Coloring Algorithms on Subcubic Graphs
Harold N. Gabow, San Skulrattanakulchai
hal@cs.colorado.edu, skulratt@cs.colorado.edu
University of Colorado at Boulder Colorado, USA
Coloring Algorithms on Subcubic Graphs – p.1/56
Coloring Algorithms on Subcubic Graphs Harold N. Gabow, San - - PowerPoint PPT Presentation
Coloring Algorithms on Subcubic Graphs Harold N. Gabow, San Skulrattanakulchai hal@cs.colorado.edu, skulratt@cs.colorado.edu University of Colorado at Boulder Colorado, USA Coloring Algorithms on Subcubic Graphs p.1/56 Introduction Graph
Harold N. Gabow, San Skulrattanakulchai
hal@cs.colorado.edu, skulratt@cs.colorado.edu
University of Colorado at Boulder Colorado, USA
Coloring Algorithms on Subcubic Graphs – p.1/56
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ℓ)
ℓ)
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ℓ = χ′?
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ℓ = χ′?
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ℓ = χ′?
ℓ = χ′′?
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ℓ = χ′?
ℓ = χ′′?
❜ ❜ ❜ ❡ ❡ ❡ ❡ ✧ ✧ ✧ ✧ ❜ ❜ ❜ ❜ ✪ ✪ ✪ ✪ ✧ ✧ ✧ ✧
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ℓ = χ′?
ℓ = χ′′?
ℓ ≤ ∆ + 1 if ∆ = 3, 4. LCC holds if graph is bipartite, or
Coloring Algorithms on Subcubic Graphs – p.4/56
ℓ = χ′?
ℓ = χ′′?
ℓ ≤ ∆ + 1 if ∆ = 3, 4. LCC holds if graph is bipartite, or
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ℓ = χ′?
ℓ = χ′′?
ℓ ≤ ∆ + 1 if ∆ = 3, 4. LCC holds if graph is bipartite, or
ℓ ≤ 5 if ∆ = 3
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ℓ ≤ 4, χ′′ ≤ 5, χ′′ ℓ ≤ 5. The first two are
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>= 2 >= 2 >= 2 >= 2 >= 2 {1,2} COLORABLE unless Odd & Same 2−list ODD >= 2 {1,2} {1,2} {1,2} {1, 2}
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>= 5 >= 3 >= 3 >= 3 >= 3 >= 1 >= 4 >= 3 >= 3 >= 3 >= 2 >= 3 Colorable
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2 3 2 1 3 1 2 1 3 2 1 2 4 4
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2 3 2 1 3 1 2 1 3 2 1 2 4 4
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2 3 2 1 3 1 2 1 3 2 1 2 4 4
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2 4 3 4 1 3 1 5
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2 4 3 4 1 3 1 5 3 2 1 2
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2 4 3 4 1 3 1 5 3 2 1 2 >=5 >=4 >=4 >=3 >=3 >=3 >=3 >=3 >=3 >=3 >=3 >=3
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2 3 1 2
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2 3 1 2 1 3 2 2
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2 3 1 2 1 3 2 2 1 2 3 1 1
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2 3 1 2 1 3 2 2 1 2 3 1 1 3
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2 1 3 3 3 3 2 1 3 2 3 1 2 1 1
doubly−forced forced unforced
Observe that in a “bad” cycle not all cycle edges are doubly-forced any cycle edge can be recolored recoloring an unforced edge makes CVCL I applicable e forced = ⇒ not both e’s neighbors are unforced = ⇒ recoloring e makes either CVCL I or CVCL II applicable. In fact, recoloring last cycle edge fixes the problem! (since it’s either forced or unforced)
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4 4 4 4 4 1 5
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4 4 4 4 4 5 1 3 2 3 1 3 2 3 2 1 2
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4 4 4 4 4 5 2
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4 4 4 4 4 5 2 1 3 2 3 1 1 3 2 3 2
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5 5 5 5 5 5 5 5 2 2 1 1 3 3 2 2 1 1 3 3 2 2 1 1 3 3 5 6k + 3
Observe that in any “bad” cycle the number of cycle vertices (and edges) is (odd and) divisible by 3 the colors of the hanging edges go in a cyclic order α, β, γ, α, β, γ, . . .; so are the colors of cycle vertices, shifted by one position any cycle vertex can be recolored recoloring a cycle vertex makes CECL II applicable So, recolor a cycle vertex to fix the problem!
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