Transcript K-1 - Rutgers University

Survey of Some Connectivity
Approximation Problems via
Survey of Techniques
Guy Kortsarz
Rutgers University,
Camden, NJ.
The talk is based on the
comprehensive survey
G. Kortsarz and Z. Nutov,
Approximating min-cost
connectivity problems,
Survey Chapter in handbook on
approximation, 2006. Chapter
58, 30 pages.
Steiner Network Problem
Steiner Network:
Instance: A complete graph with edge (or node) costs,
and connectivity requirements r(u,v) for
every pair.
Objective: Min-cost subgraph with r(u,v) edge
(vertex) disjoint uv - paths for all u,v in V.
k-edge-Connected Subgraph: r(u,v) =k for all u,v.
k-vertex-Connected Subgraph: r(u,v) =k for all u,v.
Example
b
a
c
k=2 vertex 2-connected graph
Previous Work on Steiner Network
VERTEX CASE:
Labelcover hard. [K, Krauthgamer, Lee, SICOMP]
kε approximation not possible for some universal ε>0
[Chakraborty, Chuzhoy, Khanna ,STOC 2008]
Undirected and directed problems are equivalent for
k>n/2
[Lando & Nutov, APPROX 2008]
O(log n)-approximation for metric costs. [Cheriyan &
Vetta, STOC 2005]
O(k^3log n) (k maximum demand). [Chuzhoy &
Khanna, STOC 2009]
EDGE CASE:
Edge-Connectivity: sequence of papers, until reaching a
2-approximation [Jain, FOCS 98]
Transitivity in Edge
Connectivity
If (a,b)=k and (b,c)=k then
(a,c)=k
Proof:
b
a
K-1
c
b
First Technique: Directed
Out-Connectivity
The following problem has a polynomial
solution:
Input: A directed graph G(V,E) a root r and
connectivity requirement k
Required: Min cost subgraph so that there
will be k edge disjoint paths from r to any
other vertex
Polynomial time algorithm: for the edge case by
matroids intersection (Edmonds).
Also true for k vertex disjoint paths from r
[Frank, Tardo’s] (submodular flow)
Algorithm for k-ECSG
If we have k connectivity from a vertex v
to all the rest, by transitivity the graph is
k-edge-connected
Apply the Edmonds algorithm twice:
replace every edge with two directed
edges
Once k-in-connectivity to v
Second k-out-connectivity from v
Ratio 2 guaranteed
Work on Node k-Vertex Connected
Subgraph
[Cheriyan, Vempala, Veta, STOC 2002]
O(log k)-approximation for undirected graphs with n>6k2
[K & Nutov STOC 04]
n/(n-k) O(log2 k) for any k, directed/undirected graphs.
The ratio is O(log2 k), unless k = n - o(n).
[Fackharoenphol and Laekhanukit, STOC 2008]
O(log2 k)-approximation also for k = n - o(n).
O(log k) log (n/(n-k))
[Nutov, SODA 2009]. O(log n) unless k=n-o(n)
Many excellent papers about particular cases:
– metric costs: (2+k/n) [K & Nutov]
– 1,∞-costs:
(1+1/k) [Cheriyan & Thurimella]
– small requirements:
[ADNP,DN,KN...]
Technique 2: The Cycle
Theorem of Mader
Let G(V,E) be a k-vertex connected
graph, minimal for edge deletion
and let C be a cycle in G
Then there is a vertex in C of
degree exactly k
Strange Claim?
Corolloraly
Say that (G) is at least k-1
Let F be any edge minimal
augmentation of G to a kvertex-connected subgraph
Then F is a forest
Proof
Consider a cycle in F
As all degrees are at least k-1 before F, with F
all degrees are at least k+1 which contradicts
Mader’s theorem.
Application in Minimum Power
Networks
In a power setting p(v)= max{ c(e) | eE(v)}
Reasons: transmission range.
