Approximating the Minimum Degree Spanning Tree to within One from the Optimal Degree
R96922115 陳建霖R96922055 宋彥朋B93902023 楊鈞羽R96922074 郭慶徵R96922142 林佳慶
Authors
• Martin Furer Department of Computer Science and Engineering
The Pennsylvania State University
• Balaji Raghavachari Computer Science Department, EC 3.1
University of Texas at Dallas
Application
• Distribution of mail and news on the Internet
• Designing power grids
Similar approximation properties
• Edge coloring problem
• 3-colorability of planer graphs
Theorem 1
• There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree )log( nO
Theorem 2
• There is a polynomial time approximation algorithm for the minimum degree Steiner tree problem which produces a Steiner tree of degree )log( nO
Theorem 3
• There is a polynomial time approximation algorithm for the directed version of the minimum degree spanning tree problem which produces a directed spanning tree of degree )log( nO
Theorem 4
• There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree at most 1
A Simple Approximation Algorithm
Improvement
• :the degree of vertex u in T
• If
we now introduce an “improvement” in T by adding the edge and deleting one of the edge in C incident to
)(u
2))(),(max()( vuw
),( vuw
Example
3
<=211
2
<=2
G T
Locally Optimal Tree
• Def: A locally optimal tree (LOT) is a tree in which none of the non-tree edges produce any improvements. Its maximal degree denoted by k
• Def: contains those vertices of T which have degree at least i
iS
Theorem
• For any b>1, the maximal degree k of a locally optimal tree T is less than
Hence, k = nb blog
)log( nO
Proof
• Observe the ratio of can be larger than b at most times in a row. To be more precise, there is an i with
k- +1 i k such that
Otherwise, we have
which lead to a contradiction
ii SS /1
)(log nO b
nblog ii SbS 1
log
log 1b
b
nkk nS b S n n
Proof
• Suppose we remove the vertices of from T. This split T into a forest F with t trees. And we can say that
iS
)1(2 ii SSit
Proof
• Now, we can say the average degree of vertices in in any spanning tree of G is at least
1iS
1
1
i
i
S
St
Proof
1
1
i
i
S
St
Sib
Si i 1)1(
2( 1) 1 i i i
i
i S S S
b S
b
i 1
b
nk b 1)1log(
b
nk blog
logbk b n
Algorithm (sketch)
• We start with an arbitrary spanning tree T of G
• We used the local optimality condition only on “ high”( ) degree vertices.
• The algorithm stops when no vertex in
has a local improvement.
• Each phase of the algorithm can be implemented in polynomial time.
1log nk bS
1log nk bS
Theorem 1
• There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree )log( nO
Idea
• We just need to show that the number of phases is bounded by a polynomial in n.
Proof
• Recall :the degree of vertex u in T
• Define a potential function such that
(c>2)
• Define total potential of the tree is the sum of the potentials of all vertices.
• We can say that where k is the maximal degree of T
)(u
)()( ucu
)(T
kncT )(
Proof
1 1 1
1 1
1
1 2 2
1
1 2, 2
i i a i b i i a i b
i i a i b i i a i b
i i a i b
i i i
c c c c c c
c c c c c c c
c c c c
c c c c a b
Proof
• For ,
where
2 1 2 2( 2 ) (3 ) ( 2 3 )i i i ic c c c c c nki log
2)2()1( iccc
n
c k
3
)2()1(
c
cc
2
)(
n
T
Proof
• Observe we need phases,the potential reduces by a constant factor.
• On the other hand, k cannot decrease more than n times.
• Hence, bound on the number of phases.
)( 2nO
)( 3nO
The Steiner tree version
Steiner Problem
• Consider a graph G = (V,E) and a distinguished set of vertices D V.
• Steiner spanning tree: Find a tree which spans the vertices set D, paths in the tree may go through vertices of V － D.
W1
W2W3
W4
D1 D2
D3D4
D5D6
D7D8
Steiner Problem (cont.)
• Minimum Degree Steiner (spanning) Tree– To find a tree of minimum degree, which
spans the set D.– This is a generalization of the [Minimum
Degree Spanning Tree] problem.
Steiner Problem (cont.)
• We start with an arbitrary tree T which spans D and retain only edges which separate the set D in T.
• In other words, there are no useless edges since every edge separates at least two distinguished vertices.
• Let W be the set of vertices spanned by T. (W : all vertices in the tree T)
• Non-tree path: a path between two vertices in W which goes entirely through vertices of V － W except for the end point.
Some case of non-tree paths
Steiner Problem (cont.)
