Multihop Paths and Key Predistribution in Sensor Networks Guy Rozen.
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Transcript of 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
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
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
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
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
Moving between grid typesMoving between grid types
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
k-Hop Coveragek-Hop Coverage Definition:
Theorem 5:
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
Suppose that is used in Scheme 1.
Then the k-hop coverage of the scheme is equal to k
D
C D
k-Hop Coveragek-Hop Coverage Definition:
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
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
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
First, we define a set of integer m-tuples:
Simply put, the some of elements in each tuple is zero and the sum of positive elements is k.
Some examples for m=3:
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
0 1 20,0,0 ; 1, 1,0 ; 2, 2,0 , 2, 1, 1H H H
First, we define a set of integer m-tuples:
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:
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
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:
Maximal k-hop coverageMaximal k-hop coverage
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:
Maximal k-hop coverageMaximal k-hop coverage
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.):
Maximal k-hop coverageMaximal k-hop coverage
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:
Maximal k-hop coverageMaximal k-hop coverage
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:
Maximal k-hop coverageMaximal k-hop coverage
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:
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
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:
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
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:
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
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.):
BBhh Sequences and DDC Sequences and DDC
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:
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
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:
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
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.):
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
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:
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
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 .
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
,DD m r ,r k m
What is the minimal r so that a with maximal k-hop coverage is possible? We denote this r as .
Theorem 14:
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
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
What is the minimal r so that a with maximal k-hop coverage is possible? We denote this r as .
Theorem 14:
Proof: (Upper bound proven in Theorem 12)
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
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.):
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
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.
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
* ,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:
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
2 21, , where 0.914769m o m r m m o m
We will give special attention to the case k=1. Theorem 16:
Proof:
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
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
We will give special attention to the case k=1. Theorem 16:
Proof:
Theorem 17:
Proof: Analogous hexagonal result from [2].
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
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:
Maximal k-hop coverage - Maximal k-hop coverage - boundsbounds
*
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? Theorem 20:
Minimal k-hop coverageMinimal k-hop coverage
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:
Minimal k-hop coverageMinimal k-hop coverage
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:
Minimal k-hop coverageMinimal k-hop coverage
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:
Minimal k-hop coverageMinimal k-hop coverage
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
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:
Minimal k-hop coverageMinimal k-hop coverage
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
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
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:
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
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:
Complete 2-hop coverageComplete 2-hop coverage
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 . Construction 2:
◦
Complete 2-hop coverageComplete 2-hop coverage
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
We now define a by using dots from . Construction 2:
◦
◦
Complete 2-hop coverageComplete 2-hop coverage
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
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
We now define a by using dots from . Construction 2:
◦
◦
Complete 2-hop coverageComplete 2-hop coverage
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
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
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
Lemma 24:
Complete 2-hop coverageComplete 2-hop coverage
The configuration is a 2 , with all points in a
1 2 rectangle.
DD p
p p
B
Lemma 24:
Proof:
Complete 2-hop coverageComplete 2-hop coverage
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
Proof (cont.):
Complete 2-hop coverageComplete 2-hop coverage
, , , 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:
Complete 2-hop coverageComplete 2-hop coverage
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:
Complete 2-hop coverageComplete 2-hop coverage
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.
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:
Complete 2-hop coverageComplete 2-hop coverage
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.
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
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.):
Complete 2-hop coverageComplete 2-hop coverage
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.):
Complete 2-hop coverageComplete 2-hop coverage
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
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 D44
We use the vector , 1 to cross quadrents, then use another
difference vector from to reach our destination (this second one
exists since is doubly periodic.p
p p
B
R
1
3 4
2
DD33 to D to D22
We use the vector , 1 to cross quadrents, then use another
difference vector from to reach our destination (this second one
exists since is doubly periodic.p
p p
B
R
1
3 4
2
Lemma 26:
Complete 2-hop coverageComplete 2-hop coverage
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?
Complete 2-hop coverageComplete 2-hop coverage
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:
Complete 2-hop coverageComplete 2-hop coverage
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
Lemma 26 motivation:
Complete 2-hop coverageComplete 2-hop coverage
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.
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
Lemma 26 motivation:
Complete 2-hop coverageComplete 2-hop coverage
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.
has difference vectors
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
Lemma 26 motivation:
Complete 2-hop coverageComplete 2-hop coverage
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.
has difference vectors
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
Lemma 26 motivation:
Complete 2-hop coverageComplete 2-hop coverage
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.
has difference vectors
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 .
'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 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
Lemma 26 motivation:
Complete 2-hop coverageComplete 2-hop coverage
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.
has difference vectors
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 .
's 1, vectors can't all satisfy 0.
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.
Lemma 26 motivation:
Complete 2-hop coverageComplete 2-hop coverage
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.
has difference vectors
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 .
's 1, vectors can't all satisfy 0.
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
Lemma 26 motivation:
We will now face insurmountable suspense…
Complete 2-hop coverageComplete 2-hop coverage
By similar means, we can show Lemma 26 works for vectors of the
form ,1 when .x t p
Proof (of Lemma 26):
Complete 2-hop coverageComplete 2-hop coverage
\ 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.):
Complete 2-hop coverageComplete 2-hop coverage
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:
Complete 2-hop coverageComplete 2-hop coverage
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:
Complete 2-hop coverageComplete 2-hop coverage
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