A Note On Vertex Distinguishing Edge colorings of Trees

0.12 MB·

by  Songling Shan

A Note On Vertex Distinguishing Edge colorings of Trees
Songling Shana, Bing Yaob,1,
a. Department of Mathematics and Statistics of Georgia State University,Atlanta,30303-3083, USA.
E-mail: [email protected]
6 b. College of Mathematics and Statistics, Northwest Normal University,Lanzhou,730070, P.R.China
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Abstract
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A proper edge coloring of a simple graph G is called a vertex distinguishing edge coloring
n
a (vdec) if for any two distinct vertices u and v of G, the set of the colors assigned to the edges
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incidenttoudiffersfromthesetofthecolorsassignedtotheedgesincidenttov. Theminimum
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number of colors required for all vdecs of G is denoted by χ′(G) called the vdec chromatic
s

] number of G. Let nd(G) denote the number of vertices of degree d in G. In this note, we show
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thatatreeT withn2(T)≤n1(T)holdsχs′(T)=n1(T)+1ifitsdiameterD(T)=3oroneoftwo
C
particular trees with D(T) = 4, and χs′(T) = n1(T) otherwise; furthermore χe′s(T) = χ′s(T)
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h when |E(T)|≤2(n1(T)+1), where χ′es(T) is the equitable vdec chromatic number of T.
t
a
m AMS Subject Classification (2000): 05C15
[ Keywords: vertex distinguishing coloring, edge coloring, trees

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1 1 Introduction and concepts
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2 Labeled graphs are becoming an increasingly useful family of mathematical models for a broad
0 range of applications, suchas time tabling and scheduling, frequency assignment, register allocation,
.
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computer security and so on. In [2], Burris and Schelp introduced that a proper edge k-coloring
0
6 of a simple graph G is called a vertex distinguishing edge k-coloring (k-vdec, or vdec for short) if
1
for any two distinct vertices u and v of G, the set of the colors assigned to the edges incident to
:
v
u differs from the set of the colors assigned to the edges incident to v. The minimum number of
i
X colors required for a vertex distinguishing coloring of G is denoted by χ′(G). The maximum and
s
r
a minimum degrees of G are denoted by ∆(G) and δ(G), respectively. Let nd(G) denote the number
of vertices of degree d with respect to δ(G) ≤ d ≤ ∆(G) in G, or write n = n (G) if there is no
d d
confusion. Burris and Schelp [2] presented the following conjecture:
Conjecture 1. Let G be a simple graph having no isolated edges and at most one isolated vertex,
and let k be the smallest integer such that (k) ≥ n for all d with respect to δ(G) ≤ d ≤ ∆(G).
d d
Then k ≤ χ′(G) ≤ k+1.
s
It seems very difficult to settle down Conjecture 1, since no more results to verify it (cf. [1],
[2], [3]). We show some results on trees to confirm positively Conjecture 1, and try to approximate
χ′(G) of a graph G by χ′(T) of some tree T generated from G. Graphs mentioned here are finite,
s s
1Corresponding author, Email: [email protected]
simple and undirected. We use standard terminology and notation of graph theory, and write
[m,n] = {m,m+1,...,n} for integers m,n with respect to 0 ≤ m < n. Let π be a k-vdec of a
graph G, and S = {uv : π(uv) = i,uv ∈ E(G)} for i ∈ [1,k]. We call π an equitable k-vdec if S
i i
and S differ in size at most one for distinct i,j ∈ [1,k]. The smallest value of k such that G has an
j
equitable k-vdec is denoted by χ′ (G). Clearly, χ′(G) ≤ χ′ (G). The set of neighbors of a vertex
es s es
u of G is denoted as N (u), or N(u) if no confusion. The diameter of G is written as D(G). A leaf
G
is a vertex of degree one, and a k-degree vertex is one of degree k ≥ 2. P is a path of length n−1.
