On the Complexity of
Distributed Network
Decomposition
Alessandro Panconesi
and
Aravind Srinivasan
JOURNAL OF ALGORITHMS 20, 356-374 (1996)
What we saw last week
β’ Awerbuch, Goldberg, Luby, Plotkin (1989): A network decomposition
can be computed in time:
ππ(π π )
where π π =
πππππππ
ππππ
.
Today
β’ A network decomposition can be computed in time:
nO π(π)
where π π =
πππππππ
ππππ
=
1
ππππ
β’ Near-optimal decomposition.
β’ Completeness result.
.
Setting
β’ A message-passing distributed network is an undirected graph πΊ =
(π, πΈ) where vertices correspond to processors and edges to
bidirectional communication links.
β’ The network is synchronous (=rounds).
β’ We do not charge for local computations.
β’ Messages are unbounded in size.
Definitions - reminder
β’ π: π β [π] is a called a π-coloring of πΊ if for every π’, π£ β πΈ it holds
that π π’ β π(π£).
β’ πΌ β π is called an independent set if for every π’, π£ β πΌ, π’, π£ β πΈ.
β’ πΌ β π Is called a maximal independent set if it is not a strict subset of
any other independent set.
Definitions - Network Decomposition
β’ We will assume from now on that π = π.
β’ Given a graph πΊ and a partition π =βͺπ πΆπ , the cluster graph πΊ is
defined:
π πΊ = {πΆπ }
πΈ(πΊ) = {(πΆπ , πΆπ )|π β π, βπ’ β πΆπ , βπ£ β πΆπ π . π‘. π’, π£ β πΈ)
β’ The cluster graph is said to be a (π π , π π )-decomposition of πΊ if:
β’ Every πΊ πΆπ (the graph induced by πΆπ ) is connected and is of π(π π )
diameter;
β’ The cluster graph πΊ is vertex colored with π(π π ) colors.
β’ Note that a π-coloring of πΊ is a 0, π -decomposition.
Why network decomposition?
β’ Problems like MIS, (Ξ + 1)-coloring and others can be solved in
π(π π β
π π ) time, given a (π π , π π )-decomposition of πΊ.
β’ The generic algorithm for such problems, given a cluster
decomposition, will iterate through the color classes, clusters of color
1 being processed first in parallel, clusters of color 2 being processed
next, and so on.
β’ In each cluster, πΆπ , the trivial π(π π ) algorithm (βglobal
informationβ) can be used.
Example β Computing MIS
β’ Suppose we are given an (π, π)-decomposition, πΊ = π, πΈ , of πΊ.
β’ For simplicity, assume that π = {πΆπ } are colored differently. In
particular πΆπ is of color π where π = 1, β¦ π.
β’ For π = 1, compute ππΌπ on πΆ1 and denote this set by πΌ1 .
β’ For π = 2, β¦ , π compute ππΌπ on πΆπ excluding all vertices from βͺπβ1
π=1 πΌπ
and their neighbors.
β’ The set πΌ = βͺ πΌπ is a ππΌπ of the original graph, πΊ.
Example β Computing MIS
Example β Computing MIS
compute ππΌπ on πΆ1 and denote this set by πΌ1
Example β Computing MIS
Example β Computing MIS
compute ππΌπ on πΆ2 , excluding πΌ1 and their neighbors, and denote this set by πΌ2
Example β Computing MIS
compute ππΌπ on πΆ3 , excluding πΌ1 βͺ πΌ2 and their neighbors, and denote this set by πΌ3
Example β Computing MIS
compute ππΌπ on πΆ4 , excluding πΌ1 βͺ πΌ2 βͺ πΌ3 and their neighbors, and denote this set by πΌ4
Example β Computing MIS
The πΌ =βͺπ πΌπ is a ππΌπ of πΊ
More definitions
β’ Given a graph πΊ = π, πΈ and a set π β π, denote by πΊ[π, π] the
following graph:
π(πΊ π, π ) = π
πΈ(πΊ π, π ) = π’, π£ π’, π£ β π, ππΊ π’, π£ β€ π
β’ Example:
πΊ = (π, πΈ)
πΊ[2, π]
More definitions (2)
β’ An πΌ, π½ -ruling set, S, with respect to πΊ = π, πΈ and π β π is a set
of vertices such that
β’ π β π;
β’ For any two π’ β π£ β π, ππΊ π’, π£ β₯ πΌ;
β’ For any π’ β π there is π£ β π such that ππΊ π’, π£ β€ π½.
