Minimum-Cost Bounded-Degree Spanning Trees

Minimum-cost Bounded-degree Spanning Trees

Feliciano Colella

17th December 2014

Introduction Approximation factor bv + 2 Approximation factor bv + 1 Further proofs and analysis

Outline

1 Definition of the problem;

2 Algorithm with approximation factor bv + 2;

3 Algorithm with approximation factor bv + 1;

4 Further analysis and some proofs.

Uncovered It is NP-complete to decide whether or not a givengraph has a minimum-degree spanning tree ofmaximum degree two.

Feliciano Colella Minimum-cost Bounded-degree Spanning Trees


The minimum-cost bounded-degree spanning tree problem

Input

• Undirected graph G = (V ,E ),

• Costs ce ≥ 0 for all e ∈ E ,

• A set W ⊆ V , with bounds bv ≥ 1 for all v ∈W .

Output

The minimum spanning tree (MST) such that the degree of eachnode v ∈W is no more than bv , it cannot exists.Let OPT be the cost of the MST.



Linear Programming formulation

Given a vertex set S ⊆ V , let:

• E (S) subset of edges in E that have both endpoints in S ,

• δ(S) subset of edges that have only one endpoint in S ,

xe ∈ {0, 1} indicate whether e is in the ST or not.



Linear Programming formulation

minimize∑e∈E

cexe

subject to∑e∈E

xe = |V | − 1,∑e∈E(S)

xe ≤ |V | − 1, ∀S ⊆ V , |S | ≥ 2,∑e∈δ(v)

xe ≤ bv , ∀v ∈W ,

xe ≥ 0, ∀e ∈ E .

Can be solved with ellipsoid method by giving a polynomial-timeseparation oracle.



A sequence of Max-Flow problems

Create a new graph G ′ where we add two vertices s and t, and weadd edges (s, v) and (v , t) for each v ∈ V .The capacity of e ∈ G is 1

2xe , 12

∑e∈δ(v) xe for every edge (s, v)

and 1 for every edge (v , t), the total is:

|S |+ 1

2

∑e∈δ(S)

xe +1

2

∑v /∈S

∑e∈δ(v)

xe = |S |+∑

e∈δ(S)

xe +∑

e∈E(V−S)

xe

Using that∑

e∈E xe = |V | − 1 we have:

|S |+ |V | − 1−∑

e∈E(S)

xe = |V |+ (|S |+ 1)−∑

e∈E(S)

xe

The capacity of this cut is at least |V | ⇐⇒∑

e∈E(S)

xe ≥ |S | − 1

We want |S | ≥ 2 for each pair x , y ∈ V , how ?



An algorithm for the case W = ∅

Theorem

Algorithm below yields a spanning tree of no more than the valueof the linear program.

F ← ∅while |V | > 1 do

Solve LP on (V ,E ), basic solution is xE ← E (x)Find v ∈ V s.t. there is only one edge (u, v) incident on vF ← F ∪ {(u, v)}V ← V − {v}E ← E − {(u, v)}

return F



Returning to our problem ...

... we show two main results:

Lemma

For any basic feasible solution x to the linear program, either thereis some v ∈ V such that there is at most one edge of E (x)incident on v , or there is some v ∈W such that there are at mostthree edges of E (x) incident on v .

Theorem

The next algorithm produces a spanning tree F such that thedegree of v ∈ F is at most bv + 2 for v ∈W , and such that thecost of F is at most the value of the linear program.



Algorithm with bv + 2 approximation factor

F ← ∅while |V | > 1 do

Solve LP on (V ,E ), basic solution is xE ← E (x)if ∃v ∈ V s.t. there is e = (u, v) ∈ E (x) incident on v then

F ← F ∪ {(u, v)}V ← V − {v}E ← E − {(u, v)}if u ∈W then

bu ← bu − 1

else∃v ∈W s.t. there are at most 3 edges (u, v) ∈ E (x) on vW ←W − {v}

return F



Proof of the previous Theorem

• Each step add edge to F or remove vertex from W , thus(n − 1) + 1 = 2n − 1 iterations.