n
A tree Q of diameter four has its own vertex set V(Q) = Wr ∪Xm ∪(cid:0)Sni=1Yi(cid:1), where Wr =
{w ,w : i ∈ [1,r]} if r ≥ 1, X = {s ,s′ : i ∈ [1,m]} if m ≥ 1, Y = {t ,t′ : j ∈ [1,r ]} for r ≥ 2
0 i m i i i i i,j i i
and i∈ [1,n] if n ≥ 1; and Q has its own edge set E(Q) =E(Wr)∪E(Xm)∪(cid:0)Sni=1E(Yi)(cid:1), where
E(W )= {w w : i∈ [1,r]}, E(X ) = {w s ,s s′ : i∈ [1,m]}, E(Y ) = {w t ,t t′ :j ∈ [1,r ]} for
r 0 i m 0 i i i i 0 i i i,j i
r ≥ 2andi∈ [1,n]. Thereby,wecanwritethistreeQhavingdiameterD(Q) = 4asQ = Q(r,m,n)
i
withthecenterw hereafter. Especially,Q(0,2,0) = P suchthatχ′(Q(0,2,0)) = n (Q(0,2,0))+2.
0 5 s 1
let U = {Q(0,2,0), Q(r,2,0), Q(0,1,1)}. We will show the following results in this note.
Theorem1. LetT beatreewithn (T) ≤ n (T)andatleastthree vertices. Thenχ′(T) = n (T)+1
2 1 s 1
if D(T) = 3 and T ∈ U \{Q(0,2,0)}; χ′(T) = n (T)+2 if T = Q(0,2,0); and χ′(T) = n (T) if
s 1 s 1
D(T) ≥4 and T 6∈ U.
Theorem2. LetT beatree withn (T) ≤ n (T)andatleastthree vertices. If|E(T)| ≤ 2[n (T)+1],
2 1 1
then χ′ (T)= χ′(T).
es s
2 Lemmas and proofs
Lemma 3. Let T be a tree with n (T)≥ n (T). Then there exists a 2-degree vertex such that one
2 1
of its neighbors is either a leaf or a vertex of degree 2.
Proof. Write n = n (T) for d ∈ [δ(T),∆(T)], here n = n (T) = 0 if T has no vertices of degree
d d d d
d. Let n = |V(T)|−n −n −···−n for k ≥ 2. By contradiction. Assume that each 2-degree
≥k 1 2 k−1
vertex x has its neighborhood N(x) = {x ,x } such that degree deg (x ) ≥ 3 for i = 1,2. So, we
1 2 T i
have n2 ≤ n≥3 by the assumption. Applying the formula n1 = 2+P3≤d≤∆(T)(d−2)nd shown in
the article [4], we have n ≥ 2+n for d ≥ 3, and n ≥ 2+n +2n = 2+2n −n , which
1 d 1 3 ≥4 ≥3 3
shows that n ≤ 1(n +n )−1. Thereby, the following inequalities
≥3 2 1 3
1
n ≤ n ≤ (n +n )−1 ≤ n −2
2 ≥3 1 3 1
2
show a contradiction.
Lemma 4. Let T be a tree with diameter three. Then χ′(T)= n (T)+1, and χ′ (T) = χ′(T).
