Post on 26-Jun-2020
2-DISTANCE-BALANCED GRAPHS
Bostjan FrelihUniversity of Primorska, Slovenia
Joint work with Stefko Miklavic
May 27, 2015
Motivation: Distance-balanced graphs
A graph Γ is said to be distance-balanced if for any edge uv of Γ,the number of vertices closer to u than to v is equal to the numberof vertices closer to v than to u.
A distance partition of a graph with diameter 4 with respect toedge uv .
Motivation: Distance-balanced graphs
I These graphs were first studied (at least implicitly) by K.Handa who considered distance-balanced partial cubes.
I The term itself is due to J. Jerebic, S. Klavzar and D. F. Rallwho studied distance-balanced graphs in the framework ofvarious kinds of graph products.
Motivation: Distance-balanced graphs
I These graphs were first studied (at least implicitly) by K.Handa who considered distance-balanced partial cubes.
I The term itself is due to J. Jerebic, S. Klavzar and D. F. Rallwho studied distance-balanced graphs in the framework ofvarious kinds of graph products.
References
K. Handa, Bipartite graphs with balanced (a, b)-partitions, ArsCombin. 51 (1999), 113-119.
K. Kutnar, A. Malnic, D. Marusic, S. Miklavic,Distance-balanced graphs: symmetry conditions, DiscreteMath. 306 (2006), 1881-1894.
J. Jerebic, S. Klavzar, D. F. Rall, Distance-balanced graphs,Ann. comb. (Print. ed.) 12 (2008), 71-79.
References
K. Kutnar, A. Malnic, D. Marusic, S. Miklavic, The stronglydistance-balanced property of the generalized Petersen graphs.Ars mathematica contemporanea. (Print. ed.) 2 (2009), 41-47.
A. Ilic, S. Klavzar, M. Milanovic, On distance-balanced graphs,Eur. j. comb. 31 (2010), 733-737.
S. Miklavic, P. Sparl, On the connectivity of bipartitedistance-balanced graphs, Europ. j. Combin. 33 (2012),237-247.
Distance-balanced graphs - example
Non-regular bipartite distance-balanced graph H.
Generalization: n-distance-balanced graphs
A graph Γ is said to be n-distance-balanced if there exist at leasttwo vertices at distance n in Γ and if for any two vertices u and vof Γ at distance n, the number of vertices closer to u than to v isequal to the number of vertices closer to v than to u.
We are intrested in 2-distance-balanced graphs.
Generalization: n-distance-balanced graphs
A graph Γ is said to be n-distance-balanced if there exist at leasttwo vertices at distance n in Γ and if for any two vertices u and vof Γ at distance n, the number of vertices closer to u than to v isequal to the number of vertices closer to v than to u.
We are intrested in 2-distance-balanced graphs.
2-distance-balanced graphs
A distance partition of a graph with diameter 4 with respect tovertices u and v at distance 2.
2-distance-balanced graphs: example
Distance-balanced and 2-distance-balanced non-regular graph.
2-distance-balanced graphs
Question:Are there any 2-distance-balanced graphs that are notdistance-balanced?
Theorem (Handa, 1999):
Every distance-balanced graph is 2-connected.
A graph Γ is k-vertex-connected (or k-connected) if it has morethan k vertices and the result of deleting any set of fewer than kvertices is a connected graph.
Question:Are there any connected 2-distance-balanced graphs that are not2-connected?
2-distance-balanced graphs
Question:Are there any 2-distance-balanced graphs that are notdistance-balanced?
Theorem (Handa, 1999):
Every distance-balanced graph is 2-connected.
A graph Γ is k-vertex-connected (or k-connected) if it has morethan k vertices and the result of deleting any set of fewer than kvertices is a connected graph.
Question:Are there any connected 2-distance-balanced graphs that are not2-connected?
2-distance-balanced graphs
Question:Are there any 2-distance-balanced graphs that are notdistance-balanced?
Theorem (Handa, 1999):
Every distance-balanced graph is 2-connected.
A graph Γ is k-vertex-connected (or k-connected) if it has morethan k vertices and the result of deleting any set of fewer than kvertices is a connected graph.
Question:Are there any connected 2-distance-balanced graphs that are not2-connected?
Family of graphs Γ(G , c) - construction
Let G be an arbitrary graph (not necessary connected) and c anadditional vertex.
Then Γ(G , c) is a graph constructed in such a way that
V (Γ(G , c)) = V (G ) ∪ {c},
and
E (Γ(G , c)) = E (G ) ∪ {cv | v ∈ V (G )}.
I Γ(G , c) is connected.
I G is not connected ⇐⇒ Γ(G , c) is not 2-connected.
I Diameter of Γ = Γ(G , c) is at most 2.
Family of graphs Γ(G , c) - construction
Let G be an arbitrary graph (not necessary connected) and c anadditional vertex.
Then Γ(G , c) is a graph constructed in such a way that
V (Γ(G , c)) = V (G ) ∪ {c},
and
E (Γ(G , c)) = E (G ) ∪ {cv | v ∈ V (G )}.
I Γ(G , c) is connected.
I G is not connected ⇐⇒ Γ(G , c) is not 2-connected.
I Diameter of Γ = Γ(G , c) is at most 2.
