Strong edge-coloring of (3, ∆)-bipartite graphs
Julien Bensmaila , Aurélie Lagouttea and Petru Valicovb
a. LIP – ENS de Lyon – France
b. LIF – Université Aix-Marseille – France
LIRMM
March 12th, 2015
1 / 20
Strong edge-coloring
G : undirected simple graph
c: edge-coloring of G
Definition: strong edge-coloring
We call c strong if every two edges at distance at most 2 in G are assigned
distinct colors by c.
2 / 20
Strong edge-coloring
G : undirected simple graph
c: edge-coloring of G
Definition: strong edge-coloring
We call c strong if every two edges at distance at most 2 in G are assigned
distinct colors by c.
2 / 20
Strong edge-coloring
G : undirected simple graph
c: edge-coloring of G
Definition: strong edge-coloring
We call c strong if every two edges at distance at most 2 in G are assigned
distinct colors by c.
2 / 20
Strong edge-coloring
G : undirected simple graph
c: edge-coloring of G
Definition: strong edge-coloring
We call c strong if every two edges at distance at most 2 in G are assigned
distinct colors by c.
2 / 20
Strong edge-coloring
G : undirected simple graph
c: edge-coloring of G
Definition: strong edge-coloring
We call c strong if every two edges at distance at most 2 in G are assigned
distinct colors by c.
2 / 20
Strong edge-coloring
G : undirected simple graph
c: edge-coloring of G
Definition: strong edge-coloring
We call c strong if every two edges at distance at most 2 in G are assigned
distinct colors by c.
Equivalently:
edge-partition giving induced matchings
proper vertex-coloring of L(G )2
2 / 20
Strong chromatic index
∆: maximum degree of an explicit graph
Definition: strong chromatic index
The least number of colors in a strong edge-coloring of G is the strong
chromatic index of G , denoted χ0s (G ).
3 / 20
Strong chromatic index
∆: maximum degree of an explicit graph
Definition: strong chromatic index
The least number of colors in a strong edge-coloring of G is the strong
chromatic index of G , denoted χ0s (G ).
Brooks-like argument: χ0s (G ) ≤ 2∆2 − 2∆ + 1 (≈ 2∆2 )
∆
∆−1
∆−1
u
v
∆−1
3 / 20
On the Brooks-like upper bound on χ0s
optimality of 2∆2 ?
Theorem [Molloy, Reed – 1997]
If ∆ is large enough, then χ0s (G ) ≤ 1.998∆2 .
4 / 20
On the Brooks-like upper bound on χ0s
optimality of 2∆2 ?
Theorem [Molloy, Reed – 1997]
If ∆ is large enough, then χ0s (G ) ≤ 1.998∆2 .
What would be a “worst graph”? C5∆ :
4 / 20
On the Brooks-like upper bound on χ0s
optimality of 2∆2 ?
Theorem [Molloy, Reed – 1997]
If ∆ is large enough, then χ0s (G ) ≤ 1.998∆2 .
What would be a “worst graph”? C5∆ :
• every Ij is an independent set,
I2
I3
I1
• two “adjacent” Ij ’s are complete to each other,
• if ∆ = 2k, then |Ij | = k,
I4
I5
• if ∆ = 2k + 1, then |I1 | = |I2 | = |I3 | = k,
and |I4 | = |I5 | = k + 1.
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Erdős and Nešetřil’s conjecture
Conjecture [Erdős, Nešetřil – 1985]
(
We have
χ0s (G )
≤
5 2
4 ∆ for ∆ even, and
1
2
4 (5∆ − 2∆ + 1) otherwise.
5 / 20
Erdős and Nešetřil’s conjecture
Conjecture [Erdős, Nešetřil – 1985]
(
We have
χ0s (G )
≤
5 2
4 ∆ for ∆ even, and
1
2
4 (5∆ − 2∆ + 1) otherwise.
Facts:
reached for C5∆ ’s only [Chung, Gyárfás, Tuza, Trotter – 1990]
5 / 20
Erdős and Nešetřil’s conjecture
Conjecture [Erdős, Nešetřil – 1985]
(
We have
χ0s (G )
≤
5 2
4 ∆ for ∆ even, and
1
2
4 (5∆ − 2∆ + 1) otherwise.