Example 1: any MIS is a (2,1)-ruling set
Example 2: any MIS on G[2,V] is a (3,2)-ruling set
More definitions (3)
β’ An (πΌ, π½)-ruling forest with respect to πΊ = (π, πΈ) and π β π is a
forest of rooted trees πΉ = {ππ }, where each tree is a subgraph of πΊ,
with the following properties:
β’
β’
β’
β’
β’
For all π, the root of ππ , denoted by ri , is in π;
For every π’ β π there is π such that π’ β ππ ;
ππ β© ππ = β
for π β π;
For every π β π, ππΊ ππ , ππ β₯ πΌ;
Tree depth is at most π½.
Ruling forest β black boxes
β’ Recall from the previous talk that we know how to compute an
(πΌ, π½)-ruling forest given an πΌ, π½ -ruling set in π π½ rounds.
β’ In addition, an π, πππππ -ruling forest can be computed in π(ππππ).
Question
β’ Assume weβre given:
β’
β’
β’
β’
A graph πΊ such that π = π΄ βͺ π΅;
A π, π -decomposition of πΊ[π΄];
We know: maximum degree of πΊ[π΅] is at most cβ1;
π-coloring of πΊ[B].
β’ Is it possible to compute a π, π -decomposition of the whole graph
πΊ?
πΆππ’π π‘πππππππ
β’ Input: (π, π)-decomposition of G[π΄], π-coloring of G[π΅]
β’ Output: (π, π)-decomposition of G
1. π πΊ β π(πΊ[π΄]) βͺ {{π}|π β π΅}
2. For i = 1, β¦ , π in parallel for each vertex with color π choose a color
in [π] not chosen by any of its neighbors.
β’ Time complexity?
πΆππ’π π‘πππππππ
πΆππ’π π‘πππππππ
c=
πΆππ’π π‘πππππππ
c=
πΆππ’π π‘πππππππ
c=
πΆππ’π π‘πππππππ
c=
πΆππ’π π‘πππππππ
c=
πΆππ’π π‘πππππππ
c=
πΆππ’π π‘πππππππ
c=
πΆππ’π π‘πππππππ
c=
πΆππ’π π‘πππππππ
πΆππ’π π‘πππππππ
Network Decomposition - Strategy
1. Choose a number π(π) and divide the vertices into high-degree and
low-degree groups (high-degree iff deg πΊ π£ β₯ π(π) etc).
2. Collapse high-degree vertices with their neighborhoods, resulting in
a cluster graph πΊβππβ . Compute the network decomposition
recursively on πΊβππβ .
3. Color the low-degree vertices on the remaining graph πΊπππ€ .
4. Call πΆππ’π π‘πππππππ to get a network decomposition of πΊ.
Network Decomposition - Strategy (2)
P(n)=6
Choose a number π(π) and divide the vertices into high-degree and low-degree groups (high-degree iff deg πΊ π£ β₯ π(π) etc).
Network Decomposition - Strategy (2)
Collapse high-degree vertices with their neighborhoods, resulting in a cluster graph πΊβππβ . Compute the network
decomposition recursively on πΊβππβ .
Network Decomposition - Strategy (2)
Color the low-degree vertices on the remaining graph(s).
Network Decomposition - Strategy (2)
Network Decomposition - Strategy (2)
Call πΆππ’π π‘πππππππ to get a network decomposition of πΊ.
Network Decomposition - Strategy (3)
β’ PROBLEM: It may happen that the neighborhoods of high-degree
vertices intersect! This is a symmetry-breaking problem.
Network Decomposition - Strategy (4)
β’ Solution: we handle the problem by computing a (3,4)-ruling forest
with respect to the set π of high-degree vertices and the graph πΊ.
β’ We collapse trees with their root, instead of collapsing
neighborhoods.
β’ The improvement of this algorithm compared to last week comes
from the computation of a (3,4)-ruling forest instead of a (3,3ππππ)ruling forest. The recursion can terminate faster.
