CS590A Distributed Network Algorithms

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• CS590A Distributed Network Algorithms
• Prof. Gopal Pandurangan

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An efficient distributed algorithm for
constructing small dominating setsLujun Jia,
Rajmohan Rajaraman, Torsten SuelJournal of
Distributed Computing 2002(15)

• Presented by Xun Zhou
• 2007.10.30

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Outline

• Overview and Related Work
• Local Randomized Greedy Algorithm (LRG)
• Analysis Time Complexity and Approximation
  Ratio
• Generalizations
• Open problems and comments

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Definition

• The dominating set problem asks for a small
  subset D of nodes in a graph such that every node
  is either in D or adjacent to a node in D.
• Result size of MDS (d), Optimal Size d
• Approximation Ratio (d)/d

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Related Work (1)

• Greedy Algorithm (A distributed implementation in
  Lecture 13 )
• Time Complexity O(log n log?)
• Approximation Ratio H(?) O(log?)
• All nodes must know n and ? first (? Max degree)
  
• Drawback
• if ? is not known, then O( log2 n) rounds

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Related Work (2)
Distributed Greedy Algorithm DDCH in reference
20 Span(v) The number of uncovered neighbors
of v (including v) White node uncovered node.
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Related Work (2)

• Drawback
• In some particular graphs where there is a
  long chain of nodes with decreasing degrees, the
  number of rounds until completion may be
  proportional to the length of the chain.
• To Solve this
• Span ? rounded span

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Example

__ Time Complexity of the above graph
O(v n )
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Example

• In this Kn/3 graph, run time is n/3 round.
• Each round add exactly 1 node Inefficient

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Main Result

• Local Randomized Greedy algorithm (LRG)
• Time Complexity O(log n log?)
• Approximation Ratio
• H(?) O(log?) of the optimal in expectation
• O(log n ) with high probability (whp)
• ? is the maximum degree 1 in the graph
• No pre-request for n and ?

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Problems and Results(2)

• Generalizations Multi-MDS (MMDS) Find a
  dominating set that covers v at least r(v) times
  for all v ? V .
• Time Complexity O(log n logR log?) whp.
• O(log?) in expectation and O(log n) whp,
  where R equals max u?V r(u).
• Weighted dominating set problem We wish to
  select a dominating set of minimum total weight
• Same approximation ratio as MDS.
• Time Complexity O(log n log(?W)), where W
  is the ratio of the maximum and minimum weights.

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LRG Span, Candidate, Support and Dominator
Selection

• Span d(v) The number of uncovered nodes that are
  adjacent to v (including v itself if uncovered).
• Let the rounded span of node v be the
  smallest power of base b that is at least d(v),
  where b 1 is a real constant.
• We say that a node v ? V is a candidate If
  is at least for all w ? V within
  distance 2 of v. For each candidate v, let cover
  C(v) denote the set of uncovered nodes that v
  covers (v and its neighbors).
• Support calculation
• For each uncovered node u, we calculate its
  support s(u), which is the number of candidates
  that cover u.
• Dominator Selection
• For each candidate v, we add v to D with
  probability 1/med(v), where med(v) is the median
  support of all the nodes in C(v).

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LRG

• Do the following steps in each round
• 1. Node v is a candidate for joining the
  dominating set if its span w(v) rounded to the
  next power of 2 is maximal within distance 2.
• 2. Each white node u computes its support s(u),
  which is the number of candidates that cover u.
• 3. For a candidate v, let m(v) be the median
  support of all white neighbors. v joins the
  dominating set with probability 1/m(v).

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LRG (cont)

• Once a node u and all of the neighbors of u are
  covered, u need not participate in the algorithm
  any longer (other than as an intermediate node
  for routing messages).
• Covered nodes still have chance to be candidates,
  while those in D dont.
• The algorithm will stop when all the nodes are
  covered.

