimension is V(G). They also determined dim f (G) where G is Petersen graph, cycles, hypercubes, stars, wheels, friendship graph, and grids. Some authors applied this topic to some product graphs. Feng et al. [5] have determined the fractional metric dimension of Cartesian product of two graphs. In other paper [6], Feng et al. also studied the fractional metric dimension of corona and lexicographic product graphs. In this paper we consider tree and unicyclic graph. Unicyclic graph is a graph obtained from a tree by adding one more edge. Note that unicyclic graph contains one cycle. The study of fractional metric dimension for a tree has been initiated by Arumugam and Mathew []. They have proved that dim f (P n ) =. They also provided an exact value for the fractional metric dimenison of star and bistar. In this paper, we are interested to determine the fractional metric dimension of any trees. We also investigate the fractional metric dimension of unicyclic graph.. Fractional metric dimension of tree Let G be a tree and v be a vertex of G. Abranch of G at v is defined as a maximal subtree of G containing v as an end point. So, if degree of v is k, then v has at most k different branches. A branch of v which is isomorphic to a path is called a leaf of v. If there is a leaf from v, then v is called a stem of G. A stem vertex v is called a node if v has more than one leaves. Those definitions are firstly introduced by Slater [9]. For i n and m i, we define a generalized star graph S n (m, m,...,m n ) as a tree graph having m + m m n + vertices and containing a vertex of degree n which has n leaves where the length of the i-th leaf is m i. Note that if v is a node in a tree graph G, then an induced subgraph of G by v and all of its leaves is isomorphic to a generalized star graph. Lemma. Let G be a generalized star graph and w be a node in G. Let u V(G) where degree of u is. If two distinct vertices x, y V(G) satisfy w R{x, y}, then u R{x, y}. Proof. Suppose that x and y are two distinct vertices in G such that w R{x, y} but u R{x, y}. Let L and L are leaves of w which are containing x and y respectively. Since u R{x, y}, wehaved(x, u) = d(y, u). So, u V(L ) and u V(L ). Therefore, we obtain d(x, u) = d(x, w) + d(w, u) and d(y, u) = d(y, w) + d(w, u). It follows that d(x, w) = d(y, w) which implies w R{x, y}, a contradiction. Lemma. For n 3, let G be a generalized star graph S n (m, m,, m n ). Then dim f (G) = n. Proof. First, we will prove that dim f (G) n. We define the function f : V(G) [0, ] where f (v) =, if degree of v is, For any two different vertices u, v V(G) and also considering Lemma, R{u, v} contains two distinct vertices of degree. So, we have f (R{u, v}). Therefore, f is a resolving function of G. Hence, dim f (G) f (V(G)) = n. Now, we will prove dim f (G) n. Let g be a minimum resolving function of G. Let x and y be two different vertices in G. Then there exist two different vertices u, v V(G) from two different leaves of G such that u, v R{x, y}. Let L u and L v be two leaves of G containing u and v respectively. Note that V(L u ) V(L v ) R{x, y}. So, g(v(l u ) V(L v )). It follows that for every leaf L of G,wehaveg(V(L)). Hence, dim f (G) n. Using both Lemmas and, we can obtain the following theorem. Theorem 3. Let G be a tree graph with k nodes w, w,...,w k. Let L(w) be the number of leaf of node w of G. Then dim f (G) = k L(w i ) i=

3 Daniel A. Krismanto and Suhadi Wido Saputro / Procedia Computer Science 74 ( 05 ) Proof. To proof Theorem 3 above, we use the following definitions. For i k, we define S i as an induced subgraph of G by w i and all vertices of its leaves. Note that S i is isomorphic to a generalized star graph. Let S = V(S ) V(S )... V(S k ). First, we define a function f : V(G) [0, ] as f (v) =, if v S and degree of v is, For any two different vertices u, v V(G), R{u, v} contains two distinct vertices of S of degree. So, we have f (R{u, v}). Therefore, f is a resolving function of G. Hence, dim f (G) f (V(G)) = ki= Now, let g be a minimal resolving function of G. Since S i is a generalized star graph for i k, by Lemma we obtain dim f (G) g(s ) = ki= Remark 4. Theorem 3 above strengthen Arumugam and Mathew s result of star graph. 