Multihop Paths and Key
Predistribution in Sensor
Networks
Guy Rozen
Contents
Terminoligy (quick review)
 Alternate grid types and metrics
 k-hop coverage

◦ Calculation
◦ How to optimize

Complete two-hop coverage
Terminology







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 D  v1 , v2 ,..., vm  be a distinct difference
configuration. Allocate keys to notes as follows:
◦ Label each node with its position in
2
.
◦ For every ‘shift’ u 
generate a key ku and assign it to the notes labeled by
u  vi , for i  1, 2, ..., m .
2
Alternate grid types and metrics

In a regular square grid, where the plane is tiled with unit squares,
sensor coordinates are in fact 2.
 1,1
 0,0 
Alternate grid types and metrics

In a regular square grid, where the plane is tiled with unit squares,
sensor coordinates are in fact 2.
 1,1
 0,0 

In a hexagonal grid, where the plane is tiled with hexagons, seonsor
coordinates can be depicted as  1, 0   1 2, 3 2 |  ,  

1 2,
3 2

 



1
32
 0,0 
12
Moving between grid types


y 2y 

The linear bijection  :  x, y    x 
,  transitions from a
3 3

hexagonal grid to a square one.

  3
Alternatively, can be seen as doing    ,
  ,  
2 2 

Moving between grid types



y 2y 

The linear bijection  :  x, y    x 
,  transitions from a
3 3

hexagonal grid to a square one.

  3
Alternatively, can be seen as doing    ,
  ,  
2 2 

Theorem 1: If D  v1 , v2 ,..., vm  is a DD*  m  ,
then   D     v1  ,   v2  ,...,   vm  is a DD  m  .
Similarly, if D is a DD  m  then  1 is a DD*  m  .
Moving between grid types



y 2y 

The linear bijection  :  x, y    x 
,  transitions from a
3 3

hexagonal grid to a square one.

  3
Alternatively, can be seen as doing    ,
  ,  
2 2 

Theorem 1: If D  v1 , v2 ,..., vm  is a DD*  m  ,
then   D     v1  ,   v2  ,...,   vm  is a DD  m  .
Similarly, if D is a DD  m  then  1 is a DD*  m  .

Proof:
Since  is linear: vi  v j  vk  vl    vi     v j     vk     vl 
Moving between grid types

It is important to note that  does not preserve distances.

Theorem 2:
If D is a DD*  m, r  then   D  is a DD m, r 2 .




If D is a DD  m, r  then  1  D  is a DD* m, r 3 2 .
Alternate metrics

Manhattan/Lee metric: The distance between two points  i1 , j1 
and  i2 , j2  is i1  i2  j1  j2 .
For example, a sphere of radius 2:

Theorem 3: For r  , a DD  m, r  is a DD  m, r  and a


DD  m, r  is a DD m,  2r 

Alternate 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 r  , a DD*  m, r  is a DD*  m, r  and a

  2 
DD*  m, r  is a DD*  m,  r  
  3 
k-Hop Coverage

Definition: Ck  D  is the amount of vectors of the form
  v
l
i 1
i
 vi  when  i , i  1, 2,..., m ;  i  i ;0  l  k
k-Hop Coverage

Definition: Ck  D  is the amount of vectors of the form
  v
l
i 1

i
 vi  when  i , i  1, 2,..., m ;  i  i ;0  l  k
Theorem 5:
Suppose that D is used in Scheme 1.
Then the k-hop coverage of the scheme is equal to Ck  D 
k-Hop Coverage

Definition: Ck  D  is the amount of vectors of the form
  v
l
i 1

i
 vi  when  i , i  1, 2,..., m ;  i  i ;0  l  k
Theorem 5:
Suppose that D is used in Scheme 1.
Then the k-hop coverage of the scheme is equal to Ck  D 

Proof: When using Scheme 1, we know that a pair of nodes sharing
a key are located at vi  u , v j  u , hence the vector vi  v j 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.
k-Hop Coverage

Theorem 6:
Let D be a DD*  m  and let D  be a DD  m  such that D     D  .
Then the k-hop coverage of D is equal to the k-hop coverage of D .
 Proof:  is a bilinear bijection.
Maximal k-hop coverage

