Download - Multihop Paths and Key Predistribution in Sensor Networks Guy Rozen

Multihop Paths and Key Multihop Paths and Key Predistribution in Sensor Predistribution in Sensor

NetworksNetworks

Guy Rozen

ContentsContentsTerminoligy (quick review)Alternate grid types and metricsk-hop coverage

◦Calculation◦How to optimize

Complete two-hop coverage

TerminologyTerminology DD(m) – Distinct distribution set of m points DD(m,r) – DD(m) with maximal Euclidian distance of r DD*(m)/ DD*(m,r) – DD(m)/ DD(m,r) on a hexagonal grid DD(m,r) – Denotes use of the Manhattan metric DD*(m,r) – Denotes use of the Hexagonal metric Ck(D) – Maximal value of a k-hop coverage for some DDS D

Scheme 1: Let be a distinct difference configuration. Allocate keys to notes as follows:◦ Label each node with its position in .

◦ For every ‘shift’ generate a key and assign it to the notes labeled by , for .

1 2, ,..., mD v v v

22

u uk

iu v 1, 2, ...,i m

Alternate grid types and Alternate grid types and metricsmetrics In a regular square grid, where the plane is tiled with unit squares,

sensor coordinates are in fact .

2

0,0 1,1

Alternate grid types and Alternate grid types and metricsmetrics In a regular square grid, where the plane is tiled with unit squares,

sensor coordinates are in fact .

In a hexagonal grid, where the plane is tiled with hexagons, seonsor coordinates can be depicted as

2

0,0 1,1

1,0 1 2, 3 2 | ,

0,0

1 2, 3 2 1

1 2

3 2

Moving between grid typesMoving between grid types

The linear bijection transitions from a

hexagonal grid to a square one. Alternatively, can be seen as doing

2: , ,

3 3

y yx y x

3, ,

2 2




Theorem 1:

2: , ,

3 3

y yx y x

3, ,

2 2

*1 2

1 2

1 *

If , ,..., is a ,

then , ,..., is a .

Similarly, if is a then is a .

m

m

D v v v DD m

D v v v DD m

D DD m DD m




Theorem 1:

Proof:

2: , ,

3 3

y yx y x

3, ,

2 2

*1 2

1 2

1 *

If , ,..., is a ,

then , ,..., is a .

Similarly, if is a then is a .

m

m

D v v v DD m

D v v v DD m

D DD m DD m

Since is linear: i j k l i j k lv v v v v v v v


It is important to note that does not preserve distances.

Theorem 2:

*

1 *

If is a , then is a , 2 .

If is a , then is a , 3 2 .

D DD m r D DD m r

D DD m r D DD m r

Alternate metricsAlternate metrics

Manhattan/Lee metric: The distance between two points and is .

For example, a sphere of radius 2:

Theorem 3:

1 1,i j 2 2,i j 1 2 1 2i i j j

For , a , is a , and a

, is a , 2

r DD m r DD m r

DD m r DD m r

Alternate metricsAlternate metrics

Hexagonal metric: The distance between two points is the amount of hexagons on the shortest path between the points.

For example, a sphere of radius 2:

Theorem 4:

* *

* *

For , a , is a , and a

2, is a ,

3

r DD m r DD m r

DD m r DD m r

k-Hop Coveragek-Hop Coverage Definition:

1

is the amount of vectors of the form

when , 1,2,..., ; ;0i i

k

l

i i i ii

C D

v v m l k


Theorem 5:

1


when , 1,2,..., ; ;0i i

k

l

i i i ii

C D

v v m l k

Suppose that is used in Scheme 1.

Then the k-hop coverage of the scheme is equal to k

D

C D


Theorem 5:

Proof: When using Scheme 1, we know that a pair of nodes sharing a key are located at , hence the vector is both a difference vector of D and a one hop path when using Scheme 1. Hence, an l-hop path between paths is composed of difference vectors from D.

1


when , 1,2,..., ; ;0i i

k

l

i i i ii

C D

v v m l k

Suppose that is used in Scheme 1.

Then the k-hop coverage of the scheme is equal to k

D

C D

,i jv u v u i jv v

k-Hop Coveragek-Hop Coverage Theorem 6:

Proof:

*Let be a and let be a such that .