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Analysis Time Complexity

• Lemma 3.1. If v1 and v2 are two candidates in any
  connected component of H, then
• v1
  v2
• Candidate Node
• Node covered by candidates

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Proof 3.1

• Fixed round, H (V ,E) is the subgraph of G
  where V is the union of the set of all
  candidates in the round and the set of nodes
  covered by the candidates, and E is the set of
  edges (v,w) where v is a candidate in the round
  and w ? C(v).
• Proof. Consider any path p connecting v1 and v2.
  From the definition of subgraph H, at least every
  alternate node on p is a candidate. From the
  definition of a candidate, two candidates within
  distance 2 have the same rounded span. Hence, the
  desired claim holds.

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Time Complexity

• Lemma 3.2. If m is the maximum rounded span of
  any node at the start of a round and F and F are
  them-potentials at the start and end of the
  round, then EF dF for some positive
  constant d
• Lemma establishes a bound on the expected
  decrease in the m-potential in any round.
• To prove Lemma 3.2, we need to introduce some
  definitions, and use Lemma 3.3 and Lemma 3.4.

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Proof 3.2 - definition

• i-potential
• m maximum rounded span of all the nodes at
  the start of the round.
• For each v, sort nodes in C(v) in non-decreasing
  order, according to s(u) of each node in C(v).
• T(v) Top half of the C(v). B(v) Bottom half
• v is good for u if u is in T(v)
• u is nice if at least s(u)/4 nodes are good for
  u.
• Ps(v) The probability that candidate v will be
  added to D at the end of the round.
• Pc(u) denote the probability that node u will be
  covered at the end of the round.

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Proof 3.2 Lemma 3.3

• Lemma 3.3.

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Proof 3.2 Definition (for 3.4)

• Each element in C(v) for any v with rounded span
  m, is an entry. Thus, F is the total number of
  entries (a node may have entries in several sets
  C(v)).
• Any entry that occurs in T(v), for some v, is
  referred to as a top entry. Finally, any
  occurrence of a nice node in T(v) is referred to
  as a nice top entry.

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Proof 3.2 Lemma 3.4

• Lemma 3.4. At least one-third of the total
  entries are nice top entries.
• Proof 3.4 Fix a connected component S in which
  the maximum rounded span is m. By Lemma 3.1, all
  candidates in S have the same rounded span.
• Let x (resp., y) denote the total number of
  entries (resp., number of non-nice top entries)
  in S. It follows from the definition of nice
  nodes that for any nice node u the number of
  candidates v such that T(v) contains u is at
  least s(u)/4.

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Proof 3.2 Lemma 3.4

• Therefore, there exist at least 3y non-nice node
  entries in B(v). (Non-nice nodes appear in less
  than ¼ of all T(v) )
• Clearly, x/2 - 3y 0, y x/6.
• Thus, the total number of nice node entries in
  T(v) is x/2 - y x/3. Adding over all connected
  components with rounded span m, we obtain the
  desired claim.

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Proof 3.2

• Now we go back to prove 3.2 using the result of
  3.3 and 3.4.
• v Any candidate in S
• y,z The number of nice top entries in C(v) at
  the start of/covered in the round.
• By Lemma 3.3
• Adding all Components with span m and invoke
  Lemma 3.4 . 3.2
  proved.

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Theorem 1

• Theorem 1 LRG terminates in O(log n log?) rounds
  whp.
• Proof Divide the running time of the algorithm
  into phases. A phase consists of a maximal
  sequence of rounds with the same maximum rounded
  span. The number of phases is at most the total
  number of distinct values for the rounded span,
  logb?. We now show that the number of rounds in
  each phase is O(log n) whp.

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Theorem 1 (cont)

• Let T(P) denote the number of rounds remaining in
  the phase when the m-potential at the start of a
  round is P.
• Establish Probabilistic Recurrence for T.
• For P , we have T(P) a(P) T(P ),
  where a(P) 1 for all P and P is the random
  variable denoting the m-potential at the end of
  the round.
• For P , we have T(P) 0.
• Invoke a result on probabilistic recurrence
  relations due to Karp 18, Theorem 1.3 to obtain
  the following claim
• Let w , where d is the
  constant in Lemma 3.2 .

