2-DISTANCE-BALANCED GRAPHS Bo stjan Frelih University of - - PowerPoint PPT Presentation

2-DISTANCE-BALANCED GRAPHS

Boˇ stjan Frelih University of Primorska, Slovenia Joint work with ˇ Stefko Miklaviˇ c 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 number

• f vertices closer to v than to u.

A distance partition of a graph with diameter 4 with respect to edge uv.

Motivation: Distance-balanced graphs

◮ These graphs were first studied (at least implicitly) by K.

Handa who considered distance-balanced partial cubes.

◮ The term itself is due to J. Jerebic, S. Klavˇ

zar and D. F. Rall who studied distance-balanced graphs in the framework of various kinds of graph products.

Motivation: Distance-balanced graphs

◮ These graphs were first studied (at least implicitly) by K.

Handa who considered distance-balanced partial cubes.

◮ The term itself is due to J. Jerebic, S. Klavˇ

zar and D. F. Rall who studied distance-balanced graphs in the framework of various kinds of graph products.

References

• K. Handa, Bipartite graphs with balanced (a, b)-partitions, Ars
• Combin. 51 (1999), 113-119.
• K. Kutnar, A. Malniˇ

c, D. Maruˇ siˇ c, ˇ

• S. Miklaviˇ

c, Distance-balanced graphs: symmetry conditions, Discrete

• Math. 306 (2006), 1881-1894.
• J. Jerebic, S. Klavˇ

zar, D. F. Rall, Distance-balanced graphs,

• Ann. comb. (Print. ed.) 12 (2008), 71-79.

References

• K. Kutnar, A. Malniˇ

c, D. Maruˇ siˇ c, ˇ

• S. Miklaviˇ

c, The strongly distance-balanced property of the generalized Petersen graphs. Ars mathematica contemporanea. (Print. ed.) 2 (2009), 41-47.

• A. Ili´

c, S. Klavˇ zar, M. Milanovi´ c, On distance-balanced graphs,

• Eur. j. comb. 31 (2010), 733-737.

ˇ

• S. Miklaviˇ

c, P. ˇ Sparl, On the connectivity of bipartite distance-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 least two vertices at distance n in Γ and if for any two vertices u and v

• f Γ at distance n, the number of vertices closer to u than to v is

equal 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 least two vertices at distance n in Γ and if for any two vertices u and v

• f Γ at distance n, the number of vertices closer to u than to v is

equal 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 to vertices 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 not distance-balanced?

Theorem (Handa, 1999):

Every distance-balanced graph is 2-connected. A graph Γ is k-vertex-connected (or k-connected) if it has more than k vertices and the result of deleting any set of fewer than k vertices is a connected graph.

Question:

Are there any connected 2-distance-balanced graphs that are not 2-connected?

2-distance-balanced graphs

Question:

Are there any 2-distance-balanced graphs that are not distance-balanced?

Theorem (Handa, 1999):

Every distance-balanced graph is 2-connected. A graph Γ is k-vertex-connected (or k-connected) if it has more than k vertices and the result of deleting any set of fewer than k vertices is a connected graph.

Question:

Are there any connected 2-distance-balanced graphs that are not 2-connected?

2-distance-balanced graphs

Question:

Are there any 2-distance-balanced graphs that are not distance-balanced?

Theorem (Handa, 1999):

Every distance-balanced graph is 2-connected. A graph Γ is k-vertex-connected (or k-connected) if it has more than k vertices and the result of deleting any set of fewer than k vertices is a connected graph.

Question:

Are there any connected 2-distance-balanced graphs that are not 2-connected?

Family of graphs Γ(G, c) - construction

Let G be an arbitrary graph (not necessary connected) and c an additional 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)}.

◮ Γ(G, c) is connected. ◮ G is not connected ⇐

⇒ Γ(G, c) is not 2-connected.

◮ 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 an additional 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)}.

◮ Γ(G, c) is connected. ◮ G is not connected ⇐

⇒ Γ(G, c) is not 2-connected.

◮ 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 not 2-connected?

Theorem (B.F., ˇ

• S. Miklaviˇ

c)

Graph Γ is a connected 2-distance-balanced graph that is not 2-connected iff Γ ∼ = Γ(G, c) for some not connected regular graph G.

2-distance-balanced graphs

Question:

Are there any connected 2-distance-balanced graphs that are not 2-connected?