2-distance-balanced graphs
Question:Are there any connected 2-distance-balanced graphs that are not2-connected?
Theorem (B.F., S. Miklavic)
Graph Γ is a connected 2-distance-balanced graph that is not2-connected iff Γ ∼= Γ(G , c) for some not connected regular graphG .
2-distance-balanced graphs
Question:Are there any connected 2-distance-balanced graphs that are not2-connected?
Theorem (B.F., S. Miklavic)
Graph Γ is a connected 2-distance-balanced graph that is not2-connected iff Γ ∼= Γ(G , c) for some not connected regular graphG .
2-distance-balanced graphs
Lemma (B.F., S. Miklavic)
If G is regular but not a complete graph, then Γ(G , c) is2-distance-balanced.
Proof:
Let G be a regular graph with valency k and construct Γ = Γ(G , c).
Let G1,G2, . . .Gn be connected components of G for some n ≥ 1.
Two essentially different types of vertices at distance two in Γ:
1. both from V (Gi ),
2. one from V (Gi ), the other from V (Gj) for i 6= j .
2-distance-balanced graphs
Lemma (B.F., S. Miklavic)
If G is regular but not a complete graph, then Γ(G , c) is2-distance-balanced.
Proof:
Let G be a regular graph with valency k and construct Γ = Γ(G , c).
Let G1,G2, . . .Gn be connected components of G for some n ≥ 1.
Two essentially different types of vertices at distance two in Γ:
1. both from V (Gi ),
2. one from V (Gi ), the other from V (Gj) for i 6= j .
2-distance-balanced graphs
Lemma (B.F., S. Miklavic)
If G is regular but not a complete graph, then Γ(G , c) is2-distance-balanced.
Proof:
Let G be a regular graph with valency k and construct Γ = Γ(G , c).
Let G1,G2, . . .Gn be connected components of G for some n ≥ 1.
Two essentially different types of vertices at distance two in Γ:
1. both from V (Gi ),
2. one from V (Gi ), the other from V (Gj) for i 6= j .
2-distance-balanced graphs
Lemma (B.F., S. Miklavic)
If G is regular but not a complete graph, then Γ(G , c) is2-distance-balanced.
Proof:
Let G be a regular graph with valency k and construct Γ = Γ(G , c).
Let G1,G2, . . .Gn be connected components of G for some n ≥ 1.
Two essentially different types of vertices at distance two in Γ:
1. both from V (Gi ),
2. one from V (Gi ), the other from V (Gj) for i 6= j .
Proof
1. Pick arbitrary v1, v2 ∈ V (Gi ) s.t. d(v1, v2) = 2.
W Γv1v2
= {v1} ∪ (NGi(v1) \ (NGi
(v1) ∩ NGi(v2)))
|W Γv1v2| = 1 + |NGi
(v1)| − |NGi(v1) ∩ NGi
(v2)|
|W Γv2v1| = 1 + |NGi
(v2)| − |NGi(v2) ∩ NGi
(v1)|
⇒ |W Γv1v2| = |W Γ
v2v1|
Proof
1. Pick arbitrary v1, v2 ∈ V (Gi ) s.t. d(v1, v2) = 2.
W Γv1v2
= {v1} ∪ (NGi(v1) \ (NGi
(v1) ∩ NGi(v2)))
|W Γv1v2| = 1 + |NGi
(v1)| − |NGi(v1) ∩ NGi
(v2)|
|W Γv2v1| = 1 + |NGi
(v2)| − |NGi(v2) ∩ NGi
(v1)|
⇒ |W Γv1v2| = |W Γ
v2v1|
Proof
1. Pick arbitrary v1, v2 ∈ V (Gi ) s.t. d(v1, v2) = 2.
W Γv1v2
= {v1} ∪ (NGi(v1) \ (NGi
(v1) ∩ NGi(v2)))
|W Γv1v2| = 1 + |NGi
(v1)| − |NGi(v1) ∩ NGi
(v2)|
|W Γv2v1| = 1 + |NGi
(v2)| − |NGi(v2) ∩ NGi
(v1)|
⇒ |W Γv1v2| = |W Γ
v2v1|
Proof
1. Pick arbitrary v1, v2 ∈ V (Gi ) s.t. d(v1, v2) = 2.
W Γv1v2
= {v1} ∪ (NGi(v1) \ (NGi
(v1) ∩ NGi(v2)))
|W Γv1v2| = 1 + |NGi
(v1)| − |NGi(v1) ∩ NGi
(v2)|
|W Γv2v1| = 1 + |NGi
(v2)| − |NGi(v2) ∩ NGi
(v1)|
⇒ |W Γv1v2| = |W Γ
v2v1|
Proof
1. Pick arbitrary v1, v2 ∈ V (Gi ) s.t. d(v1, v2) = 2.
W Γv1v2
= {v1} ∪ (NGi(v1) \ (NGi
(v1) ∩ NGi(v2)))
|W Γv1v2| = 1 + |NGi
(v1)| − |NGi(v1) ∩ NGi
(v2)|
|W Γv2v1| = 1 + |NGi
(v2)| − |NGi(v2) ∩ NGi
(v1)|
⇒ |W Γv1v2| = |W Γ
v2v1|
Proof
2. Pick arbitrary v ∈ V (Gi ), u ∈ V (Gj).
W Γvu = {v} ∪ NGi
(v)
W Γuv = {u} ∪ NGj
(u)
⇒ |W Γvu| = 1 + k = |W Γ
uv |
So Γ = Γ(G , c) is 2-distance-balanced.