s 1 es s
2
Proof. Let two complete graphs K and K have their own vertex and edge sets as follows:
1,m 1,n
V(K ) = {s,s ,s ,...,s } and E(K ) = {ss ,ss ,...,ss }, V(K ) = {t,t ,t ,...,t } and
1,m 1 2 m 1,m 1 2 m 1,n 1 2 n
E(K ) = {tt ,tt ,...,tt }, where s,t are the centers of K and K . Each tree of diameter 3,
1,n 1 2 n 1,m 1,n
denoted as S , can be obtained by joining two centers of K and K with an edge. So,
m+1,n+1 1,m 1,n
S has its own vertex set V(S ) = V(K )∪V(K ) and edge set E(S ) =
m+1,n+1 m+1,n+1 1,m 1,n m+1,n+1
E(K )∪E(K )∪{st}. Supposethatf isak-vdecof S suchthatk = χ′(S ). Let
1,m 1,n m+1,n+1 s m+1,n+1
C(u,f) be the set of the colors assigned to the edges incident to u of S . Since C(s ,f) 6=
m+1,n+1 i
C(s ,f), C(s ,f) 6= C(t ,f), C(t ,f)6= C(t ,f),andC(s,f)6= C(t,f),soweknow k ≥ m+n+1 =
j i l i j
n (S )+1. This k-vdec f can be exactly defined as: f(ss )= i for i∈ [1,m]; f(st)= m+1;
1 m+1,n+1 i
f(tt )= m+1+j for j ∈ [1,n]. Thereby, k = m+n+1. By the definition of the k-vdec f, we are
j
not hard to see χ′ (T)= χ′(T).
es s
Lemma 5. Let U = {Q(0,2,0), Q(r,2,0), Q(0,1,1)}. For all trees Q = Q(r,m,n) of diameter
four we have χ′(Q) = n (Q)+1 if Q ∈ U \{Q(0,2,0)}; χ′(Q) = n (Q)+2 if Q = Q(0,2,0); and
s 1 s 1
χ′(Q) = n (Q) for Q 6∈ U. Furthermore, χ′ (Q) = χ′(Q).
s 1 es s
Proof. UsingthedescriptionofatreeQ = Q(r,m,n)ofdiameterfourinSection1. Letn = n (Q),
1 1
and f be a k-vdec of Q such that k = χ′(Q). Note that D(Q)= 4.
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Case A. (r,m,n) = (r,m,0) with m ≥ 2 and r ≥ 0. Clearly, n = r+m.
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Case A1. (r,m,0) = (r,2,0), n = r+2. We can easily see χ′(Q) = n +1 for r ≥ 0, since we
1 s 1
have to color one of two edges w s and w s with one color that is not in {f(w w ),f(s s′) : s ∈
0 1 0 2 0 s i i
[1,r],i ∈ [1,2]}. As r = 0, Q is a path with 4 vertices, so χ′(Q) = n +2= 4.
s 1
Case A2. (r,m,0) with m ≥ 3, n = r + m. We show f in the following: f(s s′) = i for
1 i i
i ∈ [1,m]; f(w s ) = i+1 for i ∈ [1,m−1], and f(w s ) = 1; f(w w ) = m+j for j ∈ [1,r] if
0 i 0 m 0 j
r 6= 0. Clearly, C(u,f) 6= C(v,f) for any two vertices u,v ∈ V(Q), which means χ′(Q) = n .
s 1
Case B. (r,0,n) with n≥ 2, n1 = r+Pni=1ri. It is easy to show χ′s(Q)= n1 in this case, since
n ≤ 1.
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Case C. (0,m,n) with m ≥ 1 and n ≥ 1, n1 = m+Pni=1ri.
Case C1. (0,m,n) = (0,1,1), n = 1+r . Since f(s s′) = f(w t ) if χ′(Q) = n in this case,
1 1 1 1 0 1 s 1
we can see C(w ,f)= C(s ,f); a contradiction. So χ′(Q(0,1,1)) = 2+r = n +1.
0 1 s 1 1
Case C1. (0,m,n) = (0,1,n) with n ≥ 2, n1 = 1+Pni=1ri. To show χ′s(Q) = n1, we have f
defined as: f(s s′) = 1; f(t t′ ) = 1+j for j ∈ [1,r ]; f(t t′ ) = f(t t′ )+j for j ∈ [1,r ]
1 1 1 1,j 1 i i,j i−1 i−1,j i
and i∈ [2,n]; f(w s ) =f(t t′ ); f(w t ) = f(t t′ ) for j ∈ [1,n−1]; f(w t ) = f(s s′).