• Pseudo optimal Steiner tree (POST): – Every edge in the tree separates at least two
distinguished vertices.– None of the non-tree paths produce any
improvement for any vertex in Si for i = k- log n⌈ ⌉ , where k denotes the maximal degree of the tree.
– If this property is true for all i, then we call it locally optimal Steiner tree (LOST).
Steiner Problem (cont.)
• Similar to the theorem 2.1, we have theorem 3.1: for any b>1, the maximal degree k of a LOST T is at most bΔ* + log n⌈ ⌉ . Hence k = O(Δ* + log n ).
• This idea can be converted into a polynomial time algorithm which produces a Steiner tree of degree O(Δ* + log n ).
Steiner Problem (cont.)
• Theorem 1.2. There is a polynomial time approximation algorithm for the minimum degree Steiner tree problem which produces a Steiner tree of degree O(Δ* + log n ).
Directed spanning trees version
• Input : A directed graph G and anyone
spanning tree together with r
which is the root of the tree.
What is a directed spanning tree ?
• Def : A rooted spanning tree T of G is a
subgraph of G with the following
properties:–T does not contain any cycles.–The outdegree of every vertex except r
is exactly one.–There is a path in T from every vertex to
the root r.
An example
Recall
• The improvement in the undirected case
3
<=2
11
2
<=2
Where is the cycle in directed case ?
New “improvement”
Idea : Consider a vertex v of indegree i.
Try to reduce the indegree of the
vertex by attaching one of the i
subtrees of v to another vertex of
smaller degree.
Step1 : Move the root of the subtree T’ that is
being removed from v to a “convenient”
vertex in that subtree.
Step2 : Attach T’ to another vertex outside
the tree to which the new root has a
connection.
What is a convient vertex ?
• The set of convenient vertices are those in the
strongly connected part of the root of T’ in
the graph induced by the vertices of T’ with all
non-tree edges removed from vertices of
degree i – 1 or greater.• We pick a convenient vertex from convenient
set
T’
Algorithm in directed version
• The algorithm tries to decrease the degrees of
vertices in Si for i =k – using the
improvement step above. nlog
A Lemma
Let T be a directed spanning tree of degree k.
Let Si consist of those vertices whose indegree
is at least i. Suppose we remove the vertices of
Si from T, breaking T into a forest F.
Then there are at least t = ISiI× (i – 1) + 1 trees
of F whose vertices do not have descendants of
degree i or greater in T.
Let Y be “ a tree of F whose vertices
do not have descendants of degree i
or greater in T “ ?
What is a Y ?
i
descendent
ancestor
…
Why ISiI× (i – 1) + 1 ?
• We use induction on the size of |Si|
– Basis : if |Si| =1, then there are (i-1) +1 = i subtrees like Y (o.k)
Note : Each addition of a vertex to the set S i
can remove at most one tree from the
set of candidate trees, but adds at
least i more.
– suppose that |Si| = k holds, i.e : there are k ×(i-1)+1 subtrees like Y.
– then if |Si| = k+1, we will have at least
[k ×(i-1)+1]+i-1 = k×i-k+i = (k+1) ×(i-1)+1
• By math induction, we complete the proof.
THEOREM 1.3
There is a polynomial time approximation algorithm for the directed version of the minimum degree spanning tree problem which produces a directed spanning tree of degree O(∆* + logn).
illustration
As in the proof of the undirected case, we look for an i in the range k to k – where the
set Si does not “expand” by more than a constant factor.We will have at least [ ISiI× (i – 1) + 1 ]+ ISiI−1 edges connecting these components and each one of these edges is incident to at least one
vertex in Si-1
nlog
illustration
Hence the average degree of vertices in Si-1 in
any spanning tree of G is at least
The maximal degree ∆* is at least the average
degree of these vertices.
1
1]1)1(|[|
i
ii
S
SiS
illustration
Hence,
nbkbi
S
iS
S
SiS
i
i
i
ii
log
11)1(
*
1
1
*
The △*+1 algorithm
Definition
k-1k
kk-1
≦k-2
u v
w
≦k-2
k
k-1 k-1
k-1
w
u v
If ρ(u) k – 1, we say that u ≧ blocks w from (u, v).
If neither u nor v blocks w, then w benefits from (u, v).
T
T
The algorithm works in phases
• (Recall Def.) Suppose k were the maximal degree of a spanning tree T, Let Si be the set of vertices of degree at least i in T.
• In each phase we try to reduce the size of Sk by one. If successful, we move on to the next phase.
• So we also update k after each phase.
• Even in a phase the number of vertices of degree at most k-1 increases so many, but it’s ok!
• We use the local optimality property.
many phases
…
try to improve
Max degree = k
but |Sk| remains the same…
Local Optimality Property
Our solution
How many phases?
