Transcript Graphs with tiny vector chromatic numbers and huge

Approximating Maximum
Subgraphs Without
Short Cycles
Guy Kortsarz
Join work with Michael Langberg and Zeev Nutov
Max-g-Girth
Girth: A graph G is said to have girth g if its shortest
cycle is of length g.
Max-g-Girth: Given G, find a subgraph of G of girth at
least g with the maximum number of edges.
g=4
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Max-g-Girth: context
• Max-g-Girth:
• Used in study of “Genome Sequencing” [Pevzner Tang Tesler].
• Mentioned in [Erdos Gallai Tuza] for g=4 (triangle free).
• Complementary problem of “covering” all small cycles (size ≤ g)
with minimum number of edges was studied in past.
• [Krivelevich] addressed g=4 (covering triangles).
• Approximation ratio of 2 was achieved (ratio of 3 is easy).
• Problem is NP-Hard (even for g=4).
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Max-g-Girth on cliques
• The Max-g-Girth problem on cliques = densest graph on
n vertices with girth g.
• Has been extensively studied
• Known that Max-g-Girth has size between (n
[Erd¨os, Bondy Simonovits, …]].
1+4/(3g-12))
and O(n1+2/(g-2)).
• There is a polynomial gap! Long standing open problem.
• Implies that approximation ratio O(n ) will solve open
-
problem.
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First steps
• Positive:
• Trivial by previous bounds approximation ratio of ~ n .
• For g=5,6  n
• If g>4 part of input: ratio n .
• If g=4 (maximum triangle free graph): return random cut and
-2/(g-2)
-1/2
.
-1/2
obtain ½|EG| edges  ratio ½.
• g = 4: constant ratio, g ≥ 5 polynomial ratio!
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Our results
• Max-g-Girth: positive and negative.
• Positive:
• Improve on trivial n for general g to n .
• For g=4 (triangle free) improve from ½ to 2/3.
• For instances with n edges: ratio ~ n .
-1/2
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• Negative:
• Max-g-Girth is APX hard (any g).
-1/3
-2/3g
Large gap!
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Our results
• Covering triangles by edges.
• [Krivelevich] presented LP based 2 approx. ratio.
• Posed open problem of tightness of integrality gap.
• We solve open problem: present family of graphs in
which the gap is 2-.
• Moreover: 2- approximation implies 2- for
Vertex Cover (<1/2).
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Positive
• Theorem: Max-g-Girth admits ratio ~ n .
• Outline of proof:
• Consider optimal subgraph H.
• Remove all odd cycles in G by randomly partitioning G and
removing edges on each side.
• ½ the edges of optimal H remain  Opt. value “did not” change.
• Now G is bipartite, need to remove even cycles of size < g.
• If g=5: only need to remove cycles of length 4.
• If g=6: only need to remove cycles of length 4.
• If g>6: as any graph of girth g=2r+1 or 2r+2 contains at most ~
-1/3
n1+1/r edges, trivial algorithm gives ratio n-1/3.
• Goal: Approximate Max-5-Girth within ratio ~ n
-1/3.
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Max-5-Girth
• Goal: App. Max-5-Girth on bipartite graphs within ratio ~ n .
• Namely: given bipartite G find max. HG without 4-cycles.
• Algorithm has 2 steps:
• Step I: Find G’G that is almost regular (in both parts) such
that Opt(G’)~Opt(G).
• Step II: Find HG’ for which |E | ≥ Opt(G’)n .
-1/3
H
-1/3
Step I:
•General procedure that may be useful elsewhere.
•Let G=(A,B;E) – want G’ almost regular on A and B & Opt(G’)~Opt(G).
•Starting point: easy to make A regular (bucketing).
•Now we can make B regular, however A becomes irregular.
•Iterate …
•Can show: if we do not converge after constant # steps then it must
be the case that Opt(G) is small (in each iteration degree decreases).
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Max-5-girth
• Goal: App. Max-5-girth on bipartite graphs within ratio ~ n .
• Namely: given bipartite G find max. HG without 4-cycles.
• Algorithm has 2 steps:
• Step I: Find G’G that is almost regular (in both parts) such
that Opt(G’)~Opt(G).
• Step II: Find HG’ for which |E |≥ Opt(G’)n .
-1/3
H
-1/3
Step II:
•Now G’ is regular.
•Enables us to tightly analyze the maximum amount of 4 cycles in G’.
Regularity connects # of edges |EG’| with number of 4-cycles.
•Remove edges randomly as to break 4-cyles (“alteration method”).
•Using comb. upper bound on Opt [NaorVerstraete] yields n-1/3 ratio.
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Covering k cycles
• Our algorithm actually gives an approximation for the
problem of finding a maximum edge subgraph of G
without cycles of length exactly k.
• Trivial algorithm (return spanning tree) gives ratio of
n-2/k
• Our algorithm gives ~ n
• Significant for small values of k.
-2/k (1+1/(k-1))
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Some interesting open prob.
• LP for g=4 (maximum triangle free graph):
Max: ex(e)
st:
For every triangle C, eCx(e)2
• Max-Cut: integrality gap = 2.
• Complete graph: IG = 4/3 (x(e)=2/3).
• Conjecture: NP-Hard to obtain 2/3+ approx.
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Some interesting open prob.
Max-5-girth:
• Large gap between upper and lower bounds.
• We suspect that for some  a ratio of n is
-
NP-Hard.
• Obvious open problem: give strong lower
bound for g=5.
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Some interesting open prob.
Set Cover in which each element appears in k
sets.
• Upper bound: k.
Thanks!
• Lower bound: k-1-
• If sets are “k cycles” in given graph G we show a ratio
[Dinur et al.]
of k-1 (for odd k).
• Open problem: is k-1 possible for general set cover
(large k).
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