First, we define a set of integer m-tuples:
m
m


m
H k   a1 , a2 ,..., am  
|  ai  0,  ai  k 
i 1
i:ai  0


Maximal k-hop coverage



First, we define a set of integer m-tuples:
m
m


m
H k   a1 , a2 ,..., am  
|  ai  0,  ai  k 
i 1
i:ai  0


Simply put, the some of elements in each tuple is zero and the sum
of positive elements is k.
Some examples for m=3:
 0, 0, 0   H 0 ; 1, 1, 0   H1 ;  2, 2, 0  ,  2, 1, 1  H 2
Maximal k-hop coverage



First, we define a set of integer m-tuples:
m
m


m
H k   a1 , a2 ,..., am  
|  ai  0,  ai  k 
i 1
i:ai  0


Simply put, the some of elements in each tuple is zero and the sum
of positive elements is k.
Some examples for m=3:
 0, 0, 0   H 0 ; 1, 1, 0   H1 ;  2, 2, 0  ,  2, 1, 1  H 2

Lemma 7:
 i  w  H k1 , z  H k2  w  z  H k3 where 0  k3  k1  k2
 ii  w  H k , z  H k , w  z  w  z  H k where 1  k3  k1  k2
 iii  Any element of H k may be written as the sum of k1 elements
1
2
3
1
from H1 .
Maximal k-hop coverage

Theorem 8:
The k-hop coverage of a DD  m  is at most
k
H
i 1
m
if and only if all the vectors
a v
i 1
are distinct.
i i
i
, with equality
with  a1 , a2 ,..., am  
k
Hj
j 0
Maximal k-hop coverage

Theorem 8:
The k-hop coverage of a DD  m  is at most
k
H
i 1
m
if and only if all the vectors
a v
i 1
i i
i
, with equality
with  a1 , a2 ,..., am  
k
Hj
j 0
are distinct.
m

Proof: The difference vectors in D are all of the form  ai vi
i 1
with  a1 ,..., am   H1 .
Hence:
V is a sum of at most k difference vectors 
k
m

V   ai vi |  a1 ,..., am   H j 
j 0
 i 1

Maximal k-hop coverage

Proof (cont.):
We can now say:
k
k
k
m

Ck  D   1   ai vi |  a1 ,..., am   H j    H i  1   H i
i 1
j 0
 i 1
 i 0
Maximal k-hop coverage

Proof (cont.):
We can now say:
k
k
k
m

Ck  D   1   ai vi |  a1 ,..., am   H j    H i  1   H i
i 1
j 0
 i 1
 i 0

Corollary 9:
The two-hop coverage of a DD  m  is at most:
1
m  m  1 m  2  m  3  m  m  1 m  2   2m  m  1 
4
1
m  m  1  m 2  m  6 
4
Maximal k-hop coverage

Proof:
We proved that the two-hop coverage is at most H1  H 2 .
H1 contains m  m  1 elements.
As for H 2 we have this lovely table:
Maximal k-hop coverage - bounds


We would like to show that Theorem 8’s bound is tight.
i
Naïve approach: vi   2k  1 , 0 , i  1, 2,..., m


Maximal k-hop coverage - bounds



We would like to show that Theorem 8’s bound is tight.
i
Naïve approach: vi   2k  1 , 0 , i  1, 2,..., m
Lemma 10:
The k-hop coverage of a DD  m  given by D  v1 , v2 ,..., vm 


meets the bound of Theorem 8 
m
c v
i 1
i i
 0 for all c 
2k
Hi
i 1
Maximal k-hop coverage - bounds



We would like to show that Theorem 8’s bound is tight.
i
Naïve approach: vi   2k  1 , 0 , i  1, 2,..., m
Lemma 10:
The k-hop coverage of a DD  m  given by D  v1 , v2 ,..., vm 


meets the bound of Theorem 8 
m
c v
i i
i 1
 0 for all c 
2k
Hi
i 1
Proof:
2k
m
m
m
   Bound not met   ai vi   bi vi   ci vi  0;  c1 ,..., cm   H i

i 1
i 1
i 1
i 1
m
   Assume  ci vi  0 when  c1 ,..., cm   H l for some l  1, 2,..., 2k  
i 1
vector c is a  b where a  H  l  , b  H  l
 2

 2
 Bound not met
Bh Sequences

Definition 1:
A - an abelian group.
D  v1 , v2 ,...vm   A - a sequence of A's elements.
D is a Bh sequence over A if all sums
vi1  vi2  ...  vih with 1  i1  ...  ih  m are distinct.