Facts:
reached for C5∆ ’s only [Chung, Gyárfás, Tuza, Trotter – 1990]
verified for ∆ = 3 [Andersen – 1992]
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Erdős and Nešetřil’s conjecture
Conjecture [Erdős, Nešetřil – 1985]
(
We have
χ0s (G )
≤
5 2
4 ∆ for ∆ even, and
1
2
4 (5∆ − 2∆ + 1) otherwise.
Facts:
reached for C5∆ ’s only [Chung, Gyárfás, Tuza, Trotter – 1990]
verified for ∆ = 3 [Andersen – 1992]
for ∆ = 4, we know χ0s (G ) ≤ 22 [Cranston – 2006]
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Beyond Erdős and Nešetřil’s construction
Less dependencies for graphs with no small cycles
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Beyond Erdős and Nešetřil’s construction
Less dependencies for graphs with no small cycles
Theorem [Mahdian – 2000]
2
If G is C4 -free, then χ0s (G ) ≤ (2 + o(1)) ln∆∆ .
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Beyond Erdős and Nešetřil’s construction
Less dependencies for graphs with no small cycles
Theorem [Mahdian – 2000]
2
If G is C4 -free, then χ0s (G ) ≤ (2 + o(1)) ln∆∆ .
What for C3 - and C5 -free graphs?
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What for bipartite graphs?
Bipartite graphs are C3 and C5 -free...
Conjecture [Faudree, Gyárfás, Schelp, Tuza – 1990]
If G is bipartite, then χ0s (G ) ≤ ∆2 .
Reached e.g. for any complete bipartite graph Ka,a
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What for bipartite graphs?
Bipartite graphs are C3 and C5 -free...
Conjecture [Faudree, Gyárfás, Schelp, Tuza – 1990]
If G is bipartite, then χ0s (G ) ≤ ∆2 .
Reached e.g. for any complete bipartite graph Ka,a
G = (A, B, E ): bipartite graph with bipartition A and B
(∆A , ∆B )-bipartite graph: A and B have maximum degree ∆A and ∆B , resp.
Conjecture [Brualdi, Quinn Massey – 1993]
If G is (∆A , ∆B )-bipartite, then χ0s (G ) ≤ ∆A ∆B .
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Refined conjecture for bipartite graphs
Conjecture [Brualdi, Quinn Massey – 1993]
If G is (∆A , ∆B )-bipartite, then χ0s (G ) ≤ ∆A ∆B .
Verified when:
∆A = ∆B = 3 [Steger and Yu – 1993]
∆A = 2 [Nakprasit – 2008]
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Refined conjecture for bipartite graphs
Conjecture [Brualdi, Quinn Massey – 1993]
If G is (∆A , ∆B )-bipartite, then χ0s (G ) ≤ ∆A ∆B .
Verified when:
∆A = ∆B = 3 [Steger and Yu – 1993]
∆A = 2 [Nakprasit – 2008]
We confirm the conjecture when ∆A = 3 and ∆B ≥ 4
Theorem [B., Lagoutte, Valicov – 2014+]
If G is (3, ∆B )-bipartite, then χ0s (G ) ≤ 4∆B .
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(3, ∆B )-bipartite graphs – a proof
Theorem [B., Lagoutte, Valicov – 2014+]
If G is (3, ∆B )-bipartite, then χ0s (G ) ≤ 4∆B .
Proof. G = (A, B, E ), where all vertices in A have degree 3
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(3, ∆B )-bipartite graphs – a proof
Theorem [B., Lagoutte, Valicov – 2014+]
If G is (3, ∆B )-bipartite, then χ0s (G ) ≤ 4∆B .
Proof. G = (A, B, E ), where all vertices in A have degree 3
Idea: we produce a strong 4∆B edge-coloring c of G by combining an incidence
coloring cA of the incidences involving A and one cB of the incidences involving B
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(3, ∆B )-bipartite graphs – a proof
Theorem [B., Lagoutte, Valicov – 2014+]
If G is (3, ∆B )-bipartite, then χ0s (G ) ≤ 4∆B .