b
7
a
5
4
5
f
2
h
8
8
6
c
9
d
3
g
The power of G is v p(v)
The Min-Power Vertex kConnectivity Problem
We are given a graph G(V,E) edge costs
and an integer k
Design a min-power subgraph G(V, E)
so that every u,v V admits at least k
vertex-disjoint paths from u to v
May seem unrelated to min cost
vertex k-connectivity
Previous Work for Min-Power
Vertex k - Connectivity
Min-Power 2 Vertex-connectivity, heurisitic study
[Ramanathan, Rosales-Hain, 2000]
11/3 approximation for k =2 [K, Mirrokni, Nutov,
Tsano, 2006]
Cone-Based Topology Control for Unit-Disk
Graphs
[M. Bahramgiri, M. Hajiaghayi and V. Mirrokni,
2002]
O(k)- approximation Algorithm and a Distributed
Algorithm for Geometric Graphs
[M. Hajiaghayi, N. Immorlica, V. Mirrokni, 2003]
Comparing Power And Cost:
Spanning Tree Case
The case k = 1 is the spanning tree case
Hence the min-cost version is the
minimum spanning tree problem
Min-power spanning tree: even this simple
case is NP-hard [Clementi, Penna,
Silvestri, 2000]
Best known approximation ratio: 5/3
[E. Althaus, G. Calinescu, S.Prasad,
N. Tchervensky, A. Zelikovsky, 2004]
The Case k = 1: Spanning Tree
The minimum cost spanning tree is a
ratio 2 approximation for min-power.
Due to: L. M. Kerousis, E. Kranakis,
D. Krizank and A. Pelc, 2003
Spanning Tree (cont’)
c(T)  p(T):
Assign the parent edge ev to v
Clearly, p(v)  c(ev)
Taking the sum, the claim follows
p(G)  2c(G) (on any graph):
Assign to v its power edge ev
Every edge is assigned at most twice
The cost is at least

v
c(e v )
2
The power is at exactly  c ( e v )
v
k vertex-conn: Power, Cost
Equivalent For Approx’(!)
K, Mirrokni, Nutov, Tsano show that the
vertex k - connectivity problem is
essentially equivalent with respect to
approximation for cost and power
(somewhat surprising).
In all other problem variants, almost, the
two problems behave quite differently.
Based on a paper by
[M. Hajiaghayi, K, V. Mirrokni and Z.
Nutov, IPCO 2005].
Reduction to a Forest Solution
Say that we know how to approximate
by ratio  the following problem:
The Min-Power Edge-Cover problem:
Input: G(V, E), c(e), degree
requirements r(v) for every v V
Required: A subgraph G(V, E) of
minimum power so that degG(v)  r(v)
Remark: polynomial problem for cost
version
Reduction to Forest (cont’)
Clearly, the min power for getting
(G’)  k-1, bounds the optimum power
for k-connectivity, from below
Say that we have a 
approximation for the above
problem
Hence at cost at most opt we may
start with minimum degree k -1
Reduction to Forest (cont’)
Let H be any feasible solution for the
Edge-Multicover problem with
r(v)  k-1 for all v
Recall: let F any minimal augmentation
of H into a k vertex-connected subgraph.
Then F is a forest
Comparing the Cost and
the Power
Theorem: If MCKK admits an  approximation then
MPKK admits  + 2  approximation.
Similarly:  approximation for min-power kconnectivity gives  +  approximation for min-cost
k – connectivity.
Proof: Start with a β approximation H for the minpower vertex r(v) = k-1 cover problem
Apply the best min-cost approximation to turn H to a
minimum cost vertex k - connected subgraph H + F, F
minimal
Comparing the Cost and
the Power (cont’)
Since F is minimal, by Mader’s theorem F is a forest
Let F* be the optimum augmentation. Then the
following inequalities hold:
1) c(F)   c(F*) (this holds because  approximation)
2) p(F)  2c(F) (always true)
3) c(F*)  p(F*) (F* is a forest);
4) p(F)  2c(F)  2c(F*)  2 p(F*) QED
Approximating the Min-Power
(H) k-1 Problem
Very hard technical difficulty: Any edge adds
power to both sides.
Because of that: take k-1 best edges, ratio k-1
Admits an O(log n) ratio (Mirrokni et al).
Proof omited the (quite hard)
By The [Nutov 2009] result on min-cost edge kconnectivity O(log n) ratio (almost). SO DOES
THE POWER VARIANT
We conjecture (log n) hardness.
A Result of Khuller and
Ragavachari
There exists a 2+2(k-1)/n ratio for
minimum cost vertex k-connected
subgraph in the metric case
At most 4 always and tends to 2 for
k=o(n)
K, Nutov: 2+(k-1)/n ratio
At most 3 and tends to 2 for k=o(n)
Combines the two techniques shown
The Algorithm
Let Jk(v0) be cheapest star for any v
and its k cheapest edges. Let leaves be
{v1,…..,vk}
Averaging gives that best star has cost at
most Jk(v0)  2OPT/n
v0
v1
v2
vk
The Algorithm Continued
Let R={v0,….,vk-1}
Note, that vk is absent from R
As in [KR] add a new node s that
does not belong to V
Similar to [KR] define a new graph
Gs from G with 0 cost edges for svi
for any vertex vi
The Algorithm continued
Compute a k - outconnected graph from s
in Gs. Let Hs be this graph.