• Proof of the cost of the spanning tree still holds from theprevious theorem, we need to consider the set of bounds W .

• Since x was feasible for the constraint∑

e∈δ(u) xe ≤ bu, thenif we remove e = (v , u) with xe = 1 from the graph, we havethat x is still feasible on new graph with bound bu − 1.

Now consider a vertex v initially in W , its degree in F will be atmost bv + 2.



Proof of the previous Theorem

By cases:

1 The edge is incident on v , bv decrease by 1;

2 The edge is incident on v and v is removed from the graph:• we must have had bv ≥ 1 to have a feasible solution, so the

statement holds.

3 The node v is removed from W :• we must have had bv ≥ 1 for there to be any edges in E (x)

incident on v ;• there are at most 3 edges in E (x) incident on v ;• we can add at most all of the 3 edges, so the degree will be at

most bv + 2

�



A better algorithm

We give now an algorithm that finds a spanning tree of cost atmost OPT such that each node v ∈W has degree at most bv + 1.

while W 6= ∅ doSolve LP on (V ,E ),W , get basic optimal solution is xE ← E (x)Select v ∈W such that there are at most bv + 1 edges in Eincident on vW ←W − {v}

Compute minimum-cost ST F on (V ,E )return F



The key of the proof

The key idea is to state and prove a stronger lemma than theprevious one:

Lemma

For any basic feasible solution x to the LP in which W 6= ∅, thereis some v ∈W such that there are at most bv + 1 edges incidenton v in E (x).

Which leads to the theorem:

Theorem

The previous algorithm produces a spanning tree F such that thedegree of v ∈ F is at most bv + 1 for v ∈W , and such that thecost of F is at most the value of the linear program.



Proof of the algorithm with approx factor bv + 1

Proof.

• By induction on iterations, if xe is feasible for (V ,E ),W itwill still be on (V ,E ′),W ′;

• We remove e from E if xe = 0, or we remove v from W ,imposing fewer constraint.

• Those removals does not affect the feasibility of the solution;

• Since we can remove a vertex from W , eventually W = ∅;• When we modify E and W , the value of LP never increase;

• The cost of the MST is at most the value of the LP;

• We removed v from W only if there were at most bv + 1edges remaining (in E on v);

• The spanning tree can have at most degree bv + 1,∀v ∈W .



Starting definitions

From now on, assume that the edges e s.t. xe = 0, has beenremoved, so E = E (x).

Definition 11.7 given a subset F , we define x(F ) =∑

e∈F xe ;

Definition 11.8 given x , a constraint is tight if x(E (S)) = |S | − 1or x(δ(v)) = bv ;

Definition 11.9 A and B are intersecting if A ∩ B, A− B andB − A are all non empty;

Definition 11.10 S is laminar if no pair of sets A,B ∈ S areintersecting;

Definition 11.11 given a subset F ⊆ E , the characteristic vector isχF ∈ {0, 1}|E |, where χF (e) = 1 if e ∈ F .



Some useful Results

Lemma 11.13

Let L be a laminar collection of subset of V , where each S ∈ Lhas |S | ≥ 2. Then |L| ≤ |V | − 1.

Proof.

Induction on |V |. Base |V | = 2, L only contains one set of size 2.For |V | ≥ 2:

• pick R ∈ L with minimum cardinality

• let V ′ be V \R and L′ the sets restricted to V ′

• L′ is still laminar, so |L′| ≤ |V ′| − 1 by induction

• since |L′| = |L| − 1 and |V ′| = |V | − 1 the statement follows



Some useful Results

Lemma 11.14

If S and T are tight sets such that S and T are intersecting, thenS ∪ T and S ∩ T are also tight sets. Furthermore,

χE(S) + χE(T ) = χE(S∪T ) + χE(S∩T )

Let T = {S ⊆ V : x(E (S)) = |S | − 1, |S | ≥ 2} be the collection ofall the tight sets for the LP solution x .Let span(T ) = {χE(S) : S ∈ T } be the span of that sets of vectors.