Proof. G = (A, B, E ), where all vertices in A have degree 3
Idea: we produce a strong 4∆B edge-coloring c of G by combining an incidence
coloring cA of the incidences involving A and one cB of the incidences involving B
a1
(1, 2)
b1
(1, 3)
a2
(2, 1)
b2
(1, 1)
a3
(2, 2)
b3
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(3, ∆B )-bipartite graphs – a proof
Theorem [B., Lagoutte, Valicov – 2014+]
If G is (3, ∆B )-bipartite, then χ0s (G ) ≤ 4∆B .
Proof. G = (A, B, E ), where all vertices in A have degree 3
Idea: we produce a strong 4∆B edge-coloring c of G by combining an incidence
coloring cA of the incidences involving A and one cB of the incidences involving B
a1
(1, 2)
b1
(1, 3)
a2
(2, 1)
b2
(1, 1)
a3
(2, 2)
b3
9 / 20
(3, ∆B )-bipartite graphs – a proof
Theorem [B., Lagoutte, Valicov – 2014+]
If G is (3, ∆B )-bipartite, then χ0s (G ) ≤ 4∆B .
Proof. G = (A, B, E ), where all vertices in A have degree 3
Idea: we produce a strong 4∆B edge-coloring c of G by combining an incidence
coloring cA of the incidences involving A and one cB of the incidences involving B
a1
(1, 2)
b1
(1, 3)
a2
(2, 1)
b2
(1, 1)
a3
(2, 2)
b3
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Mixing cA and cB to get c
Three steps:
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Mixing cA and cB to get c
Three steps:
1. choose cB as a specific ∆B -incidence coloring
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Mixing cA and cB to get c
Three steps:
1. choose cB as a specific ∆B -incidence coloring
2. according to cB , define cA using at most 4 colors
10 / 20
Mixing cA and cB to get c
Three steps:
1. choose cB as a specific ∆B -incidence coloring
2. according to cB , define cA using at most 4 colors
3. mix cA and cB , i.e. set c(e) = (cA (e), cB (e)) for every e ∈ E
10 / 20
Mixing cA and cB to get c
Three steps:
1. choose cB as a specific ∆B -incidence coloring
2. according to cB , define cA using at most 4 colors
3. mix cA and cB , i.e. set c(e) = (cA (e), cB (e)) for every e ∈ E
Remark: we have c(e) 6= c(f ) as soon as cA (e) 6= cA (f ) or cB (e) 6= cB (f )
10 / 20
Mixing cA and cB to get c
Three steps:
1. choose cB as a specific ∆B -incidence coloring
2. according to cB , define cA using at most 4 colors
3. mix cA and cB , i.e. set c(e) = (cA (e), cB (e)) for every e ∈ E
Remark: we have c(e) 6= c(f ) as soon as cA (e) 6= cA (f ) or cB (e) 6= cB (f )
As cB , just consider a proper ∆B -incidence coloring
(?, 1)
(?, 2)
b1
(?, 3)
(adjacent incidences are of the form (b1 , e) and (b1 , f ))
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Coloring procedure
cB yields a characterization of the vertices in A as follows:
11 / 20
Coloring procedure
cB yields a characterization of the vertices in A as follows:
Type 1: all three incident edges have the same color by cB
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Coloring procedure
cB yields a characterization of the vertices in A as follows:
Type 1: all three incident edges have the same color by cB
Type 2: two incident edges (= paired) have the same color by cB , which is
different from the color of the third one (= lonely )
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Coloring procedure
cB yields a characterization of the vertices in A as follows:
Type 1: all three incident edges have the same color by cB
Type 2: two incident edges (= paired) have the same color by cB , which is
different from the color of the third one (= lonely )
Type 3: all three incident edges have distinct colors by cB
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Coloring procedure
cB yields a characterization of the vertices in A as follows:
Type 1: all three incident edges have the same color by cB
Type 2: two incident edges (= paired) have the same color by cB , which is
different from the color of the third one (= lonely )
Type 3: all three incident edges have distinct colors by cB
As cB , choose the one maximizing the number of Type 1 vertices, and then
maximizing the number of Type 2 vertices
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Coloring procedure
cB yields a characterization of the vertices in A as follows:
Type 1: all three incident edges have the same color by cB
Type 2: two incident edges (= paired) have the same color by cB , which is
different from the color of the third one (= lonely )
Type 3: all three incident edges have distinct colors by cB