Elements may be used more than once.
Bh Sequences and DDC

Theorem 11:
Let k  2 be a fixed integer and D  v1 , v2 ,..., vm  
2
. Then D is a
DD  m  with maximal k-hop coverage  D is a B2 k over
2
Bh Sequences and DDC

Theorem 11:
Let k  2 be a fixed integer and D  v1 , v2 ,..., vm  
2
. Then D is a
DD  m  with maximal k-hop coverage  D is a B2 k over
 Proof:
   Assume D is a B2 k over 2
2
 i  If vi  v j for i  j we get a contradiction  vi  v j
 ii  If vi  v j  vi  v j  for i  j , i   j  we get a contradiction 
vi  v j  vi   v j   or i  i  and j  j  
 iii  If D does not have maximal k-hop coverage  by Lemma 10
there is a c 
2k
m
H i so that
i 1
c v
i i
i 1
 0  a is c's positives
m
m
i 1
i 1
and b  a - c   2k  t  v1   ai vi   2k  t  v1   bi vi 
we get a contradiction  D has maximal k-hop coverage
Bh Sequences and DDC
Proof (cont.):
   Assume D is a DD  m  with maximal k-hop coverage

If D is not a B2 k sequence then there are two sums as seen in the Bh
definition which are equal 
m
m
m
m
 a v   b v ;1   a ,  b
i 1
i i
i 1
i i
c  a  b  H t gives
i 1
i
i 1
m
c v
i 1
i i
i
 t  2k 
 0  we get a contradiction.
Using Bh sequences to build a DDC

Construction 1:
Let k  2 be a fixed integer. Let q be a prime power so that
q 2 k  1  ab where a, b are coprime. Then there exists a set of
dots X 
2
which is doubly periodic with periods a, b and that for
each rectangle R of size a  b, R  X is a DD  q  with maximal
k-hop coverage.
Using Bh sequences to build a DDC

Construction 1:
Let k  2 be a fixed integer. Let q be a prime power so that
q 2 k  1  ab where a, b are coprime. Then there exists a set of
dots X 
2
which is doubly periodic with periods a, b and that for
each rectangle R of size a  b, R  X is a DD  q  with maximal

k-hop coverage.
Proof:
Example 1 gave a B2 k sequence over
there is an isomorphism
sequence v1 , v2 ,..., vq 
q 2 k 1
a


a
q 2 k 1

b
with q elements. Also,
 We get a B2 k
b
Define    x, y     x mod a, y mod b  , so we have  :

and can define X as v 
2
2


a

|   v   v1 , v2 ,..., vq   Every a  b
rectangle R will give us X  R  v1 , v2 ,..., vq  , a B2 k sequence
b
Maximal k-hop coverage - bounds

Theorem 12:
Let k  2 be a fixed integer, and c   16  21 k . Then there exists
a DD  m, r  with maximal k-hop coverage such that m
cr1 k .
Maximal k-hop coverage - bounds

Theorem 12:
Let k  2 be a fixed integer, and c   16  21 k . Then there exists
a DD  m, r  with maximal k-hop coverage such that m cr1 k .
 Proof:
S  2 set of points in a  r / 2  radius from the origin. Note that
S   4  r 2  O  r  . Let q be the smallest prime power for which
q k  2r and we get q
 2r 
1k
We now define:
q k  1
when q is even
 k
a   q  1 2 when q k  3mod 4
 k
k
 1mod 4
q
hen
w
2
1

q



b   q 2 k  1 a
Maximal k-hop coverage - bounds

Proof (cont.):
As we saw in Construction 1, a a  b rectangle contains q dots.
A shift T of S contains S q  ab  dots at average  T so that
T  X  S q  ab  . We define D  T  X where
D  m   S q  ab    4  r 2 q  2r   16  21 k r 1 k 
D is contained in a sphere of radius  r 2  , and in a a  b rectangle
2
which is a DD  q  coverage  D is a DD  m, r  with maximal
k-hop coverage.
Maximal k-hop coverage - bounds

Proof (cont.):
As we saw in Construction 1, a a  b rectangle contains q dots.
A shift T of S contains S q  ab  dots at average  T so that
T  X  S q  ab  . We define D  T  X where
D  m   S q  ab    4  r 2 q  2r   16  21 k r 1 k 
D is contained in a sphere of radius  r 2  , and in a a  b rectangle
2
which is a DD  q  coverage  D is a DD  m, r  with maximal
k-hop coverage.

