1. The MST and the upper box dimension
Gady Kozma - Weizmann Institute
Zvi Lotker -Ben Gurion University
Gideon Stupp - MIT

2. Today Power in Rd Introduction to upper box dimension
The connection between them

3. Different deployment topologiesGG
MST
RNG
Set of connected graphs where each is a sub graph of the previous

4. Introduction to Sensor Networks
One of the main problems is power.
Common assumption is that the power of transmission is
E(r)=C·r Where 24

5. First we do the broadcast on the MST
The Euclidean case
Phase tran.
=d
Theorem [SNYDER-STEELE87]
If nodes are placed in [0,1]d
then the energy needed for broadcast is E<C ·nmax{0,1- /d}
First we do the broadcast on the MST

6. Planar MST The opposite of a big side in triangle is a big angle60.
Lemma: Deg vi <7
Proof:
The opposite of a big side in triangle is a big angle60.

7. Question E<C ·nmax{0,1- /d}
When is the broadcast energy bounded by a constant ?
These type of Questions are answered by the dimension theory.

8. Topological dimension
point has d=0
line is divided by point with
d=0 -> line has d=1
surface is divided by line with
d=1 -> surface has d=2
interior of closed surface is divided by surface with
d=2 -> it has d=3
Finite set of points d=0
Smooth line d=1
Smooth surface d=2
Interior of the d=3
closed surface

9. Dimension the Combinatorics approach (upper box dimension)
D=1 N parts, scaled by ratio r=1/N(r), N(r)r1=1
N(r)=1/r
Dimension the Combinatorics approach (upper box dimension)
N(r) is the max # of disjoint balls(box) in M with radius r
D=2 N parts, scaled by ratio r=1/N(r)1/2, N(r)r2=1
N(r)=(1/r)2
D=3 N parts, scaled by ratio r=1/N(r)1/3, N(r)r3=1
N(r)=(1/r)3

10. Example upper box dimension
Suggested definition
Cantor set:
n=0 1
n=1
1/3
n=2
N(1/3n)=2n
Koch curve:

11. Example

12. More Example

13. More Example

14. More Example

15. More Example

16. More Example

17. Fractals and Self Similarity
Mountain is fractals

18. Related work
E. N. Gilbert, H. O. Pollak, Steiner random minimal trees, SIAM J. Appl. Math. 16 (1968),
T. L. Snyder and J. Michael Steele, Worst case growth rates for minimal and greedy matchings with power weighted edges (1987).
P. Jaillet, Cube versus torus models for combinatorial optimization problems and the Euclidean minimum spanning tree constant}, Annals of Applied Probability, 3 (1993),
Andrea E. F. Clementi, Pierluigi Crescenzi, Paolo Penna, Gianluca Rossi and Paola Vocca, On the complexity of computing minimum energy consumption broadcast subgraphs, STACS 2001,

19. MST dimension
Example
dimMST(Rd)=d
We are looking
for Phase tran d

20. Our Results for Rd Theorm dimMST(M)=dimBox(M)
We estimate the total energy necessary for broadcasts in sensor networks on a fractal terrain
Theorm dimMST(M)=dimBox(M)

21. All the balls are disjoint
Our Results for Rd w1 v1 17 8 13 13 9 v2 w2 13 3 17 3

22. Simple Propreity Lemma
Let w1,w2Rd satisfy ||w1||1, ||w2|| 1 and ||w1-w2|| 1. Then ||w1+w2||/2 3/2

23. Note that the conclusion of the lemma 1 is fails
Proof of the theorem
Assume B1B2 ||e1||||e2||, by the triangle inequality we get that d(v1,(w1+w2)/2) 
0.5||e1||+0.1||e1||+0.1||e2||<
3/2||e1||
Note that the conclusion of the lemma 1 is fails

24. Proof of the theorem
we know that one of it's assumptions is fails. Since
||w1-w2||=||e2||||e1|| the failing assumption must be one of the first two
which means that for some i{1,2}, d(v1,wi)<||e1||.
Similarly, for some j, d(v2,wj)<||e1||.

25. Proof of the theorem Assume that v2 is in the same component as w1.
Using d(v1,wi)<||e1|| we get a contradiction.

26. All the balls are disjoint
Proof of the theorem
Since the center of the balls is in [0,1]d we get the: v1 v2 w2 w1 3 8 9 13 17
All the balls are disjoint

27. Proof of the theorem 1/p+1/q=1 ( ||e||d)/d n(1-d/ ) C n(1-/d)
Now we use Holders inequality
|fg| |f|1/p |g|1/q
1/p+1/q=1
||e||=||e||·1
|| ||e|| ||d/  || 1|| d/ (d-) =
( ||e||d)/d n(1-d/ ) C n(1-/d)

28. Proof of the theorem Lemma
For every metric space M, dimBox(M)dimMST(M)
Proof:
If dimBox(M)=0
Assume that <dimBox(M) it is possible to find sequences of disjoint balls B(v1, ),B(v2, ),…,B(vN(), ) s.t
Now look on the MST of the points vi.
We get that every edge of T is   therefore ET=||e||> N() and we get dimBox(M)> 

29. Proof of the theorem Lemma
For every metric space M, dimBox(M)dimMST(M)
Proof:
W.l.o.g we can assume that diam(M)=1
If >dimBox(M) so we have N()<C (1/)
We use Prim's algorithm For every edge eT satisfying ||e||> define B(e) to be a ball of radius /3 centered at the edge of e which entered last into T

30. Proof of the theorem 

31. Proof of the theorem It follows that these balls are disjoint.
Lemma
For every metric space M, dimBox(M)dimMST(M)
Proof:
It follows that these balls are disjoint.
Since >dimBox(M) we know that there are no more than C·- edges of length >  in T. Let  <. Then

32. Proof of the theorem

33. Conclusion The power assumption E(r)=C·r Where 24
The space assumpion “The dimension is 2d3
These two assumption don’t get well