0 1 n n,1 0 j j+1 j+1,1 0 n 1 1
Case C2. (0,m,n) = (0,m,1) with m ≥ 2, n = m + r . For showing χ′(Q) = n we
1 1 s 1
define: f(s s′) = i for i ∈ [1,m]; f(w s ) = i + 1 for i ∈ [1,m − 1], and f(w s ) = f(t t′ );
i i 0 i 0 m 1 1,1
f(t t′ ) = m+j for j ∈ [1,r ] and f(w t ) = m.
1 1,j 1 0 1
Case C3. (0,m,n) = (0,m,n) with m ≥ 2 and n ≥ 1, n1 = m+Pni=1ri. We have χ′s(Q) = n1
by defining f appropriately by the methods showing in Case C1 and Case C2.
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Case D. (r,m,n) with r ≥ 1, m ≥ 1 and n ≥ 1, n1 = r+m+Pni=1ri. By the techniques use
in the above cases, we can define f to show χ′(Q)= n .
s 1
Through the above Cases A, B, C and D, we conclude χ′ (Q) = χ′(Q), since one color is used
es s
at most twice under the vdec f. The lemma is covered.
The proof of Theorem 1. Let n = n (T) for i = 1,2, and neighborhoods N(w) = N (w)
i i T
for w ∈ V(T). Because T = K when D(T) = 2, so χ′(T) = n = n−1. For D(T) = 3,4, by
1,n−1 s 1
Lemmas 4 and 5, the theorem holds true. So, we show χ′(T)= n by induction on vertex numbers
s 1
of trees T with diameter D(T) ≥ 5. We will use the description of trees having diameter four in
the following discussion.
Case 1. There exists a leaf v having a neighbor u with degree deg (u) ≥ 4. Let T ′ = T −v, so
T
n (T ′)= n −1. Clearly, D(T ′) ≥ 5.
1 1
Case 1.1. n2(T ′) ≤ n1(T ′). Then by induction hypothesis, there is an edge coloring ξT′ :
E(T ′) → {1,2,...,b′} such that χ′(T ′) = b′ = n (T ′) = n −1. It is straightforward to define
s 1 1
a vdec ξT: E(T) → {1,2,...,b′,b′ + 1} such that ξT(e) = ξT′(e) for e ∈ E(T) \ {uv}, and
ξ (uv) = b′+1.
T
Case 1.2. n (T ′) = n (T ′)−1. By Lemma 3, there exists a 2-degree vertex x ∈ V(T ′) with
1 2
NT′(x) = {x1,x2} such that degT(x1) = degT′(x1) ≤ 2. Notice that degT′(x) = degT(x) and
degT′(x1) = degT(x1). Let T1 be the tree obtained from T ′ by suppressing the vertex x, that is,
T = T ′−x+x x . Clearly, D(T )≥ 4, n (T ) = n (T ′) and n (T )= n (T ′)−1.
1 1 2 1 1 1 1 2 1 2
If D(T ) = 4, we have T = Q(0,m,0) with m ≥ 3, since n (T ) = n (T ) and deg (u) ≥ 4.
1 1 1 1 2 1 T
Thus, u is the center of T = Q(0,m,0), the leaf v is the leaf w of Q(1,m,0), and T is the tree
1 1
obtained by subdivision of the edge s s′. It is not difficult to see χ′(Q(1,m,0)) = n (Q(1,m,0)),
1 1 s 1
so χ′(T) = n .
s 1
If D(T ) ≥ 5, again by induction hypothesis, there is a vdec ξ : E(T ) → {1,2,...,b }
1 T1 1 1
such that χ′(T ) = b = n (T ) = n − 1. Hence, T has a proper edge coloring ξ defined
s 1 1 1 1 1 T
as: ξ (e) = ξ (e) for e ∈ E(T) \ {uv,xx ,xx }, and ξ (uv) = b + 1, ξ (xx ) = b + 1 and
T T1 1 2 T 1 T 1 1
ξ (xx )= ξ (x x ). Itiseasy tocheck ξ tobeadesiredn -vdecofT,sinceC(ξ ,u) 6= C(ξ ,x ).