• As the size of Sk reduces by one in each phase (except the last one), there are at most O (n / k) phases when the maximal degree is k.
• At most phases... ( log )1 3
n n nO n n
k k
k=1
k=2
maximal degree from k to k-1
…
Theorem
• Let * be the degree of a MDST. △• Let S be the set Sk.• Let B be an arbitrary subset of vertices of degree k–1 in T.
S U B
suppose no such edges exist then k *+1≦△
F
Try to prove * something△ ≧
1
1
1
| | 1*
| |
| | 2(| | 1) | | 1
| |
( 1) | | 1
| |
k
k
k k k
k
k
k
t S
S
k S S S
S
k S
S
△
△* ≧ average degree of any subset
we want the vertices of high degree
take Sk
Sk
|F| = t
take S U BS U B
S U B may be either connected or disconnected (Case 1) (Case 2)
( 1) | | 1
| |
k
k
k S
b S
S U B is connected
• Case 1
S U B is connected (a tree), then F contains exactly |S|k+|B|(k-1)-2(|S|+|B|-1) subtrees.
Exactly |S|k+|B|(k-1)-2(|S|+|B|-1)+(|S|+|B|-1) edges connect to S U B.
The average degree of vertices in S U B is exactly
|S|k+|B|(k-1)-(|S|+|B|-1) |S|(k-1)+|B|(k-1)-(|B|-1) |B|-11
|S|+|B| |S|+|B| |S|+|B|k
S U B
S U B is not connected
• Case 2
S U B is not connected (not a tree), then F contains more than |S|k+|B|(k-1)-2(|S|+|B|-1) subtrees.
More than |S|k+|B|(k-1)-2(|S|+|B|-1)+(|S|+|B|-1) edges connect to S U B.
S U B
The average degree of vertices in S U B is more than
|S|k+|B|(k-1)-(|S|+|B|-1) |S|(k-1)+|B|(k-1)-(|B|-1) |B|-11
|S|+|B| |S|+|B| |S|+|B|k
Finally we proved * k-1△ ≧
• △* ≧ average degree of S U B =
• Therefore * k-1 △ ≧ k *+1≦ △
| | 11
| | | |
Bk
S B
0 1 1* 1 1
|S| 0 |S|
*
k k
k
△
△
• Note ： what if |B|=0?
Observation
Let B = Vk-1, apply Theorem 5.1, we are done.
Vk U Vk-1
≦k-2
≦k-2
≦k-2
≦k-2
If no any Sk on the cycle, just union and mark good
kk
k-1k-1k-2
kk-1
Case 1Case 2aCase 2b
k-2k-1…
≦k-1≦k-1
Algorithm• Step 1. Given a SPT (the degree of all vertices is known),
remove Vk U Vk-1 (marked as bad) from T and mark all the connected components as good.
Vk U Vk-1
F
≦k-2≦k-2 ≦k-2
good
bad
Algorithm• Step 2. Choose any edge between 2 components and find
the cycle generated, also check all bad vertices on the cycle. If no such edges exist, stop.
Vk U Vk-1
F
uv≦k-2
≦k-2
≦k-2
≦k-2
Algorithm• Step 3. (Case 1) At least a vertex in Vk on the cycle,
reduce the size of Vk by one, go to Step 1.
Vk U Vk-1
F
uv
≦k-2
≦k-2 ≦k-2
≦k-2
kk-1
Algorithm• Step 3. (Case 2) No any vertex in Vk on the cycle,
(imply) at least a bad vertex in Vk-1 on the cycle, make a union of all components along with all bad vertices on the cycle, also mark the bad vertices as good, go to Step 2.
Vk U Vk-1
Fuv≦k-2
≦k-2 ≦k-2
≦k-2
k-1k-1 k-1
good
Illustration of example 1
≦k-2
k
k-1k-1
good good
k-2
≦k-1
k-1
Illustration of example 2
k or k-1
k-1 ≦k-2
Must not be choosed
Illustration of example 3
k-1k k
Time analysis
• Step 1. Given a SPT (the degree of all vertices is known), remove Vk U Vk-1 (marked as bad) from T and mark all the connected components as good.
• Step 2. choose any edge between 2 components and find
the cycle generated , also check all bad vertices on the cycle. If no such edges exist, stop.
• Step 3. (Case 1) At least a vertex in Vk on the cycle, reduce the size of Vk by one, go to Step 1.
• Step 3. (Case 2) No any vertex in Vk on the cycle, (imply) at least a bad vertex in Vk-1 on the cycle, make a union of all components along with all bad vertices on the cycle, also mark the bad vertices as good, go to Step 2.
( ( ))O m n
(1)O
( ( ))O n n
)(nO
( )O n