As cB , choose the one maximizing the number of Type 1 vertices, and then
maximizing the number of Type 2 vertices
For every edge e, we assign a color to cA (e) in such a way that no conflict appears
11 / 20
Coloring procedure
cB yields a characterization of the vertices in A as follows:
Type 1: all three incident edges have the same color by cB
Type 2: two incident edges (= paired) have the same color by cB , which is
different from the color of the third one (= lonely )
Type 3: all three incident edges have distinct colors by cB
As cB , choose the one maximizing the number of Type 1 vertices, and then
maximizing the number of Type 2 vertices
For every edge e, we assign a color to cA (e) in such a way that no conflict appears
Coloring procedure:
Step 1: color the
Step 2: color the
Step 3: color the
Step 4: color the
edges incident to Type 1 vertices
paired edges incident to Type 2
edges incident to Type 3 vertices
lonely edges incident to Type 2 vertices
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Step 1: color the edges incident to Type 1 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
12 / 20
Step 1: color the edges incident to Type 1 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
Proof. Follows from the properness of cB
(−, j 0 )
a1
(−, j 0 )
(−, j 0 )
b
(?, j)
a2
(?, j)
(?, j)
12 / 20
Step 1: color the edges incident to Type 1 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
Proof. Follows from the properness of cB
(−, j 0 )
a1
(−, j 0 )
(−, j 0 )
b
(1, j)
a2
(2, j)
(3, j)
12 / 20
Step 2: color the paired edges incident to Type 2 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
13 / 20
Step 2: color the paired edges incident to Type 2 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
Proof. There may be a forbidden color (−, j) adjacent to the bottom-most edge
– this is the only one since cB is proper and “maximum”
(−, `)
a1
(−, `)
(−, `)
b
(?, j)
a2
(?, j)
(−, k)
if a1 is Type 1, then the colors are different
13 / 20
Step 2: color the paired edges incident to Type 2 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
Proof. There may be a forbidden color (−, j) adjacent to the bottom-most edge
– this is the only one since cB is proper and “maximum”
(1, j)
a1
(2, j)
(−, `)
b
(?, j)
a2
(?, j)
(−, k)
if a1 is Type 2...
13 / 20
Step 2: color the paired edges incident to Type 2 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
Proof. There may be a forbidden color (−, j) adjacent to the bottom-most edge
– this is the only one since cB is proper and “maximum”
(−, j)
a1
(−, j)
(−, j)
b
(−, `)
a2
(−, j)
(−, k)
... we could just switch two colors and make a1 Type 1
13 / 20
Step 2: color the paired edges incident to Type 2 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
Proof. There may be a forbidden color (−, j) adjacent to the bottom-most edge
– this is the only one since cB is proper and “maximum”
(−, j)
a1
(−, j)
(−, j)
b
(−, `)
a2
(−, j)
(−, k)
... we could just switch two colors and make a1 Type 1
13 / 20
Step 3: color the edges incident to Type 3 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
14 / 20
Step 3: color the edges incident to Type 3 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
Proof. There may be up to two forbidden colors (−, j) near the bottom-most
edges – these are the only ones since cB is proper and “maximum”
(−, m)
a1
(−, m)
(−, m)
b
(?, j)
a2
(−, k)
(−, `)
if a1 is Type 1, then the colors are different
14 / 20
Step 3: color the edges incident to Type 3 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
Proof. There may be up to two forbidden colors (−, j) near the bottom-most
edges – these are the only ones since cB is proper and “maximum”
(1, j)
a1
(2, j)
(−, m)
b
(?, j)
a2
(−, k)
(−, `)
if a1 is Type 2...
14 / 20
Step 3: color the edges incident to Type 3 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
Proof. There may be up to two forbidden colors (−, j) near the bottom-most
edges – these are the only ones since cB is proper and “maximum”
(−, j)
a1
(−, j)
(−, j)
b
(−, m)
a2
(−, k)
(−, `)
... we could switch two colors and make a1 Type 1
14 / 20
Step 3: color the edges incident to Type 3 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
Proof. There may be up to two forbidden colors (−, j) near the bottom-most
edges – these are the only ones since cB is proper and “maximum”
(−, n)
a1
(1, j)
(−, m)
b
(?, j)
a2
(−, k)
(−, `)
if a1 is Type 3...