Then the k-hop coverage of is equal to the k-hop coverage of .

D DD m D DD m D D

D D

is a bilinear bijection.

First, we define a set of integer m-tuples:

Maximal k-hop coverageMaximal k-hop coverage

1 21 : 0

, ,..., | 0,i

m mm

k m i ii i a

H a a a a a k


Simply put, the some of elements in each tuple is zero and the sum of positive elements is k.

Some examples for m=3:


1 21 : 0

, ,..., | 0,i

m mm

k m i ii i a

H a a a a a k

0 1 20,0,0 ; 1, 1,0 ; 2, 2,0 , 2, 1, 1H H H


Simply put, the some of elements in each tuple is zero and the sum of positive elements is k.

Some examples for m=3:

Lemma 7:


1 21 : 0

, ,..., | 0,i

m mm

k m i ii i a

H a a a a a k

0 1 20,0,0 ; 1, 1,0 ; 2, 2,0 , 2, 1, 1H H H

1 2 3

1 2 3

1

3 1 2

3 1 2

1

1

, where 0

, , where 1

Any element of may be written as the sum of elements

from .

k k k

k k k

k

w H z H w z H k k ki

w H z H w z w z H k k kii

H kiii

H

Theorem 8:


1

1 21 0

The k-hop coverage of a is at most , with equality

if and only if all the vectors with , ,...,

are distinct.

k

ii

km

i i m ji j

DD m H

a v a a a H

Theorem 8:

Proof:


1

1 21 0

The k-hop coverage of a is at most , with equality

if and only if all the vectors with , ,...,

are distinct.

k

ii

km

i i m ji j

DD m H

a v a a a H

1

1 1

The difference vectors in are all of the form

with ,..., .

m

i ii

m

D a v

a a H

11 0

Hence:

is a sum of at most difference vectors

| ,...,km

i i m ji j

V k

V a v a a H

Proof (cont.):


11 0 10

We can now say:

1 | ,..., 1km k k

k i i m j i ii i ij

C D a v a a H H H

Proof (cont.):

Corollary 9:


11 0 10

We can now say:

1 | ,..., 1km k k

k i i m j i ii i ij

C D a v a a H H H

2

The two-hop coverage of a is at most:

11 2 3 1 2 2 1

41

1 64

DD m

m m m m m m m m m

m m m m

Proof:


1 2

1

2

We proved that the two-hop coverage is at most .

contains 1 elements.

As for we have this lovely table:

H H

H m m

H

We would like to show that Theorem 8’s bound is tight. Naïve approach:

Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds

2 1 ,0 , 1,2,...,i

iv k i m

We would like to show that Theorem 8’s bound is tight. Naïve approach: Lemma 10:


2 1 ,0 , 1,2,...,i

iv k i m

1 2

2

1 1

The k-hop coverage of a given by , ,...,

meets the bound of Theorem 8 0 for all

m

km

i i ii i

DD m D v v v

c v c H

We would like to show that Theorem 8’s bound is tight. Naïve approach: Lemma 10:

Proof:


2 1 ,0 , 1,2,...,i

iv k i m

1 2

2

1 1

The k-hop coverage of a given by , ,...,

meets the bound of Theorem 8 0 for all

m

km

i i ii i

DD m D v v v

c v c H

2

11 1 1 1

Bound not met 0; ,...,km m m

i i i i i i m ii i i i

a v b v c v c c H

11

2 2

Assume 0 when ,..., for some 1,2,..., 2

vector is where , Bound not met

m

i i m li

l l

c v c c H l k

c a b a H b H

Definition 1:

Elements may be used more than once.

BBhh Sequences Sequences

1 2

1 2

1

- an abelian group.