3,4 -ruling forest how?
We will see in a few slidesβ¦
Network Decomposition - Algorithm outline
β’ The algorithm uses two procedures:
β’ πππ‘π€ππππ·πππππππ π(πΊ) to compute a (ππ π , ππ π )-decomposition of
some graph πΊ with π π = 1/ππππ.
β’ π
π’πππππΉππππ π‘(π, πΊ) to compute a (3,4)-ruling forest with respect to set π
and graph πΊ.
β’ We will show the algorithm achieves π ππ
β’ Definition: π π π
=π π π
π 1
.
π
running time.
πππ‘π€ππππ·πππππππ π
β’ Input: a graph H = (π, πΈ) with π β€ π vertices
β’ Output: an (8logπ π , π)-network decomposition of π».
1. Define π β {π£| deg π» π£ β₯ π} and call πΉ = {ππ } β π
π’πππππΉππππ π‘(P, H)
2. Construct the graph HF in the following manner:
β’ π(HF ) = {ππ }
β’ E(HF ) = {(Ti , Tj )|i β π, βπ’ β ππ , π£ β ππ : π’, π£ β πΈ(π»)}
3. Call πππ‘π€ππππ·πππππππ π(HF ).
4. Let π β π\F. Compute a π-coloring of H π .
5. Call πΆππ’π π‘πππππππ on HF and H[π].
πππ‘π€ππππ·πππππππ π β Correctness &
complexity
β’ Claim 1:
π(π»)
π(π»)
π(HF ) β€
β€
π+1
π
πππ‘π€ππππ·πππππππ π - Correctness &
complexity (2)
β’ Claim 1:
π(π»)
π(π»)
π(HF ) β€
β€
π+1
π
β’ Claim 2:
πππ‘π€ππππ·πππππππ π computes a network decomposition with
clusters of at most 8logπ π diameter.
πππ‘π€ππππ·πππππππ π - Correctness &
complexity (3)
β’ Claim 1:
π(π»)
π(π»)
π(HF ) β€
β€
π+1
π
β’ Claim 2:
πππ‘π€ππππ·πππππππ π computes a network decomposition with
clusters of at most 8logπ π diameter.
β’ Claim 3:
πππ‘π€ππππ·πππππππ π computes a network decomposition that uses
at most π colors.
πππ‘π€ππππ·πππππππ π - Correctness &
complexity (4)
β’ What is the time complexity?
πππ‘π€ππππ·πππππππ π - Correctness &
complexity (5)
β’ The time complexity πππ· satisfies the following recursive inequality:
πππ· π β€ ππ
πΉ π + 8 β
πππ·
π
+ π(π β
ππππ)
π
Computing a (3,4)-ruling forest
β’ In comparison to a (3,3ππππ)-ruling forest, it is not known how to
compute (3, π)-ruling forest in π(ππππ) time.
β’ However, a (3,4)-ruling forest can be computed using network
decomposition, in (hopefully) running time π ππ(π) .
β’ We will see that this modification improves the overall time
complexity.
π
π’πππππΉππππ π‘
β’ Input: a graph π» with π β€ π vertices and a set π β π(π»).
β’ Output: a (3,4)-ruling forest with respect to π and π».
1. Compute a (3,3ππππ)-ruling forest πΉ = {ππ } with respect to π and
π».
2. Within each tree ππ , the root ππ will assign numbers 1,2, β¦ , π
cyclically to the vertices in ππ β© π. Let ππ be the set of all vertices in
π with color π.
π
π’πππππΉππππ π‘
3. From each ππ we obtain a new set ππ β ππ by marking some
elements of ππ ; ππ is the set of unmarked vertices in ππ . Let π1 = π1 .
In phases 2, β¦ , π repeat the following: in phase π, any vertex π’ β ππ
such that ππ» π’, π£ β€ 2 for some vertex π£ ββͺπ<π ππ will be marked.
π
π’πππππΉππππ π‘
4. Consider π»[2, ππ ]. In parallel, for all π, invoke
πππ‘π€ππππ·πππππππ π(π» 2, ππ ) to compute a network
decomposition and use it to get an ππΌπ πΌπ in π»[2, ππ ].