Theorem (B.F., ˇ

• S. Miklaviˇ

c)

Graph Γ is a connected 2-distance-balanced graph that is not 2-connected iff Γ ∼ = Γ(G, c) for some not connected regular graph G.

2-distance-balanced graphs

Lemma (B.F., ˇ

• S. Miklaviˇ

c)

If G is regular but not a complete graph, then Γ(G, c) is 2-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 = j.

2-distance-balanced graphs

Lemma (B.F., ˇ

• S. Miklaviˇ

c)

If G is regular but not a complete graph, then Γ(G, c) is 2-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 = j.

2-distance-balanced graphs

Lemma (B.F., ˇ

• S. Miklaviˇ

c)

If G is regular but not a complete graph, then Γ(G, c) is 2-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 = j.

2-distance-balanced graphs

Lemma (B.F., ˇ

• S. Miklaviˇ

c)

If G is regular but not a complete graph, then Γ(G, c) is 2-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 = j.

Proof

• 1. Pick arbitrary v1, v2 ∈ V (Gi) s.t. d(v1, v2) = 2.

W Γ

v1v2 = {v1} ∪ (NGi(v1) \ (NGi(v1) ∩ NGi(v2)))

|W Γ

v1v2| = 1 + |NGi(v1)| − |NGi(v1) ∩ NGi(v2)|

|W Γ

v2v1| = 1 + |NGi(v2)| − |NGi(v2) ∩ NGi(v1)|

⇒ |W Γ

v1v2| = |W Γ v2v1|

Proof

• 1. Pick arbitrary v1, v2 ∈ V (Gi) s.t. d(v1, v2) = 2.

W Γ

v1v2 = {v1} ∪ (NGi(v1) \ (NGi(v1) ∩ NGi(v2)))

|W Γ

v1v2| = 1 + |NGi(v1)| − |NGi(v1) ∩ NGi(v2)|

|W Γ

v2v1| = 1 + |NGi(v2)| − |NGi(v2) ∩ NGi(v1)|

⇒ |W Γ

v1v2| = |W Γ v2v1|

Proof

• 1. Pick arbitrary v1, v2 ∈ V (Gi) s.t. d(v1, v2) = 2.

W Γ

v1v2 = {v1} ∪ (NGi(v1) \ (NGi(v1) ∩ NGi(v2)))

|W Γ

v1v2| = 1 + |NGi(v1)| − |NGi(v1) ∩ NGi(v2)|

|W Γ

v2v1| = 1 + |NGi(v2)| − |NGi(v2) ∩ NGi(v1)|

⇒ |W Γ

v1v2| = |W Γ v2v1|

Proof

• 1. Pick arbitrary v1, v2 ∈ V (Gi) s.t. d(v1, v2) = 2.

W Γ

v1v2 = {v1} ∪ (NGi(v1) \ (NGi(v1) ∩ NGi(v2)))

|W Γ

v1v2| = 1 + |NGi(v1)| − |NGi(v1) ∩ NGi(v2)|

|W Γ

v2v1| = 1 + |NGi(v2)| − |NGi(v2) ∩ NGi(v1)|

⇒ |W Γ

v1v2| = |W Γ v2v1|

Proof

• 1. Pick arbitrary v1, v2 ∈ V (Gi) s.t. d(v1, v2) = 2.

W Γ

v1v2 = {v1} ∪ (NGi(v1) \ (NGi(v1) ∩ NGi(v2)))

|W Γ

v1v2| = 1 + |NGi(v1)| − |NGi(v1) ∩ NGi(v2)|

|W Γ

v2v1| = 1 + |NGi(v2)| − |NGi(v2) ∩ NGi(v1)|

⇒ |W Γ

v1v2| = |W Γ v2v1|

Proof

• 2. Pick arbitrary v ∈ V (Gi), u ∈ V (Gj).

W Γ

vu = {v} ∪ NGi(v)

W Γ

uv = {u} ∪ NGj(u)

⇒ |W Γ

vu| = 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 Γ

vu = {v} ∪ NGi(v)

W Γ

uv = {u} ∪ NGj(u)

⇒ |W Γ

vu| = 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 Γ

vu = {v} ∪ NGi(v)

W Γ

uv = {u} ∪ NGj(u)

⇒ |W Γ

vu| = 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 Γ

vu = {v} ∪ NGi(v)

W Γ

uv = {u} ∪ NGj(u)

⇒ |W Γ

vu| = 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 Γ

vu = {v} ∪ NGi(v)

W Γ

uv = {u} ∪ NGj(u)

⇒ |W Γ

vu| = 1 + k = |W Γ uv|

So Γ = Γ(G, c) is 2-distance-balanced.

