Applied Mathematical Sciences, Vol. 9, 2015, no. 60, 2957 - 2964
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ams.2015.5291
Locating Sets in a Graph
Stephanie A. Omega1 and Sergio R. Canoy, Jr.
Department of Mathematics and Statistics
Mindanao State University-Iligan Institute of Technology
Tibanga Highway, Iligan City, Philippines
c 2015 Stephanie A. Omega and Sergio R. Canoy, Jr. This article is
Copyright distributed under the Creative Commons Attribution License, which permits unrestricted
use, distribution, and reproduction in any medium, provided the original work is properly
cited.
Abstract
In this paper, we investigate the concept locating set in a graph. In
particular, we characterize the locating sets in the join and corona of
graphs and determine the locating numbers of these graphs.
Mathematics Subject Classification: 05C69
Keywords: locating set, strictly locating set, join, corona
1
Introduction
Let G = (V (G), E(G)) be a connected graph and u ∈ V (G). The neighborhood
of u is the set NG (u) = N (u) = {v ∈ V (G) : uv ∈ E(G)}. The distance dG (u, v)
in G of two vertices u and v is the length of the shortest u − v path in G. A
subset S of V (G) is a locating set in a connected graph G if every two vertices
u and v of V (G)\S, NG (u) ∩ S =
6 NG (v) ∩ S. It is a strictly locating set if it
is locating and NG (u) ∩ S 6= S for all u ∈ V (G)\S. The minimum cardinality
of a locating set in G, denoted by ln(G), is called the locating number of G.
The minimum cardinality of a strictly locating set in G, denoted by sln(G), is
called the strictly locating number of G.
In a given network, locating set can be viewed as a set of monitors which
can determine the exact location of an intruder (e.g. burglar, fire, etc.). The
concept of locating set is studied in [1], [2], [4] and [5].
1
This research is funded by the Department of Science and Technology-Accelerated Science
and Technology Human Resource Development Program (DOST-ASTHRDP), Philippines
2958
2
Stephanie A. Omega and Sergio R. Canoy, Jr.
Results
Remark 2.1 For any connected graph G of order n ≥ 2, 1 ≤ ln(G) ≤ n − 1.
Lemma 2.2 For any complete graph Kn of order n ≥ 2, ln(Kn ) = n − 1.
Theorem 2.3 Let G be a connected non-trivial graph. Then ln(G) = 1 if and
only if G = P2 or G = P3 .
Proof : Suppose ln(G) = 1, say S = {v} is a minimum locating set of G.
Since G is connected and non-trivial, there exists x ∈ V (G)\ {v} such that
xv ∈ E(G). If |V (G)| = 2, then G = K2 = P2 . Suppose that |V (G)| ≥ 3.
If y ∈ V (G)\ {x, v}, then yv ∈
/ E(G) since S is a locating set in G. Hence,
NG (y) ∩ S = ∅. If there exists z ∈ V (G)\ {x, y, v}, then NG (z) ∩ S = ∅.
This implies that S is not a locating set, contrary to our assumption. Thus,
|V (G)| = 3. Since G is connected, xy ∈ E(G). This shows that G = P3 =
[v, x, y].
The converse is easy. Theorem 2.4 Let G be a connected graph of order n ≥ 3. If ln(G) = 2, then
3 ≤ |V (G)| ≤ 6.
Proof : Supppose that ln(G) = 2. Clearly, 3 ≤ |V (G)|. Suppose that |V (G)| > 6.
Let S = {x, y} be a minimum locating set in G and let vi ∈ V (G)\S, where
i = 1, 2, 3, 4, 5. Since NG (vi ) ∩ S is either ∅, {x}, {y} or S, for each i =
1, 2, 3, 4, 5, there exist distinct vertices k, j ∈ {1, 2, 3, 4, 5} such that NG (vk ) ∩
S = NG (vj ) ∩ S, contrary to our assumption that S is a locating set in G. Thus,
|V (G)| ≤ 6. Therefore, 3 ≤ |V (G)| ≤ 6. Theorem 2.5 Let G be a non-trivial connected graph. Then ln(G) = n − 1 if
and only if G = Kn .
