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

Previous Results…(continued) Degree-3 MST in R d, d > 2 5/3-approximation [Khuller, Raghavachari & Young ‘96] approximation [Chan ‘03] Degree-δ MST in R d QPTAS – (n O(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 < 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 A B 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 ≤ 90 o. Then 2|BC|cosθ ≤ |AB| ≤ |BC| 2cosθ C B 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 b c d 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 b c d θ

p v a b c d Ingredients…(continued)

p v a b c d

p v a b c da b v

v p v a b c da b

Chan’s degree-4 recipe v 2 biological child & 1 foster child

Chan’s degree-4 recipe recurse v 2 biological child & 1 foster child

Chan’s degree-4 recipe v 3 biological child & 1 foster child

Chan’s degree-4 recipe v 3 biological child & 1 foster child

Chan’s degree-4 recipe v 3 biological child & 1 foster child

Chan’s degree-4 recipe v 3 biological child & 1 foster child recurse

Chan’s degree-4 recipe 4 biological child & 1 foster child v

Chan’s degree-4 recipe 4 biological child & 1 foster child recurse v

Snippets of our recipe v vertex under consideration v 1 -v 4 v’s biological children v’v’s foster child Obj.Reduce the degree of v to 3 Notemax{θ 1,θ 2,θ 3,θ 4 +θ 5 } ≥ 120 o Solve it by case-by-case analysis. v1v1 v v2v2 v3v3 v4v4 v’ θ1θ1 θ2θ2 θ3θ3 θ4θ4 θ5θ5

Secrets behind our recipe’s success Let θ 5 ≤ 60 o. Then |vv’| ≥ |vv 1 |. Otherwise |v 1 v’| ≥ |vv 1 |, which contradicts the fact that vv 1 was chosen over v 1 v’ to be the MST edge. Removal of vv’ and adding v 1 v’ results in savings, which is used for future extra charge accounting. Smarter charging scheme. v1v1 v v2v2 v3v3 v’ θ1θ1 θ2θ2 θ3θ3 θ4θ4 θ5θ5 v4v4

A stone we turned Case θ 2 ≥ 120 o, θ 4 ≥ 60 o, θ 5 ≤ 60 o, θ 3 > 90 o v v2v2 v3v3 v4v4 v’ θ1θ1 ≥ 120 o θ4θ4 θ5θ5 θ3θ3 θ2θ2 ≤ 60 o ≥ 60 o v1v1 > 90 o

A stone we turned Case θ 2 ≥ 120 o, θ 4 ≥ 60 o, θ 5 ≤ 60 o, θ 3 > 90 o For simplicity, let |vv 1 | = x 1 ; … ; |vv 4 | = x 4 and |vv’| = x 5 v θ1θ1 ≥ 120 o θ4θ4 θ5θ5 θ3θ3 θ2θ2 ≤ 60 o ≥ 60 o x1x1 x2x2 x3x3 x4x4 x5x5 > 90 o

A stone we turned Case θ 2 ≥ 120 o, θ 4 ≥ 60 o, θ 5 ≤ 60 o, θ 3 > 90 o For simplicity, let |vv 1 | = x 1 ; … ; |vv 4 | = x 4 and |vv’| = x 5 Subcase x 3 or x 4 is the 2 nd smallest among {x 1, x 2, x 3, x 4 } AND x 1 ≤ x 2, x 3, x 4 v θ1θ1 ≥ 120 o θ4θ4 θ5θ5 θ3θ3 θ2θ2 ≤ 60 o ≥ 60 o > 90 o x1x1 x2x2 x3x3 x4x4 x5x5

A stone we turned Case θ 2 ≥ 120 o, θ 4 ≥ 60 o, θ 5 ≤ 60 o, θ 3 > 90 o For simplicity, let |vv 1 | = x 1 ; … ; |vv 4 | = x 4 and |vv’| = x 5 Subcase x 3 or x 4 is the 2 nd smallest among {x 1, x 2, x 3, x 4 } AND x 1 ≤ x 2, x 3, x 4 Savings = 2(sin(90-θ 5 )-1) x 1 v θ1θ1 ≥ 120 o θ4θ4 θ5θ5 θ3θ3 θ2θ2 ≤ 60 o ≥ 60 o > 90 o x1x1 x2x2 x3x3 x4x4 x5x5

A stone we turned Case θ 2 ≥ 120 o, θ 4 ≥ 60 o, θ 5 ≤ 60 o, θ 3 > 90 o For simplicity, let |vv 1 | = x 1 ; … ; |vv 4 | = x 4 and |vv’| = x 5 Subcase x 3 or x 4 is the 2 nd smallest among {x 1, x 2, x 3, x 4 } AND x 1 ≤ x 2, x 3, x 4 Savings = 2(sin(90-θ 5 )-1) x 1 It is as if we have at least an additional 2(sin(90-θ 5 )-1) x 1 / to charge. v θ1θ1 ≥ 120 o θ4θ4 θ5θ5 θ3θ3 θ2θ2 ≤ 60 o ≥ 60 o > 90 o x1x1 x2x2 x3x3 x4x4 x5x5

A stone we turned W.l.o.g., let x 3 ≤ x 4 Savings = 2(sin(90-θ 5 )-1) x 1 /0.1381) Additional cost due to transformations is ≤ F(θ 3 ) * (x 1 + x 2 + x 3 + x 4 + Savings ) 3 + 2cos(120 o –θ 5 /2) * (1+Savings) which is bounded by (x 1 + x 2 + x 3 + x 4 + Savings ) v θ1θ1 ≥ 120 o θ4θ4 θ5θ5 θ3θ3 θ2θ2 ≤ 60 o ≥ 60 o x1x1 x2x2 x3x3 x4x4 x5x5 > 90 o

A stone we turned W.l.o.g., let x 3 ≤ x 4 Savings = 2(sin(90-θ 5 )-1) x 1 /0.1381) Additional cost due to transformations is ≤ F(θ 3 ) * (x 1 + x 2 + x 3 + x 4 + Savings ) 3 + 2cos(120 o –θ 5 /2) * (1+Savings) which is bounded by (x 1 + x 2 + x 3 + x 4 + Savings ) v θ1θ1 ≥ 120 o θ4θ4 θ5θ5 θ3θ3 θ2θ2 ≤ 60 o ≥ 60 o x1x1 x2x2 x3x3 x4x4 x5x5 > 90 o

A stone we turned There are cases where the ratio is bounded by (√2 – 1 )/3 < That makes the approximation ratio of our algorithm to be (√2 + 2 )/3 <

Dessert Our ratio of (√2 + 2 )/3 < 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 (2sin36 o + 4)/5 < times the weight of MST.

Future Problems Points in the plane Is degree-4 MST problem NP-hard? Improve the ratio for degree-4 trees. Improve the ratio for degree-3 trees. Is there a PTAS for degree-δ MST problem? Points in higher dimensions Improve the ratio for degree-3 trees. Approximation of degree-δ trees in general metric spaces (no triangle inequality), within ratios better than 2.