Corollary 13:
1 2k
Using theorem 2, 6, and 12 we can say that for c   16  21 k  2 3
there exists a DD*  m, r  with maximal k-hop coverage so that
m
cr1 k
Maximal k-hop coverage - bounds

What is the minimal r so that a DD  m, r  with maximal k-hop
coverage is possible? We denote this r as r  k , m  .
Maximal k-hop coverage - bounds


What is the minimal r so that a DD  m, r  with maximal k-hop
coverage is possible? We denote this r as r  k , m  .
Theorem 14:
For an integer k  2,
k
mk
1  16 
 o  mk   r  k , m     mk  o  mk 
2  
  k ! k
Maximal k-hop coverage - bounds


What is the minimal r so that a DD  m, r  with maximal k-hop
coverage is possible? We denote this r as r  k , m  .
Theorem 14:
For an integer k  2,
k
mk
1  16 
 o  mk   r  k , m     mk  o  mk 
2  
  k ! k

Proof: (Upper bound proven in Theorem 12)
k
Remember Ck  D    H i .
i 1
Define B   a1 , a2 ,..., am   H k : i : ai  0  2k .
We get B 
m!
.
2
 m  2k ! k !
k
So we can say 
i 1
m!
m2k
2k
2k
Hi 

o
m


o
m




k !2
 m  2k ! k !2
Maximal k-hop coverage - bounds

Proof (cont.):
Vectors in Ck  D  are at most k difference vectors. All difference
vectors are r at most. All vectors Ck  D  are inside a kr sphere,
which contains at most   kr   O  r  vectors.
2
m2k
2
We get Ck  D   2  o  m 2 k     kr   O  r  .
k!
Maximal k-hop coverage - bounds

Proof (cont.):
Vectors in Ck  D  are at most k difference vectors. All difference
vectors are r at most. All vectors Ck  D  are inside a kr sphere,
which contains at most   kr   O  r  vectors.
2
m2k
2
We get Ck  D   2  o  m 2 k     kr   O  r  .
k!


For a hexagonal grid we present an equivalent term r *  k , m  .
Theorem 15:
For an integer k  2,
k

2
mk
2 1  16  k

 o  mk   r *  k , m  
   m  o  mk 
3   k ! k
3 2  
Proof: Theorem 2 & 14.
Maximal k-hop coverage - bounds


We will give special attention to the case k=1.
Theorem 16:
2
2
m  o  m   r 1, m   m  o  m  , where   0.914769


Maximal k-hop coverage - bounds



We will give special attention to the case k=1.
Theorem 16:
2
2
m  o  m   r 1, m   m  o  m  , where   0.914769


Proof:
 2  For DD  m, r 
we have m 
 2  DD  m, r  : m  


2
r  O  r 2 3   Lower bound
 r  o  r   Upper bound
2
Maximal k-hop coverage - bounds



We will give special attention to the case k=1.
Theorem 16:
2
2
m  o  m   r 1, m   m  o  m  , where   0.914769


Proof:
 2  For DD  m, r 
we have m 
 2  DD  m, r  : m  



2
r  O  r 2 3   Lower bound

 r  o  r   Upper bound
2
Theorem 17:
231 4

m  o  m   r 1, m  
*
21 2 31 4

m  o m
Proof: Analogous hexagonal result from [2].
Maximal k-hop coverage - bounds


Finally, using results in [2] we can prove:
Theorem 19:
r 1, m   2  m  o  m 
 2 3 m  o m  r
*
1, m    2   m  o  m  ,   1.58887
Minimal k-hop coverage

What is the smallest value for a k-hop coverage?
Minimal k-hop coverage


What is the smallest value for a k-hop coverage?
Theorem 20:
The k-hop coverage of a DD  m  is at least km  m  1
Minimal k-hop coverage



What is the smallest value for a k-hop coverage?
Theorem 20:
The k-hop coverage of a DD  m  is at least km  m  1
Proof:
The 1-hop coverage of a DD  m  is m  m  1 , the amount of
difference vectors. Define S1 the set of initial difference vectors.
We pick u=  d , e   S1 with maximal d
 and if need be, e  .
We can assume d  0, e  0 and can say that S1 is composed of:
S1   x, y  |  x, y   S1 , x  0 or  x  0 and y  0  ,
S1    x, y  |  x, y   S1 
For i  1: Si  w   i  1 u | w  S1    w   i  1 u | w  S1 
Si  S j   for i  j , Si  m  m  1 .
Minimal k-hop coverage

Lemma 21:
For an integer k  2, suppose D is a DD  m  where two differences
are not parallel. Then the k-hop coverage of D is more than km  m  1 .
Minimal k-hop coverage

Lemma 21:
For an integer k  2, suppose D is a DD  m  where two differences
are not parallel. Then the k-hop coverage of D is more than km  m  1 .