2-distance-balanced graphs

Lemma (B.F., ˇ

• S. Miklaviˇ

c)

If Γ is a connected 2-distance-balanced graph that is not 2-connected then Γ ∼ = Γ(G, c) for some not connected regular graph G. Proof: Let Γ connected 2-distance-balanced graph that is not 2-connected and c a cut vertex in Γ. If we delete vertex c, we get some subgraph G with connected components G1, G2, . . . Gn for some n ≥ 2.

2-distance-balanced graphs

Lemma (B.F., ˇ

• S. Miklaviˇ

c)

If Γ is a connected 2-distance-balanced graph that is not 2-connected then Γ ∼ = Γ(G, c) for some not connected regular graph G. Proof: Let Γ connected 2-distance-balanced graph that is not 2-connected and c a cut vertex in Γ. If we delete vertex c, we get some subgraph G with connected components G1, G2, . . . Gn for some n ≥ 2.

2-distance-balanced graphs

Lemma (B.F., ˇ

• S. Miklaviˇ

c)

If Γ is a connected 2-distance-balanced graph that is not 2-connected then Γ ∼ = Γ(G, c) for some not connected regular graph G. Proof: Let Γ connected 2-distance-balanced graph that is not 2-connected and c a cut vertex in Γ. If we delete vertex c, we get some subgraph G with connected components G1, G2, . . . Gn for some n ≥ 2.

2-distance-balanced graphs

Lemma (B.F., ˇ

• S. Miklaviˇ

c)

If Γ is a connected 2-distance-balanced graph that is not 2-connected then Γ ∼ = Γ(G, c) for some not connected regular graph G. Proof: Let Γ connected 2-distance-balanced graph that is not 2-connected and c a cut vertex in Γ. If we delete vertex c, we get some subgraph G with connected components 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 Γ