T 2 T1 1 2 T 1 T T 1
Case 2. There is a leaf v having a 3-degree neighbor u, and Case 1 is false.
Case 2.1. v′ isanother leaf intheneighborhoodN(u) = {v,v′,u′}, andT hasa2-degreevertex
x having its neighborhood {x ,x } such that x is a leaf of T.
1 2 1
We have a tree T = T −{v,v′,x}+x x . Clearly, D(T ) ≥ 3 since D(T) ≥ 5, n (T )= n −1
1 1 2 1 1 1 1
and n (T )= n −1.
2 1 2
If D(T ) = 3, T is a 3-diameter tree S (we use the description shown in the proof of
1 1 m+1,n+1
Lemma 4) obtained by joining two centers of K and K with an edge. Since D(T) ≥ 5, so
1,m 1,n
T can be obtained by subdividing the edge ss (= x x ) to form a path sxs = x xx of T and
1 1 2 1 1 2
joining t (= u) to two new vertices v and v′ to S . It is not hard to make a desired n -vdec
1 m+1,n+1 1
of T based on the structure of T.
4
If D(T ) = 4, then T = Q(r,m,n). If T = Q(0,2,0), thus it goes to T such that n = 3 and
1 1 1 1
n = 4; a contradiction. As T = Q(r,m,n) 6= Q(0,2,0), we can show χ′(T) = n by the edge
2 1 s 1
colorings used in the proof of Lemma 5.
For D(T ) ≥ 5, T , by induction hypothesis, has a vdec ξ : E(T ) → {1,2,...,b } such that
1 1 T1 1 1
χ′(T ) = b = n (T ) = n −1. Notice that u is a leaf of T . So, we can extend ξ to a vdec
s 1 1 1 1 1 1 T1
ξ of T as follows: ξ (e) = ξ (e) for e ∈ E(T) \ {uv,uv′,xx ,xx }; if ξ (x x ) 6∈ C(ξ ,u′),
T T T1 1 2 T1 1 2 T1
we set ξ (uu′) = b + 1, ξ (uv′) = ξ (uu′), ξ (uv) = ξ (x x ), ξ (xx ) = ξ (x x ) and
T 1 T T1 T T1 1 2 T 2 T1 1 2
ξ (xx ) = b + 1; if ξ (x x ) ∈ C(ξ ,u′), we define ξ (uu′) = ξ (uu′), ξ (uv) = ξ (x x ),
T 1 1 T1 1 2 T1 T T1 T T1 1 2
ξ (uv′) = b +1, ξ (xx )= b +1 and ξ (xx )= ξ (uu′).
T 1 T 2 1 T 1 T1
Case 2.2. deg (v′) = 1 for v′ ∈ N(u) = {v,v′,u′}, and T has no a 2-degree vertex that is
T
adjacent to a leaf of T. Let P = p p ···p p be a longest path of T, where D is the diameter
1 2 D−1 D
of T. Here, deg (p ) = 3 = deg (p ).
T 2 T D−1
Without loss of generality, p 6= u, so p is adjacent to two leaves p ,p′ ∈ N(p ) = {p ,p′,p }
2 2 1 1 2 1 1 3
with deg (p ) 6= 1. Suppose that x,y are two 2-degree vertices of T. Let N(x) = {x ,x } and
T 3 1 2
N(y) = {y ,y }, then we know that deg (x ) ≥ 2 and deg (y ) ≥ 2 for i = 1,2 by the hypothesis
1 2 T i T i
of Case 2.2. Make a tree T = T −{v,v′,p ,p′,x,y} + {x x ,y y } with n (T ) = n −2 and
2 1 1 1 2 1 2 1 2 1
n (T ) =n −2.