5. πΌ =βͺπ πΌπ is a (3,4)-ruling set; use it to compute a 3,4 -ruling forest.
π
π’πππππΉππππ π‘ - example
P(n)=2
Everyone is in π!
π
π’πππππΉππππ π‘ - example
Compute a (3,3ππππ)-ruling forest πΉ = {ππ } with respect to π and π».
π
π’πππππΉππππ π‘ - example
Within each tree ππ , the root ππ will assign numbers 1,2, β¦ , π cyclically to the vertices in ππ β© π.
π
π’πππππΉππππ π‘ - example
π1 = π1 , π2 = β
π
π’πππππΉππππ π‘ - example
In parallel, compute ππΌπ πΌπ in π»[2, ππ ].
π
π’πππππΉππππ π‘ - example
πΌ =βͺπ πΌπ is a (3,4)-ruling set; use it to compute a 3,4 -ruling forest.
π
π’πππππΉππππ π‘ - Correctness & complexity
β’ Claim 1:
for all π β [π], we have
2π
ππ β€ .
π
π
π’πππππΉππππ π‘ - Correctness & complexity (2)
β’ Claim 1:
for all π β [π], we have
2π
ππ β€ .
π
β’ Claim 2:
The set πΌ =βͺπ πΌπ is a (3,4)-ruling set with respect to π and π».
π
π’πππππΉππππ π‘ - Correctness & complexity (3)
β’ What is the time
complexity?
π
π’πππππΉππππ π‘ - Correctness & complexity (4)
β’ The complexity of ππ
πΉ satisfies the following inequality:
ππ
πΉ π β€ π ππππ + π π + 2πππ· (2π/π)
π
π’πππππΉππππ π‘ - Correctness & complexity (4)
β’ The complexity of ππ
πΉ satisfies the following inequality:
ππ
πΉ π β€ π ππππ + π π + 2πππ· (2π/π)
πππ· π β€ ππ
πΉ π + 8 β
πππ·
π
+ π(π β
ππππ)
π
Time complexities combined
β’ Combining the inequalities for πππ· , TRF we get (assuming π β₯ ππππ)
πππ· π β€ π πππππ + 10πππ·
2π
π
β€ π πππππ 10logπ/2 π
= π(π, π)
β’ The minimum is attained when π = 2π(βππππ) = π(ππ
π
).
THEOREM 1:
Given a graph πΊ with π vertices, procedure
πππ‘π€ππππ·πππππππ π computes an
(ππ π , ππ π )-decomposition of πΊ in π ππ π
time, where π π = 1/ ππππ.
Part 2
Near-optimal decomposition
Near-optimal decomposition
β’ A (ππππ, ππππ)-decomposition is called a near-optimal
decomposition. We would be interested in computing a near-optimal
decomposition in π ππ π .
β’ Suppose there is a procedure π΄πΏπΊ which computes a (π π , π π )decomposition in time π(π‘ π ).
β’ We will prove one can compute a near-optimal decomposition in time
π(π‘ π log π + π π π π log 2 π)
A sequential scheme
β’ For any index π and vertex π£ β π define (βclusterβ and βboundaryβ)
πΆπ£ β π’ ππΊ π£, π’ β€ π
π΅π£ β π’ ππΊ π£, π’ = π + 1
β’ Now, either there exists an index π,
0 β€ π β€ log 2 π β 1
such that π΅π£ β€ πΆπ£ or not.
β’ If there exists no such index, then πΊ has π(ππππ) diameter, so we can
take the whole graph to be one cluster.
β’ Otherwise let ππ£ = π(ππππ) always be the smallest such index.
A sequential scheme (2)
1. For the procedure, start with any vertex π£.
2. Choose i = ππ£ to be the smallest such that π΅π£ β€ |πΆπ£ |.
3. Proceed by removing the vertices πΆπ£ βͺ π΅π£ and the edges incident at
them and repeat this process on the remaining graph.
4. Each set πΆπ£ now becomes a cluster and gets color 1.
5. Repeat the above process on the remaining graph (union of π΅π£ ) to
assign color classes 2,3, β¦.
A sequential scheme - correctness
β’ Claim 1:
The diameter of each cluster πΆπ£ is at most π(ππππ).
β’ Claim 2:
The coloring of the cluster graph is valid.