, ,... - a sequence of 's elements.

is a sequence over if all sums

... with 1 ... are distinct.h

m

h

i i i h

A

D v v v A A

D B A

v v v i i m

Theorem 11:

BBhh Sequences and DDC Sequences and DDC

21 2

22

Let 2 be a fixed integer and , ,..., . Then is a

with maximal k-hop coverage is a over

m

k

k D v v v D

DD m D B

Theorem 11:

Proof:


21 2

22

Let 2 be a fixed integer and , ,..., . Then is a

with maximal k-hop coverage is a over

m

k

k D v v v D

DD m D B

22Assume is a over

If for we get a contradiction

If for , we get a contradiction

or and

If does not have maximal k-hop coverag

k

i j i j

i j i j

i j i j

D B

v v i j v vi

v v v v i j i jii

v v v v i i j j

Diii

2

11

1 11 1

e by Lemma 10

there is a so that 0 is 's positives

and - 2 2

we get a contradiction has maximal k-hop coverage

k m

i i iii

m m

i i i ii i

c H c v a c

b a c k t v a v k t v b v

D

Proof (cont.):


2

1 1 1 1

Assume is a with maximal k-hop coverage

If is not a sequence then there are two sums as seen in the

definition which are equal

;1 , 2

gives

k h

m m m m

i i i i i ii i i i

t i

D DD m

D B B

a v b v a b t k

c a b H c

1

0 we get a contradiction.m

ii

v

Construction 1:

Using BUsing Bhh sequences to build a sequences to build a DDCDDC

2

2

Let 2 be a fixed integer. Let be a prime power so that

1 where , are coprime. Then there exists a set of

dots which is doubly periodic with periods , and that for

each rectangle of

k

k q

q ab a b

X a b

R

size , is a with maximal

k-hop coverage.

a b R X DD q

Construction 1:

Proof:

Using BUsing Bhh sequences to build a sequences to build a DDCDDC

2

2

Let 2 be a fixed integer. Let be a prime power so that

1 where , are coprime. Then there exists a set of

dots which is doubly periodic with periods , and that for

each rectangle of

k

k q

q ab a b

X a b

R

size , is a with maximal

k-hop coverage.

a b R X DD q

2

2

2 1

21

1 2

2

Example 1 gave a sequence over with elements. Also,

there is an isomorphism We get a

sequence , ,...,

Define , mod , mod , so we have :

and can define

k

k

k q

a b kq

q a b

a b

B q

B

v v v

x y x a y b

21 2

1 2 2

as | , ,..., Every

rectangle will give us , ,..., , a sequence

q

q k

X v v v v v a b

R X R v v v B

Theorem 12:


1

1

Let 2 be a fixed integer, and 16 2 . Then there exists

a DD , with maximal k-hop coverage such that .

k

k

k c

m r m cr

Theorem 12:

Proof:


1

1

Let 2 be a fixed integer, and 16 2 . Then there exists

a DD , with maximal k-hop coverage such that .

k

k

k c

m r m cr

2

2

1

set of points in a / 2 radius from the origin. Note that

4 . Let be the smallest prime power for which

2 and we get 2

We now define:

1 when is even

1 2 when 3mod 4

1 2 w

kk

k

k k

k

S r

S r O r q

q r q r

q q

a q q

q

2

hen 1mod 4

1

k

k

q

b q a

Proof (cont.):


22 1 1

As we saw in Construction 1, a rectangle contains dots.

A shift of contains dots at average so that

. We define where

4 2 16 2

is contained in a s

k k

a b q

T S S q ab T

T X S q ab D T X

D m S q ab r q r r

D

phere of radius 2 , and in a rectangle

which is a coverage is a , with maximal

k-hop coverage.

r a b

DD q D DD m r

Proof (cont.):

Corollary 13:


22 1 1

As we saw in Construction 1, a rectangle contains dots.

A shift of contains dots at average so that

. We define where

4 2 16 2

is contained in a s

k k

a b q

T S S q ab T

T X S q ab D T X

D m S q ab r q r r

D

phere of radius 2 , and in a rectangle

which is a coverage is a , with maximal

k-hop coverage.

r a b

DD q D DD m r

1 21

*

1

Using theorem 2, 6, and 12 we can say that for 16 2 2 3

there exists a , with maximal k-hop coverage so that

kk

k

c

DD m r

m c r

What is the minimal r so that a with maximal k-hop coverage is possible? We denote this r as .