Proof : Suppose that ln(G) = n − 1 and suppose further that G 6= Kn . Then
there exists x, y ∈ V (G) such that dG (x, y) = 2. Let z ∈ NG (x) ∩ NG (y) and
let S = V (G)\ {x, z}. Since y ∈ NG (z)\NG (x), NG (z) ∩ S 6= NG (x) ∩ S. Thus,
S is a locating set in G. It follows that ln(G) ≤ |S| = n − 2, contrary to our
assumption. Therefore, G = Kn .
The converse follows from Lemma 2.2. Theorem 2.6 Let G be a connected graph of order n = 4. Then ln(G) = 2 if
and only if G 6= K4 .
Locating sets in a graph
2959
Proof : Suppose that ln(G) = 2. Then by Theorem 2.5, G 6= K4 .
For the converse, suppose that G =
6 K4 . Since n = 4, ln(G) ≥ 2 by Theorem
2.3. Choose x, y ∈ V (G) such that dG (x, y) = 2. Let z ∈ NG (x) ∩ NG (y) and
let w ∈ V (G)\ {x, y, z}. Consider the following cases:
Case 1.
Suppose that w ∈ NG (x) ∩ NG (y). Take S = {x, z}. Then S is a locating set
in G.
Case 2.
Suppose that w ∈ NG (x)\NG (y) or w ∈ NG (y)\NG (x), say w ∈ NG (x)\NG (y).
Let S = {x, y}. Then S is a locating set in G.
Therefore, ln(G) = 2. Theorem 2.7 Let G be a connected graph of order n = 5. Then ln(G) = 2
if and only if there exist distinct vertices x and y of G satisfying one of the
following properties:
(i) |NG (x) ∩ NG (y)| = 0 and |NG (x)\ {y}| = |NG (y)\ {x}| = 1;
(ii) |NG (x) ∩ NG (y)| = 1 and |NG (x)\ {y}| = |NG (y)\ {x}| = 2 or
|NG (x)\ {y}| = 2 and |NG (y)\ {x}| = 1 or |NG (y)\ {x}| = 2 and
|NG (x)\ {y}| = 1.
Proof : Suppose that ln(G) = 2. Then there exist distinct vertices x and y of G
such that S = {x, y} is a minimum locating set of G. Hence, |NG (x) ∩ NG (y)| ≤
1. Suppose |NG (x) ∩ NG (y)| = 0. Since S is a locating set, |NG (x)\ {y}| ≤ 1.
Suppose |NG (x)\ {y}| = 0. Then |NG (y)\ {x}| = 1 since S is a locating set.
This implies that there exist at least two vertices say z, w such that z, w ∈
/
NG (x) ∪ NG (y). Consequently, NG (z) ∩ S = NG (w) ∩ S = ∅, contrary to
our assumption that S is a locating set. Thus, |NG (x)\ {y}| = 1. Similarly,
|NG (y)\ {x}| = 1. Hence, (i) holds.
Suppose that |NG (x) ∩ NG (y)| = 1. Let a ∈ V (G) such that ax, ay ∈ E(G)
and let b, c ∈ V (G)\ {x, y, a}. Then b, c ∈
/ NG (x) ∩ NG (y). Since the subsets
of S are ∅, {x, y} , {x} and {y} and since NG (a) ∩ S is {x, y}, the remaining
two sets NG (b) ∩ S and NG (c) ∩ S are {x} and {y} or {x} and ∅ or {y} and
∅, respectively. Thus, |NG (x)\ {y}| = |NG (y)\ {x}| = 2 or |NG (x)\ {y}| = 2
and |NG (y)\ {x}| = 1 or |NG (y)\ {x}| = 2 and |NG (x)\ {y}| = 1. Therefore,
(ii) holds.
For the converse, suppose there exist distinct vertices x, y ∈ V (G) satisfying
(i) or (ii). Let S = {x, y}. Then S is a minimum locating set in G. Therefore,
ln(G) = 2. Theorem 2.8 Let G be a connected graph of order n = 6. Then ln(G) = 2 if
and only if there exist distinct vertices x, y ∈ V (G) such that |NG (x) ∩ NG (y)| =
1 and |NG (x)\ {y}| = |NG (y)\ {x}| = 2.
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Stephanie A. Omega and Sergio R. Canoy, Jr.