3. Fractional metric dimension of unicyclic graph In this section, let G be a unicyclic graph. Let R be an induced subgraph of G which is isomorphic to a cycle. A vertex v V(R) is called a root if degree of v is more than. If a root v has degree 3 and G v has a path component, then v is called a grass root. We also consider a vertex set Q V(R) which is a set of all vertices of degree at least 3 in R and vertex set A Q which is a maximal subset of Q such that for every u, v A, d(u, v). For vertex v V(R), we still use the definitions of branch, leaf, stem, and node as stated in Section. Let A, B, C, D, E be collections of unicyclic graphs G where:. A is a set of G which is isomorphic to a cycle graph.. B is a set of G having only grass root(s). Furthermore, we decompose B into 5 partitions B, B,...,B 5 where: (a) G B if G Band R of G is isomorphic to C 4. (b) G B if G Band R of G is an even cycle of order at least 6 and A 3. (c) G B 3 if G Band R of G is an even cycle of order at least 6 and A 4. (d) G B 4 if G Band R is isomorphic to C 3. (e) G B 5 if G Band R of G is an odd cycle of order at least C is a set of G having exactly one root which is not grass root. 4. D is a set of G having exactly two roots which are not grass root. Furthermore, we decompose D into partitions D and D where: (a) G D if G D, R is isomorphic to even cycle, and for two distinct roots u, v V(G), d(u, v) =. (b) G D if G Dbut G D. 5. E is the set of G which is not in class A, B, C,orD. In theorem below, we give an exact values of fractional metric dimension of a unicyclic graph G in some collections. Theorem 5. Let G be a unicyclic graph of order n 3 having k nodes w, w,...,w k. Let L(w) be the number of leaf of node w of G. Let R be an induced subgraph of G which is isomorphic to a cycle. Let Q V(R) be a set of all vertices of degree at least 3 in R and A Q be a maximal subset of Q such that for every u, v A, d(u, v).if G B 5, then, if G Aand R is even cycle,, if G Aand R is odd cycle,, if G B, 3 dim f (G) =, if G B B 4, A A +, if G B 3, + ki= L(w i ), if G C D, ki= L(w i ), if G D E.

4 50 Daniel A. Krismanto and Suhadi Wido Saputro / Procedia Computer Science 74 ( 05 ) 47 5 Proof. Note that all unicyclic graphs in A are isomorphic to cycle graph. The fractional metric dimension of cycle graph has been determined by Arumugam and Mathew []. Now, we assume that G A B 5. Case 5.. G B First, we define a function f : V(G) [0, ] such that for v V(G), we have, if degree of v is, f (v) =, if degree of v is and v V(R), Let X be vertex set of G containing all vertices of degree. Let X be vertex set of R containing all vertices of degree. Note that for any two different vertices u, v V(G), there exist two distinct vertices x, y X X such that x, y R{u, v}. So, we have f (R{u, v}). Therefore, f is a resolving function of G. Hence, dim f (G) f (V(G)) =. Now, let g be a minimal resolving function of G. Note that there exist two different pair of vertices (a, b) and (c, d) in V(G) such that R{a, b} R{c, d} =. It follows that dim f (G) g (R{a, b} R{c, d}). Case 5.. G B B 3 We define vertex set X as the set of all vertices of R where for every x X there exists a A such that d(x, a) =. If A X 6, we define vertex set C V(R) \ (A X) such that C = 6 A X and for every x C there exists y C such that d(x, y) =. We also define B as the subset of X where degree of every vertex of B is. Note that A, B, and C are disjoint. Let V be a subset of V(G) containing all vertices of degree, and S = V B C. First, we define a function f : V(G) [0, ] such that for v V(G), we have 4, if v S and G B, f (v) = A +, if v S and G B 3, Note that for any two different vertices u, v V(G), if G B, then R{u, v} S 4, otherwise R{u, v} S A +. So, we have f (R{u, v}). Therefore, f is a resolving function of G. Hence, dim f (G) f (V(G)) = 3 for G B and dim f (G) f (V(G)) = A A + for G B 3. For i A, let w i A, x i and y i be two different vertices