Proof:
Define u and Si as in Theorem 20. Let v be a difference vector not
parallel to u , and the most perpendicular.
The k-hop coverage of D is at least S1  S 2  ...  S k  2v
We saw that Si  m  m  1 , and we can see that 2v  S1  ...  S k ,
since p  w   i  1 u   p  w   p  v   2 p  v   p  2v  .
Minimal k-hop coverage

Lemma 21:
For an integer k  2, suppose D is a DD  m  where two differences
are not parallel. Then the k-hop coverage of D is more than km  m  1 .

Proof:
Define u and Si as in Theorem 20. Let v be a difference vector not
parallel to u , and the most perpendicular.
The k-hop coverage of D is at least S1  S 2  ...  S k  2v
We saw that Si  m  m  1 , and we can see that 2v  S1  ...  S k ,
since p  w   i  1 u   p  w   p  v   2 p  v   p  2v  .

Lemma 21 can be used to prove Theorem 21:
For an integer k  2, suppose D is a DD  m  . D meets the bound of
Theorem 20 if and only if it is equivalent to a perfect Golomb ruler.
Complete 2-hop coverage


For a prime p  5 , we will show a construction of a DD  m  with
complete 2-hop coverage.
That ensures a two-hop path between a point x and any other grid
point within a  2 p  3   2 p  1 rectangle centered at x.
Height
Width
Complete 2-hop coverage


For a prime p  5 , we will show a construction of a DD  m  with
complete 2-hop coverage.
That ensures a two-hop path between a point x and any other grid
point within a  2 p  3   2 p  1 rectangle centered at x.
Height

Width
Definition 2 (Welch Periodic Array):
Let  be a primitive root modulo a prime p.
Define the Welch periodic array: R p   i, j  
2
|  j  i mod p.
R p is double periodic:  i, j   R p   i   p, j    p  1   R p
for  ,   .

Equivalent points:
A   i, j    i , j    A  i   i   p, j   j    p  1 .
j
Example of an array

10

9
8




6



5
4




3



2


1
1
 3


7
p5

2
3

4
5
6
7
8
9
10
11
i
Complete 2-hop coverage

Lemma 23:
Let d and e be integers such that d  0 mod p and e  0 mod  p  1 .
Suppose that R p contains dots A   i1 , j2  , B  i1  d , j2  e  as well as
A   i2 , j2  , B  i2  d , j2  e  . Then A  A, B  B .
Complete 2-hop coverage

Lemma 23:
Let d and e be integers such that d  0 mod p and e  0 mod  p  1 .
Suppose that R p contains dots A   i1 , j2  , B  i1  d , j2  e  as well as
A   i2 , j2  , B  i2  d , j2  e  . Then A  A, B  B .

Proof:
i1   j1 mod p


j2
i2   mod p

j1
j2
e



1



 0 mod p





j1  e
i1  d  
mod p 
i2  d   j2  e mod p 
Since e  0 mod  p  1 we get j1  j2 mod  p  1
and from that i1  i2 mod p
Complete 2-hop coverage

From Lemma 23 we conclude:
if R p contains  i, j  ,  i  d , j  then d  0 mod p.
if R p contains  i, j  ,  i, j  e  then e  0 mod  p  1 .
A vector  d , e  can occur at most once as a difference between
two points in R p , within a particular  p  1  p rectangle.
Complete 2-hop coverage

We now define a DD  m  by using dots from R p .
Complete 2-hop coverage


We now define a DD  m  by using dots from R p .
Construction 2:
p an odd prime,  i, j   2 is such that  i, j  ,  i  1, j  1  R p
j
◦ Do such points exist? Why yes!   i  1   1 mod p
and also  j 1     1  1   1  1   i  1 mod p
Complete 2-hop coverage


We now define a DD  m  by using dots from R p .
Construction 2:
p an odd prime,  i, j   2 is such that  i, j  ,  i  1, j  1  R p
j
◦ Do such points exist? Why yes!   i  1   1 mod p
and also  j 1     1  1   1  1   i  1 mod p
S is a  p  1  p rectangle bounded by  i, j  ,  i  p  1, j  ,
 i, j  p  2  and  i  p  1, j  p  2  . There are  p  1 dots
in S .
◦ Why p  1 dots? Remember what  is!
Complete 2-hop coverage