Proof : Suppose that ln(G) = 2. Let S = {x, y} be a minimum locating set
in G. Then |NG (x) ∩ NG (y)| ≤ 1. Suppose that |NG (x) ∩ NG (y)| = 0. Then
|NG (x)\ {y}| = |NG (y)\ {x}| = 1. Let a ∈ NG (x)\ {y}, b ∈ NG (y)\ {x} and
u, v ∈ V (G)\ {x, y, a, b}. Then NG (u) ∩ S = NG (v) ∩ S = ∅, contrary to
our assumption that S is a locating set. Hence, |NG (x) ∩ NG (y)| = 1. Let
v ∈ V (G) such that xv, yv ∈ E(G) and let v1 , v2 , v3 ∈ V (G)\ {x, y, v}. Then
v1 , v2 , v3 ∈
/ NG (x)∩NG (y). Since S is a locating set and NG (v)∩S = S, it follows
that NG (vi ) ∩ S is either ∅, {x} or {y} for each i = 1, 2, 3. Without loss of
generality assume that NG (v1 )∩S = ∅, NG (v2 )∩S = {x} and NG (v3 )∩S = {y}.
Then |NG (x)\ {y}| = 2 and |NG (y)\ {x}| = 2.
For the converse, suppose there exist distinct vertices x, y ∈ V (G) satisfying
the given properties. Let S = {x, y}. Then it is easy to show that S is a
minimum locating set in G. Therefore, ln(G) = 2. 3
Locating Sets in the Join of Graphs
Let A and B be sets which are not necessarily disjoint. The disjoint union of A
•
and B, denoted by A ∪ B, is the set obtained by taking the union of A and B
treating each element in A as distinct from each element in B. The join G + H
•
of two graphs G and H is the graph with vertex-set V (G + H) = V (G) ∪ V (H)
•
and edge-set E(G + H) = E(G) ∪ E(H) ∪ {uv : u ∈ V (G), v ∈ V (H)}.
Theorem 3.1 Let G and H be connected non-trivial graphs. A set S ⊆
V (G + H) is a locating set in G + H if and only if S1 = V (G) ∩ S and
S2 = V (H) ∩ S are locating sets in G and H, respectively, where S1 or S2 is a
strictly locating set.
Proof : Let S ⊆ V (G + H) be a locating set in G + H. Let S1 = V (G) ∩ S
and S2 = V (H) ∩ S. Suppose S1 = ∅. Then for any two distinct vertices
x, y ∈ V (G), NG+H (x) ∩ S = NG+H (y) ∩ S = S contrary to our assumption
that S is a locating set. Thus, S1 =
6 ∅. Similarly, S2 =
6 ∅. Next, suppose
that S1 or S2 , say S1 is not a locating set in G. Then there exist distinct
vertices u, v ∈ V (G) such that NG (u) ∩ S1 = NG (v) ∩ S1 . Since S2 ⊆ NG+H (u)
and S2 ⊆ NG+H (v), it follows that NG+H (u) ∩ S = (NG (u) ∩ S1 ) ∪ S2 =
(NG (v) ∩ S1 ) ∪ S2 = NG+H (v) ∩ S. Thus, S is not a locating set in G + H,
contrary to our assumption. Therefore, S1 and S2 are locating sets in G and H,
respectively. Now, suppose that both S1 and S2 are not strictly locating sets.
Then there exist u ∈ V (G)\S1 and v ∈ V (H)\S2 such that NG (u) ∩ S1 = S1
and NG (v) ∩ S2 = S2 . Hence, NG+H (u) ∩ S = S = NG+H (v) ∩ S, contrary to
our assumption that S is a locating set. Hence, S1 is a strictly locating set in
G or S2 is a strictly locating set in H.