,DD m r ,r k m


Theorem 14:


For an integer 2,

1 16,

2!

kkk k k

k

mo m r k m m o m

k k

,DD m r ,r k m


Theorem 14:

Proof: (Upper bound proven in Theorem 12)


For an integer 2,

1 16,

2!

kkk k k

k

mo m r k m m o m

k k

,DD m r ,r k m

1

1 2

2

22 2

2 21

Remember .

Define , ,..., : : 0 2 .

!We get .

2 ! !

!So we can say

2 ! ! !

k

k ii

m k i

kkk k

ii

C D H

B a a a H i a k

mB

m k k

m mH o m o m

m k k k

Proof (cont.):


2

222

2

Vectors in are at most difference vectors. All difference

vectors are at most. All vectors are inside a sphere,

which contains at most vectors.

We get .!

k

k

kk

k

C D k

r C D kr

kr O r

mC D o m kr O r

k

Proof (cont.):

For a hexagonal grid we present an equivalent term . Theorem 15:

Proof: Theorem 2 & 14.


* ,r k m

*

For an integer 2,

2 2 1 16,

3 3 2!

kkk k k

k

mo m r k m m o m

k k

2

222

2

Vectors in are at most difference vectors. All difference

vectors are at most. All vectors are inside a sphere,

which contains at most vectors.

We get .!

k

k

kk

k

C D k

r C D kr

kr O r

mC D o m kr O r

k

We will give special attention to the case k=1. Theorem 16:


2 21, , where 0.914769m o m r m m o m


Proof:


2 32 For , we have Lower bound2

2 , : Upper bound2

DD m r m r O r

DD m r m r o r

2 21, , where 0.914769m o m r m m o m


Proof:

Theorem 17:

Proof: Analogous hexagonal result from [2].


2 32 For , we have Lower bound2

2 , : Upper bound2

DD m r m r O r

DD m r m r o r

2 21, , where 0.914769m o m r m m o m

1 4 1 2 1 4

*23 2 31,m o m r m m o m

Finally, using results in [2] we can prove: Theorem 19:


*

1, 2

2 3 1, 2 , 1.58887

r m m o m

m o m r m m o m

What is the smallest value for a k-hop coverage?

Minimal k-hop coverageMinimal k-hop coverage

What is the smallest value for a k-hop coverage? Theorem 20:


The k-hop coverage of a is at least 1DD m km m

What is the smallest value for a k-hop coverage? Theorem 20:

Proof:


The k-hop coverage of a is at least 1DD m km m

1

1

The 1-hop coverage of a is 1 , the amount of

difference vectors. Define the set of initial difference vectors.

We pick u= , with maximal and if need be, .

We can assume 0, 0 and can sa

DD m m m

S

d e S d e

d e

1

1 1

1 1

1 1

y that is composed of:

, | , , 0 or 0 and 0 ,

, | ,

For 1: 1 | 1 |

for , 1 .

i

i j i

S

S x y x y S x x y

S x y x y S

i S w i u w S w i u w S

S S i j S m m

Lemma 21:


For an integer 2, suppose is a where two differences

are not parallel. Then the k-hop coverage of is more than 1 .

k D DD m

D km m

Lemma 21:

Proof:




k D DD m

D km m

1 2

Define and as in Theorem 20. Let be a difference vector not

parallel to , and the most perpendicular.

The k-hop coverage of is at least ... 2

We saw that 1 , and we can see that

i

k

i

u S v

u

D S S S v

S m m

12v ... ,

since 1 2 2 .

kS S

p w i u p w p v p v p v

Lemma 21:

Proof:

Lemma 21 can be used to prove Theorem 21:




k D DD m

D km m

1 2

Define and as in Theorem 20. Let be a difference vector not

parallel to , and the most perpendicular.

The k-hop coverage of is at least ... 2

We saw that 1 , and we can see that

i

k

i

u S v

u

D S S S v

S m m

12v ... ,

since 1 2 2 .

kS S

p w i u p w p v p v p v

For an integer 2, suppose is a . meets the bound of

Theorem 20 if and only if it is equivalent to a perfect Golomb ruler.

k D DD m D

For a prime , we will show a construction of a with complete 2-hop coverage.

That ensures a two-hop path between a point x and any other grid point within a rectangle centered at x.

Complete 2-hop coverageComplete 2-hop coverage5p DD m

2 3 2 1p p

Height Width

For a prime , we will show a construction of a with complete 2-hop coverage.