14 / 20
Step 3: color the edges incident to Type 3 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
Proof. There may be up to two forbidden colors (−, j) near the bottom-most
edges – these are the only ones since cB is proper and “maximum”
(−, n)
a1
(−, j)
(−, j)
b
(−, m)
a2
(−, k)
(−, `)
... we could switch two colors and make a1 Type 2
14 / 20
Step 3: color the edges incident to Type 3 vertices
Just assign the colors among {1, 2, 3} greedily
Lemma
There is at least one available color for every edge to color.
Proof. There may be up to two forbidden colors (−, j) near the bottom-most
edges – these are the only ones since cB is proper and “maximum”
(−, n)
a1
(−, j)
(−, j)
b
(−, m)
a2
(−, k)
(−, `)
... we could switch two colors and make a1 Type 2
14 / 20
Step 4: color the lonely edges incident to Type 2 vertices
Cj : (connected) subgraph induced by the j-lonely edges (i.e. with cB = j)
Alternate cycle of Cj : edges alternate between j-lonely and non-j-lonely
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Step 4: color the lonely edges incident to Type 2 vertices
Cj : (connected) subgraph induced by the j-lonely edges (i.e. with cB = j)
Alternate cycle of Cj : edges alternate between j-lonely and non-j-lonely
Lemma
Every cycle of Cj is alternate.
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Step 4: color the lonely edges incident to Type 2 vertices
Cj : (connected) subgraph induced by the j-lonely edges (i.e. with cB = j)
Alternate cycle of Cj : edges alternate between j-lonely and non-j-lonely
Lemma
Every cycle of Cj is alternate.
Proof. Assume C is a cycle of Cj – if C is not alternate, then there are two
adjacent non-lonely edges e and e 0 both incident to a vertex in B
?
?
?
?
?
?
hidden text
15 / 20
Step 4: color the lonely edges incident to Type 2 vertices
Cj : (connected) subgraph induced by the j-lonely edges (i.e. with cB = j)
Alternate cycle of Cj : edges alternate between j-lonely and non-j-lonely
Lemma
Every cycle of Cj is alternate.
Proof. Assume C is a cycle of Cj – if C is not alternate, then there are two
adjacent non-lonely edges e and e 0 both incident to a vertex in B
?
?
?
?
?
Type 2 + 2 adjacent j-lonely = new Type 1
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Step 4: color the lonely edges incident to Type 2 vertices
Cj : (connected) subgraph induced by the j-lonely edges (i.e. with cB = j)
Alternate cycle of Cj : edges alternate between j-lonely and non-j-lonely
Lemma
Every cycle of Cj is alternate.
Proof. Assume C is a cycle of Cj – if C is not alternate, then there are two
adjacent non-lonely edges e and e 0 both incident to a vertex in B
?
?
?
?
No two adjacent j-lonely
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Step 4: color the lonely edges incident to Type 2 vertices
Cj : (connected) subgraph induced by the j-lonely edges (i.e. with cB = j)
Alternate cycle of Cj : edges alternate between j-lonely and non-j-lonely
Lemma
Every cycle of Cj is alternate.
Proof. Assume C is a cycle of Cj – if C is not alternate, then there are two
adjacent non-lonely edges e and e 0 both incident to a vertex in B
?
?
?
hidden text
15 / 20
Step 4: color the lonely edges incident to Type 2 vertices
Cj : (connected) subgraph induced by the j-lonely edges (i.e. with cB = j)
Alternate cycle of Cj : edges alternate between j-lonely and non-j-lonely
Lemma
Every cycle of Cj is alternate.
Proof. Assume C is a cycle of Cj – if C is not alternate, then there are two
adjacent non-lonely edges e and e 0 both incident to a vertex in B
?