For the converse, suppose that S1 and S2 are locating sets in G and H,
Locating sets in a graph
2961
respectively, and S1 or S2 is a strictly locating set. Let x, y ∈ V (G + H)\S with
x 6= y. If x, y ∈ V (G), then NG (x) ∩ S1 6= NG (y) ∩ S1 . Hence, NG+H (x) ∩ S =
(NG (x) ∩ S1 ) ∪ S2 6= (NG (y) ∩ S1 ) ∪ S2 = NG+H (y) ∩ S. Similarly, if x, y ∈ V (H),
then NG+H (x) ∩ S 6= NG+H (y) ∩ S. Suppose that x ∈ V (G) and y ∈ V (H)
and suppose that S1 is a strictly locating set in G. Then S1 * NG+H (x). Since
S1 ⊆ NG+H (y), it follows that NG+H (x) ∩ S =
6 NG+H (y) ∩ S. Therefore, S is a
locating set in G + H. Corollary 3.2 Let G and H be connected non-trivial graphs. Then
ln(G + H) = min {sln(H) + ln(G), sln(G) + ln(H)} .
Proof : Let S be a minimum locating set in G + H. Let S1 = V (G) ∩ S
and S2 = V (H) ∩ S. By Theorem 3.1, S1 and S2 are locating sets in G and
H, respectively, where S1 or S2 is a strictly locating set. If S1 is strictly
locating set, then sln(G) + ln(H) ≤ |S1 | + |S2 | = |S| = ln(G + H). If S2 is
strictly locating set, then sln(H) + ln(G) ≤ |S2 | + |S1 | = |S| = ln(G + H).
Thus, ln(G + H) ≥ min {sln(H) + ln(G), sln(G) + ln(H)}. Next suppose that
sln(G) + ln(H) ≤ sln(H) + ln(G). Let S1 be a minimum strictly locating set
in G and S2 be a minimum locating set in H. Then S = S1 ∪ S2 is a locating
set by Theorem 3.1. Hence, ln(G + H) ≤ |S| = |S1 | + |S2 | = sln(G) + ln(H).
Therefore, ln(G + H) = min {sln(H) + ln(G), sln(G) + ln(H)} . Theorem 3.3 [1] Let G be a connected graph of order n ≥ 2. If ln(G) <
sln(G), then 1 + ln(G) = sln(G).
Corollary 3.4 Let G be a connected non-trivial graph and let Kn be a complete
graph of order n ≥ 2. Then ln(G + Kn ) = sln(G) + n − 1.
Proof : Note that ln(Kn ) = n − 1 and sln(Kn ) = n. From Corollary 3.2,
ln(G+Kn ) = min {sln(G) + n − 1, ln(G) + n}. By Theorem 3.3, sln(G)−1 ≤
ln(G). Therefore,
ln(G + Kn ) = min {sln(G) + n − 1, ln(G) + n} = sln(G) + n − 1.
Theorem 3.5 Let H be connected non-trivial graph and let K1 = hvi. Then
S ⊆ V (H + K1 ) is a locating set in H + K1 if and only if either v ∈
/ S and S
is a strictly locating set in H or S = {v} ∪ S1 , where S1 is a locating set in H.
Proof : Let S ⊆ V (H + K1 ) be a locating set in H + K1 . If v ∈
/ S, then
S ⊆ V (H) is a locating set in H. Suppose there exists u ∈ V (H)\S such that
NH (u) ∩ S = S. Then NH+K1 (u) ∩ S = S = NH+K1 (v) ∩ S, contrary to our
assumption that S is a locating set in H + K1 . Hence, S is a strictly locating set
in H. Next, suppose that S = {v} ∪ S1 , where S1 = V (H) ∩ S. Then S1 6= ∅.
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Stephanie A. Omega and Sergio R. Canoy, Jr.
Let w, z ∈ V (H)\S1 with z 6= w. Then (NH (z) ∩ S1 ) ∪ {v} = NH+K1 (z) ∩ S 6=
NH+K1 (w) ∩ S = (NH (w) ∩ S1 ) ∪ {v}. Thus, NH (z) ∩ S1 =
6 NH (w) ∩ S1 . Hence,
S1 is a locating set in H.
For the converse, assume first that v ∈
/ S and S is a strictly locating set in H.
Let x, y ∈ V (H + K1 )\S. If x, y ∈ V (H), then NH+K1 (x) ∩ S = NH (x) ∩ S 6=
NH (y) ∩ S = NH+K1 (y) ∩ S. Suppose that x ∈ V (H) and y = v. Then
NH+K1 (v) ∩ S = S. Since S is a strictly locating set in H, NH (x) ∩ S 6= S.