cv2 ⊇ {c} ∪ V (Gj) ⇒ 1 + |V (Gj)| ≤ |W Γ cv2|

W Γ

v2c ⊆ V (Gi) \ {v1} ⇒ |W Γ v2c| ≤ |V (Gi)| − 1

⇒ |V (Gj)| ≤ |V (Gi)| − 2 W Γ

cu2 ⊇ {c} ∪ V (Gi) ⇒ 1 + |V (Gi)| ≤ |W Γ cu2|

W Γ

u2c ⊆ 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 Γ

cv2 ⊇ {c} ∪ V (Gj) ⇒ 1 + |V (Gj)| ≤ |W Γ cv2|

W Γ

v2c ⊆ V (Gi) \ {v1} ⇒ |W Γ v2c| ≤ |V (Gi)| − 1

⇒ |V (Gj)| ≤ |V (Gi)| − 2 W Γ

cu2 ⊇ {c} ∪ V (Gi) ⇒ 1 + |V (Gi)| ≤ |W Γ cu2|

W Γ

u2c ⊆ 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 Γ

cv2 ⊇ {c} ∪ V (Gj) ⇒ 1 + |V (Gj)| ≤ |W Γ cv2|

W Γ

v2c ⊆ V (Gi) \ {v1} ⇒ |W Γ v2c| ≤ |V (Gi)| − 1

⇒ |V (Gj)| ≤ |V (Gi)| − 2 W Γ

cu2 ⊇ {c} ∪ V (Gi) ⇒ 1 + |V (Gi)| ≤ |W Γ cu2|

W Γ

u2c ⊆ 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 Γ

cv2 ⊇ {c} ∪ V (Gj) ⇒ 1 + |V (Gj)| ≤ |W Γ cv2|

W Γ

v2c ⊆ V (Gi) \ {v1} ⇒ |W Γ v2c| ≤ |V (Gi)| − 1

⇒ |V (Gj)| ≤ |V (Gi)| − 2 W Γ

cu2 ⊇ {c} ∪ V (Gi) ⇒ 1 + |V (Gi)| ≤ |W Γ cu2|

W Γ

u2c ⊆ 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 Γ

cv2 ⊇ {c} ∪ V (Gj) ⇒ 1 + |V (Gj)| ≤ |W Γ cv2|

W Γ

v2c ⊆ V (Gi) \ {v1} ⇒ |W Γ v2c| ≤ |V (Gi)| − 1

⇒ |V (Gj)| ≤ |V (Gi)| − 2 W Γ

cu2 ⊇ {c} ∪ V (Gi) ⇒ 1 + |V (Gi)| ≤ |W Γ cu2|

W Γ

u2c ⊆ 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 Γ

cv2 ⊇ {c} ∪ V (Gj) ⇒ 1 + |V (Gj)| ≤ |W Γ cv2|

W Γ

v2c ⊆ V (Gi) \ {v1} ⇒ |W Γ v2c| ≤ |V (Gi)| − 1

⇒ |V (Gj)| ≤ |V (Gi)| − 2 W Γ

cu2 ⊇ {c} ∪ V (Gi) ⇒ 1 + |V (Gi)| ≤ |W Γ cu2|

W Γ

u2c ⊆ 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 Γ

cv2 ⊇ {c} ∪ V (Gj) ⇒ 1 + |V (Gj)| ≤ |W Γ cv2|

W Γ

v2c ⊆ V (Gi) \ {v1} ⇒ |W Γ v2c| ≤ |V (Gi)| − 1

⇒ |V (Gj)| ≤ |V (Gi)| − 2 W Γ

cu2 ⊇ {c} ∪ V (Gi) ⇒ 1 + |V (Gi)| ≤ |W Γ cu2|

W Γ

u2c ⊆ 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 Γ

cv2 ⊇ {c} ∪ V (Gj) ⇒ 1 + |V (Gj)| ≤ |W Γ cv2|

W Γ

v2c ⊆ V (Gi) \ {v1} ⇒ |W Γ v2c| ≤ |V (Gi)| − 1

⇒ |V (Gj)| ≤ |V (Gi)| − 2 W Γ

cu2 ⊇ {c} ∪ V (Gi) ⇒ 1 + |V (Gi)| ≤ |W Γ cu2|

W Γ

u2c ⊆ 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.

◮ d(u, v) = 2 for every v ∈ V (G1) ◮ |W Γ ux| = |W Γ uy| for arbitrary x, y ∈ V (G1) ◮ W Γ vu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)

⇒ |W Γ

vu| = 1 + |NΓ(v)| − 1 = |NG1(v)| + 1 for every

v ∈ V (G1) ⇒ |NG1(x)| + 1 = |NΓ(x)| = |W Γ

xu| = |W Γ ux| = |W Γ uy| = |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.

◮ d(u, v) = 2 for every v ∈ V (G1) ◮ |W Γ ux| = |W Γ uy| for arbitrary x, y ∈ V (G1) ◮ W Γ vu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)

⇒ |W Γ

vu| = 1 + |NΓ(v)| − 1 = |NG1(v)| + 1 for every

v ∈ V (G1) ⇒ |NG1(x)| + 1 = |NΓ(x)| = |W Γ

xu| = |W Γ ux| = |W Γ uy| = |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.

◮ d(u, v) = 2 for every v ∈ V (G1) ◮ |W Γ ux| = |W Γ uy| for arbitrary x, y ∈ V (G1) ◮ W Γ vu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)

⇒ |W Γ

vu| = 1 + |NΓ(v)| − 1 = |NG1(v)| + 1 for every

v ∈ V (G1) ⇒ |NG1(x)| + 1 = |NΓ(x)| = |W Γ

xu| = |W Γ ux| = |W Γ uy| = |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.

◮ d(u, v) = 2 for every v ∈ V (G1) ◮ |W Γ ux| = |W Γ uy| for arbitrary x, y ∈ V (G1) ◮ W Γ vu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)

⇒ |W Γ

vu| = 1 + |NΓ(v)| − 1 = |NG1(v)| + 1 for every

v ∈ V (G1) ⇒ |NG1(x)| + 1 = |NΓ(x)| = |W Γ

xu| = |W Γ ux| = |W Γ uy| = |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.

◮ d(u, v) = 2 for every v ∈ V (G1) ◮ |W Γ ux| = |W Γ uy| for arbitrary x, y ∈ V (G1) ◮ W Γ vu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)

⇒ |W Γ

vu| = 1 + |NΓ(v)| − 1 = |NG1(v)| + 1 for every

v ∈ V (G1) ⇒ |NG1(x)| + 1 = |NΓ(x)| = |W Γ

xu| = |W Γ ux| = |W Γ uy| = |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.