We now define a DD  m  by using dots from R p .
Construction 2:
p an odd prime,  i, j   2 is such that  i, j  ,  i  1, j  1  R p
j
◦ Do such points exist? Why yes!   i  1   1 mod p
and also  j 1     1  1   1  1   i  1 mod p
S is a  p  1  p rectangle bounded by  i, j  ,  i  p  1, j  ,
 i, j  p  2  and  i  p  1, j  p  2  . There are  p  1 dots
in S .
◦ Why p  1 dots? Remember what  is!
Since R p is periodic, it also contains dots at  i, j   p  1  ,  i  p, j 
and  i  p  1, j  p  .
Our, actual construction is a configuration B , which is S and the
above dots.
Meet 
j p
B
j  p 1
A
p2
p 3
S
Central region
j 1
B
j
A
A
i
i 1
i  p i  p 1
 - Vital statistics





Contained in a  p  1   p  2  square.
Has a border region of width 2 which contains exactly 5 points.
Has a central region which is a  p  3   p  2  rectangle.
The central region contains p  3 dots. One column is empty.
A  A  A and B  B and there are no other equivalent points.
j
Example of 

10

9
8




6

B’

A’
5
4



3


2

B
1
A
1
 3


7
p5

2
3
A’’
4
5
6
7
8
9
10
11
i
Complete 2-hop coverage

Lemma 24:
The configuration B is a DD  p  2  , with all points in a
 p  1   p  2  rectangle.
Complete 2-hop coverage

Lemma 24:
The configuration B is a DD  p  2  , with all points in a
 p  1   p  2  rectangle.

Proof:
We have already shown that B contains the dots as needed.
Suppose that X and Y and X  and Y  are distinct dot pairs with the
same difference vector  d , e  .
Suppose d  0,  p, p and/or e  0,   p  1 ,  p  1 .
It is impossible for one of X , Y , X , Y  to be in the central region.
Why?
This is why
j p
B
j  p 1
A
p2
p2
p 3
j 1
B
j
p 1
A
i
i 1
A
i  p i  p 1
Complete 2-hop coverage

Proof (cont.):
X , Y , X , Y  must be outside the central region, but no two pairs have
the same difference vector.
So d  0,  p, p and e  0,   p  1 ,  p  1 and since all dots are in
a  p  1   p  2  rectangle d  0 mod p and e  0 mod  p  1 .
Lemma 23  X  X , Y  Y .
X  X   Y  Y   point pairs are not distinct  X  X 
X , X  and Y , Y  must be on configuration border, but we've already
seen that leads to a contradiction, and so we have our proof.
Complete 2-hop coverage

Motivational boost:
In order to show a  2 p  3   2 p  1 rectangle with complete 2-hop
coverage, we will show that any vector  d , e  where d   p  1
and e   p  2  can be expressed as a difference vector of B..
Complete 2-hop coverage

Motivational boost:
In order to show a  2 p  3   2 p  1 rectangle with complete 2-hop
coverage, we will show that any vector  d , e  where d   p  1
and e   p  2  can be expressed as a difference vector of B..

Lemma 25:
Any vector of the form  d , e  , where 0  d  p  1 and 0  e  p  2
 where d , e are integers  is a sum of two difference vectors from B.
Complete 2-hop coverage

Motivational boost:
In order to show a  2 p  3   2 p  1 rectangle with complete 2-hop
coverage, we will show that any vector  d , e  where d   p  1
and e   p  2  can be expressed as a difference vector of B..

Lemma 25:
Any vector of the form  d , e  , where 0  d  p  1 and 0  e  p  2
 where d , e are integers  is a sum of two difference vectors from B.

Proof:
Remember the  p  1  p rectangle S from Construction 2.
Define A as a restriction of R p to the  2 p  2   2 p sub-array
whose lower left corner coincides with S .
illustrated
3
4
1
2
S
illustrated
D1  S  B.
Since R p is doubly
periodic with S 's size,
the points in D1 occur
in the other quadrents
as well.
3
4
1
2
Complete 2-hop coverage

Proof (cont.):
We want to show that any vector  d , e  as defined is a difference of
two points in A . We assume d  0.
Assume e  0 and can say that there is a  i , j    D1 so that
d
mod p
 e 1
It is easy to see that  i , j   ,  i   d , j   e   R p , and since d , e  0 and
j
  i 
are small enough,  i , j   ,  i   d , j   e   A .
In the case where e  0 we can just pick  i , j    D3 and follow use
same concept.
Complete 2-hop coverage