Proof
2. Pick arbitrary v ∈ V (Gi ), u ∈ V (Gj).
W Γvu = {v} ∪ NGi
(v)
W Γuv = {u} ∪ NGj
(u)
⇒ |W Γvu| = 1 + k = |W Γ
uv |
So Γ = Γ(G , c) is 2-distance-balanced.
Proof
2. Pick arbitrary v ∈ V (Gi ), u ∈ V (Gj).
W Γvu = {v} ∪ NGi
(v)
W Γuv = {u} ∪ NGj
(u)
⇒ |W Γvu| = 1 + k = |W Γ
uv |
So Γ = Γ(G , c) is 2-distance-balanced.
Proof
2. Pick arbitrary v ∈ V (Gi ), u ∈ V (Gj).
W Γvu = {v} ∪ NGi
(v)
W Γuv = {u} ∪ NGj
(u)
⇒ |W Γvu| = 1 + k = |W Γ
uv |
So Γ = Γ(G , c) is 2-distance-balanced.
Proof
2. Pick arbitrary v ∈ V (Gi ), u ∈ V (Gj).
W Γvu = {v} ∪ NGi
(v)
W Γuv = {u} ∪ NGj
(u)
⇒ |W Γvu| = 1 + k = |W Γ
uv |
So Γ = Γ(G , c) is 2-distance-balanced.
2-distance-balanced graphs
Lemma (B.F., S. Miklavic)
If Γ is a connected 2-distance-balanced graph that is not2-connected then Γ ∼= Γ(G , c) for some not connected regulargraph G .
Proof:
Let Γ connected 2-distance-balanced graph that is not 2-connectedand c a cut vertex in Γ.
If we delete vertex c , we get some subgraph G with connectedcomponents G1,G2, . . .Gn for some n ≥ 2.
2-distance-balanced graphs
Lemma (B.F., S. Miklavic)
If Γ is a connected 2-distance-balanced graph that is not2-connected then Γ ∼= Γ(G , c) for some not connected regulargraph G .
Proof:
Let Γ connected 2-distance-balanced graph that is not 2-connectedand c a cut vertex in Γ.
If we delete vertex c , we get some subgraph G with connectedcomponents G1,G2, . . .Gn for some n ≥ 2.
2-distance-balanced graphs
Lemma (B.F., S. Miklavic)
If Γ is a connected 2-distance-balanced graph that is not2-connected then Γ ∼= Γ(G , c) for some not connected regulargraph G .
Proof:
Let Γ connected 2-distance-balanced graph that is not 2-connectedand c a cut vertex in Γ.
If we delete vertex c , we get some subgraph G with connectedcomponents G1,G2, . . .Gn for some n ≥ 2.
2-distance-balanced graphs
Lemma (B.F., S. Miklavic)
If Γ is a connected 2-distance-balanced graph that is not2-connected then Γ ∼= Γ(G , c) for some not connected regulargraph G .
Proof:
Let Γ connected 2-distance-balanced graph that is not 2-connectedand c a cut vertex in Γ.
If we delete vertex c , we get some subgraph G with connectedcomponents G1,G2, . . .Gn for some n ≥ 2.
Proof - Claim 1
Claim 1: c is adjacent to every vertex in Gl for at least one l ,1 ≤ l ≤ n.
Suppose this statement is not true. Then for arbitrary Gi and Gj :
∃v2 ∈ V (Gi ) s.t. d(c, v2) = 2
⇒ ∃v1 ∈ V (Gi ) s.t. d(c , v1) = d(v1, v2) = 1, and
∃u2 ∈ V (Gj) s.t. d(c, u) = 2
⇒ ∃u1 ∈ V (Gi ) s.t. d(c , u1) = d(u1, u2) = 1
Proof - Claim 1
Claim 1: c is adjacent to every vertex in Gl for at least one l ,1 ≤ l ≤ n.
Suppose this statement is not true. Then for arbitrary Gi and Gj :
∃v2 ∈ V (Gi ) s.t. d(c, v2) = 2
⇒ ∃v1 ∈ V (Gi ) s.t. d(c , v1) = d(v1, v2) = 1, and
∃u2 ∈ V (Gj) s.t. d(c, u) = 2
⇒ ∃u1 ∈ V (Gi ) s.t. d(c , u1) = d(u1, u2) = 1
Proof - Claim 1
W Γcv2⊇ {c} ∪ V (Gj)⇒ 1 + |V (Gj)| ≤ |W Γ
cv2|
W Γv2c ⊆ V (Gi ) \ {v1} ⇒ |W Γ
v2c | ≤ |V (Gi )| − 1
⇒ |V (Gj)| ≤ |V (Gi )| − 2
W Γcu2⊇ {c} ∪ V (Gi )⇒ 1 + |V (Gi )| ≤ |W Γ
cu2|
W Γu2c ⊆ V (Gj) \ {u1} ⇒ |W Γ
u2c | ≤ |V (Gj)| − 1
⇒ |V (Gi )|+ 2 ≤ |V (Gj)|
So: |V (Gi )|+ 2 ≤ |V (Gj)| ≤ |V (Gi )| − 2, contradiction.
