A PrimalDual Approach to the Feedback Vertex Set Problem

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A Primal-Dual Approach to the Feedback Vertex Set
Problem

• Presented by
• Yuval Mishory
• Oren Ben Zwi

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A Primal-Dual interpretation of two
2-approximation algorithms for the Feedback
Vertex Set problem in undirected graphs

• Written by
• Fabian A. Chudak, Michel X. Goemans,
• Dorit S. Hochbaum, David P. Williamson.

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Outline

• Problem definition
• Related work
• New linear programs
• A primal-dual algorithm
• Equivalence to two previous algorithms

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The (fvs) Feedback Vertex Set problem for
undirected graphs

• Given an undirected graph G(V, E) and nonnegative
  weights wv for the vertices, find a subset F of
  V s.t. F meets every cycle of G.
• The minimum fvs (Feedback Vertex Set) problem is
  finding a minimum weight fvs.

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The classic IP formulation of FVS

• The classic formulation of the problem
• (CYC) Min. ?v?V wv xv s.t.
• ?v?C xv ? 1 , ?C??
• xv ? ?0,1? , v?V
• is the set of all cycles(node sets) C in the
  graph

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Related work

• Garey Johnson 79 decision - NP-complete
• Bar-Yehuda Geiger Naor Roth 94 O(logN)
  approximation
• Becker Geiger 94, 96 2-approximation
• Bafna Berman Fujito 95 2-approximation
• Even Naor Schiber Zosin 96 Integrality Gap of
  CYC is ?(logN)
• CGHW 96 A p-d interpretation of BBF BG
• Bar-Yehuda Ravitz 01 local ratio primal dual

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Some notations

• Given S a subset of V, let ES denote the subset
  of edges that have both endpoints in S.
• Let GS denote the subgraph (S, ES).
• Let ds(v) denote the degree of v in GS.
• Let b(s) ES - S 1.
• Let t denote the cardinality of the smallest fvs.

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Inequalities

• Let F denote any fvs of a graph G(V, E) where E
  is not empty. Then

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proof

• If FV then the hypothesis that E is not empty
  ensures (1).
• Else we split E into the edges that have at
  least one vertex in F and the edges that dont.
  in the first set there are not more than the sum
  of the vertexs degrees, while in the other since
  GV-F is a forest there are less than the number
  of vertices.

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A new formulation for FVS
Observe that if F is an fvs for G then F?S is an
fvs for GS. from (1) we say that for any
(edges)nonempty set S?V ?v?F?S (dS(v) 1) ?
b(S) (IP) Min. ?v?V wvxv s.t. ?v?S (dS(v)
1)xv ? b(S) , S?V ES ?? xv ? ?0,1? ,
v?V
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An LP relaxation of FVS
(LP) min. ?v?V wvxv s.t. ?v?S (dS(v) 1)xv ?
b(S) , S?V ES ?? xv ? 0 , v?V
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The dual problem of the LP relaxation of FVS
(D) max. b(s) yS s.t. ?Sv?S (dS(v) 1)yS ?
wv , v?V yS ? 0 , S?V ES ??
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Another IP for fvs
By inequality (2) we say that for any (edges)
nonempty set S?V and F any fvs ?v?F?S dS(v) ?
b(S) ?(S) (IP(2)) Min. ?v?V wvxv s.t. ?v?S
dS(v)xv ? b(S) ?(S) , S?V ES ?? xv ?
?0,1? , v?V
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The dual problem of the 2nd LP relaxation of FVS
(D(2)) max. ( b(s) ?(S)) yS s.t. ?Sv?S
dS(v)yS ? wv , v?V yS ? 0 , S?V
ES ??
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The algorithm
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The algorithm (cont.)

• 7. Recursively remove degree 1 vertices
  and incident edges from V and E.
• 8. S ? Violation (V, E)
• 9. Increase ys until
• 10. F ? F?vl
• 11. Remove vl from V and adjacent edges from E

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The algorithm (end)

• 12. For j i to 1
• 13. If F vj is an fvs F ? F vj
• 14. F ? F
• 15. output F and y.

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More notations

• We say that a semi-disjoint cycle is a cycle in
  the graph, with no more than one vertex of degree
  grater than 2.
• We call a 2-connected component, an endblock, if
  it has at most one cutvertex (there are 2
  vertices s.t. this vertex belongs to every path
  between them)

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Primal-dual version of the Bafna-Berman-Fujito
algorithm

• Violation(V, E)
• If (V, E) contains a semi-disjoint cycle
  return the vertex set of it.
• Else return V

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The Bafna-Berman-Fujito algorithm

• Start with an empty set F
• Look first for a semi-disjoint cycle S. If none
  is found set S to be the remaining vertices.
• Pick the vertex v from S that achieves the min.
• for all
• set

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The Bafna-Berman-Fujito algorithm (cont.)