β’ Claim 3:
There are at most π(ππππ) colors.
A distributed scheme
β’ Any sequential algorithm can be computed trivially in π(π) rounds,
where π is the diameter of the given graph.
β’ We seek a faster approach by exploiting the idea of network
decomposition, in a similar manner MIS, for instance, is computed.
β’ Technical assumption: π π , π π β₯ Ξ©(logn).
A distributed scheme
1. Use π΄πΏπΊ to compute a (π(π), π(π))-decomposition of πΊ[2ππππ, π]
2. For πππ€πππππ = 1, β¦ , ππππ do
β’ For π = 1, β¦ , π(π(π)), each cluster with color π in πΊ[2ππππ, π] proceeds as
follows. Each cluster π simulates the sequential algorithm to compute the sets
πΆπ£ , makes them new clusters with color πππ€πππππ (vertices are allowed to
βeat intoβ old clusters which are not of color π). Newly colored clusters are
removed from the network.
A distributed scheme - correctness
β’ Since the initial (π(π), π(π))-decomposition was computed on
πΊ 2ππππ, π , vertices with the same old color that belong to different
clusters are at least 2 πππ π + 1 apart.
β’ In addition, if two clusters receive the same πππ€πππππ there are two
possibilities:
β’ Theyβre in the same old cluster β by the correctness of the sequential algorithm the
clusters will not have an edge between them.
β’ They're in different clusters β because of the above bullet and that clusters are at
most ππππ in diameter, they donβt interfere with each other.
β’ Hence the created cluster graph has a valid coloring. There are at most
ππππ new colors. Additionally, each cluster is at most πππ π in diameter as
in the sequential case.
β’ We have obtained a (ππππ, ππππ)-decomposition.
A distributed scheme - time complexity?
1. Use π΄πΏπΊ to compute a (π(π), π(π))-decomposition of πΊ[2ππππ, π]
2. For πππ€πππππ = 1, β¦ , ππππ do
β’ For π = 1, β¦ , π(π(π)), each cluster with color π in πΊ[2ππππ, π] proceeds as
follows. Each cluster π simulates the sequential algorithm to compute the sets
πΆπ£ , makes them new clusters with color πππ€πππππ (vertices are allowed to
βeat intoβ old clusters which are not of color π). Newly colored clusters are
removed from the network.
THEOREM 2:
Suppose there is a procedure π΄πΏπΊ which
computes a (π π , π π )-decomposition in time
π(π‘ π ). Then we can compute a near-optimal
decomposition in
2
π(π‘ π ππππ + π π π π log π).
Applications
β’ Corollary: Given a graph πΊ with π vertices, the following functions can
be computed in π(ππ π ) time in the distributed model of
computation:
β’
β’
β’
β’
β’
Maximal independent set
(Ξ + 1)-vertex coloring
(2Ξ β 1)-edge coloring
Maximal matching
(ππππ, ππππ)-decomposition
Part 3
Completeness
Completeness
β’ In the previous parts we gave an algorithm for computing a network
decomposition in time π ππ π .
β’ One can ask a question from a different angle: suppose that we know
how to compute a decomposition in π(ππππ) but only for some
graphs. Can we use it in conjunction with other methods to compute
a decomposition for all graphs?
β’ We call this a completeness result: we seek a class of π-vertex graphs
which is complete for the task of computing a network
decomposition.
Completeness (2)
β’ We will see a result of the following nature
In order to have a (ππππ¦ππππ, ππππ¦ππππ)-decomposition running in π(ππππ) time,
we just need to look at graphs of maximum degree Ξ β€ π(π1/πππππππ ).
Completeness (3)
β’ In general, let β(π) be any non-decreasing function and let
π π = π π 1/β(π)
π π = π π β(π)
where π π = π(π1/ ππππ ).
β’ Suppose we have an algorithm, π΄πΏπΊ, that computes a (π π , π π )decomposition of graphs with maximum degree Ξ β€ π(π) in time
π(π π ). We wish to compute a decomposition on all graphs.
β’ For π π = π(ππππ), we have
β π = ππππ/πππππππ
Completeness - strategy
β’ To compute a network decomposition for all π-vertex graphs, we will
use a modified version of the mutually recursive procedures
πππ‘π€ππππ·πππππππ π and π
π’πππππΉππππ π‘.