From now on (w.l.o.g.): c is adjacent to every vertex in V (G1).
Proof - Claim 1
W Γcv2⊇ {c} ∪ V (Gj)⇒ 1 + |V (Gj)| ≤ |W Γ
cv2|
W Γv2c ⊆ V (Gi ) \ {v1} ⇒ |W Γ
v2c | ≤ |V (Gi )| − 1
⇒ |V (Gj)| ≤ |V (Gi )| − 2
W Γcu2⊇ {c} ∪ V (Gi )⇒ 1 + |V (Gi )| ≤ |W Γ
cu2|
W Γu2c ⊆ V (Gj) \ {u1} ⇒ |W Γ
u2c | ≤ |V (Gj)| − 1
⇒ |V (Gi )|+ 2 ≤ |V (Gj)|
So: |V (Gi )|+ 2 ≤ |V (Gj)| ≤ |V (Gi )| − 2, contradiction.
From now on (w.l.o.g.): c is adjacent to every vertex in V (G1).
Proof - Claim 1
W Γcv2⊇ {c} ∪ V (Gj)⇒ 1 + |V (Gj)| ≤ |W Γ
cv2|
W Γv2c ⊆ V (Gi ) \ {v1} ⇒ |W Γ
v2c | ≤ |V (Gi )| − 1
⇒ |V (Gj)| ≤ |V (Gi )| − 2
W Γcu2⊇ {c} ∪ V (Gi )⇒ 1 + |V (Gi )| ≤ |W Γ
cu2|
W Γu2c ⊆ V (Gj) \ {u1} ⇒ |W Γ
u2c | ≤ |V (Gj)| − 1
⇒ |V (Gi )|+ 2 ≤ |V (Gj)|
So: |V (Gi )|+ 2 ≤ |V (Gj)| ≤ |V (Gi )| − 2, contradiction.
From now on (w.l.o.g.): c is adjacent to every vertex in V (G1).
Proof - Claim 1
W Γcv2⊇ {c} ∪ V (Gj)⇒ 1 + |V (Gj)| ≤ |W Γ
cv2|
W Γv2c ⊆ V (Gi ) \ {v1} ⇒ |W Γ
v2c | ≤ |V (Gi )| − 1
⇒ |V (Gj)| ≤ |V (Gi )| − 2
W Γcu2⊇ {c} ∪ V (Gi )⇒ 1 + |V (Gi )| ≤ |W Γ
cu2|
W Γu2c ⊆ V (Gj) \ {u1} ⇒ |W Γ
u2c | ≤ |V (Gj)| − 1
⇒ |V (Gi )|+ 2 ≤ |V (Gj)|
So: |V (Gi )|+ 2 ≤ |V (Gj)| ≤ |V (Gi )| − 2, contradiction.
From now on (w.l.o.g.): c is adjacent to every vertex in V (G1).
Proof - Claim 1
W Γcv2⊇ {c} ∪ V (Gj)⇒ 1 + |V (Gj)| ≤ |W Γ
cv2|
W Γv2c ⊆ V (Gi ) \ {v1} ⇒ |W Γ
v2c | ≤ |V (Gi )| − 1
⇒ |V (Gj)| ≤ |V (Gi )| − 2
W Γcu2⊇ {c} ∪ V (Gi )⇒ 1 + |V (Gi )| ≤ |W Γ
cu2|
W Γu2c ⊆ V (Gj) \ {u1} ⇒ |W Γ
u2c | ≤ |V (Gj)| − 1
⇒ |V (Gi )|+ 2 ≤ |V (Gj)|
So: |V (Gi )|+ 2 ≤ |V (Gj)| ≤ |V (Gi )| − 2, contradiction.
From now on (w.l.o.g.): c is adjacent to every vertex in V (G1).
Proof - Claim 1
W Γcv2⊇ {c} ∪ V (Gj)⇒ 1 + |V (Gj)| ≤ |W Γ
cv2|
W Γv2c ⊆ V (Gi ) \ {v1} ⇒ |W Γ
v2c | ≤ |V (Gi )| − 1
⇒ |V (Gj)| ≤ |V (Gi )| − 2
W Γcu2⊇ {c} ∪ V (Gi )⇒ 1 + |V (Gi )| ≤ |W Γ
cu2|
W Γu2c ⊆ V (Gj) \ {u1} ⇒ |W Γ
u2c | ≤ |V (Gj)| − 1
⇒ |V (Gi )|+ 2 ≤ |V (Gj)|
So: |V (Gi )|+ 2 ≤ |V (Gj)| ≤ |V (Gi )| − 2, contradiction.
From now on (w.l.o.g.): c is adjacent to every vertex in V (G1).
Proof - Claim 1
W Γcv2⊇ {c} ∪ V (Gj)⇒ 1 + |V (Gj)| ≤ |W Γ
cv2|
W Γv2c ⊆ V (Gi ) \ {v1} ⇒ |W Γ
v2c | ≤ |V (Gi )| − 1
⇒ |V (Gj)| ≤ |V (Gi )| − 2
W Γcu2⊇ {c} ∪ V (Gi )⇒ 1 + |V (Gi )| ≤ |W Γ
cu2|
W Γu2c ⊆ V (Gj) \ {u1} ⇒ |W Γ
u2c | ≤ |V (Gj)| − 1
⇒ |V (Gi )|+ 2 ≤ |V (Gj)|
So: |V (Gi )|+ 2 ≤ |V (Gj)| ≤ |V (Gi )| − 2, contradiction.
