Low Degree Spanning Trees of Small Weight

1
Low Degree Spanning Trees of Small Weight

• Samir Khuller,
• Balaji Raghavachari,
• and Neal Young

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2
Outline

• Introduction
• Spanning tree of degree 3
• Algorithm
• Analysis theorem proof
• Spanning tree of degree 4
• Points in Higher dimensions

3
Introduction
Euclidean degree-K spanning tree problem Given
n points in the plane, find a minimum spanning
tree of degree at most K. When K3 and 4, it was
shown to be NP-hard by Papadimitriou and
Vazirani. On two geometric problems related to
the traveling salesman problem, J. Algorighms, 5
(1984) When K5, the problem can be solved in
polynomial time. Monma and Suri, Transitions in
geometric minimum spanning trees, Discrete
Computational Geometry, 8 (1992) In this paper,
an 1.5 approximation algorithm for degree-3
spanning tree and an 1.25 approximation algorithm
for degree-4 spanning tree are constructed. (Impro
vements (2003 SoCG) degree-3 1.402degree-4
1.143)
4
Some definitions

1. Degree of a vertex
2. Degree of a tree deg(T) max deg(v)
   v ? T
3. VIVJ the length of the segment connecting VI
   and VJ
4. V1V2V3Vk V1V2 V2V3 Vk-1Vk VkV1
5. Tk A spanning tree of degree at most K

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Algorithm
V is a set of points in the plane, T is a minimum
spanning tree of V with degree at most 5

• TREE-3 ( V, T ) - Find a degree 3 tree of V
• Root the MST T at a leaf vertex r.
• For each vertex v ? V do
• Compute Pv, the shortest path starting at v
  and visiting all the children of v.
• Return T3, the tree formed by the union of the
  paths Pv

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Feasibility
Claim T3 is a spanning tree of degree at
most 3 sketch of the proof ( i )
spanning tree Induction shows that the union of
the paths forms a tree. ( ii ) of
degree at most 3 Observe that each vertex is on
at most two paths and is an interior
vertex of at most one path.
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Quality
Theorem Let T be a minimum spanning tree of
V, T3 be the spanning tree output by
TREE-3(V,T). Then w (T3) lt 1.5 w (T)
To see the theorem, we need two addition lemmas.
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lemma 1
lemma 2
Let v be a vertex in an MST T of a set of points
in R2, Pv be a shortest path starting at v and
visiting childT(v). Then w (Pv) lt 1.5 ?
v vi vi?childT(v)
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Theorem Let T be a minimum spanning tree of
V, T3 be the spanning tree output by
TREE-3(V,T). Then w (T3) lt 1.5 w (T)
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Proof of lemma 1
Proof
Let B and C be points on XB and XC such that XA
XB XC . By triangle inequality, we haveABC
lt ABC 2 BB 2 CC
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Proof of lemma 1
By scaling, we can assume that XA 1, then the
right hand side becomes 3 3 . Thus, it suffices
to show that the maximum perimeter achieved by
any triangle whose vertices lie on a unit circle
is 3 3 . This is easily proved.
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Proof of lemma 2
Let v be a vertex in an MST T of a set of points
in R2, Pv be a shortest path starting at v and
visiting childT(v). Then w (Pv) lt 1.5 ?
v vi vi?childT(v)
Proof
The proof is based on the number of children of v.
v
Case 1 v has two children, v1 and v2. There are
two possible paths for Pv, namely P1v,v1,v2
and P2v,v2,v1. Clearly, we have w(Pv) lt min
w(P1), w(P2) lt 0.5 ( w(P1)
w(P2) ) 0.5 ( v v1 v v2 ) v1v2
lt 1.5 ( v v1 v v2 )
2
1
v
2
1
13
v
v
Proof of lemma 2
v
2
1
3
2
3
1
2
1
v
Case 2v has 3 children, v1, v2, and v3.
3
1
3
2
Let v1 be the child that is nearest to
v. Consider the four pathsP1 v,v1,v2,v3 ,
P2 v,v1,v3,v2 , P3 v,v2,v1,v3 , P4
v,v3,v1,v2 By lemma 1, we have
v1v2 v2v3 v3v1 lt ( 3 3 - 4 ) v v1
2 ( v v2 v v3 ) lt 1.25 v v1 2 ( v
v2 v v3 ) ? w(P1) w(P2) / 3 w(P3)
w(P4) / 6 lt 1.5( v v1 v v2 v v3 ) So, w
(Pv) lt min w(P1), w(P2), w(P3), w(P4)
lt w(P1) w(P2) / 3 w(P3) w(P4) /
6 lt 1.5 ( v v1 v v2 v v3 )
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Proof of lemma 2
Case 3v has 4 children, v1, v2, v3, and v4,
ordered clockwise around v.
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Proof of lemma 2
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Proof of lemma 2
If we shrink v v3 by a,then the left side
decreases by at most 2a, whereas the right side
decrease by exactly 3a. So if we shrink v v3 ,
the inequality still holds. Suppose v v3
shrinks and becomes equal to another edge v vi
for some i ? 1,2,4 . We now shrink v v3 and
v vi simultaneously. Again it is easy to show
that the inequality still holds by the same
argument.
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Proof of lemma 2
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Proof of lemma 2
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Proof of lemma 2
Case 3b. v v2 b . In this case, we have
v1v2v3v4 ( v1v2 v1v4 ) gt 8 2b . Define
F(b) v1v2v3v4 ( v1v2 v1v4 ) - 8 - 2b
0ltblt1 As the same reason in 3a, we see that F
is also a convex function of b and reaches its
maximum at either b0 or b1. The case b1 leads
to the same configuration as in case 3a. When
b0, F can be written as a function of a single
variable and reaches a maximum value of 5 0.8
5 , which is negative. So, we have F(b) lt 0 for
0ltblt1, a contradiction. From case 3a and 3b,
we have proved the claim and this lemma.
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Spanning Trees for Degree 4

• Given the MST which has degree at most 5
• High Level Description
• Explain Pv

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Spanning Trees for Degree 4

• Done.
• This algorithm runs in linear time.
• The only problem left is to prove the bound on
  the approximation ratio.