Thus, NH+K1 (v) ∩ S 6= NH (x) ∩ S = NH+K1 (x) ∩ S. Therefore, S is a locating
set in H + K1 . Next, suppose that S = {v} ∪ S1 where S1 is a locating
set in H. Let x, y ∈ V (H + K1 )\S with x 6= y. Then x, y ∈ V (H)\S1 and
NH (x) ∩ S1 =
6 NH (y) ∩ S1 . Thus, NH+K1 (x) ∩ S = (NH (x) ∩ S1 ) ∪ {v} 6=
(NH (y) ∩ S1 ) ∪ {v} = NH+K1 (y) ∩ S. Hence, S is a locating set in H + K1 . Corollary 3.6 Let H be a connected non-trivial graph. Then
ln(H + K1 ) = sln(H).
Proof : By Theorem 3.5, ln(H + K1 ) = min {sln(H), ln(H) + 1}. By Theorem
3.3, sln(H) − 1 ≤ ln(H). Thus, sln(H) ≤ ln(H) + 1. Therefore, ln(H + K1 ) =
sln(H). 4
Locating Sets in the Corona of Graphs
Let G and H be graphs of order m and n, respectively. The corona of two
graphs G and H is the graph G ◦ H obtained by taking one copy of G and m
copies of H, then joining the ith vertex of G to every vertex of the ith copy
of H. For every v ∈ V (G), denote by H v the copy of H whose vertices are
attached one by one to the vertex v. Denote by v + H v the subgraph of the
corona G ◦ H corresponding the join h{v}i + H v .
Theorem 4.1 Let G and H be non-trivial connected graphs. Then S ⊆
V (G ◦ H) is a locating set in G ◦ H if and only if (V (G ◦ H)\S) admits
at most a single element x with NG◦H (x) ∩ S = ∅ and S = A ∪ B ∪ C ∪ D,
S
where A ⊆ V (G), B = {Bv : v ∈ A and Bv is a locating set in H v } , C =
S
{Ew : w ∈
/ A, NG (w) ∩ A 6= ∅ and Ew is a locating set in H w } and
S
D = {Dw : w ∈
/ A, NG (w) ∩ A = ∅ and Dw is strictly locating set in H w } .
Proof : Suppose that S is a locating set in G ◦ H. Let A = V (G) ∩ S and
let v ∈ A. Let Bv = V (H v ) ∩ S and let x, y ∈ V (H v )\Bv with x 6= y. Then
x, y ∈
/ S. Since S is a locating set in G◦H, (NH v (x)∩Bv )∪{v} = NG◦H (x)∩S 6=
NG◦H (y) ∩ S = (NH v (y) ∩ Bv ) ∪ {v}. Hence, Bv is a locating set in H v . Next,
let w ∈
/ A. Consider the following cases:
Case 1. Suppose that NG (w) ∩ A 6= ∅.
Locating sets in a graph
2963
Let Ew = V (H w ) ∩ S and x, y ∈ V (H w )\Ew with x 6= y. Then x, y ∈
/ S. Since
S is a locating set and w ∈
/ S, NH w (x) ∩ Ew = NG◦H (x) ∩ S 6= NG◦H (y) ∩ S =
NH w (y) ∩ Ew . Thus, Ew is a locating set in H w .
Case 2. Suppose that NG (w) ∩ A = ∅.
Let Dw = V (H w ) ∩ S. As in Case 1, Dw is a locating set in H w . Suppose
there exists x ∈ V (H w ) such that NH w (x) ∩ Dw = NG◦H (x) ∩ S = Dw . Since
w∈
/ S and NG (w) ∩ A = ∅, NG◦H (x) ∩ S = NG◦H (w) ∩ S = Dw . Thus, S is
not a locating set in G ◦ H, contrary to our assumption. Thus, Dw is a strictly
locating set in H w .
S
S
Let B = {Bv : v ∈ A and Bv is a locating set in H v } , C = {Ew : w ∈
/
S
A, NG (w) ∩ A 6= ∅ and Ew is a locating set in H w } and D = {Dw : w ∈
/
A, NG (w) ∩ A = ∅ and Dw is strictly locating set in H w }. Then S = A ∪ B ∪
C ∪ D. Moreover, since S is a locating set, V (G ◦ H)\S admits at most a single
element whose neighborhood does not intersect with S.
