Coloring problem
It is highly recommended to read the Independent set problem chapter before reading this one.
Problem definition
A vertex coloring is an assignment of labels or colors to each vertex of a graph such that no edge connects two identically colored vertices. In the following, we are going to defined a 3-coloring problem for the Petersen graph.
using GenericTensorNetworks, Graphs
graph = Graphs.smallgraph(:petersen)
{10, 15} undirected simple Int64 graph
We can visualize this graph using the following function
rot15(a, b, i::Int) = cos(2i*π/5)*a + sin(2i*π/5)*b, cos(2i*π/5)*b - sin(2i*π/5)*a
locations = [[rot15(0.0, 60.0, i) for i=0:4]..., [rot15(0.0, 30, i) for i=0:4]...]
show_graph(graph, locations; format=:svg)
Generic tensor network representation
We can define the 3-coloring problem with the Coloring
type as
coloring = Coloring{3}(graph)
Coloring{3, UnitWeight}(Graphs.SimpleGraphs.SimpleGraph{Int64}(15, [[2, 5, 6], [1, 3, 7], [2, 4, 8], [3, 5, 9], [1, 4, 10], [1, 8, 9], [2, 9, 10], [3, 6, 10], [4, 6, 7], [5, 7, 8]]), UnitWeight())
The tensor network representation of the 3-coloring problem can be obtained by
problem = GenericTensorNetwork(coloring)
GenericTensorNetwork{Coloring{3, UnitWeight}, OMEinsum.DynamicNestedEinsum{Int64}, Int64}(Coloring{3, UnitWeight}(Graphs.SimpleGraphs.SimpleGraph{Int64}(15, [[2, 5, 6], [1, 3, 7], [2, 4, 8], [3, 5, 9], [1, 4, 10], [1, 8, 9], [2, 9, 10], [3, 6, 10], [4, 6, 7], [5, 7, 8]]), UnitWeight()), 3∘7∘5∘6, 3∘6∘5∘7 ->
├─ 3∘7∘5∘6, 3∘5∘6∘7 -> 3∘7∘5∘6
│ ├─ 1∘3∘7, 5∘6∘1 -> 3∘7∘5∘6
│ │ ├─ 1∘2, 3∘7∘2 -> 1∘3∘7
│ │ │ ├─ 2, 2∘1 -> 1∘2
│ │ │ │ ├─ 2
│ │ │ │ └─ 1, 1∘2 -> 2∘1
│ │ │ │ ⋮
│ │ │ │
│ │ │ └─ 2∘3, 2∘7 -> 3∘7∘2
│ │ │ ├─ 3, 2∘3 -> 2∘3
│ │ │ │ ⋮
│ │ │ │
│ │ │ └─ 7, 2∘7 -> 2∘7
│ │ │ ⋮
│ │ │
│ │ └─ 1∘5, 1∘6 -> 5∘6∘1
│ │ ├─ 5, 1∘5 -> 1∘5
│ │ │ ├─ 5
│ │ │ └─ 1∘5
│ │ └─ 6, 1∘6 -> 1∘6
│ │ ├─ 6
│ │ └─ 1∘6
│ └─ 3∘9∘5, 6∘7∘9 -> 3∘5∘6∘7
│ ├─ 3∘9∘4, 4∘5 -> 3∘9∘5
│ │ ├─ 3∘4, 4∘9 -> 3∘9∘4
│ │ │ ├─ 4, 3∘4 -> 3∘4
│ │ │ │ ⋮
│ │ │ │
│ │ │ └─ 9, 4∘9 -> 4∘9
│ │ │ ⋮
│ │ │
│ │ └─ 4∘5
│ └─ 6∘9, 7∘9 -> 6∘7∘9
│ ├─ 6∘9
│ └─ 7∘9
└─ 3∘6∘10, 5∘7∘10 -> 3∘6∘5∘7
├─ 3∘8, 6∘10∘8 -> 3∘6∘10
│ ├─ 8, 3∘8 -> 3∘8
│ │ ├─ 8
│ │ └─ 3∘8
│ └─ 6∘8, 8∘10 -> 6∘10∘8
│ ├─ 6∘8
│ └─ 8∘10
└─ 5∘10, 7∘10 -> 5∘7∘10
├─ 10, 5∘10 -> 5∘10
│ ├─ 10
│ └─ 5∘10
└─ 7∘10
, Dict{Int64, Int64}())
Theory (can skip)
Type Coloring
can be used for constructing the tensor network with optimized contraction order for a coloring problem. Let us use 3-coloring problem defined on vertices as an example. For a vertex $v$, we define the degrees of freedom $c_v\in\{1,2,3\}$ and a vertex tensor labelled by it as
\[W(v) = \left(\begin{matrix} 1\\ 1\\ 1 \end{matrix}\right).\]
For an edge $(u, v)$, we define an edge tensor as a matrix labelled by $(c_u, c_v)$ to specify the constraint
\[B = \left(\begin{matrix} 1 & x & x\\ x & 1 & x\\ x & x & 1 \end{matrix}\right).\]
The number of possible coloring can be obtained by contracting this tensor network by setting vertex tensor elements $r_v, g_v$ and $b_v$ to 1.
Solving properties
counting all possible coloring
num_of_coloring = solve(problem, CountingMax())[]
(15.0, 120.0)ₜ
finding one best coloring
single_solution = solve(problem, SingleConfigMax())[]
is_vertex_coloring(graph, single_solution.c.data)
vertex_color_map = Dict(0=>"red", 1=>"green", 2=>"blue")
show_graph(graph, locations; format=:svg, vertex_colors=[vertex_color_map[Int(c)]
for c in single_solution.c.data])
Let us try to solve the same issue on its line graph, a graph that generated by mapping an edge to a vertex and two edges sharing a common vertex will be connected.
linegraph = line_graph(graph)
show_graph(linegraph, [0.5 .* (locations[e.src] .+ locations[e.dst])
for e in edges(graph)]; format=:svg)
Let us construct the tensor network and see if there are solutions.
lineproblem = Coloring{3}(linegraph);
num_of_coloring = solve(GenericTensorNetwork(lineproblem), CountingMax())[]
(28.0, 1800.0)ₜ
You will see the maximum size 28 is smaller than the number of edges in the linegraph
, meaning no solution for the 3-coloring on edges of a Petersen graph.
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