• Remove v from the graph
• Recursively remove all degree one vertices and
  incident edges.
• Repeat until F is an fvs.
• Perform the reverse delete step to obtain F

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A new version

• Violation(V, E)
• Return the vertices of an endblock of(V, E)

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Primal-dual version of the Becker-Geiger algorithm

• Violation(V, E)
• Return V
• Replace line 9 with

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The Becker-Geiger primal-dual again

• (LP(2)) Min. ?v?V wvxv s.t.
• ?v?S dS(v)xv ? b(S) ?(S) , S?V ES ??
• xv ? 0 v?V
• (D(2)) Max. ?S( b(s) ?(S)) yS s.t.
• ?Sv?S dS(v)yS ? wv , v?V
• yS ? 0 , S?V ES ??

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The Becker-Geiger algorithm

• Start with an empty set F
• Recursively remove all degree one vertices and
  incident edges.
• Pick the vertex v that achieves the minimum
• for all
• set

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Becker-Geiger algorithm (end)

• Remove v from the graph
• Repeat until F is an fvs.
• Perform the reverse delete step to obtain F

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????? ??????

• ????? ?? ????? ?- fvs ????? ??? ???? ??? ?????
  ????, ????? ??? ????? ??? ?????????? ??????-2
  ??????? ????? ????? local-ratio.
• ????? 3 ??????? ?????????, ??????? ??? ???? ???
  ????? ???????? ???? ??????? ????? ??????? ????
  ????? ???? 2
• ????? ???????? ?? ??????? ??? 3 ?????? ?????, 2
  ??????? ??????????? ??????? ???? ????.
• ???? ????? ???? ??????????? ?????? ?????-2 ?

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More inequalities

• If every vertex of G has degree at least two and
  FM is any minimal fvs, then

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and another...

• If every vertex of G has degree at least two,
  the graph contains no semi-disjoint cycles or is
  itself a cycle, and FAM is any almost minimal
  fvs, (in almost minimal we mean that we cant
  remove more than one vertex) then

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The Bafna-Berman-Fujito version of the Algorithm
is 2-approximation

• The algorithm with the BBF Violation function
  constructs an fvs F and a solution y feasible
  for the dual plan s.t.
• Hence it is a 2-approximation algorithm

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Proof
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Proof (cont.)

• In order to apply inequality (4) it is sufficient
  to argue that S?F is a minimal fvs for GS, and
  that GS has the requisite properties. if it is
  so then

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Proof (end)

• When S is selected, none of its vertices is in F.
  Therefore (because of the deletion step) F-F is
  minimal w.r.t. V.
• In other words, since no vertex of S is in F when
  S is selected, if we assume a spare vertex v in
  F?S there is a vertex u in F?S that entered F
  after v and it is the vertex that met vs
  cycle, now since the reverse delete stage will
  check v after u it cant be the case that v is in
  F.
• Hence

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The integrality gap

• We saw a 2-approximation algorithm that uses the
  primal dual plan we have so we conclude that the
  integrality gap is less than 2
• Unfortunately this gap is tight, as we can see
  if we take a clique with n vertices with the
  trivial weight function of 1 to each vertex, the
  solution xv ½ is a feasible solution to the LP
  relaxation but the value of a minimum fvs is n-2.

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Not today...

• The other two algorithms the Becker-Geiger one
  and the new one are also 2-approximation (the two
  proofs have the same structure and use the two
  inequalities.)
• The second Linear plan has also an integrality
  gap of 2.
• We will leave inequality (4) and only proove (3)

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Proof of (3)
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Proof (3) cont.

• Define the weighted bipartite graph H obtained by
  shrinking in G every connected component of
  GV-FM and by removing all edges of GFM. The
  weight of an edge is the number of edges from the
  respective node in FM to the nodes in the
  respective connected component of GV-FM. We
  need to show that the total weight of H is at
  least 2FM2k-2?.

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Proof (3) cont.

• We first observe that for each vertex v of FM
  there must be some witness cycle cv in G s.t. cv?
  FM v otherwise FM would not be minimal. Thus
  in H there must be at least one edge of weight at
  least 2 incident to every vertex of FM. We
  designate one such edge for each vertex and call
  it a primary edge.

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Yet more notations...

• Let T be a maximum collection of vertex disjoint
  witness cycles, each corresponding to a certain
  primary edge. Let A denote the vertices of FM
  whose witness cycle is not in T.
• Notice that ? ? T, so that the inequality is
  implied if the weight of H is at least 2A2k

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Proof (3) (end)

• By the properties of T, for every connected
  component of GV-FM adjacent to a vertex of A
  via a primary edge, it must also be adjacent to a
  vertex of FM-A via a primary edge.
• Thus if we remove the primary edge adjacent to
  the vertices of A from H there still must be
  weight two adjacent to each connected component.
  Hence the weight of H is at least 2A 2k.

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?????

• ????? ?? ????? fvs , ????? ????? ????? ?????
  ????? ?? ?? ??????? ???? ????. ????? NP ??? ???
  ????????? ????? ??????-2.
• ??????????? ??????? ?????? ????? local ratio.
  ????? ???? ???? ?? ??? ?????????? ??? ???????????
  ??????? ????? primal dual.
• ????? ??????? ????????? ????? ?? ??? ????? 2.
• ????? ???? ??? ???? ???? ?????????? ????? ???
  ?????? ?????? ????? ????.
• ??? ??????? ?????? ?????? ???? 2001 ??? ????
  ???? ????? ???? ?? ???? ?"? ???? ???? ??????
  ??-????? ??????? ????.