Lemma 11.15

There exists a laminar collection of sets L such thatspan(L) ⊇ span(T ), where each S ∈ L is tight and has |S | ≥ 2,and the vectors χE(S) for S ∈ L are linearly independent.



Some useful Results

Theorem 11.12

For any basic feasible solution x to the LP, there is a set Z ⊆Wand a collection L of subsets of vertices with the followingproperties:

1 For all S ∈ L, |S | ≥ 2 and S is tight, and for all v ∈ Z , v istight.

2 The vectors χE(S) for S ∈ L and χδv) ∈ Z are linearlyindependent.

3 |L| = |Z |+ |E |.4 The collection L is laminar.



Some useful Results

Lemma 11.18

For any e ∈ E such that xe = 1, χe ∈ span(L).Proof Follows from the previous theorem.

Lemma 11.16

For any basic feasible solution x to the LP in which W 6= ∅, thereis some v ∈W such that there are at most bv + 1 edges incidenton v in E (x).



Proof of Lemma 11.16

Proof

By contradiction, there are at least bv + 2 edges for each v ∈ Zwith W 6= ∅.We derive the contradiction with a charging scheme, in which weassign each v ∈ Z and each S ∈ L a charge.

We will show that:

• each v ∈ Z and each S ∈ L receives a charge of at least one

• total charge is |Z |+ |L| = |E |• total charge must be strictly less than |E |

Assign a charge of 12(1− xe) to each endpoint of e ∈ Z , and assign

of xe to the smallest S ∈ L that contains both endpoints of e, ifexists. Notice that the total charge is at most 1 per edge e ∈ E .




Proof

By contradiction, there are at least bv + 2 edges for each v ∈ Zwith W 6= ∅.We derive the contradiction with a charging scheme, in which weassign each v ∈ Z and each S ∈ L a charge.We will show that:

• each v ∈ Z and each S ∈ L receives a charge of at least one

• total charge is |Z |+ |L| = |E |• total charge must be strictly less than |E |

Assign a charge of 12(1− xe) to each endpoint of e ∈ Z , and assign

of xe to the smallest S ∈ L that contains both endpoints of e, ifexists. Notice that the total charge is at most 1 per edge e ∈ E .




Proof

Each v ∈ Z receives a charge of 12(1− xe) from each edge of E

incident. By hypothesis, there are at least bv + 2 edges on v .Since v ∈ Z implies v is tight, we know that

∑e∈δ(v) xe = bv , thus

the total is:

∑e∈δ(v)

1

2(1− xe) =

1

2

|δ(v)| −∑

e∈δ(v)

xe

≥ 1

2[(bv + 2)− bv ]

= 1.




Proof

Consider a set S ∈ L. If S contains no other set C ∈ L, since S istight and |S | ≥ 2 then the total charge is at least 1. Suppose thatS has a child C ∈ L which is strictly contained in S . Let C1 . . .Ck

be the children of S , they are all tight and x(E (Ci )) = |Ci | − 1. Bylaminarity of L and for Theorem 11.12, the total charge to S is:

x(E (S))−k∑

i=1

x(E (Ci )) > 0

The following is integral, which leads to a charge of at least one

x(E (S))−k∑

i=1

x(E (Ci )) = (|S | − 1)−k∑

i=1

(|Ci | − 1)




Proof.

We show that the total charge is less than |E | by cases:

1 If V /∈ L2 Exists some e = (u, v) ∈ E with xe < 1 s.t. one of its

endpoint is not in Z

3 Suppose V ∈ L and for any e ∈ E with xe < 1, bothendpoints of e are in Z .

1 Exists e /∈ E (S) ∀S ∈ L, charge xe > 0 is not made to anyset, so the total is strictly less than |E |.

2 Let u ∈ Z , his charge is not made to 12(1− xe) > 0.

3 This case lead to a contradiction (for 11.12), one of theprevious cases must occur (proved using lemma 11.18).


Questions?

Minimum-Cost Bounded-Degree Spanning Trees

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