β’ As one could expect, we will divide the vertices into high-degree and
low-degree ones. This will allow us to use algorithm π΄πΏπΊ which
operates only on graphs of degree Ξ β€ π(π).
β’ In addition, π
π’πππππΉππππ π‘ will compute a (3,6)-ruling forest.
πππ‘π€ππππ·πππππππ π
β’ Input: a graph H = (π, πΈ) with π β€ π vertices
β’ Output: an (π(π), π π log π π π)-network decomposition of π».
1. Define π β {π£| deg π» π£ β₯ π(π)} and call πΉ = {ππ }
β π
π’πππππΉππππ π‘(P, H)
2. Construct the graph π»πΉ in the following manner:
β’ π(π»πΉ ) = {ππ }
β’ E(HF ) = {(Ti , Tj )|i β π, βπ’ β ππ , π£ β ππ : π’, π£ β πΈπ» }
3. Call πππ‘π€ππππ·πππππππ π(π»πΉ ).
4. Let π β π\F. Compute a (π π , π π )-decomposition of H π using π΄πΏπΊ
with fresh colors.
πππ‘π€ππππ·πππππππ π - Correctness &
complexity
β’ Claim 1:
π(π»)
π(π»)
π(π»πΉ ) β€
β€
π(π) + 1
π(π)
β’ Claim 2:
πππ‘π€ππππ·πππππππ π computes a network decomposition with
clusters of at most π(π) diameter.
β’ Claim 3:
πππ‘π€ππππ·πππππππ π uses at most π π π log π / log π π
colors.
πππ‘π€ππππ·πππππππ π - Correctness &
complexity (2)
β’ The complexity of πππ· satisfies the following inequality:
πππ· π β€ ππ
πΉ π + 12 β
πππ·
π
+ π(π π )
π
π
π’πππππΉππππ π‘
β’ Input: a graph π» with π β€ π vertices and a set π β π(π»).
β’ Output: a (3,6)-ruling forest with respect to π and π».
1. Partition π into disjoint sets ππ , π β π , as before, by computing a
(3,3ππππ)-ruling forest and assigning numbers 1,2, . . , π cyclically
inside each tree.
2. Let π»π = π» 4, ππ . In parallel, for all π β [π], compute an MIS πΌπ of π»π
by calling πππ‘π€ππππ·πππππππ π(π»π ). Let πΌ =βͺπ πΌπ .
π
π’πππππΉππππ π‘
3. Let πΉ = π»[2, πΌ]. Invoke π΄πΏπΊ to compute a MIS, π½, of πΉ. (Ξ πΉ β€ π)
4. The set π½ is a (3,6)-ruling set; use it to compute a (3,6)-ruling
forest.
π
π’πππππΉππππ π‘ - Correctness & complexity
β’ Claim 1:
for all π β [π], we have
2π
ππ β€ .
π
β’ Claim 2:
We have
Ξ πΉ β€π π .
β’ Claim 3:
The set π½ is a (3,6)-ruling set with respect to π and π».
π
π’πππππΉππππ π‘ - Correctness & complexity (2)
β’ The complexity of ππ
πΉ satisfies the following inequality:
ππ
πΉ π β€ π ππππ + 4πππ· (2π/π)+ π π(π)
π
π’πππππΉππππ π‘ - Correctness & complexity (3)
β’ The complexity of ππ
πΉ satisfies the following inequality:
ππ
πΉ π β€ π ππππ + 4πππ· (2π/π)+ π π(π)
πππ· π β€ ππ
πΉ π + 12 β
πππ·
π
+ π(π π )
π
Time complexities combined
β’ Combining the inequalities for πππ· , TRF we get (assuming π β₯ ππππ)
πππ·
2π
π β€ 16πππ·
+ π(π π )
π
β€ π(π π )
THEOREM 3:
Let π π = 2 ππππ . Suppose we want an π(π n 1/β π )
time algorithm to distributively compute an
(π(π n 1/β π ), π(π n 1/β π ))-decomposition, for any
non-decreasing function β β
. This is self-reducible to
graphs of maximum degree O(p n β π ).
Thank you!
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