That ensures a two-hop path between a point x and any other grid point within a rectangle centered at x.

Definition 2 (Welch Periodic Array):

Equivalent points:

Complete 2-hop coverageComplete 2-hop coverage5p

2

Let be a primitive root modulo a prime .

Define the Welch periodic array: , | mod .

is double periodic: , , 1

for , .

jp

p p p

p

i j i p

i j i p j p

R

R R R

DD m

2 3 2 1p p

Height Width

, , , 1 .A i j i j A i i p j j p

Example of an arrayExample of an array

5

3

p

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

j

i11

10

Lemma 23:

Complete 2-hop coverageComplete 2-hop coverage

1 2 1 2

2 2 2 2

Let and be integers such that 0mod and 0mod 1 .

Suppose that contains dots , , , as well as

, , , . Then , .

p

d e d p e p

A i j B i d j e

A i j B i d j e A A B B

R

Lemma 23:

Proof:


1 2 1 2

2 2 2 2

Let and be integers such that 0mod and 0mod 1 .

Suppose that contains dots , , , as well as

, , , . Then , .

p

d e d p e p

A i j B i d j e

A i j B i d j e A A B B

R

1

2

1 2

1

2

1

2

1

2

1 2

1 2

mod

mod1 0mod

mod

mod

Since 0mod 1 we get mod 1

and from that mod

j

jj je

j e

j e

i p

i pp

i d p

i d p

e p j j p

i i p

From Lemma 23 we conclude:


if contains , , , then 0mod .

if contains , , , then 0mod 1 .

A vector , can occur at most once as a difference between

two points in , within a particular 1 rectangle.

p

p

p

i j i d j d p

i j i j e e p

d e

p p

R

R

R

We now define a by using dots from .

Complete 2-hop coverageComplete 2-hop coverage DD m pR

We now define a by using dots from . Construction 2:

◦


2 an odd prime, , is such that , , 1, 1 pp i j i j i j R

DD m pR

1

Do such points exist? Why yes! 1 1 mod

and also 1 1 1 1 1 mod

j

j

i p

i p


◦

◦



DD m pR

1



j

j

i p

i p

is a 1 rectangle bounded by , , 1, ,

, 2 and 1, 2 . There are 1 dots

in .

S p p i j i p j

i j p i p j p p

S

Why 1 dots? Remember what is!p


◦

◦



DD m pR

1



j

j

i p

i p

is a 1 rectangle bounded by , , 1, ,

, 2 and 1, 2 . There are 1 dots

in .

S p p i j i p j

i j p i p j p p

S

Why 1 dots? Remember what is!p

Since is periodic, it also contains dots at , 1 , ,

and 1, .

Our, actual construction is a configuration , which is and the

above dots.

p i j p i p j

i p j p

S

R

B

Meet Meet j p

1j p

1j

j

i 1i i p 1i p

A

B

A

B

A

2p

3p S

Central region

Contained in a square. Has a border region of width 2 which contains exactly 5 points. Has a central region which is a rectangle. The central region contains dots. One column is empty. and there are no other equivalent points.

- Vital statistics- Vital statistics 1 2p p

3 2p p 3p

and A A A B B

Example of Example of

B’

A’

B

AA’’

5

3

p

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

j

i11

10

Lemma 24:


The configuration is a 2 , with all points in a

1 2 rectangle.

DD p

p p

B

Lemma 24:

Proof:


The configuration is a 2 , with all points in a

1 2 rectangle.

DD p

p p

B

We have already shown that contains the dots as needed.

Suppose that and and and are distinct dot pairs with the

same difference vector , .

Suppose 0, , and/or 0, 1 , 1 .

It is impossi

X Y X Y

d e

d p p e p p

B

ble for one of , , , to be in the central region.

Why?

X Y X Y

This is whyThis is why

j p

1j p

1j

j

i 1i i p 1i p

A

B

A

B

A

2p

3p

2p

1p

Proof (cont.):


, , , must be outside the central region, but no two pairs have

the same difference vector.

So 0, , and 0, 1 , 1 and since all dots are in

a 1 2 rectangle 0mod and 0mod 1 .

