Tight lower bound for the Channel Assignment
problem
Arkadiusz Socała
University of Warsaw
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
Graph Coloring
Recall the Graph Coloring problem.
2
1
3
3
1
2
1
coloring c : V → {1, . . . , s}
c(u) 6= c(v ) for every edge uv ∈ E
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
Graph Coloring
Recall the Graph Coloring problem.
2
1
1
1
3
1
1
1
3
1
1
1
1
1
1
2
1
coloring c : V → {1, . . . , s}
|c(u) − c(v )| ≥ 1 for every edge uv ∈ E
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
Channel Assignment
Weights on edges mean minimum allowed color differences.
9
4
1
2
3
5
3
4
4
2
2
3
7
1
5
6
1
coloring c : V → {1, . . . , s}
|c(u) − c(v )| ≥ w (uv ) for every edge uv ∈ E
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
The Channel Assignment problem.
Input
graph G (V , E )
weight function w : E → N+ .
Definitions
An assignment c : V → {1, . . . , s} is called proper when
∀uv ∈E |c(u) − c(v )| ≥ w (uv ).
The number s is called the span of c.
Problem
Find a proper assignment of minimum span.
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
The Channel Assignment problem.
Motivation
Assignment of the radio
frequencies:
n radio emitters
they interfere each
other
minimize the range
of the used
frequencies (span)
Introduced by
Hale in 1980.
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
Algorithms for Channel Assignment
General version
O ∗ (n!)-time (McDiarmid, 2003)
Our result: There is no 2o(n log n) -time algorithm (under ETH)
`-bounded version
Assume ∀uv ∈E w (uv ) ≤ `.
O ∗ ((2` + 1)n )-time (McDiarmid, 2003)
O ∗ ((` + 2)n )-time (Kral, 2005)
O ∗ ((` + 1)n )-time (Cygan and Kowalik, 2011)
√
O ∗ ((2 ` + 1)n )-time (Kowalik and S, 2014)
These are all dynamic programming.
All algorithms above can be modified to count proper colorings.
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
CSP Hierarchy
(s, 2)-CSP
Generalized T -Coloring
Channel Assignment
T -Coloring
Graph Coloring
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
T -Coloring
List of forbidden differences.
T = {0, 1, 2, 5}
7
1
4
4
1
7
1
coloring c : V → {1, . . . , s}
|c(u) − c(v )| 6∈ T for every edge uv ∈ E
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
Generalized T -Coloring
Lists of forbidden differences.
7
0, 1, 2
1
0, 2
6
0, 1, 2, 3
0, 1, 7
0, 1, 3, 4
3
0, 1, 2
0, 2, 5
2, 4
4
0, 1, 3
0, 3, 5
2
6
coloring c : V → {1, . . . , s}
|c(u) − c(v )| 6∈ T (uv ) for every edge uv ∈ E
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
(s, 2)-CSP (Constraint Satisfaction Problem)
Lists of forbidden pairs of colors.
3
(1, 1), (1, 2)
(3, 1), (3, 3)
1
2
(3, 3)
(1, 2), (2, 1)
(2, 1), (2, 3)
4
(1, 1)
(2, 4)
(1, 4), (2, 2)
5
(1, 2), (1, 4)
1
(1, 1), (2, 2)
2
coloring c : V → {1, . . . , s}
(c(u), c(v )) 6∈ C (uv ) for every edge uv ∈ E
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
CSP Hierarchy
(s, 2)-CSP
no 2O(n) under ETH (Traxler)
Generalized T -Coloring √
o( n)
under ETH
no 22
(Kowalik and S)
Channel Assignment
O ∗ (n!)
T -Coloring
(McDiarmid)
Graph Coloring
O ∗ (2n )
(Björklund et al.)
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
CSP Hierarchy
(s, 2)-CSP
no 2O(n) under ETH (Traxler)
Generalized T -Coloring √
o( n)
no 22
under ETH
(Kowalik and S)
T -Coloring
Channel Assignment
O ∗ (n!) (McDiarmid)
Our Result:
no 2o(n log n) under ETH
Graph Coloring
O ∗ (2n ) (Björklund et al.)
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
Summary of our results
Channel Assignment cannot be solved in 2o(n log n) under ETH
(tight!)
nor in 2n·o(log log `)
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
From 3-CNF-SAT to Family Intersection
From 3-CNF-SAT to Family Intersection
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
From 3-CNF-SAT to Family Intersection
Family Intersection:
Input: We are given two matrices.
We have to pick exactly one element from every row.
Question: Is it possible to pick elements in such a way that
both sums are equal?
Choice
Choice
Choice
Choice
1:
2:
3:
4:
1
7
6
2
14
6
4
0
1
10
7
14
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Choice 1:
Choice 2:
Choice 3:
5
2
3
7
8
8
5
7
8
4
7
9
Tight lower bound for the Channel Assignment problem
From 3-CNF-SAT to Family Intersection
Family Intersection:
Input: We are given two matrices.
We have to pick exactly one element from every row.
Question: Is it possible to pick elements in such a way that
both sums are equal?
Choice
Choice
Choice
Choice
1:
2:
3:
4:
1
7
6
2
14
6
4
0
1
10
7
14
Choice 1:
Choice 2:
Choice 3:
5
2
3
7
8
8
5
7
8
4
7
9
e.g. 1 + 7 + 7 + 0 = 5 + 7 + 3
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
From 3-CNF-SAT to Family Intersection
An example for a 2-CNF-SAT formula ϕ = (α ∨ β) ∧ (¬α ∨ γ):
We number all the occurrences (α1 ∨ β2 ) ∧ (¬α3 ∨ γ4 )
We interpret 4-bit numbers as the valuations of the
occurrences
α:
β:
γ:
False
00002
00002
00002
True
10102
01002
00012
α1 ∨ β 2 :
¬α3 ∨ γ4 :
Arkadiusz Socała
1st
10002
00002
2nd
01002
00112
1st and 2nd
11002
00012
Tight lower bound for the Channel Assignment problem
From 3-CNF-SAT to Family Intersection
An example for a 2-CNF-SAT formula ϕ = (α ∨ β) ∧ (¬α ∨ γ):
We number all the occurrences (α1 ∨ β2 ) ∧ (¬α3 ∨ γ4 )
We interpret 4-bit numbers as the valuations of the
occurrences
α:
β:
γ:
False
00002
00002
00002
True
10102
01002
00012
α1 ∨ β 2 :
¬α3 ∨ γ4 :
1st
10002
00002
2nd
01002
00112
1st and 2nd
11002
00012
Left matrix represents consistent valuations of the occurrences.
Right matrix represents valuations of the occurrences such
that every clause is satisfied.
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
From Family Intersection to Common Matching Weight
From Family Intersection to Common Matching
Weight
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
From Family Intersection to Common Matching Weight
For a matrix with n rows and c columns we can build a weighted
full bipartite graph such that:
set of weights of all perfect matchings = set of all possible
sums of choices
number of the vertices is O logn n (for fixed c)
Common Matching Weight:
Input: We are given two weighted bipartite graphs.
Question: Is it possible to pick two perfect matchings in such
a way that both sum of weights are equal?
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
From Family Intersection to Common Matching Weight
How do we compress? A nice recursive idea behind:
Assume that we already have matchings that represents
”something”.
By merging them we can additionally represent a function ρ on
the left vertices of the first matching.
h1̂i
h2̂i
h3̂i
h1i
h2i
h3i
h1̂i
h2̂i
h3̂i
h1̂i
h2̂i
h3̂i
h1i
h2i
h3i
h1i
h2i
h3i
(ρ)
h1̂, 1̂i
h1̂, 2̂i
h1̂, 3̂i
h1, 1i
h1, 2i
h1, 3i (ρ(h2̂i) = 1)
h2̂, 1̂i
h2̂, 2̂i
h2̂, 3̂i
h3̂, 1̂i
h3̂, 2̂i
h3̂, 3̂i
h2, 1i
h2, 2i
h2, 3i
Arkadiusz Socała
h3, 1i (ρ(h3̂i) = 3)
h3, 2i (ρ(h1̂i) = 3)
h3, 3i
Tight lower bound for the Channel Assignment problem
From Family Intersection to Common Matching Weight
We can use this idea to build our weighted graph recursively. When
merging bipartite graphs:
We copy the weights that for every vertex of the left side there
is no difference to which of the merged parts it will be
matched.
Then for every vertex of the first part we can add to the
weights of its edges a function depending only on the part to
which it is connected.
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
From Family Intersection to Common Matching Weight
For the matrix with 4 rows and 2 columns we get:
0 0 0 0
A1,1 A1,2 A2,1 A2,2
0
0
0
0