2 2 2
IfD(T ) = 2,soT = K withV(K ) = {s,s ,s ,...,s }andE(K )= {ss ,ss ,...,ss }
2 2 1,m 1,m 1 2 m 1,m 1 2 m
for m ≥ 2. Without loss of generality, u = s , p = s , ss = x x and ss = y y for j 6= 1 since
1 2 2 1 1 2 j 1 2
D(T) ≥ 5. So, D(T) = 5 or 6, n ≥ 4 and n = 2. The structure of T enables us to show
1 2
χ′(T) =n by defining a desired n -vdec of T.
s 1 1
Consider the case D(T ) = 3. As T = P , T is obtained by replacing by a path w xx yt the
2 2 4 0 1 1
edge w t of Q(0,0,2) with r = r = 2 (t = u, t = p ), so n = n = 4. It is easy to define a
0 1 1 2 2 1 2 1 2
desired4-vdecof T bymeans ofthetechniques usedintheproofof Lemma5. WhenT = S
2 m+1,n+1
with m+n ≥ 3, the structure of T is clear such that we can show easily χ′(T)= n .
s 1
For D(T ) = 4, thus T = Q(r,0,n) with 0≤ r ≤ 1 and n ≥2 according to T has no a 2-degree
2 2
vertex that is adjacent to a leaf of T. We have two subcases T = Q(1,0,2) and T = Q(0,0,n)
2 2
with n ≥ 3. So, x x = w t and y y = w t , and u and p both are the leaves of T = Q(1,0,2)
1 2 0 1 1 2 0 2 2 2
or T = Q(0,0,n) with n ≥ 3. We claim χ′(T) = n by the structure of T in this situation.
2 s 1
When D(T ) ≥ 5, by induction hypothesis, we take a vdec ξ : E(T ) → {1,2,...,b } having
2 T2 2 2
χ′(T ) = b = n (T ) = n − 2, and extend ξ to a proper edge coloring ξ of T by setting
s 2 2 1 2 1 T2 T
ξ (e) = ξ (e)fore∈ E(T)\{uv,uv′,uu′,xx ,xx ,p p ,p p′,p p ,yy ,yy },andξ (uu′)= b +2,
T T2 1 2 2 1 2 1 2 3 1 2 T 2
ξ (uv′) = ξ (uu′), ξ (uv) = b +1, ξ (xx ) = ξ (x x ), and ξ (xx ) = b +1; ξ (p p ) = b +1,
T T2 T 2 T 2 T2 1 2 T 1 2 T 2 3 2
ξ (p p′) = ξ (p p ), ξ (p p ) = b + 2, ξ (yy ) = ξ (y y ), and ξ (yy ) = b + 2. Since
T 2 1 T2 2 3 T 2 1 2 T 2 T2 1 2 T 1 2
C(ξ ,p )\{ξ (p p )} is not equal to one of C(ξ ,x )\{ξ (x x )} for i = 1,2, and C(ξ ,u)\
T2 3 T2 2 3 T2 i T2 1 2 T2
{ξ (uu′)} is not equal to one of C(ξ ,y )\{ξ (y y )} for i = 1,2, so C(ξ ,p ) 6= C(ξ ,x ) and
T2 T2 i T2 1 2 T 3 T 1
C(ξ ,u′) 6= C(ξ ,y ), without loss of generality. Hence, it is not hard to verify that ξ is a vdec
T T 1 T
of T such that χ′(T) = b +2= n .
s 2 1
5
Case 2.3. The above Case 2.1 and Case 2.2 are false, simultaneously. So, any leaf of T is either
adjacent to a 2-degree vertex or a 3-degree vertex having two neighbors of non-leaves. Thereby,
deg (v) = 1, deg (v′) ≥ 2 and deg (u′) ≥ 2 in N(u) = {v,v′,u′}. Let P = x x ···x x be
T T T 1 2 D−1 D
a longest path of T with deg (x ) = 2 or deg (x ) = 2, where D = D(T). Without loss of
T 2 T D−1
generality, x has its two neighbors x and x holding deg (x ) = 1 and deg (x )≥ 2.
