Examples of Mixed Hypergraphs

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Transcript Examples of Mixed Hypergraphs

V.Voloshin: Examples of Mixed
Hypergraphs
“pictures speak louder than
words...”
Drawing a graph G=(V,E):
V={1,2,3,4,5}
E={{1,2},{2,3},{3,4},{4,5},{5,1},{1,4},{3,5}}
1
5
2
adjacent vertices
4
3
Coloring of G=(V,E):
V={1,2,3,4,5}
E={{1,2},{2,3},{3,4},{4,5},{5,1},{1,4},{3,5}}
1
5
2
Χ(G)=3
4
3
Drawing a hypergraph
H=(X,D), X={1,2,3,4},
D={{1,2,3},{2,3,4},{1,4}}={D1,D2,D3,}
2
D1
3
1
D3
4
D2
Coloring a hypergraph:
H=(X,D), X={1,2,3,4},
D={{1,2,3},{2,3,4},{1,4}}={D1,D2,D3,}
2
D1
3
1
D3
4
D2
Χ(H)=2
Drawing a mixed hypergraph
H=(X,C,D):
X={1,2,3,4}, C={{1,2,4}}={C1},
D={{2,3,4},{2,4}}={D1,D2}
3
C1
2
D2
D1
1
4
Uncolorable mixed hypergraphs:
D
1
2
the smallest uncolorable mixed hypergraph
C
violated C-edge
1
D1
C1
C2
2
uncolorable mixed hypergraph H=(X,C,D)
violated D-edge
D4
D2
1
1
D3
Coloring mixed hypergraph
H=(X,C,D): numbers = colors
Proper 2-,3-,…λ-coloring:
Strict 2-coloring:
C1
1
1
1
D2
D1
2
Chromatic numbers:
1
C
1
C
1
1
1
3
2
2
D
Χ(H)=2
Χ(H)=3
D
Uniquely colorable mixed
hypergraphs:
H2=(X,C,D)
H1=(X,Ø,D)
1
1
D
3
C
2
D
D
2
D
D
D
D
C
1
1
R(H1)=(0,0,1)
Χ = Χ =3
2
R(H2)=(0,1,0,0,0)
Χ=Χ=2
C
Chromatic polynomial:
1
3
C
D
2
H
C
1
2
3
D
C
1
3
H
D
2
H
R (H)=(0,3,0)
3 feasible partitions
3 2! strict 2-colorings
If we have λ≥2 colors, then the number P (H, λ) of proper λ-colorings is:
3 2! λ(λ-1)/2! = 3 λ(λ-1)=3λ(2)
How to get a gap? – Step 1:
• Construct a “can”:
R (H )=(0,r2,0,0)
C
C
C
C
Feasible set S={2}
D
How to get a gap? - Step 2:
Construct:
R (H ) = (0, r2, r3, 0,0,0)
can
can
can
Feasible set S={2,3}
How to get a gap? - Step 3:
Each subset = “can”
R (H)=(0,r2, 0,r4,0,0,0,0)
Feasible set S={2,4}
We have broken it!!!
The smallest mixed hypergraph
with a gap, H2,4 (Jiang, Mubayi,
Tuza, Voloshin, West):
D
Bi-edges
R(H2,4)=(0,4,0,1,0,0)
D
C
S(H)={2,4}
C
C
The smallest 3-uniform bihypergraph
with a gap (L.Gionfriddo, V.Voloshin,
2000):
R(H)=(0,12,0,3,0,0,0), S(H)={2,4}
Planar mixed hypergraph with a
gap, H ’2,4 (Kobler, Kundgen, 2001)
D
D
C
C
D
D
D
D
C
R(H’2,4)=(0,1,0,1,0,0), S(H)={2,4}
C
THANK YOU!