◮ d(u, v) = 2 for every v ∈ V (G1) ◮ |W Γ ux| = |W Γ uy| for arbitrary x, y ∈ V (G1) ◮ W Γ vu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)

⇒ |W Γ

vu| = 1 + |NΓ(v)| − 1 = |NG1(v)| + 1 for every

v ∈ V (G1) ⇒ |NG1(x)| + 1 = |NΓ(x)| = |W Γ

xu| = |W Γ ux| = |W Γ uy| = |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.

◮ d(u, v) = 2 for every v ∈ V (G1) ◮ |W Γ ux| = |W Γ uy| for arbitrary x, y ∈ V (G1) ◮ W Γ vu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)

⇒ |W Γ

vu| = 1 + |NΓ(v)| − 1 = |NG1(v)| + 1 for every

v ∈ V (G1) ⇒ |NG1(x)| + 1 = |NΓ(x)| = |W Γ

xu| = |W Γ ux| = |W Γ uy| = |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.

◮ d(u, v) = 2 for every v ∈ V (G1) ◮ |W Γ ux| = |W Γ uy| for arbitrary x, y ∈ V (G1) ◮ W Γ vu = {v} ∪ (NΓ(v) \ {c}) for every v ∈ V (G1)

⇒ |W Γ

vu| = 1 + |NΓ(v)| − 1 = |NG1(v)| + 1 for every

v ∈ V (G1) ⇒ |NG1(x)| + 1 = |NΓ(x)| = |W Γ

xu| = |W Γ ux| = |W Γ uy| = |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 Γ

vu1 = {v} ∪ (NΓ(v) \ {c})

|W Γ

vu1| = 1 + k + 1 − 1 = k + 1

W Γ

cu2 ⊇ V (G1) ∪ {c}

|W Γ

cu2| ≥ |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 Γ

vu1 = {v} ∪ (NΓ(v) \ {c})

|W Γ

vu1| = 1 + k + 1 − 1 = k + 1

W Γ

cu2 ⊇ V (G1) ∪ {c}

|W Γ

cu2| ≥ |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 Γ

vu1 = {v} ∪ (NΓ(v) \ {c})

|W Γ

vu1| = 1 + k + 1 − 1 = k + 1

W Γ

cu2 ⊇ V (G1) ∪ {c}

|W Γ

cu2| ≥ |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 Γ

vu1 = {v} ∪ (NΓ(v) \ {c})

|W Γ

vu1| = 1 + k + 1 − 1 = k + 1

W Γ

cu2 ⊇ V (G1) ∪ {c}

|W Γ

cu2| ≥ |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 Γ

vu1 = {v} ∪ (NΓ(v) \ {c})

|W Γ

vu1| = 1 + k + 1 − 1 = k + 1

W Γ

cu2 ⊇ V (G1) ∪ {c}

|W Γ

cu2| ≥ |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 Γ

vu1 = {v} ∪ (NΓ(v) \ {c})

|W Γ

vu1| = 1 + k + 1 − 1 = k + 1

W Γ

cu2 ⊇ V (G1) ∪ {c}

|W Γ

cu2| ≥ |V (G1)| + 1 ≥ k + 2

Proof - Claim 3

Define: U = d

i=1 =

• Di−1

i

∪ Di

i

• W Γ

u2c ⊆ U ⊆ W Γ u1v

⇒ |W Γ

u2c| ≤ |W Γ u1v| = k + 1

⇒ k + 2 ≤ |W Γ

cu2| = |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 =

• Di−1

i

∪ Di

i

• W Γ

u2c ⊆ U ⊆ W Γ u1v

⇒ |W Γ

u2c| ≤ |W Γ u1v| = k + 1

⇒ k + 2 ≤ |W Γ

cu2| = |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 =

• Di−1

i

∪ Di

i

• W Γ

u2c ⊆ U ⊆ W Γ u1v

⇒ |W Γ

u2c| ≤ |W Γ u1v| = k + 1

⇒ k + 2 ≤ |W Γ

cu2| = |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 =

• Di−1

i

∪ Di

i

• W Γ

u2c ⊆ U ⊆ W Γ u1v

⇒ |W Γ

u2c| ≤ |W Γ u1v| = k + 1

⇒ k + 2 ≤ |W Γ

cu2| = |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 =

• Di−1

i

∪ Di

i

• W Γ

u2c ⊆ U ⊆ W Γ u1v

⇒ |W Γ

u2c| ≤ |W Γ u1v| = k + 1

⇒ k + 2 ≤ |W Γ

cu2| = |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 =

• Di−1

i

∪ Di

i

• W Γ

u2c ⊆ U ⊆ W Γ u1v

⇒ |W Γ

u2c| ≤ |W Γ u1v| = k + 1

⇒ k + 2 ≤ |W Γ

cu2| = |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 Γ

uv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)