2 1 3 T 1 T 3
We have a tree T = T−{v,u,x }+{x x ,v′u′}. Clearly, D(T ) ≥ 3 since D(T) ≥ 5, n (T ) =
3 2 1 3 3 1 3
n −1andn (T )= n −1. IfD(T ) = 3,thepossiblecaseisT = P = x x x x = x u′v′x ,which
1 2 3 2 3 3 4 1 3 4 5 1 5
implies T = x x u′uv′x +uv. Thereby, we have χ′(T) = n . IfD(T )= 4, thenT = Q(1,2,0) or
1 2 5 s 1 3 3
T = Q(0,m,0) with m ≥ 3 (T = Q(0,2,0) will induce T has n = 3 and n = 4; a contradiction).
3 3 1 2
Therefore, s = u′, w = v′, and x is a leaf of T = Q(1,2,0) or T = Q(0,m,0) with m ≥ 3. The
1 0 1 3 3
structure of T is clear when D(T ) = 4, so we can show a n -vdec of T by the methods used in the
3 1
proof of Lemma 5.
ForD(T )≥ 5, byinductionhypothesis,T admitsavdecξ : E(T )→ {1,2,...,b }suchthat
3 3 T3 3 3
χ′(T )= b = n (T )= n −1. It is straightforward to definea n -vdec ξ of T as follows: ξ (e) =
s 3 3 1 3 1 1 T T
ξ (e) for e ∈ E(T) \ {uv,uv′,uu′,x x ,x x }, and ξ (uu′) = b + 1 if ξ (x x ) 6∈ C(ξ ,u′),
T3 2 1 2 3 T 3 T3 1 3 T3
ξ (uv′) = ξ (u′v′), ξ (uv) = ξ (x x ), ξ (x x ) = ξ (x x ), and ξ (x x ) = b +1.
T T3 T T3 1 3 T 2 3 T3 1 3 T 2 1 3
Case 3. The above Case 1 and Case 2 do not appear simultaneously, and every leaf is adjacent
to a 2-degree vertex in T, which means n = n .
2 1
Case 3.1. Each 2-degree vertex x has its neighborhood N(x) = {x ,x } such that deg (x )= 1
1 2 T 1
and deg (x ) ≥ 3, and no two 2-degree vertices have a common neighbor. Let R = p p ···p be
T 2 1 2 l
a longest path in T, where l = D(T). Thereby, deg (p ) ≥ 3 and deg (p ) ≥ 3 since n = n .
T 3 T l−2 2 1
Notice that deg (x) ≤ 2 for x ∈ N(p ) \ {p ,p } (resp. x ∈ N(p ) \ {p ,p }) because
T 3 2 4 l−2 l−3 l−1
x 6∈ V(R), and furthermore this vertex x is neither a leaf (Case 2 has been assumed to disappear)
nor a 2-degree vertex (by the hypothesis of this subcase and n = n ). We claim that this subcase
2 1
does not exist.
Case 3.2. Case 3.1 does not exist at all. We have a subgraph H with V(H) = {w,u}∪V ′ with
m ≥ 2 and V ′ = {y ,x : i ∈ [1,m]} and E(H) = {wu,uy ,y x : i ∈ [1,m]}, where deg (w) ≥ 2,
i i i i i T
every y is a 2-degree vertex and every x is a leaf for i∈ [1,m]. Then we have a tree T = T −V ′
i i 1
such that D(T )≥ 3 because D(T) ≥ 5.
1
If D(T )= 3, the possible structure is T = P , but, which implies n = n +1; a contradiction.
1 1 4 2 1
For D(T )= 4, thus T = Q(1,m,0) with m ≥ 2 under the restriction of Case 3.2, and s′ = u.