From now on (w.l.o.g.): c is adjacent to every vertex in V (G1).
Proof - Claim 1
W Γcv2⊇ {c} ∪ V (Gj)⇒ 1 + |V (Gj)| ≤ |W Γ
cv2|
W Γv2c ⊆ V (Gi ) \ {v1} ⇒ |W Γ
v2c | ≤ |V (Gi )| − 1
⇒ |V (Gj)| ≤ |V (Gi )| − 2
W Γcu2⊇ {c} ∪ V (Gi )⇒ 1 + |V (Gi )| ≤ |W Γ
cu2|
W Γu2c ⊆ V (Gj) \ {u1} ⇒ |W Γ
u2c | ≤ |V (Gj)| − 1
⇒ |V (Gi )|+ 2 ≤ |V (Gj)|
So: |V (Gi )|+ 2 ≤ |V (Gj)| ≤ |V (Gi )| − 2, contradiction.
From now on (w.l.o.g.): c is adjacent to every vertex in V (G1).
Proof - Claim 2
Claim 2: Induced subgraph G1 is regular.
Observations: Pick arbitrary u ∈ V (G ) \ V (G1) adjacent to c .
I d(u, v) = 2 for every v ∈ V (G1)
I |W Γux | = |W Γ
uy | for arbitrary x , y ∈ V (G1)
I W Γvu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)
⇒ |W Γvu| = 1 + |NΓ(v)| − 1 = |NG1(v)|+ 1 for every
v ∈ V (G1)
⇒ |NG1(x)|+ 1 = |NΓ(x)| = |W Γxu| = |W Γ
ux | = |W Γuy | = |W Γ
yu| =|NΓ(y)| = |NG1(y)|+ 1 for arbitrary x , y ∈ V (G1)
From now on: Induced subgraph G1 is regular with valency k .(Every vertex in V (G1) has valency k + 1 in Γ.)
Proof - Claim 2
Claim 2: Induced subgraph G1 is regular.
Observations: Pick arbitrary u ∈ V (G ) \ V (G1) adjacent to c .
I d(u, v) = 2 for every v ∈ V (G1)
I |W Γux | = |W Γ
uy | for arbitrary x , y ∈ V (G1)
I W Γvu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)
⇒ |W Γvu| = 1 + |NΓ(v)| − 1 = |NG1(v)|+ 1 for every
v ∈ V (G1)
⇒ |NG1(x)|+ 1 = |NΓ(x)| = |W Γxu| = |W Γ
ux | = |W Γuy | = |W Γ
yu| =|NΓ(y)| = |NG1(y)|+ 1 for arbitrary x , y ∈ V (G1)
From now on: Induced subgraph G1 is regular with valency k .(Every vertex in V (G1) has valency k + 1 in Γ.)
Proof - Claim 2
Claim 2: Induced subgraph G1 is regular.
Observations: Pick arbitrary u ∈ V (G ) \ V (G1) adjacent to c .
I d(u, v) = 2 for every v ∈ V (G1)
I |W Γux | = |W Γ
uy | for arbitrary x , y ∈ V (G1)
I W Γvu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)
⇒ |W Γvu| = 1 + |NΓ(v)| − 1 = |NG1(v)|+ 1 for every
v ∈ V (G1)
⇒ |NG1(x)|+ 1 = |NΓ(x)| = |W Γxu| = |W Γ
ux | = |W Γuy | = |W Γ
yu| =|NΓ(y)| = |NG1(y)|+ 1 for arbitrary x , y ∈ V (G1)
From now on: Induced subgraph G1 is regular with valency k .(Every vertex in V (G1) has valency k + 1 in Γ.)
Proof - Claim 2
Claim 2: Induced subgraph G1 is regular.
Observations: Pick arbitrary u ∈ V (G ) \ V (G1) adjacent to c .
I d(u, v) = 2 for every v ∈ V (G1)
I |W Γux | = |W Γ
uy | for arbitrary x , y ∈ V (G1)
I W Γvu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)
⇒ |W Γvu| = 1 + |NΓ(v)| − 1 = |NG1(v)|+ 1 for every
v ∈ V (G1)
⇒ |NG1(x)|+ 1 = |NΓ(x)| = |W Γxu| = |W Γ
ux | = |W Γuy | = |W Γ
yu| =|NΓ(y)| = |NG1(y)|+ 1 for arbitrary x , y ∈ V (G1)
From now on: Induced subgraph G1 is regular with valency k .(Every vertex in V (G1) has valency k + 1 in Γ.)
Proof - Claim 2
Claim 2: Induced subgraph G1 is regular.
Observations: Pick arbitrary u ∈ V (G ) \ V (G1) adjacent to c .
