multihop paths and key predistribution in sensor networks guy rozen
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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