# Leave No Stone Unturned: Improved Approximation Algorithm for Degree-Bounded MSTs

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Leave No Stone Unturned: Improved Approximation Algorithm

for Degree-Bounded MSTs

Raja Jothi

University of Texas at Dallas

Joint work with Balaji Raghavachari

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Appetizer On-tray Dessert

Degree-Bounded MSTs

Problem Definition

Given n points in the Euclidean plane, the degree-δ MST asks

for a minimum weight spanning tree in which the degree of each

vertex is at most δ.

Complexity δ=5: Polynomial time solvable [Monma & Suri ’92] NP-hard for δ=3 [Papadimitriou & Vazirani ’84] NP-hard for δ=4?

Previous Results

Degree-2 MST PTASs [Arora ’96 & Mitchell ‘97]

Degree-3 MST 1.5-approximation [Khuller, Raghavachari & Young ‘96] 1.402-approximation [Chan ‘03]

Degree-4 MST 1.25-approximation [Khuller, Raghavachari & Young ‘96] 1.143-approximation [Chan ‘03]

Previous Results…(continued)

Degree-3 MST in Rd, d > 2 5/3-approximation [Khuller, Raghavachari & Young ‘96] 1.633-approximation [Chan ‘03]

Degree-δ MST in Rd

QPTAS – (nO(log n)) [Arora & Chang ‘03]

Our Result

For any arbitrary collections of points in the plane,

there always exists a degree-4 MST of weight at most

(√2 + 2 )/3 < 1.1381 times the weight of an MST.

Preliminaries

Input: Degree-5 MST T Root T at an arbitrary vertex Every vertex, but root, has at most 4 children

Khuller, Raghavachari & Young’s recipe

Degree-3 MST

Khuller, Raghavachari & Young’s recipe

Degree-3 MST

Khuller, Raghavachari & Young’s recipe

Degree-3 MST

Khuller, Raghavachari & Young’s recipe

Degree-3 MST

Khuller, Raghavachari & Young’s recipe

Degree-4 MST

Khuller, Raghavachari & Young’s recipe

Degree-4 MST

Khuller, Raghavachari & Young’s recipe

Degree-4 MST

Khuller, Raghavachari & Young’s recipe

Degree-4 MST

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Ingredients

Strengthened triangle inequality

If |AB| ≤ |BC|, then |AC| ≤ F(θ)|AB| + |BC|

Where F(θ) = sqrt(2(1-cosθ)) – 1

AB

C

B’

θ

Ingredients…(continued)

Bounds on edge weights of an MST

Let AB and BC be two edges that intersect

at point B in an MST. Let θ = ABC ≤ 90o.

Then

2|BC|cosθ ≤ |AB| ≤ |BC| 2cosθ

CB

A

θ

Ingredients…(continued)

Charging Scheme

Let |av| ≤ |bv|. Then, cost of the new

tree formed by replacing bv by ab is

at most

(|av| + |bv| + |cv| + |dv|) * F(θ) / k

p

v a

bc

dθ

Ingredients…(continued)

Charging Scheme

Let |av| ≤ |bv|. Then, cost of the new

tree formed by replacing bv by ab is

at most

(|av| + |bv| + |cv| + |dv|) * F(θ) / k

p

v a

bc

dθ

p

v a

bc

d

Ingredients…(continued)

Ingredients…(continued)

p

v a

bc

d

Ingredients…(continued)

p

v a

bc

d a

b v

v

Ingredients…(continued)

p

v a

bc

d a

b

Chan’s degree-4 recipe

v2 biological child & 1 foster child

Chan’s degree-4 recipe

recurserecurse

v

recurse

2 biological child & 1 foster child

Chan’s degree-4 recipe

v3 biological child & 1 foster child

Chan’s degree-4 recipe

v3 biological child & 1 foster child

Chan’s degree-4 recipe

v3 biological child & 1 foster child

Chan’s degree-4 recipe

v3 biological child & 1 foster child

recurserecurse

recurse

Chan’s degree-4 recipe

4 biological child & 1 foster childv

Chan’s degree-4 recipe

4 biological child & 1 foster child

recurse recurse

v

recurse

Snippets of our recipe

v vertex under consideration

v1-v4 v’s biological children

v’ v’s foster child

Obj. Reduce the degree of v to 3

Note max{θ1,θ2,θ3,θ4+θ5} ≥ 120o

Solve it by case-by-case analysis.

v1 v

v2 v3

v4

v’

θ1

θ2

θ3

θ4θ5

Secrets behind our recipe’s success

Let θ5 ≤ 60o.

Then |vv’| ≥ |vv1|.