Proof (cont.):
Now that any vector  d , e  is a difference of two points in A . All we
need to show is that every two points in A can be connected using
two difference vectors from B.
You'll see this more easily if I show you.
D1 to D1 (or any Dx to Dx)
By definition, all such vectors are a single difference vector from B.
3
4

1
2
D1 to D3 (or D2 to D4)
We use the vector  0, p  1 to cross quadrents, then use another
difference vector from B to reach our destination (this second one
exists since R p is doubly periodic.
3
4
1
2
D1 to D3 (or D2 to D4)
We use the vector  p, 0  to cross quadrents, then use another
difference vector from B to reach our destination (this second one
exists since R p is doubly periodic.
3
4
1
2
D1 to D4
We use the vector  p, p  1 to cross quadrents, then use another
difference vector from B to reach our destination (this second one
exists since R p is doubly periodic.
3
4
1
2
D3 to D2
We use the vector  p,   p  1  to cross quadrents, then use another
difference vector from B to reach our destination (this second one
exists since R p is doubly periodic.
3
4
1
2
Complete 2-hop coverage

Lemma 26:
For an integer t  3, let F be a set of integers where:
a
b
c
d 
e
F  t 1
F    t  1 ,   t  2  ,..., 1  1, 2,..., t  1  t  1
1,   t  1 ,  t  1  F
i  F \ 1,   t  1 , t  1 with i  0
if i  0 and i  F \ 1,   t  1 , t  1 then i  t  F
Then for each positive integer  where 1    t  1 there are
i, j  F so that   j  1.
Complete 2-hop coverage

Lemma 26:
For an integer t  3, let F be a set of integers where:
a
b
c
d 
e
F  t 1
F    t  1 ,   t  2  ,..., 1  1, 2,..., t  1  t  1
1,   t  1 ,  t  1  F
i  F \ 1,   t  1 , t  1 with i  0
if i  0 and i  F \ 1,   t  1 , t  1 then i  t  F
Then for each positive integer  where 1    t  1 there are
i, j  F so that   j  1.

Why do we need this?
Complete 2-hop coverage

Lemma 26 motivation:
Let us look at all vectors of the form 1, y  where y  0 present in B.
We will show how the right-sided coordinates of these vectors
satisfy the conditions of Lemma 26 for t  p  1.
Complete 2-hop coverage

 difference vectors 
Lemma 26 motivation:
Let us look at all vectors of the form 1, y  where y  0 present in B.
We will show how the right-sided coordinates of these vectors
satisfy the conditions of Lemma 26 for t  p  1.
 a  B has p difference vectors as defined above.
Proof of (a)
j p
B
j  p 1
A
p2
p 3
One of the p  2 columns
is empty, so we have
p  3 vectors
j 1
B
j
A
A
i
i 1
i  p i  p 1
Complete 2-hop coverage

Lemma 26 motivation:
Let us look at all vectors of the form 1, y  where y  0 present in B.
We will show how the right-sided coordinates of these vectors
satisfy the conditions of Lemma 26 for t  p  1.
 a  B has p difference vectors as defined above.
 b  Except for 1, p  , all of B 's 1, y  vectors satisfy y  p  2.
Proof of (b)
j p
B
j  p 1
A
p2
y  p  1,
p 3
but this
y p
vector is
not legitimate
j 1
B
j
A
A
i
i 1
i  p i  p 1
Complete 2-hop coverage

Lemma 26 motivation:
Let us look at all vectors of the form 1, y  where y  0 present in B.
We will show how the right-sided coordinates of these vectors
satisfy the conditions of Lemma 26 for t  p  1.
 a  B has p difference vectors as defined above.
 b  Except for 1, p  , all of B 's 1, y  vectors satisfy y  p  2.
 c  The vectors 1,1 , 1,   p  2   , 1, p  are all in B.
Proof of (c)
j p
B
j  p 1
A
1,   p  2  
p2
p 3
1, p 
j 1
B
j
A
i
1,1
i 1
A
i  p i  p 1
Complete 2-hop coverage