I d(u, v) = 2 for every v ∈ V (G1)
I |W Γux | = |W Γ
uy | for arbitrary x , y ∈ V (G1)
I W Γvu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)
⇒ |W Γvu| = 1 + |NΓ(v)| − 1 = |NG1(v)|+ 1 for every
v ∈ V (G1)
⇒ |NG1(x)|+ 1 = |NΓ(x)| = |W Γxu| = |W Γ
ux | = |W Γuy | = |W Γ
yu| =|NΓ(y)| = |NG1(y)|+ 1 for arbitrary x , y ∈ V (G1)
From now on: Induced subgraph G1 is regular with valency k .(Every vertex in V (G1) has valency k + 1 in Γ.)
Proof - Claim 2
Claim 2: Induced subgraph G1 is regular.
Observations: Pick arbitrary u ∈ V (G ) \ V (G1) adjacent to c .
I d(u, v) = 2 for every v ∈ V (G1)
I |W Γux | = |W Γ
uy | for arbitrary x , y ∈ V (G1)
I W Γvu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)
⇒ |W Γvu| = 1 + |NΓ(v)| − 1 = |NG1(v)|+ 1 for every
v ∈ V (G1)
⇒ |NG1(x)|+ 1 = |NΓ(x)| = |W Γxu| = |W Γ
ux | = |W Γuy | = |W Γ
yu| =|NΓ(y)| = |NG1(y)|+ 1 for arbitrary x , y ∈ V (G1)
From now on: Induced subgraph G1 is regular with valency k .(Every vertex in V (G1) has valency k + 1 in Γ.)
Proof - Claim 2
Claim 2: Induced subgraph G1 is regular.
Observations: Pick arbitrary u ∈ V (G ) \ V (G1) adjacent to c .
I d(u, v) = 2 for every v ∈ V (G1)
I |W Γux | = |W Γ
uy | for arbitrary x , y ∈ V (G1)
I W Γvu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)
⇒ |W Γvu| = 1 + |NΓ(v)| − 1 = |NG1(v)|+ 1 for every
v ∈ V (G1)
⇒ |NG1(x)|+ 1 = |NΓ(x)| = |W Γxu| = |W Γ
ux | = |W Γuy | = |W Γ
yu| =|NΓ(y)| = |NG1(y)|+ 1 for arbitrary x , y ∈ V (G1)
From now on: Induced subgraph G1 is regular with valency k .(Every vertex in V (G1) has valency k + 1 in Γ.)
Proof - Claim 2
Claim 2: Induced subgraph G1 is regular.
Observations: Pick arbitrary u ∈ V (G ) \ V (G1) adjacent to c .
I d(u, v) = 2 for every v ∈ V (G1)
I |W Γux | = |W Γ
uy | for arbitrary x , y ∈ V (G1)
I W Γvu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)
⇒ |W Γvu| = 1 + |NΓ(v)| − 1 = |NG1(v)|+ 1 for every
v ∈ V (G1)
⇒ |NG1(x)|+ 1 = |NΓ(x)| = |W Γxu| = |W Γ
ux | = |W Γuy | = |W Γ
yu| =|NΓ(y)| = |NG1(y)|+ 1 for arbitrary x , y ∈ V (G1)
From now on: Induced subgraph G1 is regular with valency k .(Every vertex in V (G1) has valency k + 1 in Γ.)
Proof - Claim 3
Claim 3: c is adjacent to every vertex in V (G ).
Suppose this statement is not true. Then:
∃u2 ∈ V (G2) s.t. d(c , u2) = 2
⇒ ∃u1 ∈ V (G2) s.t. d(c , u1) = (d(u1, u2) = 1
We know: |NΓ(v)| = k + 1 for arbitrary v in V (G1).
W Γvu1
= {v} ∪ (NΓ(v) \ {c})
|W Γvu1| = 1 + k + 1− 1 = k + 1
W Γcu2⊇ V (G1) ∪ {c}
|W Γcu2| ≥ |V (G1)|+ 1 ≥ k + 2
Proof - Claim 3
Claim 3: c is adjacent to every vertex in V (G ).
Suppose this statement is not true. Then:
∃u2 ∈ V (G2) s.t. d(c , u2) = 2
⇒ ∃u1 ∈ V (G2) s.t. d(c , u1) = (d(u1, u2) = 1
We know: |NΓ(v)| = k + 1 for arbitrary v in V (G1).
W Γvu1
= {v} ∪ (NΓ(v) \ {c})
|W Γvu1| = 1 + k + 1− 1 = k + 1
W Γcu2⊇ V (G1) ∪ {c}
|W Γcu2| ≥ |V (G1)|+ 1 ≥ k + 2
Proof - Claim 3
Claim 3: c is adjacent to every vertex in V (G ).
Suppose this statement is not true. Then:
∃u2 ∈ V (G2) s.t. d(c , u2) = 2
⇒ ∃u1 ∈ V (G2) s.t. d(c , u1) = (d(u1, u2) = 1
We know: |NΓ(v)| = k + 1 for arbitrary v in V (G1).
W Γvu1
= {v} ∪ (NΓ(v) \ {c})
|W Γvu1| = 1 + k + 1− 1 = k + 1
W Γcu2⊇ V (G1) ∪ {c}
|W Γcu2| ≥ |V (G1)|+ 1 ≥ k + 2
Proof - Claim 3
Claim 3: c is adjacent to every vertex in V (G ).