A1,1 + A3,1 A1,2 + A3,1 A1,1 + A3,2 A1,2 + A3,2
 A4,1

A4,1
A4,2
A4,2



 A2,1
A2,2
A2,1
A1,2
0
0
0
0
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
From Common Matching Weight to Channel Assignment
From Common Matching Weight to Channel
Assignment
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
From Common Matching Weight to Channel Assignment
We encode a bipartite graph and its matchings into:
vL = v1
2M
M
v1
2M
v2 M w1M v3
aπ(1)
8M
vM = w2
2M
M
2M
v4M w2M v5
aπ(2)
vR = v8
8M
2M
M
2M
v6 M w3M v7
bπ(1)
2M
M
v8
bπ(2)
The position (color) of the middle vertex represent the weight of
the matching corresponding to the permutation π.
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
From Common Matching Weight to Channel Assignment
We can encode two bipartite graphs with the common middle
vertex. Its position (color) needs to represent the common weight
of both matchings.
(1)
(1)
(1)
wL
(2)
wL
(2)
vL
vR
vL
vM
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(1)
wR
(2)
wR
(2)
vR
Tight lower bound for the Channel Assignment problem
Reductions
3-CNFFamily InSAT → tersection →
O(n)
O(n)
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Common
Matching
Weight
O logn n
→
Channel
Assignment
O logn n
Tight lower bound for the Channel Assignment problem
Summary
Our results
Channel Assignment cannot be solved in 2o(n log n) under ETH
(tight!)
nor in 2n·o(log log `)
Further research
Faster algorithm for `-bounded version? e.g. O(log `)n ?
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem
The end.
Arkadiusz Socała
Tight lower bound for the Channel Assignment problem