in V(R) such that d(x i, w i ) = d(y i, w i ) =, and z i V(R) such that d(w i, z i ) =. Let H i = R{x i, z i } R{y i, z i }. We distinguish two cases.. G B Let g be a minimal resolving function of G. Since V(G) = (R{x i, z i } R{y i, z i })\H i, we obtain g (V(G))+g (H i ). Since R{x i, y i } V(G) \ H i, we obtain g (V(G)) g (H i ). Therefore, we have dim f (G) g (V(G)) 3.. G B 3 Suppose that g be a minimal resolving function of G where g (V(G)) < A A +. Since V(G) = (R{x i, z i } R{y i, z i }) \ H i, we obtain g (V(G)) + g (H i ). Since R{x i, y i } V(G)\ H i, we obtain g (H i ) V(G) = V(H ) V(H )... V(H A ). Therefore, A g (V(G)) = g (H i ) A A +. So, it is impossible to have a minimal resolving function g of G where g (V(G)) < A A +. i= A +. Note that Case 5.3. G B 4 Let S be a vertex set of G containing all vertices of G of degree and all vertices of R of degree. First, we define a function f 3 : V(G) [0, ] such that for v V(G), we have f 3 (v) = Note that for any two different vertices u, v V(G), R{u, v} S. So, we have f 3 (R{u, v}). Therefore, f 3 is a resolving function of G. Hence, dim f (G) f 3 (V(G)) = 3.

5 Daniel A. Krismanto and Suhadi Wido Saputro / Procedia Computer Science 74 ( 05 ) Now, let g 3 be a minimal resolving function of G. Let V(R) = {x, x, x 3 } and L i be a leaf of x i for i 3. Note that for distinct i, j {,, 3}, V(L i ) V(L j ) R{x i, x j }. Without loss of generality, let g 3 (V(L )) = t for t [0, ]. So, we obtain g 3 (V(L )) = t and g 3 (V(L 3 )) = max{t, t}. It follows that t =. So, we have that dim f (G) g 3 (V(G)) = 3. Case 5.4. G C D For i k, we define a vertex set S i as the set of all vertices of degree one in leaf of node w i. Choose a pair of vertices x, y V(R) such that d(x, y) = and for a {x, y}, either degree of a is or a is a grass root of G. We define S = S S... S k and T = S {x, y}. First, we define a function f 4 : V(G) [0, ] such that for v V(G), we have f 4 (v) =, if v T, Note that for any two different vertices u, v V(G), R{u, v} T. So, we have f 4 (R{u, v}). Therefore, f 4 is a resolving function of G. Hence, dim f (G) f 4 (V(G)) = + ki= For the lower bound, let g 4 be a minimal resolving function of G. For i k, let F i be an induced subgraphs of G by S i. Note that F i is a generalized star graph. By Lemma, we obtain dim f (F i ) = L(w i). So, we can say that dim f (S ) = ki= However, we also always can find a pair of vertices u, v V(G) such that R{u, v} S =. Therefore, we obtain that dim f (G) g 4 (V(G)) + ki= Case 5.5. G D E For i k, we define a vertex set S i as the set of all vertices of degree one in leaf of node w i. We define S = S S... S k. First, we define a function f 5 : V(G) [0, ] such that for v V(G), we have f 5 (v) = Note that for any two different vertices u, v V(G), R{u, v} S. So, we have f 5 (R{u, v}). Therefore, f 5 is a resolving function of G. Hence, dim f (G) f 5 (V(G)) = ki= For the lower bound, let g 5 be a minimal resolving function of G. For i k, let F i be an induced subgraphs of G by S i. Note that F i is a generalized star graph. By Lemma, we obtain dim f (F i ) = L(w i). So, we can say that dim f (S ) = ki= Therefore, we obtain that dim f (G) g 5 (V(G)) ki= In Theorem 6, we give the upper bound of dim f (G) for G B 5. We also show that the bound is sharp. In Theorem 7, we provide an existence of unicyclic graph G B 5 such that dim f (G) is not equal to the upper bound in Theorem 6. Theorem 6. Let G be a unicyclic graph in class B 5. Let R be an induced subgraph of G which is isomorphic to cycle graph. Then dim f (G) + 3. The bounds is sharp. Proof. Let S = {v V(G) degree of v is } {v V(R) degree of v is }. We define a function f : V(G) [0, ] such that for v V(G), we have { f (v) = +3 Note that for any two different vertices u, v V(G), R{u, v} S +3. So, we have f (R{u, v}). Therefore, f is a resolving function of G. Hence, dim f (G) f (V(G)) = +3. Now, we consider a unicyclic G which is obtained from C n where n 5 is odd and every vertex of C n has one leaf. We will prove that dim f (G) = +3. Now, we need to show that dim f (G) +3.