Suppose this statement is not true. Then:
∃u2 ∈ V (G2) s.t. d(c , u2) = 2
⇒ ∃u1 ∈ V (G2) s.t. d(c , u1) = (d(u1, u2) = 1
We know: |NΓ(v)| = k + 1 for arbitrary v in V (G1).
W Γvu1
= {v} ∪ (NΓ(v) \ {c})
|W Γvu1| = 1 + k + 1− 1 = k + 1
W Γcu2⊇ V (G1) ∪ {c}
|W Γcu2| ≥ |V (G1)|+ 1 ≥ k + 2
Proof - Claim 3
Claim 3: c is adjacent to every vertex in V (G ).
Suppose this statement is not true. Then:
∃u2 ∈ V (G2) s.t. d(c , u2) = 2
⇒ ∃u1 ∈ V (G2) s.t. d(c , u1) = (d(u1, u2) = 1
We know: |NΓ(v)| = k + 1 for arbitrary v in V (G1).
W Γvu1
= {v} ∪ (NΓ(v) \ {c})
|W Γvu1| = 1 + k + 1− 1 = k + 1
W Γcu2⊇ V (G1) ∪ {c}
|W Γcu2| ≥ |V (G1)|+ 1 ≥ k + 2
Proof - Claim 3
Claim 3: c is adjacent to every vertex in V (G ).
Suppose this statement is not true. Then:
∃u2 ∈ V (G2) s.t. d(c , u2) = 2
⇒ ∃u1 ∈ V (G2) s.t. d(c , u1) = (d(u1, u2) = 1
We know: |NΓ(v)| = k + 1 for arbitrary v in V (G1).
W Γvu1
= {v} ∪ (NΓ(v) \ {c})
|W Γvu1| = 1 + k + 1− 1 = k + 1
W Γcu2⊇ V (G1) ∪ {c}
|W Γcu2| ≥ |V (G1)|+ 1 ≥ k + 2
Proof - Claim 3
Define: U =⋃d
i=1 =(D i−1i ∪ D i
i
)W Γ
u2c ⊆ U ⊆W Γu1v
⇒ |W Γu2c | ≤ |W
Γu1v | = k + 1
⇒ k + 2 ≤ |W Γcu2| = |W Γ
u2c | ≤ k + 1, contradiction.
From now on: c is adjacent to every vertex in V (G ).
Proof - Claim 3
Define: U =⋃d
i=1 =(D i−1i ∪ D i
i
)W Γ
u2c ⊆ U ⊆W Γu1v
⇒ |W Γu2c | ≤ |W
Γu1v | = k + 1
⇒ k + 2 ≤ |W Γcu2| = |W Γ
u2c | ≤ k + 1, contradiction.
From now on: c is adjacent to every vertex in V (G ).
Proof - Claim 3
Define: U =⋃d
i=1 =(D i−1i ∪ D i
i
)W Γ
u2c ⊆ U ⊆W Γu1v
⇒ |W Γu2c | ≤ |W
Γu1v | = k + 1
⇒ k + 2 ≤ |W Γcu2| = |W Γ
u2c | ≤ k + 1, contradiction.
From now on: c is adjacent to every vertex in V (G ).
Proof - Claim 3
Define: U =⋃d
i=1 =(D i−1i ∪ D i
i
)W Γ
u2c ⊆ U ⊆W Γu1v
⇒ |W Γu2c | ≤ |W
Γu1v | = k + 1
⇒ k + 2 ≤ |W Γcu2| = |W Γ
u2c | ≤ k + 1, contradiction.
From now on: c is adjacent to every vertex in V (G ).
Proof - Claim 3
Define: U =⋃d
i=1 =(D i−1i ∪ D i
i
)W Γ
u2c ⊆ U ⊆W Γu1v
⇒ |W Γu2c | ≤ |W
Γu1v | = k + 1
⇒ k + 2 ≤ |W Γcu2| = |W Γ
u2c | ≤ k + 1, contradiction.
From now on: c is adjacent to every vertex in V (G ).
Proof - Claim 3
Define: U =⋃d
i=1 =(D i−1i ∪ D i
i
)W Γ
u2c ⊆ U ⊆W Γu1v
⇒ |W Γu2c | ≤ |W
Γu1v | = k + 1
⇒ k + 2 ≤ |W Γcu2| = |W Γ
u2c | ≤ k + 1, contradiction.
From now on: c is adjacent to every vertex in V (G ).
Proof - Claim 4
Claim 4: Graph G is regular with valency k.
W.l.o.g we prove that G2 is regular with valency k.
Pick arbitrary u ∈ V (G2) and v ∈ V (G1) (we already now:d(u, v) = 2).
W Γuv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)
W Γvu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)
⇒ |W Γuv | = 1 + |NG2(u)| in |W Γ
vu| = 1 + k .
Since |W Γuv | = |W Γ
vu|
⇒ |NG2(u)| = k for every u ∈ V (G2)�
Proof - Claim 4
Claim 4: Graph G is regular with valency k.
W.l.o.g we prove that G2 is regular with valency k.
Pick arbitrary u ∈ V (G2) and v ∈ V (G1) (we already now:d(u, v) = 2).
W Γuv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)
W Γvu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)
⇒ |W Γuv | = 1 + |NG2(u)| in |W Γ
vu| = 1 + k .