W Γ

vu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)

⇒ |W Γ

uv| = 1 + |NG2(u)| in |W Γ vu| = 1 + k.

Since |W Γ

uv| = |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 Γ

uv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)

W Γ

vu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)

⇒ |W Γ

uv| = 1 + |NG2(u)| in |W Γ vu| = 1 + k.

Since |W Γ

uv| = |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 Γ

uv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)

W Γ

vu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)

⇒ |W Γ

uv| = 1 + |NG2(u)| in |W Γ vu| = 1 + k.

Since |W Γ

uv| = |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 Γ

uv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)

W Γ

vu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)

⇒ |W Γ

uv| = 1 + |NG2(u)| in |W Γ vu| = 1 + k.

Since |W Γ

uv| = |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 Γ

uv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)

W Γ

vu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)

⇒ |W Γ

uv| = 1 + |NG2(u)| in |W Γ vu| = 1 + k.

Since |W Γ

uv| = |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 Γ

uv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)

W Γ

vu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)

⇒ |W Γ

uv| = 1 + |NG2(u)| in |W Γ vu| = 1 + k.

Since |W Γ

uv| = |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 Γ

uv = {u} ∪ (NΓ(u) \ {c}) = {u} ∪ NG2(u)

W Γ

vu = {v} ∪ (NΓ(v) \ {c}) = {v} ∪ NG1(v)

⇒ |W Γ

uv| = 1 + |NG2(u)| in |W Γ vu| = 1 + k.

Since |W Γ

uv| = |W Γ vu|

⇒ |NG2(u)| = k for every u ∈ V (G2)

2-distance-balanced cartesian product GH

Question:

Which cartesian products GH are 2-distance-balanced? Cartesian product of graphs G and H, denoted by GH, is a graph with vertex set V (G) × V (H), where (u1, v1) and (u2, v2) are adjacent if and only if either

◮ u1 = u2 and v1 is adjacent to v2 in H, or ◮ v1 = v2 and u1 is adjacent to u2 in G.

2-distance-balanced cartesian product GH

Question:

Which cartesian products GH are 2-distance-balanced? Cartesian product of graphs G and H, denoted by GH, is a graph with vertex set V (G) × V (H), where (u1, v1) and (u2, v2) are adjacent if and only if either

◮ u1 = u2 and v1 is adjacent to v2 in H, or ◮ v1 = v2 and u1 is adjacent to u2 in G.

2-distance-balanced cartesian product GH

Theorem (B.F., ˇ

• S. Miklaviˇ

c)

Cartesian product GH is 2-distance-balanced iff one of the following statements is true: (i) Both graphs G an H are 2-distance-balanced and 1-distance-balanced. (ii) G is a complete graph Kn for some n ≥ 2 and H is a connected 2-distance-balanced and 1-distance-balanced graph. (iii) H is a complete graph Kn for some n ≥ 2 and G is a connected 2-distance-balanced and 1-distance-balanced graph. (iv) G is a complete graph Kn and H is a complete graph Km for some 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 a graph with vertex set V (G) × V (H), where (u1, v1) and (u2, v2) are adjacent if and only if either

◮ u1 is adjacent to u2 in G, or ◮ 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 a graph with vertex set V (G) × V (H), where (u1, v1) and (u2, v2) are adjacent if and only if either

◮ u1 is adjacent to u2 in G, or ◮ u1 = u2 and v1 is adjacent to v2 in H.

2-distance-balanced lexicographic product G[H]

Theorem (B.F., ˇ

• S. Miklaviˇ

c)

Lexicographic product G[H] is 2-distance-balanced iff one of the following statements is true: (i) G is a connected 2-distance-balanced graph and H is a regular graph. (ii) G is a complete graph and H is a regular graph, which is not complete. (iii) G is a complete graph and H is a connected complete bipartite graph.

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