For the converse, suppose that S = A ∪ B ∪ C ∪ D, where A, B, C and
D are the sets possessing the properties described. Let x, y ∈ V (G ◦ H)\S
with x 6= y and let u, v ∈ V (G) such that x ∈ V (u + H u ) and y ∈ V (v + H v ).
Suppose u = v. Consider the following cases:
Case 1. Suppose that v ∈ S.
Then x, y ∈ V (H v )\Bv , where Bv is a locating set in H v . Hence, NH v (x)∩Bv 6=
NH v (y)∩Bv . Thus, NG◦H (x)∩S = (NH v (x)∩Bv )∪{v} =
6 (NH v (y)∩Bv )∪{v} =
NG◦H (y) ∩ S.
Case 2. Suppose that v ∈
/ S.
If x, y ∈ V (H v ), then x, y ∈
/ Sv = V (H v ) ∩ S, where Sv (Ev or Dv ) is a locating
set in H v by assumption. Thus, NG◦H (x) ∩ S = NH v (x) ∩ Sv 6= NH v (y) ∩ Sv =
NG◦H (y) ∩ S. Suppose that x = v and y ∈ V (H v ). If NG (v) ∩ S 6= ∅, say
w ∈ NG (v) ∩ S, then w ∈ [NG◦H (x) ∩ S]\[NG◦H (y) ∩ S]. Thus, NG◦H (x) ∩ S 6=
NG◦H (y)∩S. If NG (v)∩S = ∅, then Sv = V (H v )∩S = Dv is a strictly locating
set by assumption. Thus, NG◦H (x) ∩ S = Dv 6= NH v (y) ∩ Sv = NG◦H (y) ∩ S.
Suppose now that u 6= v. Consider the following cases:
Case 1. Suppose that u, v ∈ S.
Then x 6= u and y =
6 v. Since u ∈ NG◦H (x) and v ∈ NG◦H (y), we have
NG◦H (x) ∩ S 6= NG◦H (y) ∩ S.
Case 2. Supppose that u ∈
/ S or v ∈
/ S.
We may suppose that u ∈
/ S. Then Su = V (H u ) ∩ S is (equal to Eu or
Du ). If x = u, then there exists c ∈ Su such that c ∈ NG◦H (x)\NG◦H (y).
Suppose x 6= u. If v ∈ S, then v ∈ NG◦H (y)\NG◦H (x). Suppose v ∈
/ S. If
NG◦H (x) ∩ S 6= ∅ and NG◦H (y) ∩ S 6= ∅, then NG◦H (x) ∩ S =
6 NG◦H (y) ∩ S. If
one of NG◦H (x) ∩ S and NG◦H (y) ∩ S is empty, then the other is non-empty by
assumption.
Therefore, in all cases S is a locating set in G ◦ H. Corollary 4.2 Let G and H be non-trivial connected graphs, where |V (G)| =
2964
Stephanie A. Omega and Sergio R. Canoy, Jr.
m. Then mln(H) ≤ ln(G ◦ H) ≤ msln(H).
Proof : Let S be a minimum locating set in G ◦ H. Then S = A ∪ B ∪ C ∪ D,
where A, B, C and D are the sets decribed in Theorem 4.1. By Theorem 3.3,
sln(H) ≤ ln(H) + 1. Moreover, since ln(H) ≤ sln(H), it follows that
ln(G ◦ H)
= |S|
= |A| + |B| + |C| + |D|
≥ |A| + |A| ln(H) + (m − |A|)ln(H)
= |A| (1 + ln(H)) + (m − |A|)ln(H)
≥ |A| sln(H) + (m − |A|)ln(H)
≥ |A| ln(H) + (m − |A|)ln(H)
= mln(H).
Next, let F be a minimum strictly locating set in H. For each v ∈ V (G),
S
pick Fv ⊆ V (H v ) with hFv i ∼
= hF i. Then S = v∈V (G) Fv is a locating set in
G ◦ H by Theorem 4.1. Therefore, mln(H) ≤ ln(G ◦ H) ≤ |S| = msln(H). Corollary 4.3 Let G and H be non-trivial connected graphs with |V (G)| = m
such that ln(H) = sln(H). Then ln(G ◦ H) = msln(H).
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Received: February 13, 2015; Published: April 12, 2015