By [KR] the cost of Hs is at most 2opt
(remark: our R is different then the one in
[KR])
In [KR] it is shown that if we add all
edges between the R vertices to Hs ,the
resulting graph is k-connected.
Unlike [KR] we add a MINIMAL feasible
solution out of E(R) to Hs
The Approximation Ratio
k out-connectivity from s
implies (HS)  k-1
Thus F is a forest with k nodes.
We bound the cost of edges in the forest F. For
every vi,vj  v0
we upper bound
c(vi vj) c(v0 vi)+c(vj v0)
We call these costs the new costs
Upper Bounding c(F)
For vi,vjv0 we get
vivjF c(vivj) 
vivjFc(viv0)+c(v0vj)
 vivjF c(v0wk-1)+c(vj,v0)
There are k-1 edges in F but
we did not take the edges of v0
which means that c(v0wk-1) is
counted at most k-2 times.
Proof Continued
Note that according to the new costs
we got a star rooted at vk-1
The node v0 is (in the worst case)
also connected to vk-1 directly. This
adds c(v0vk-1) to the cost of F.
Thus c(F) (k-2)·c(v0vk-1)+c(Jk-1(v0))
Proof Continued
c(F)  (k-2) c(v0vk-1) + 1 ik-1 c(v0vj )
We know that c(Jk(v0))  2opt/n
Thus c(v0vk-1)+c(v0vk)  2opt/n
Thus c(v0vk-1)  opt/n
c(F)  (k-2) c(v0vk-1) + c(Jk(v0)) –c(v0vk) 
(k-3)· c(v0vk-1) + c(Jk(v0))
c(F)  (k-3)opt/n+2opt/n=(k-1)opt/n
Thus the final ratio is 2+(k-1)opt/n
Laminar Families
We present the Jain result with a
simplified proof due to Ravi et. al.
The LP: R(S) maximum demand of a
separated vertex vS, uS
d(S)=number of edges going out of S
LP= min  wexe
Subject to x((S))R(S)-d(S)
xe0
Jain: one of the xe at least
1/2
For the sake of contradiction
assume the contrary
May assume tight inequalities in a
BFS give laminar family
(folklore?).
Let L be laminar family and E’
non-zero edges. Thus |E’|=|L|
Charging
Total charging equals |E’|=|L|
1-2xe
xe
xe
All Possible Edges
All edge types.
C1
S
C2
C1
C3
S1
How Many Tokens S Owns?
Let E(S) be edges internal to S.
The sets C discussed now are children of S.
S owns a vertex in S if does not belong to
any child
e is assigned to the smallest S so that
eE(S)
Define the tokens in S:
t(S)=E(S) −E(C)+x((S))− (C)
Contribution to Both Sides
of Every Edge
t(S)=E(S) −E(C)+x((S))− (C)
An edge with no endpoint in S or an
edge that enters a child of and exits
S. Contribution 0.
An edge with both end points in S
that does not enter a child of S can
not exist.
More cases
t(S)=E(S)−E(C)+x((S))− (C)
An edge that enters S but not a
child of S contributes xe
An edge that enters a child of S
but not S contributes 1-xe
An edge between two children of S
contributes 1-2xe .
t(S) IS NOT ZERO
It can not be that all edges exit S and
enter a child of S. Namely, it can not
be that all contributions are 0.
Indeed in this case S is the sum of its
children
In all other cases the contribution is
positive.
t(S)1
Consider:
t(S)=E(S)−E(C)+x((S))− (C)
The children C belong to the
laminar family, hence they are
tight namely their (C) is integral.
Thus t(S)1.
We Charged Already |L|
Because one per S
Thus we found t(S) associated with
S only, that is at least 1
Clearly the parts associated are
disjoint
This implies that we found already
a fraction of |L|.
We are going to show that some
fraction remains, contradiction.
The Contradiction
Look at the maximum S.
Some edges must be leaving it
because its violated.
The 1-2xe of these edges is positive.
Uncharged.
This means t(S)|E’|>|L|,
contradiction.
Thank you
for attention.
Questions?