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Theorem 1 (cont)
For any instance with m-potential P, the number
of rounds T(P) until the end of the phase
satisfies the inequality

(1) for any constant c
0.
is the solution
fort (P) a(P)t (dP), Which is a deterministic
counterpart of Eq. (1), corresponding to the
case The m-potential decreases by exactly d in
each round. P bnm , so each phase will end
with at most
rounds. So combined with the number of
phases, we get the final result
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Analysis - Approximation Ratio

• Theorem 2 LRG yields a dominating set of
• expected size 4bH? OPT and size O(OPT
  log n) whp.
• Proof Assign cost (u) to node u in the round
  when u is covered. u is covered by v.
• Set cost(u) to be 1/ . Note all vs that
  cover u in the same round have the same rounded
  span.

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Lemma 3.5

• Lemma 3.5 Su?V cost (u) H? OPT.
• Proof Consider any v ? OPT. Let COPT(v) denote
  the set of nodes covered by v at the beginning of
  the algorithm, and let l be COPT(v). We sort
  all u ? COPT(v) to obtain the sequence u1, u2,
  ...,ul such that for any 1 i assigned a cost not after uj . Then we have cost
  (ui) 1/(i1).
• So,
• 
  H?
• 
  
• Since every u is covered by a node v in OPT, we
  can get

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Lemma 3.6

• Lemma 3.6 If S is the set of candidates added to
  D in any round and Z is the total cost assigned
  in the round, then ES 4bEZ.
• Proof For any u ? V , let c(u) 1/ ,
  where
• u ? C(v). For any v ? S, we have C(v)
  d(v)/b.
• Thus we have
• For any u ? V , t(u) denote the number of
  candidates v that are added to D at the end of
  this round such that u ? B(v).

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Lemma 3.6

• Note that t(u) 0 if u is not covered or u ?
  B(v) for all v ? S that cover u. Thus, we have
• Since c(u) is fixed in a given round and Et(u)
  equals Prt(u) 0 Et(u) t(u) 0, we
  obtain
• Let W v u ? B(v) and let p(v) denote the
  probability that v is added to D for a given v ?
  W. For v ? W, since u ? B(v), we have p(v)
  1/med(v) 1/s(u) 1/W. Thus,

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Lemma 3.6

• Substituting the bound on Et(u)t(u) 0 in Eq.
  (2), we get
• ES 4b u?V c(u) Prt(u) 0 4bEZ.
  Done.

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Proof of Theorem 2

• Let Si denote the set of candidates added to the
  dominating set in round i, and let Zi denote the
  cost assigned in round i. Then the expected size
  of the dominating set computed by LRG is
• This is the first half (expectation of ratio) of
  Theorem 2

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Analysis - Approximation Ratio
Now we start to prove the second part of Theorem
2 (Ratio with high probability) Construct
a tree T that captures all possible executions of
LRG on the given network. Each path in T depicts
one execution of LRG. The outcomes of rounds 1
though i-1 specify a particular path from the
root of T to a vertex x. The vertex x depicts the
coin tosses corresponding to the random dominator
selections in round i. Transform T into a binary
tree Tb. the path from root to any vertex in Tb,
x, signifies a particular execution of LRG,
E. Assign x a vertex value of px, which is the
probability that the network node corresponding
to vertex x is selected as a dominator at the
last step of the execution E.
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• Add nodes to make all the value of paths equal to
  the max. Tb is transformed into a full binary
  tree Tf
• Each non-leaf node is associated with a random
  variable Zx. which denotes the sum of the edge
  values along any path from x to a leaf vertex.
• We can conclude EZx 0.
• is the average vertex value of x.
• Finally we have
• PrGZ (ß - 1) pN

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Analysis - Approximation Ratio

• Theorem 3 Algorithm e-LRG terminates in O(log n
  log?) rounds whp and achieves an expected
  approximation ratio of
• (1 3e)H?, for any sufficiently small
  constant e 0.
• e-LRG e positive real constant. select a
  candidate with probability e/re(v), where re(v)
  is the support of rank among the
  supports of all nodes in C(v), set b to be e1 .
• Modify the definitions of the sets T(v) and B(v)
• Let T(v) and B(v) denote the set of the top
  and bottom d(v)- entries,
  respectively. a node u is nice if at least
  es(u)/2 candidates covering u are good for u.