?
hidden text
15 / 20
Step 4: color the lonely edges incident to Type 2 vertices
Cj : (connected) subgraph induced by the j-lonely edges (i.e. with cB = j)
Alternate cycle of Cj : edges alternate between j-lonely and non-j-lonely
Lemma
Every cycle of Cj is alternate.
Proof. Assume C is a cycle of Cj – if C is not alternate, then there are two
adjacent non-lonely edges e and e 0 both incident to a vertex in B
?
15 / 20
Step 4: color the lonely edges incident to Type 2 vertices
Cj : (connected) subgraph induced by the j-lonely edges (i.e. with cB = j)
Alternate cycle of Cj : edges alternate between j-lonely and non-j-lonely
Lemma
Every cycle of Cj is alternate.
Proof. Assume C is a cycle of Cj – if C is not alternate, then there are two
adjacent non-lonely edges e and e 0 both incident to a vertex in B
?
How to close the tour?
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Step 4: color the lonely edges incident to Type 2 vertices
Cj : (connected) subgraph induced by the j-lonely edges (i.e. with cB = j)
Alternate cycle of Cj : edges alternate between j-lonely and non-j-lonely
Lemma
Every cycle of Cj is alternate.
Proof. Assume C is a cycle of Cj – if C is not alternate, then there are two
adjacent non-lonely edges e and e 0 both incident to a vertex in B
?
How to close the tour?
15 / 20
On the structure of the Cj ’s
Lemma
Every two cycles of Cj are disjoint.
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On the structure of the Cj ’s
Lemma
Every two cycles of Cj are disjoint.
Proof. If two cycles C1 and C2 of Cj share a vertex without sharing an edge, then
a vertex is adjacent to two j-lonely edges
16 / 20
On the structure of the Cj ’s
Lemma
Every two cycles of Cj are disjoint.
Proof. If two cycles C1 and C2 of Cj share a vertex without sharing an edge, then
a vertex is adjacent to two j-lonely edges
C1
C2
hidden text
16 / 20
On the structure of the Cj ’s
Lemma
Every two cycles of Cj are disjoint.
Proof. If two cycles C1 and C2 of Cj share a vertex without sharing an edge, then
a vertex is adjacent to two j-lonely edges
C1
C2
Type 2 + 2 adjacent lonely = new Type 1
16 / 20
On the structure of the Cj ’s
Lemma
Every two cycles of Cj are disjoint.
Proof. If two cycles C1 and C2 of Cj share a vertex without sharing an edge, then
a vertex is adjacent to two j-lonely edges
C1
C2
Second end point of the intersecting path cannot be correct
16 / 20
On the structure of the Cj ’s
Lemma
Every two cycles of Cj are disjoint.
Proof. If two cycles C1 and C2 of Cj share a vertex without sharing an edge, then
a vertex is adjacent to two j-lonely edges
C1
C2
Second end point of the intersecting path cannot be correct
16 / 20
On the structure of the Cj ’s
Lemma
Cj cannot have two disjoint cycles joined by a path.
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On the structure of the Cj ’s
Lemma
Cj cannot have two disjoint cycles joined by a path.
Proof. Assume C1 and C2 are linked by a path u...v where u ∈ C1 and v ∈ C2
C1
?
?
?
?
C2
hidden text
17 / 20
On the structure of the Cj ’s
Lemma
Cj cannot have two disjoint cycles joined by a path.
Proof. Assume C1 and C2 are linked by a path u...v where u ∈ C1 and v ∈ C2
C1
?
?
?
?
C2
Type 2 + 2 adjacent lonely = new Type 1
17 / 20
On the structure of the Cj ’s
Lemma
Cj cannot have two disjoint cycles joined by a path.
Proof. Assume C1 and C2 are linked by a path u...v where u ∈ C1 and v ∈ C2
C1
?
?
?
?
C2
hidden text
17 / 20
On the structure of the Cj ’s
Lemma
Cj cannot have two disjoint cycles joined by a path.
Proof. Assume C1 and C2 are linked by a path u...v where u ∈ C1 and v ∈ C2
C1
?
?
?
C2
hidden text
17 / 20
On the structure of the Cj ’s
Lemma
Cj cannot have two disjoint cycles joined by a path.
Proof. Assume C1 and C2 are linked by a path u...v where u ∈ C1 and v ∈ C2
C1
?