6 5 Daniel A. Krismanto and Suhadi Wido Saputro / Procedia Computer Science 74 ( 05 ) 47 5 Let g be a resolving function of G. Let V(R) = {x 0, x,...,x n } where R = x 0 x...x n x 0. For 0 i n, let y i V(R) buty i x i E(G) and L(x i ) be a leaf of x i. Note that R{x i mod n, y (i ) mod n } = V(G) \ (V(L(x ( n+ +i) mod n))... V(L(x (n +i) mod n )) {x i }). Since g(r{a, b}) for every two distinct vertices a, b V(G), we obtain that n g(r{x i mod n, y (i ) mod n }) = n g(v(g)) n 3 i=0 Since g(x i ) 0 for i n, we will obtain that g(v(g)) n n+3. n g(v(g)) g(x i ) n Theorem 7. There exists a unicyclic graph G B 5 such that for an induced subgraph R of G which is isomorphic to cycle graph, dim f (G) < +3. Proof. Let G be a unicyclic graph having an odd cycle R with 7 and 3 grass roots such that grass roots a, b V(G) are adjacent and another grass root c V(G) satisfies d(c, a) = d(c, b) =. We will prove that dim f (G) = 5 4. Note that for 7, we have 5 5( + 3) = 4 4( + 3) < 8 4( + 3) = + 3. First, let S = {u V(G) degree of u is } {v V(R) vc E(G)}. Now, we define a function f : V(G) [0, ] such that for v V(G), we have f (v) = 4 Note that for any two different vertices u, v V(G), R{u, v} S 4. So, we have f (R{u, v}). Therefore, f is a resolving function of G. Hence, dim f (G) f (V(G)) = 5 4. For the lower bound, suppose that dim f (G) < 5 4. Let g be a minimal resolving function of G such that g(v(g)) < 5 4. Let y a, y b, y c V(R) such that for t {a, b, c}, ty t E(G). Let v, v V(R) be two distinct vertices which are adjacent to c and L w be a leaf of node w in R. Since R{v, v } = V(G) \ V(L c ), we obtain g(v(l c )) = 4 k for positive k R. Now, we consider R{y a, b} and R{y b, a}. We can say that R{y a, b} = V(L c ) X and R{y b, a} = V(L c ) Y. So, we obtain that g(x) = k = g(y). Note that X Y = and R{v, v } = V(G) \ V(L c ) = X Y. It follows that g(v(g)) = g(v(l c )) + g(x) + g(y) = k, a contradiction. i=0 Acknowledgement. This work is supported by Institut Teknologi Bandung, Indonesia by research grant Riset dan Inovasi ITB 05. The authors also are very grateful to the referees for their careful reading with corrections and useful comments. References. Arumugam S, Mathew V. The fractional metric dimension of graphs. Discrete Math. 0;3: Chartrand G, Eroh L, Johnson MA, Oellermann OR. Resolvability in graphs and the metric dimension of a graph. Discrete Appl. Math. 000;05: Chvátal V. Mastermind. Combinatorica 983;3: Currie J, Oellermann OR. The metric dimension and metric independence of a graph. J. Combin. Math. Combin. Comput. 00;39: Feng M, Lv B, Wang K. On the fractional metric dimension of graphs. Discrete Appl. Math. 04;70: Feng M, Wang K. On the fractional metric dimension of corona product graphs and lexicographic product graphs. arxiv:06.906v. 7. Khuller S, Raghavachari B, Rosenfeld A. Landmarks in graphs. Discrete Appl. Math. 996;70: Shapiro H, Soderberg S. A combinatory detection problem. Amer. Math. Monthly 963;70: Slater PJ. Leaves of trees. Congr. Numer. 975;4:

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