Lemma 26 motivation:
Let us look at all vectors of the form 1, y  where y  0 present in B.
We will show how the right-sided coordinates of these vectors
satisfy the conditions of Lemma 26 for t  p  1.
a
b
c
d 
B has p difference vectors as defined above.
Except for 1, p  , all of B 's 1, y  vectors satisfy y  p  2.
The vectors 1,1 , 1,   p  2   , 1, p  are all in B.
B 's 1, y  vectors can't all satisfy y  0.
Proof of (d) – case one
j p
B
j  p 1
A
p2
p 3
1, y 
y0
j 1
B
j
A
A
i
i 1
i  p i  p 1
Proof of (d) – case two
j p
B
j  p 1
A
p2
j 1
B
j
A
i
1,1
i 1
No dots
p 3
1,1
A
i  p i  p 1
Complete 2-hop coverage

Lemma 26 motivation:
Let us look at all vectors of the form 1, y  where y  0 present in B.
We will show how the right-sided coordinates of these vectors
satisfy the conditions of Lemma 26 for t  p  1.
a
b
c
d 
e
B has p difference vectors as defined above.
Except for 1, p  , all of B 's 1, y  vectors satisfy y  p  2.
The vectors 1,1 , 1,   p  2   , 1, p  are all in B.
B 's 1, y  vectors can't all satisfy y  0.
If 1, y  where y  1, p is a vector in B , 1, y   p  1  is not.
Complete 2-hop coverage

Lemma 26 motivation:
Let us look at all vectors of the form 1, y  where y  0 present in B.
We will show how the right-sided coordinates of these vectors
satisfy the conditions of Lemma 26 for t  p  1.
a
b
c
d 
e
B has p difference vectors as defined above.
Except for 1, p  , all of B 's 1, y  vectors satisfy y  p  2.
The vectors 1,1 , 1,   p  2   , 1, p  are all in B.
B 's 1, y  vectors can't all satisfy y  0.
If 1, y  where y  1, p is a vector in B , 1, y   p  1  is not.
 e  is proven thusly:
If 1, y   p  1  were in B then by Lemma 23 the points involved
would be equivalent to those making the vector 1, y  .
Such points would be on the border region, and that is impossible.
Complete 2-hop coverage

Lemma 26 motivation:
By similar means, we can show Lemma 26 works for vectors of the
form  x,1 when t  p.

We will now face insurmountable suspense…
Complete 2-hop coverage

Proof (of Lemma 26):
F \ 1,   t  1 , t  1 contains t  2 elements  F must contain
precisely one element of each pair i, i  t for i  2,3,..., t  1.
Suppose, for a contradiction,   t -1 that cannot be expressed as
a difference of two elements from F .
Suppose   1 and then: 1, t  1  F  1   , t  1    F
However, since 1     t  1     t one of them is in F 
contradiction
Complete 2-hop coverage

Proof (of Lemma 26, cont.):
Suppose   1  F does not contain a pair of integers which differ
by 1
If t is odd then F \ t  1 contains at most  t  1 2 positive integers
and at most  t  1 2 negative integers  F contains at most
 t  1  1  t integers, which contradicts  a  .
If t is even then in order for F to be of size t  1, F \ t  1 must
contain t 2 positive integers (which must be odd) and t 2 negative
integers (also odd). This implies that for each odd integer 1  i  t
we have i, i  t  F and this contradicts  e  .
Complete 2-hop coverage

Theorem 27:
Let p  5 be a prime. The distinct difference configuration B achieves
complete two-hop coverage on a  2 p  3   2 p  1 rectangle, relative
to the central point of the rectangle.
Complete 2-hop coverage

Theorem 27:
Let p  5 be a prime. The distinct difference configuration B achieves
complete two-hop coverage on a  2 p  3   2 p  1 rectangle, relative
to the central point of the rectangle.

Proof:
We notice that all vectors from the central point  d , e  are such
that 0  d  p  1, 0  e  p  2.
When d  0, e  0, Lemma 25 proves that the vector is composed
of two of B ' s difference vectors.
When either d  0 or e  0 we can use Lemma 26. For 0  e  p  2,
we have shown that there are two vectors in B which are 1, y  , 1, y  
so that y  y   e and so  0, e   1, y   1, y   . Similarly for  d , 0 
where 0  d  p  1.
Conclusion and open problems








We have shown maximal k-hop coverage as B2 k over 2 .
We used a construction of B2 k over to produce a DD  m, r 
with maximal k-hop coverage and of the order of r1 k dots.
We have found a bound for r  k , m  (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 DD  m, r  with complete 2-hop coverage
from the center of a rectangular region.
The rectangle’s region is of order m2. Can we find a construction
for significantly larger rectangles? For circles?
Can we find constructions for k-hop coverage where k>2?