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Spanning Trees for Degree 4

• If we can prove
• Then immediately follows.
• How to prove this?
• Since Pv is the shortest path visiting v and its
  children, we can enumerate several possible paths
  that do so.
• Pv is no greater than any of them, and is no
  greater than any convex combination of them.
• Choose a combination for each case so that it
  gives us the bound.

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Spanning Trees for Degree 4

• Cases
• Identified according to the number of children of
  v on the original MST, T (which is given as the
  input, and has the property that each non-root
  internal node has at most 4 children)
• Case 0 v has no children, one child, or two
  children ? trivial.
• Case 1 v has 3 children.
• Case 2 v has 4 children

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Spanning Trees for Degree 4

• Case 1 v has 3 children v1,v2,v3
• Let v1 be the closest to v, among its children.
• Consider the following 4 paths.

Notice all these paths contain (v,v1) , which is
the shortest
New Target
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Spanning Trees for Degree 4

• Case 1 v has 3 children v1,v2,v3

about 1.244 lt 1.25
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Spanning Trees for Degree 4

• Case 1 v has 3 children v1,v2,v3

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Spanning Trees for Degree 4

• Case 2 v has 4 children v1,v2,v3,v4
• Assume v1 is closest to v.
• Two cases
• 2.1 v3 is furthest from v
• 2.2 v4 is furthest from v (the other case is
  symmetric)

v1
v1
v2
v2
v4
v4
v3
v3
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Spanning Trees for Degree 4

• Case 2.1 v1 is closest to v, and v4 is farthest
  from v.
• Consider the following paths

shortest
longest
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Spanning Trees for Degree 4

• Case 2.1 v1 is closest to v, and v4 is farthest
  from v.

New Target
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Spanning Trees for Degree 4

• Proof Strategy
• By contradiction. Suppose there is a
  counter-example. After....
• 1.Shrinking
• 2.Scaling
• 3.Rotating
• 4.Estimating
• We prove that this counter-example is not
  possible.

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Spanning Trees for Degree 4

• 1.Shrinking
• move v4, v3 (or v2) closer to v until
• The lhs is decreased by at most 2b2a, while the
  rhs is decreased by exactly 4b4a
• 2. Scaling
• As shown
• We now wish to prove

v2
v1
e
v
1
v3
v4
b
a
a
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Spanning Trees for Degree 4

• 3.Rotating
• It is maximum when v1 and v3 are on the bisector
  of v1v v2
• 4.Estimating
• Define
• We will show that F is non-positive over the
  domain of possible e.
• F is a sum of convex functions minus a linear
  function and thus is convex.
• So the extrema happens oh the boundaries.
• On F(1), F is less than 4v25.66 . So F(1) lt0
• On the midpoint of (v1 ,v2), F is less than 3 v3
  5.2

v1
v2
v
1
v4
v3
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Spanning Trees for Degree 4
Notice We always avoid the farthest edge(v?v3 in
this case, and v?v4 in the previous case )

• Case 2.2 v1 is closest to v, and v3 is farthest
  from v.

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Spanning Trees for Degree 4

• Case 2.2 v1 is closest to v, and v3 is farthest
  from v.

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Degree 3 trees for higher dimensions

• 5 to k
• Instead of an MST of degree at most 5 in plane
  case, in higer (fixed) dimension, the MSTs
  degree is bounded by a constant.
• Replace branches to children by a path that
  starts at v and visit all children.
• Suppose the number of children is k.
• We want to prove that there exist a path
  satisfying (Lemma 4.1)

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Degree 3 trees for higher dimensions

• By induction
• Sort vs children in increasing distance from v
  as v1,...,vk.
• Induction base k0,1,2
• Induction hypothesis
• The lemma is true for all values of k up to some
  l gt 2.
• Consider when k l1
• By the induction hypothesis, the claim is true
  when v has k-3 (l-2) children.

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Degree 3 trees for higher dimensions
v
v1,...,vk-3

• Indution

vj
vk
vk-2
vk-1
v1,...,vk
Induction If P is already within 5/3 times the
edges it replaced, then we want to prove P?P
is also a 5/3 - approximation
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Degree 3 trees for higher dimensions

• P1 to P6 all six possibilities
• vj ? vk ? vk-1 ? vk-2
• vk ? vk-2 ? vk-1
• vk-1 ? vk ? vk-2
• vk-1 ? vk-2 ? vk
• vk-2 ? vk ? vk-1
• vk-1 ? vk-1 ? vk

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Degree 3 trees for higher dimensions
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Degree 3 trees for higher dimensions

• Shrinking
• Remember v1 to vk is sorted according to the
  distance from v
• Scaling

vk
vj
v
1
vk-2
vk-1
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Degree 3 trees for higher dimensions
vj
vk

• Rotating Estimating (Extremum)
• Recall our target
• can be expressed as the sum of the following
  two terms
• 1) The perimeter of a triangle, which is bounded
  by 3v3
• 2) The perimeter of a tetrahedron, which is
  bounded by 4v6
• 4v6 3v3 is about 14.994. The rhs of (9) is 15

vk-2
vk-1