C2
Type 2 + 2 adjacent lonely = new Type 1
17 / 20
On the structure of the Cj ’s
Lemma
Cj cannot have two disjoint cycles joined by a path.
Proof. Assume C1 and C2 are linked by a path u...v where u ∈ C1 and v ∈ C2
C1
?
C2
How to join the two cycles?
17 / 20
On the structure of the Cj ’s
Lemma
Cj cannot have two disjoint cycles joined by a path.
Proof. Assume C1 and C2 are linked by a path u...v where u ∈ C1 and v ∈ C2
C1
?
C2
How to join the two cycles?
17 / 20
Back to coloring Step 4
Step 4:
Phase 1: color the unique cycle C of Cj
Phase 2: color every tree T of the forest Cj − E (C )
18 / 20
Back to coloring Step 4
Step 4:
Phase 1: color the unique cycle C of Cj
Phase 2: color every tree T of the forest Cj − E (C )
Lemma
The j-lonely edges of C can be colored with {1, 2, 3, 4}.
18 / 20
Back to coloring Step 4
Step 4:
Phase 1: color the unique cycle C of Cj
Phase 2: color every tree T of the forest Cj − E (C )
Lemma
The j-lonely edges of C can be colored with {1, 2, 3, 4}.
Proof. Color the j-lonely edges consecutively
(1, j)
(−, 6= j)
C
(?, j)
(−, 6= j)
(?, j)
hidden text
18 / 20
Back to coloring Step 4
Step 4:
Phase 1: color the unique cycle C of Cj
Phase 2: color every tree T of the forest Cj − E (C )
Lemma
The j-lonely edges of C can be colored with {1, 2, 3, 4}.
Proof. Color the j-lonely edges consecutively
(1, j)
(2, j)
(−, 6= j)
(−, 6= j)
C
(?, j)
(−, 6= j)
(?, j)
hidden text
18 / 20
Back to coloring Step 4
Step 4:
Phase 1: color the unique cycle C of Cj
Phase 2: color every tree T of the forest Cj − E (C )
Lemma
The j-lonely edges of C can be colored with {1, 2, 3, 4}.
Proof. Color the j-lonely edges consecutively
(1, j)
(2, j)
(−, 6= j)
(−, 6= j)
C
(?, j)
(?, ?)
(−, k)
(−, 6= j)
(?, ?)
(?, j)
hidden text
18 / 20
Back to coloring Step 4
Step 4:
Phase 1: color the unique cycle C of Cj
Phase 2: color every tree T of the forest Cj − E (C )
Lemma
The j-lonely edges of C can be colored with {1, 2, 3, 4}.
Proof. Color the j-lonely edges consecutively
(1, j)
(2, j)
(−, 6= j)
(−, 6= j)
C
(?, j)
(3, j)
(−, k)
(−, 6= j)
(4, j)
(?, j)
Switch to create a new Type 1 vertex
18 / 20
Back to coloring Step 4
Step 4:
Phase 1: color the unique cycle C of Cj
Phase 2: color every tree T of the forest Cj − E (C )
Lemma
The j-lonely edges of C can be colored with {1, 2, 3, 4}.
Proof. Color the j-lonely edges consecutively
(1, j)
(2, j)
(−, 6= j)
(−, 6= j)
C
(?, j)
(−, k)
(−, k)
(−, 6= j)
(?, j)
(?, j)
Not yet colored
18 / 20
Back to coloring Step 4
Step 4:
Phase 1: color the unique cycle C of Cj
Phase 2: color every tree T of the forest Cj − E (C )
Lemma
The j-lonely edges of C can be colored with {1, 2, 3, 4}.
Proof. Color the j-lonely edges consecutively
(1, j)
(2, j)
(−, 6= j)
(−, 6= j)
C
(?, j)
(−, `)
(−, k)
(−, 6= j)
(3, j)
(?, j)
Switch to create a new Type 2 vertex
18 / 20
Back to coloring Step 4
Step 4:
Phase 1: color the unique cycle C of Cj
Phase 2: color every tree T of the forest Cj − E (C )
Lemma
The j-lonely edges of C can be colored with {1, 2, 3, 4}.