1 1 1
The clear structure of T enables us to show χ′(T) = n .
s 1
For D(T ) ≥ 5, n (T ) = n − m + 1 > n (T ) = n − m from n = n , and u is a leaf
1 1 1 1 2 1 2 1 2
of T . By induction hypothesis, T admits a vdec ξ : E(T ) → S = {1,2,...,b } having
1 1 T1 1 1
χ′(T ) = b = n (T ) = n −(m−1). We can define a n -vdec ξ of T as follows: ξ (e) = ξ (e)
s 1 1 1 1 1 1 T T T1
for e ∈ E(T)\E(H), and ξ (wu) = ξ (wu); ξ (uy ) = b +i for i ∈ [1,m−1], and ξ (uy ) ∈
T T1 T i 1 T m
S \{ξ (wu)}; ξ (y x )= ξ (wu), ξ (y x ) = b +j−1 for j ∈ [2,m].
T1 T 1 1 T1 T j j 1
Thereby, Theorem 1 follows from the principle of induction. (cid:3)
6
Proof of Theorem 2. It is not hard to let the number of a color be one or two in the desired
vdecs ξ of T in all cases of the proof of Theorem 1, since |E(T)| ≤ 2(n +1). Thereby, Theorem
T 1
2 follows from the principle of induction. (cid:3)
Corollary 6. Let G be a connected graph having cycles, p vertices and q edges. If n (G) ≤
2
2(q−p+1)+n (G), then χ′(G) ≤ 2(q−p+1)+n (G).
1 s 1
Proof. Let H be a spanning tree of G, so E′ = E(G) \ E(H). Hence, we have another tree T
obtained by deleting every edge uv ∈ E′, and then adding two new vertices u′,v′ by joining u′
with u and v′ with v simultaneously. Clearly, n (T) = 2(q −p+1)+n (G), and n (T) = n (G)
1 1 2 2
and ∆(T) = ∆(G). On the other hand, each k-vdec π of G with k = χ′(G) corresponds to an
s
edge coloring π′ of T such that for any two distinct non-leaf vertices u and v of T, the set of the
colors assigned to the edges incident to u differs from the set of the colors assigned to the edges
incident to v. Also, π′ is a proper edge coloring of T by setting π′(uu′) = π′(vv′) = π(uv) for
uu′,vv′ ∈ E(T) and uv ∈ E′; and π′(e) = π(e) for e ∈ E(H) ⊂ E(T). Thereby, k ≤ χ′(T) since
s
k = χ′(G). This corollary follows by Theorem 1.
s
Corollary 7. Let T be a spanning tree of a connected graph G, and G[E′] be an induced graph
over the edge subset E′ = E(G)\E(T). Then χ′(G) ≤ χ′(T)+χ′(G[E′]), where χ′(H) is the
s s
chromatic index of a graph H.
As further work we present the following
Conjecture 2. Let T be a tree with 2n (T)≤ (n (T)+k−1)2 for k ≥ 1. Then χ′(T) ≤ n (T)+k.
2 1 s 1
Acknowledgment. The authors thank sincerely two referees’ sharp opinions and helpful sug-
gestions that improve greatly the article. B. Yao was supported by the National Natural Science
Foundation of China under Grant No. 61163054 and No. 61363060; X.-en Chen was supported by
the National Natural Science Foundation of China under Grant No. 61163037.
References
[1] P.N.Balister,B.Bollob´asandR.H.Schelp.Vertexdistinguishingcoloringofgraphswith∆(G)=2.DiscreteMath.
252 (2002), 17-29.
[2] A.C.BurrisandR.H.Schelp.Vertex-Distingushingproperedge-colorings.JournalofGraphTheory,26(2)(1997),
73-82.
[3] J. Cˇerny´, M. Hornˇ´ak and R.Sota´k, Observability of a graph, Math. Slovaca 46 (1996), 21-31.
[4] Bing Yao, Zhong-fu Zhang and Jian-fang Wang. Some results on spanning trees. Acta Mathematicae Applicatae
Sinica,English Series 26 (4) (2010) 607-616.
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