Since |W Γuv | = |W Γ
vu|
⇒ |NG2(u)| = k for every u ∈ V (G2)�
Proof - Claim 4
Claim 4: Graph G is regular with valency k.
W.l.o.g we prove that G2 is regular with valency k.
Pick arbitrary u ∈ V (G2) and v ∈ V (G1) (we already now:d(u, v) = 2).
W Γuv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)
W Γvu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)
⇒ |W Γuv | = 1 + |NG2(u)| in |W Γ
vu| = 1 + k .
Since |W Γuv | = |W Γ
vu|
⇒ |NG2(u)| = k for every u ∈ V (G2)�
Proof - Claim 4
Claim 4: Graph G is regular with valency k.
W.l.o.g we prove that G2 is regular with valency k.
Pick arbitrary u ∈ V (G2) and v ∈ V (G1) (we already now:d(u, v) = 2).
W Γuv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)
W Γvu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)
⇒ |W Γuv | = 1 + |NG2(u)| in |W Γ
vu| = 1 + k .
Since |W Γuv | = |W Γ
vu|
⇒ |NG2(u)| = k for every u ∈ V (G2)�
Proof - Claim 4
Claim 4: Graph G is regular with valency k.
W.l.o.g we prove that G2 is regular with valency k.
Pick arbitrary u ∈ V (G2) and v ∈ V (G1) (we already now:d(u, v) = 2).
W Γuv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)
W Γvu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)
⇒ |W Γuv | = 1 + |NG2(u)| in |W Γ
vu| = 1 + k .
Since |W Γuv | = |W Γ
vu|
⇒ |NG2(u)| = k for every u ∈ V (G2)�
Proof - Claim 4
Claim 4: Graph G is regular with valency k.
W.l.o.g we prove that G2 is regular with valency k.
Pick arbitrary u ∈ V (G2) and v ∈ V (G1) (we already now:d(u, v) = 2).
W Γuv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)
W Γvu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)
⇒ |W Γuv | = 1 + |NG2(u)| in |W Γ
vu| = 1 + k .
Since |W Γuv | = |W Γ
vu|
⇒ |NG2(u)| = k for every u ∈ V (G2)�
Proof - Claim 4
Claim 4: Graph G is regular with valency k.
W.l.o.g we prove that G2 is regular with valency k.
Pick arbitrary u ∈ V (G2) and v ∈ V (G1) (we already now:d(u, v) = 2).
W Γuv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)
W Γvu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)
⇒ |W Γuv | = 1 + |NG2(u)| in |W Γ
vu| = 1 + k .
Since |W Γuv | = |W Γ
vu|
⇒ |NG2(u)| = k for every u ∈ V (G2)�
2-distance-balanced cartesian product G�H
Question:Which cartesian products G�H are 2-distance-balanced?
Cartesian product of graphs G and H, denoted by G�H, is agraph with vertex set V (G )× V (H), where (u1, v1) and (u2, v2)are adjacent if and only if either
I u1 = u2 and v1 is adjacent to v2 in H, or
I v1 = v2 and u1 is adjacent to u2 in G .
2-distance-balanced cartesian product G�H
Question:Which cartesian products G�H are 2-distance-balanced?
Cartesian product of graphs G and H, denoted by G�H, is agraph with vertex set V (G )× V (H), where (u1, v1) and (u2, v2)are adjacent if and only if either
I u1 = u2 and v1 is adjacent to v2 in H, or
I v1 = v2 and u1 is adjacent to u2 in G .
2-distance-balanced cartesian product G�H
Theorem (B.F., S. Miklavic)
Cartesian product G�H is 2-distance-balanced iff one of thefollowing statements is true:
(i) Both graphs G an H are 2-distance-balanced and1-distance-balanced.
(ii) G is a complete graph Kn for some n ≥ 2 and H is aconnected 2-distance-balanced and 1-distance-balanced graph.
(iii) H is a complete graph Kn for some n ≥ 2 and G is aconnected 2-distance-balanced and 1-distance-balanced graph.
(iv) G is a complete graph Kn and H is a complete graph Km forsome m, n ≥ 2.
2-distance-balanced lexicographic product G [H]
Question:Which lexicographic products G [H] are 2-distance-balanced?
Lexicographic product of graphs G and H, denoted by G [H], is agraph with vertex set V (G )× V (H), where (u1, v1) and (u2, v2)are adjacent if and only if either
I u1 is adjacent to u2 in G , or
I u1 = u2 and v1 is adjacent to v2 in H.
2-distance-balanced lexicographic product G [H]
Question:Which lexicographic products G [H] are 2-distance-balanced?
Lexicographic product of graphs G and H, denoted by G [H], is agraph with vertex set V (G )× V (H), where (u1, v1) and (u2, v2)are adjacent if and only if either
I u1 is adjacent to u2 in G , or
I u1 = u2 and v1 is adjacent to v2 in H.
2-distance-balanced lexicographic product G [H]
Theorem (B.F., S. Miklavic)
Lexicographic product G [H] is 2-distance-balanced iff one of thefollowing statements is true:
(i) G is a connected 2-distance-balanced graph and H is a regulargraph.
(ii) G is a complete graph and H is a regular graph, which is notcomplete.
(iii) G is a complete graph and H is a connected completebipartite graph.
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