Leave No Stone Unturned: Improved Approximation Algorithm for DegreeBounded MSTs

1
Leave No Stone Unturned Improved Approximation
Algorithm for Degree-Bounded MSTs

• Raja Jothi
• University of Texas at Dallas
• raja_at_utdallas.edu
• Joint work with Balaji Raghavachari

2
Menu

• Appetizer
• On-tray
• Dessert

3
Degree-Bounded MSTs

• Problem Definition
• Given n points in the Euclidean plane, the
  degree-d MST asks
• for a minimum weight spanning tree in which the
  degree of each
• vertex is at most d.
• Complexity
• d5 Polynomial time solvable Monma Suri 92
• NP-hard for d3 Papadimitriou Vazirani 84
• NP-hard for d4?

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

5
Previous Results(continued)

• Degree-3 MST in Rd, d 2
• 5/3-approximation Khuller, Raghavachari Young
  96
• 1.633-approximation Chan 03
• Degree-d MST in Rd
• QPTAS (nO(log n)) Arora Chang 03

6
Our Result

• For any arbitrary collections of points in the
  plane,
• there always exists a degree-4 MST of weight at
  most
• (v2 2 )/3

7
Preliminaries

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

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Khuller, Raghavachari Youngs recipe
Degree-3 MST
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Khuller, Raghavachari Youngs recipe
Degree-3 MST
10
Khuller, Raghavachari Youngs recipe
Degree-3 MST
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Khuller, Raghavachari Youngs recipe
Degree-3 MST
12
Khuller, Raghavachari Youngs recipe
Degree-4 MST
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Khuller, Raghavachari Youngs recipe
Degree-4 MST
14
Khuller, Raghavachari Youngs recipe
Degree-4 MST
15
Khuller, Raghavachari Youngs recipe
Degree-4 MST
16
Menu

• Appetizer
• On-tray
• Dessert

17
Ingredients

• Strengthened triangle inequality
• If AB BC, then AC F(?)AB BC
• Where F(?) sqrt(2(1-cos?)) 1

C
B
?
B
A
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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 ? ABC 90o.
• Then
• 2BCcos? AB

A
BC 2cos?
?
B
C
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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(?) / k

p
a
d
v
?
c
b
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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(?) / k
p
a
d
v
?
c
b
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Ingredients(continued)
p
a
d
v
c
b
22
Ingredients(continued)
p
a
d
v
c
b
23
Ingredients(continued)
v
p
b
a
d
a
v
c
b
24
Ingredients(continued)
v
p
b
a
d
a
v
c
b
25
Chans degree-4 recipe
2 biological child 1 foster child
v
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Chans degree-4 recipe
2 biological child 1 foster child
v
recurse
recurse
recurse
27
Chans degree-4 recipe
3 biological child 1 foster child
v
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Chans degree-4 recipe
3 biological child 1 foster child
v
29
Chans degree-4 recipe
3 biological child 1 foster child
v
30
Chans degree-4 recipe
3 biological child 1 foster child
v
recurse
recurse
recurse
31
Chans degree-4 recipe
4 biological child 1 foster child
v
32
Chans degree-4 recipe
4 biological child 1 foster child
v
recurse
recurse
recurse
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Snippets of our recipe

• v vertex under consideration
• v1-v4 vs biological children
• v vs foster child
• Obj. Reduce the degree of v to 3
• Note max?1,?2,?3,?4?5 120o
• Solve it by case-by-case analysis.

v
?5
?4
v1
v
v4
?1
?3
?2
v2
v3
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Secrets behind our recipes success

• Let ?5 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.

v
?5
?4
v1
v
v4
?1
?3
?2
v2
v3
35
A stone we turned

• Case ?2 120o, ?4 60o, ?5 60o , ?3 90o

v
60o
60o
?5
?4
v
v4
v1
?1
?3
90o
?2
120o
v2
v3
36
A stone we turned

• Case ?2 120o, ?4 60o, ?5 60o , ?3 90o
• For simplicity,
• let vv1 x1 vv4 x4 and vv x5

x5
60o
60o
?5
?4
v
x1
x4
?1
?3
90o
?2
120o
x2
x3
37
A stone we turned

• Case ?2 120o, ?4 60o, ?5 60o , ?3 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

x5
60o
60o
?5
?4
v
x1
x4
?1
?3
90o
?2
120o
x2
x3
38
A stone we turned

• Case ?2 120o, ?4 60o, ?5 60o , ?3 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-?5)-1) x1

x5
60o
60o
?5
?4
v
x1
x4
?1
?3
90o
?2
120o
x2
x3
39
A stone we turned

• Case ?2 120o, ?4 60o, ?5 60o , ?3 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-?5)-1) x1
• It is as if we have at least an additional
• 2(sin(90-?5)-1) x1/0.1381 to charge.

x5
60o
60o
?5
?4
v
x1
x4
?1
?3
90o
?2
120o
x2
x3
40
A stone we turned

• W.l.o.g., let x3 x4
• Savings 2(sin(90-?5)-1) x1/0.1381)
• Additional cost due to transformations is
• F(?3) (x1 x2 x3 x4 Savings )
• 3 2cos(120o?5/2) (1Savings)
• which is bounded by 0.0709(x1 x2 x3 x4
  Savings )

x5
60o
60o
?5
?4
v
x1
x4
?1
?3
90o
?2
120o
x2
x3
41
A stone we turned

• W.l.o.g., let x3 x4
• Savings 2(sin(90-?5)-1) x1/0.1381)
• Additional cost due to transformations is
• F(?3) (x1 x2 x3 x4 Savings )
• 3 2cos(120o?5/2) (1Savings)
• which is bounded by 0.0709(x1 x2 x3 x4
  Savings )

x5
60o
60o
?5
?4
v
x1
x4
?1
?3
90o
?2
120o
x2
x3
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A stone we turned

• There are cases where the ratio is bounded by
• (v2 1 )/3
• That makes the approximation ratio of our
  algorithm to be (v2 2 )/3

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Dessert

• Our ratio of (v2 2 )/3 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 MST.

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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-d MST problem?
• Points in higher dimensions
• Improve the 1.633 ratio for degree-3 trees.
• Approximation of degree-d trees in general metric
  spaces (no triangle inequality), within ratios
  better than 2.