Otherwise |v1v’| ≥ |vv1|, which contradicts

the fact that vv1was chosen over v1v’

to be the MST edge. Removal of vv’ and adding v1v’ results

in savings, which is used for future extra

charge accounting. Smarter charging scheme.

v1 v

v2 v3

v’

θ1

θ2

θ3

θ4θ5

v4

A stone we turned

Case θ2 ≥ 120o, θ4 ≥ 60o, θ5 ≤ 60o , θ3 > 90o

v

v2 v3

v4

v’

θ1

≥ 120o

θ4θ5

θ3

θ2

≤ 60o≥ 60o

v1

> 90o

A stone we turned

Case θ2 ≥ 120o, θ4 ≥ 60o, θ5 ≤ 60o , θ3 > 90o

For simplicity,

let |vv1| = x1; … ; |vv4| = x4 and |vv’| = x5

vθ1

≥ 120o

θ4θ5

θ3

θ2

≤ 60o≥ 60o

x1

x2 x3

x4

x5

> 90o

A stone we turned

Case θ2 ≥ 120o, θ4 ≥ 60o, θ5 ≤ 60o , θ3 > 90o

For simplicity,

let |vv1| = x1; … ; |vv4| = x4 and |vv’| = x5

Subcase x3 or x4 is the 2nd smallest among

{x1, x2, x3, x4} AND x1 ≤ x2, x3, x4

vθ1

≥ 120o

θ4θ5

θ3

θ2

≤ 60o≥ 60o

> 90o

x1

x2 x3

x4

x5

A stone we turned

Case θ2 ≥ 120o, θ4 ≥ 60o, θ5 ≤ 60o , θ3 > 90o

For simplicity,

let |vv1| = x1; … ; |vv4| = x4 and |vv’| = x5

Subcase x3 or x4 is the 2nd smallest among

{x1, x2, x3, x4} AND x1 ≤ x2, x3, x4

Savings = 2(sin(90-θ5)-1) x1

vθ1

≥ 120o

θ4θ5

θ3

θ2

≤ 60o≥ 60o

> 90o

x1

x2 x3

x4

x5

A stone we turned

Case θ2 ≥ 120o, θ4 ≥ 60o, θ5 ≤ 60o , θ3 > 90o

For simplicity,

let |vv1| = x1; … ; |vv4| = x4 and |vv’| = x5

Subcase x3 or x4 is the 2nd smallest among

{x1, x2, x3, x4} AND x1 ≤ x2, x3, x4

Savings = 2(sin(90-θ5)-1) x1

It is as if we have at least an additional

2(sin(90-θ5)-1) x1/0.1381 to charge.

vθ1

≥ 120o

θ4θ5

θ3

θ2

≤ 60o≥ 60o

> 90o

x1

x2 x3

x4

x5

A stone we turned

W.l.o.g., let x3 ≤ x4

Savings = 2(sin(90-θ5)-1) x1/0.1381)

Additional cost due to transformations is

≤ F(θ3) * (x1 + x2 + x3 + x4 + Savings )

3 + 2cos(120o–θ5/2) * (1+Savings)

which is bounded by 0.0709(x1 + x2 + x3 + x4 + Savings )

vθ1

≥ 120o

θ4θ5

θ3

θ2

≤ 60o≥ 60o

x1

x2 x3

x4

x5

> 90o

A stone we turned

W.l.o.g., let x3 ≤ x4

Savings = 2(sin(90-θ5)-1) x1/0.1381)

Additional cost due to transformations is

≤ F(θ3) * (x1 + x2 + x3 + x4 + Savings )

3 + 2cos(120o–θ5/2) * (1+Savings)

which is bounded by 0.0709(x1 + x2 + x3 + x4 + Savings )

vθ1

≥ 120o

θ4θ5

θ3

θ2

≤ 60o≥ 60o

x1

x2 x3

x4

x5

> 90o

A stone we turned

There are cases where the ratio is bounded by

(√2 – 1 )/3 < 0.1381 That makes the approximation ratio of our algorithm

to be (√2 + 2 )/3 < 1.1381.

Dessert

Our ratio of (√2 + 2 )/3 < 1.1381 CANNOT be improved by using just local changes.

Improvement of the ratio requires a global approach. There exists a degree-4 tree whose weight is

(2sin36o + 4)/5 < 1.0352 times the weight of MST.

Future Problems

Points in the plane Is degree-4 MST problem NP-hard? Improve the 1.1381 ratio for degree-4 trees. Improve the 1.402 ratio for degree-3 trees. Is there a PTAS for degree-δ MST problem?Points in higher dimensions Improve the 1.633 ratio for degree-3 trees. Approximation of degree-δ trees in general metric spaces

(no triangle inequality), within ratios better than 2.