Lemma 23

X Y X Y

d p p e p p

p p d p e p

, .

point pairs are not distinct

, and , must be on configuration border, but we've already

seen that leads to a contradiction, and so we have our proof.

X X Y Y

X X Y Y X X

X X Y Y

Motivational boost:


In order to show a 2 3 2 1 rectangle with complete 2-hop

coverage, we will show that any vector , where 1

and 2 can be expressed as a difference vector of .

p p

d e d p

e p

B.

Motivational boost:

Lemma 25:





p p

d e d p

e p

B.

Any vector of the form , , where 0 1 and 0 2

where , are integers is a sum of two difference vectors from .

d e d p e p

d e

B

Motivational boost:

Lemma 25:

Proof:





p p

d e d p

e p

B.

Any vector of the form , , where 0 1 and 0 2

where , are integers is a sum of two difference vectors from .

d e d p e p

d e

B

Remember the 1 rectangle from Construction 2.

Define as a restriction of to the 2 2 2 sub-array

whose lower left corner coincides with .

p

p p S

p p

S

A R

illustratedillustrated

S

1

3 4

2

illustratedillustrated

1

3 4

2

1

1

.

Since is doubly

periodic with 's size,

the points in occur

in the other quadrents

as well.

p

S

S

D B

R

D

Proof (cont.):


1

We want to show that any vector , as defined is a difference of

two points in . We assume 0.

Assume 0 and can say that there is a , so that

mod1

It is easy to see that , ,

je

d e

d

e i j

di p

i j i d

A

D

3

, , and since , 0 and

are small enough, , , , .

In the case where 0 we can just pick , and follow use

same concept.

pj e d e

i j i d j e

e i j

R

A

D

Proof (cont.):


Now that any vector , is a difference of two points in . All we

need to show is that every two points in can be connected using

two difference vectors from .

You'll see this more easily if I show y

d e A

A

B

ou.

DD11 to D to D11 (or any D (or any Dxx to D to Dxx))

By definition, all such vectors are a single difference vector from .B

1

3 4

2

DD11 to D to D33 (or D (or D22 to D to D44)) We use the vector 0, 1 to cross quadrents, then use another

difference vector from to reach our destination (this second one

exists since is doubly periodic.p

p

B

R

1

3 4

2

DD11 to D to D33 (or D (or D22 to D to D44)) We use the vector ,0 to cross quadrents, then use another



p

B

R

1

3 4

2

DD11 to D to D44

We use the vector , 1 to cross quadrents, then use another



p p

B

R

1

3 4

2

DD33 to D to D22

We use the vector , 1 to cross quadrents, then use another



p p

B

R

1

3 4

2

Lemma 26:


For an integer 3, let be a set of integers where:

1

1 , 2 ,..., 1 1,2,..., 1 1

1, 1 , 1

\ 1, 1 , 1 with 0

if 0 and \ 1, 1 , 1 then

Then for each positive integer

t

ta

b t t t t

c t t

d i t t i

e i i t t i t

F

F

F

F

F

F F

where 1 1 there are

, so that 1.

t

i j j

F

Lemma 26:

Why do we need this?


For an integer 3, let be a set of integers where:

1

1 , 2 ,..., 1 1,2,..., 1 1

1, 1 , 1

\ 1, 1 , 1 with 0

if 0 and \ 1, 1 , 1 then

Then for each positive integer

t

ta

b t t t t

c t t

d i t t i

e i i t t i t

F

F

F

F

F

F F

where 1 1 there are

, so that 1.

t

i j j

F

Lemma 26 motivation:


Let us look at all vectors of the form 1, where 0 present in .

We will show how the right-sided coordinates of these vectors

satisfy the conditions of Lemma 26 for 1.

y y

t p

B






has difference vectors

y y

t p

pa

B

B as defined above.

difference vectors

Proof of (a)Proof of (a)

j p

1j p

1j

j

i 1i i p 1i p

A

B

A

B

A

2p

3p

One of the 2 columns

is empty, so we have

3 vectors

p

p







y y

t p

pa

B

B

as defined above.