Proof. Color the j-lonely edges consecutively
(1, j)
(2, j)
(−, 6= j)
(−, 6= j)
C
(?, j)
(−, `)
(−, k)
(−, 6= j)
(3, j)
(?, j)
Switch to create a new Type 2 vertex
18 / 20
Step 4 – the end!
Remark: color 4 may be needed for the last edge of C
19 / 20
Step 4 – the end!
Remark: color 4 may be needed for the last edge of C
Lemma
The j-lonely edges of T can be colored with {1, 2, 3, 4}.
19 / 20
Step 4 – the end!
Remark: color 4 may be needed for the last edge of C
Lemma
The j-lonely edges of T can be colored with {1, 2, 3, 4}.
Proof. Color the j-lonely edges as given by a BFS algorithm
BFS order
(2, j)
(−, 6= j)
(1, j)
(−, 6= j)
(2, j)
(−, 6= j)
(?, j)
hidden text
19 / 20
Step 4 – the end!
Remark: color 4 may be needed for the last edge of C
Lemma
The j-lonely edges of T can be colored with {1, 2, 3, 4}.
Proof. Color the j-lonely edges as given by a BFS algorithm
BFS order
(2, j)
(−, 6= j)
(1, j)
(−, 6= j)
(2, j)
(−, 6= j)
(?, j)
(−, 6= j)
(2, j)
hidden text
19 / 20
Step 4 – the end!
Remark: color 4 may be needed for the last edge of C
Lemma
The j-lonely edges of T can be colored with {1, 2, 3, 4}.
Proof. Color the j-lonely edges as given by a BFS algorithm
BFS order
(2, j)
(−, 6= j)
(1, j)
(−, 6= j)
(2, j)
(−, 6= j)
(?, j)
(−, 6= j)
(2, j)
(−, k)
(3, j)
Switch to create a new Type 1 vertex
(4, j)
19 / 20
Step 4 – the end!
Remark: color 4 may be needed for the last edge of C
Lemma
The j-lonely edges of T can be colored with {1, 2, 3, 4}.
Proof. Color the j-lonely edges as given by a BFS algorithm
BFS order
(2, j)
(−, 6= j)
(1, j)
(−, 6= j)
(2, j)
(−, 6= j)
(?, j)
(−, 6= j)
(2, j)
(−, k)
(−, `)
Switch to create a new Type 2 vertex
(3, j)
19 / 20
Step 4 – the end!
Remark: color 4 may be needed for the last edge of C
Lemma
The j-lonely edges of T can be colored with {1, 2, 3, 4}.
Proof. Color the j-lonely edges as given by a BFS algorithm
BFS order
(2, j)
(−, 6= j)
(1, j)
(−, 6= j)
(2, j)
(−, 6= j)
(?, j)
(−, 6= j)
(2, j)
(−, k)
(−, `)
Switch to create a new Type 2 vertex
(3, j)
19 / 20
Conclusions and open questions
The refined conjecture says 3∆B ...
... can our proof be improved?
20 / 20
Conclusions and open questions
The refined conjecture says 3∆B ...
... can our proof be improved?
Hardly generalizable to larger values of ∆A ...
... though it might be successful for 4
20 / 20
Conclusions and open questions
The refined conjecture says 3∆B ...
... can our proof be improved?
Hardly generalizable to larger values of ∆A ...
... though it might be successful for 4
Particular construction of c...
... what for the list version?
20 / 20
Conclusions and open questions
The refined conjecture says 3∆B ...
... can our proof be improved?
Hardly generalizable to larger values of ∆A ...
... though it might be successful for 4
Particular construction of c...
... what for the list version?
Everything is done in polynomial time with cB in hand...
... it is NP-complete to choose it maximum...
... but maximality is actually not necessary
20 / 20
Conclusions and open questions
The refined conjecture says 3∆B ...
... can our proof be improved?
Hardly generalizable to larger values of ∆A ...
... though it might be successful for 4
Particular construction of c...
... what for the list version?
Everything is done in polynomial time with cB in hand...
... it is NP-complete to choose it maximum...
... but maximality is actually not necessary
Thank you for your attention.
20 / 20