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Analysis - Approximation Ratio
Lemma 3.7
Lemma 3.8. At least an (e/2)-fraction of the
total entries are nice top entries. Proof 3.8 (in
brief) Each nice node u has at least es(u)/2
entries in T(v), for each candidate v. there
exist at least 2(1 - e/2)y/e non-nice node
entries in B(v). Clearly, (1 - e)x - 2(1 -
e/2)y/e 0, from which we get y ex/2. So, the
total number of nice node entries in T(v) is at
least ex-ex/2 ex/2 Adding all the components,
we get the result.
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Analysis - Approximation Ratio

• Lemma 3.9. If S is the set of candidates added to
  D in any round and Z is the total cost assigned
  in the round, then ES (1 3e)bEZ.
• Proof 3.9 is a modification of proof 3.6
• Applying Lemma 3.9 over all rounds and invoking
• Lemma 3.5, we obtain the desired bound on the
  expected approximation ratio.

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Tightness

• The analysis of LRG is tight to within constant
• Consider the Network showed in Figure 3. 2i core
  nodes and 22i fringe nodes in level i. Number of
  nodes n (4m3 - 18)/7 2m.
• In level i, the span of each core node is 22i
  2i if 1 i (22i2i)/(22i-22i-1) and 22i/(22i-2 2i-1) are
  both at least 2, core nodes of level i-1 wont be
  covered until all fringe nodes in level i are
  covered.
• Consider a level at least 1 node in level i is
  uncovered, all nodes in level i1log m are
  covered.

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• For i (log m)/2

The probability that level i, for i (logm)/2,
is covered in r (logm)/8 rounds is at most The
probability that level i takes at least (logm)/8
rounds to be covered is at least
Adding over levels (logm)/2 through logm, the
number of rounds LRG takes for this network is
O(log2m) whp. The total running time is O(log n
log?) whp. The analysis is tight.
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4.1 Generalization AVE

• AVE The probability is the average of the
  inverse of the supports. (LRG is the inverse of
  the median of support)
• Theorem 4 ave computes a dominating set of size
  within O(log n log?) of the optimal in O(log n
  log?) rounds whp.

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4.2 Dominating sets with multiple coverage

• Lemma 4.2. If u is nice, then Eq(u) 3
  q(u)/4.
• Lemma 4.3. If F and F are the potentials at the
  start and end of a round, then EF (1 - O(1/
  logR)) F.
• Theorem 5 There exists a randomized distributed
  algorithm for MMDS that achieves an approximation
  ratio of O(log?) in expectation and O(log n) whp
  and a time complexity of O(log n log?logR) whp.

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4.3 Weights on the nodes

• mds associates a weight w(u) with every node u
  and seeks a dominating set of minimum total
  weight.
• Instead of comparing the rounded span of the
  nodes, we compare the ratio of the span to the
  weight of the node. We again round this value,
  which we refer to as the normalized span, to a
  nearest power of a constant b 1 (allowing
  negative powers).
• O(log(W?) log n) whp.

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Open Problems

• Whether there exists a distributed O(log n)-time
  O(log?)-approximation algorithm for mds.
• to determine the best approximation-time tradeoff
  achieveable by a deterministic distributed
  algorithm
• k-dominating set problem

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Questions
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References

• Lujun Jia, Rajmohan Rajaraman, Torsten Suel, An
  efficient distributed algorithm for constructing
  small dominating sets, Distrib. Comput. (2002)
  15 193205
• http//dcg.ethz.ch/lectures/ss04/distcomp/lecture/
  chapter12.pdf
• http//www.dsi.uniroma1.it/grandoni/DGP06chapter.
  pdf.
• Devdatt Dubhashi, Fabrizio Grandoni and
  Alessandro Panconesi, Distributed Approximation
  Algorithms via LP-duality and Randomization
• http//www.dsi.uniroma1.it/grandoni/DGP06chap
  ter.pdf
• CS590A Lecture Side 13 http//www.cs.purdue.edu/ho
  mes/gopal/CS590A-2007/13.pdf
• B. Liang, Z. Haas. Virtual backbone generation
  and maintenance in ad hoc network mobility
  management. In Proceedings of the 2000 IEEE
  INFOCOM, March 2000