Except for 1, , all of 's 1, vectors satisfy 2.b p y y p B

Proof of (b)Proof of (b)

j p

1j p

1j

j

i 1i i p 1i p

A

B

A

B

A

2p

3p

y p

1,

but this

vector is

not legitimate

y p







y y

t p

pa

B

B

as defined above.

Except for 1, , all of 's 1, vectors satisfy 2.

The vectors 1,1 , 1, 2 , 1, are all in .

b p y y p

c p p

B

B

Proof of (c)Proof of (c)

j p

1j p

1j

j

i 1i i p 1i p

A

B

A

B

A

2p

3p

1, p

1,1

1, 2p







y y

t p

pa

B

B

as defined above.



's 1, vectors can't all satisfy 0.

b p y y p

c p p

d y y

B

B

B

Proof of (d) – case oneProof of (d) – case one

j p

1j p

1j

j

i 1i i p 1i p

A

B

A

B

A

2p

3p

1,

0

y

y

Proof of (d) – case twoProof of (d) – case two

j p

1j p

1j

j

i 1i i p 1i p

A

B

A

B

A

2p

3p

No dots

1,1

1,1







y y

t p

pa

B

B

as defined above.




If 1, where 1, is a vector in , 1, 1 is

b p y y p

c p p

d y y

e y y p y p

B

B

B

B not.







y y

t p

pa

B

B

as defined above.




If 1, where 1, is a vector in , 1, 1 is

b p y y p

c p p

d y y

e y y p y p

B

B

B

B not.

is proven thusly:

If 1, 1 were in then by Lemma 23 the points involved

would be equivalent to those making the vector 1, .

Such points would be on the border region, and that is impossible.

e

y p

y

B


We will now face insurmountable suspense…


By similar means, we can show Lemma 26 works for vectors of the

form ,1 when .x t p

Proof (of Lemma 26):


\ 1, 1 , 1 contains 2 elements must contain

precisely one element of each pair , for 2,3,..., 1.

Suppose, for a contradiction, -1 that cannot be expressed as

a difference of two elements

t t t

i i t i t

t

F F

from .

Suppose 1 and then: 1, 1 1 , 1

However, since 1 1 one of them is in

contradiction

t t

t t

F

F F

F

Proof (of Lemma 26, cont.):


Suppose 1 does not contain a pair of integers which differ

by 1

If is odd then \ 1 contains at most 1 2 positive integers

and at most 1 2 negative integers contains at most

1 1 integers, w

t t t

t

t t

F

F

F

hich contradicts .

If is even then in order for to be of size 1, \ 1 must

contain 2 positive integers (which must be odd) and 2 negative

integers (also odd). This implies that for each odd int

a

t t t

t t

F F

eger 1

we have , and this contradicts .

i t

i i t e

F

Theorem 27:


Let 5 be a prime. The distinct difference configuration achieves

complete two-hop coverage on a 2 3 2 1 rectangle, relative

to the central point of the rectangle.

p

p p

B

Theorem 27:

Proof:


Let 5 be a prime. The distinct difference configuration achieves

complete two-hop coverage on a 2 3 2 1 rectangle, relative

to the central point of the rectangle.

p

p p

B

We notice that all vectors from the central point , are such

that 0 1,0 2.

When 0, 0, Lemma 25 proves that the vector is composed

of two of ' difference vectors.

When either 0 or 0 we c

d e

d p e p

d e

s

d e

B

an use Lemma 26. For 0 2,

we have shown that there are two vectors in which are 1, , 1,

so that and so 0, 1, 1, . Similarly for ,0

where 0 1.

e p

y y

y y e e y y d

d p

B

We have shown maximal k-hop coverage as We used a construction of to produce a

with maximal k-hop coverage and of the order of We have found a bound for (verifying the order above). Could we find tighter bounds? What is the exact value for small k

and m? The questions above also hold for the hexagonal grid and the

alternate metrics. We have constructed a with complete 2-hop

coverage from the center of a rectangular region. The rectangle’s region is of order . Can we find a construction

for significantly larger rectangles? For circles? Can we find constructions for k-hop coverage where k>2?

Conclusion and open Conclusion and open problemsproblems

22 over .kB

2 over kB ,DD m r1 dots.kr

,r k m

,DD m r

2m

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