
 
Definition 2: The  area of the i-subrelation 
(area_i_sub) is defined from  its size (s_sub
i
) and 
the number of the relations (n_rel) that are selected. 
It is given by (s_sub
i
) x (n_rel). 
Example 2: From a relation with s_rel=5 and if 
n_rel=5  then the area of this relation is  5x 5 . 
Hence there are 25 locations that have to be 
completed. 
Example 3: If two relations are: (1,2,3,5,6,7), 
(1,3,4,8,9,10) with s_sub
3
=3, s_sub
2
=2, then : 2-
subrelation
1
 =(2,3) and 2-subrelation
2
=(3,4). The last 
two subrelations ((2,3),(3,4)) comes from S
2
 
={2,3,4} having 3 as repeated item. 
Definition 3: An BP is full if it provides at least 2 
repetitions of items  and it does not include empty 
slots in the area_rel 
Definition 4: The number of items that can be 
repeated in a subrelation is called item multiplicity 
(it_mu) or number of repetitions (n-rep). 
Definition 5: The optimal BP for S
1s
 < S
2s
 < S
3s
 , is 
the full BP taken with the maximum effective n_rel ( 
providing also the maximum items multiplicity  for 
the subrelations). For optimal BP the most popular 
items are transmitted more often. Our full BP is also 
optimal BP since the items of S
2 
and S
1
 are repeated 
more than one times. 
Definition 6: Integrated relations (or integrated 
grouping)  are when after the grouping, each group 
contains relations with all the data of S
2
 and S
1
. This 
happens when: ( (2_subrelation) = S
2 
) 
 
 (  
(1_subrelation) = S
1
). See example 7 for details.  
Definition 7: An FBP is direct when k S
div  
and S
2s
 | 
k (S
2s
<k). It is indirect when k S
div  
and k | S
2s
  
(k>S
2s
) 
It is considered that a|b (a divides b) only when b 
mod a =0 (f.e. 14 mod 2=0). The relation with  the 
maximum value  of n_rel provides the opportunity 
of  maximum multiplicity for all the items of  S
2
 and 
S
1
 and finally creates the minor cycle of a full BP. 
The major cycle is obtained by  placing  the minor 
cycles on line. The S 
div
 contain all the divisors of 
S
3s
. Hence S
dil
={d
1
,d
2
,..,d
n
}.  
3 SOME ANALYTICAL RESULTS 
FOR BP AND RBP CREATION 
A set of Lemmas can discover the possibility of 
having a full equal spacing BP. from the sets (S
is
 / 
i=1,2,3). 
Lemma 1: Let us be k any integer divisor of S
3s
. If 
k≥S
is
 (i=2,3) and S
is
 | k then we can take a full direct  
BP. 
Proof: If  k≥ S
is
 and S
is
 | k =>  k= S
is
 * m (m I ) 
and any item of S
2s 
 can be repeated  for m times  
Hence it_mu 
i
 = k / S
is
 . Since this happens for all the 
sets, a full BP can be produced using just the items 
of  the S
is
. 
Example 4: (full BP)Consider the case of: S
1
 = {1}, 
S
2
 ={2,3},  S
3
 = { 4,5,6,7,8,9, 10, 11}. Finding the 
integer divisions of  S
3s 
 (=8) which are 4(8/2) and 
2(8/4). The n_rel could be 4(8 /2) or 2(8/4). Hence 
S
div
 = {d1,d2}= {4,2}. If n_rel=4 the format of  the 
four relations with S
1
 could be: 
( * * * ..* 4,5), ( * * * ..* 6,7), ( * * * ..* 8,9), ( * 
* * ..* 10,11). For n_rel=  k =4  then 4>2  and it_mu 
i
 =2=4/2  it means that there is  a  full BP for S
2 
. 
Using again the same for the S
1
 we take   4>1 and 
it_mu 
i
 =2=4/1 it means that there is a  full BP for 
S
1
. One relation of the full, direct could be: (1,2,4,5). 
Lemma 2: If k<S
is
 (i=2,3) and  k |  S
is
  then we can 
take a full indirect BP. In this case the total number 
of items (t_n_i 
i
) that transferred and the s_sub
i
 can 
be easily computed. 
Proof: If k<S
is
 (i=2,3) and k | S
is
 then S
is
 = k *m (m 
I ) and this gives again it_mu
i
=S
is
/k. 
Additionally, a predefined it_mu for S
is
 can be 
defined so that t_n_i 
i
 = S
is
 * it_mu 
i
 and s_sub = 
t_n_i / n_rel. 
Example 5: Let’s consider S
1
 = 1, S
2
 = {2,…,13}, S
3
 
= {15,…,32} with: S
1s
 = 1, S
2s
 = 12, S 
3s
 = 18. 
Finding the integer divisors of S
3s 
 (=18) which are 
9(18/2), 6(18/3), 3(18/6). The decreasing order is: 
9,6,3. (a) For n_rel=k=9, since 9<12 and 912 only 
empty slot BP possibility. (b) Taking the next k 
value (k=6),  since 6<12  and 6|12  there is a FBP 
with it_mu=2, t_n_i=12*2=24, s_sub=t_n_i / k = 
24/6=4. Hence  the 2- subrelation for the 6 relations 
can be: (..,2,3,4,5,..), (..,6,7,8,9..), (..,10,11,12,13,..), 
(..,2,3,4,5,..), (..,6,7,8,9,..), (..,10,11,12,13,. ..) having  
two repetitions for each item. Hence 1-subrelation = 
6, 2-subrelation=4. 
Lemma 3: If k<S
is
 (i=2,3) and k    S
is
  then it is not 
possible to take a full BP.  
Proof: Because it_mu
i
 = k/S
is
  I. 
Example 6: Let us consider S
1
 = 1, S
2
 = {2,3,4}, S
3
 = 
{5,…,22} with: S
1s
 = 1, S
2s
 = 3, S 
3s
 = 18. Finding 
the integer divisors of S
3s 
(=18) which are 9(18/2), 
6(18/3), 3(18/6). Decreasing order 9,6,3. For 
n_rel=k=9,since 9>3,(from (1),(2)) with it_mu
3
 =3 
=9/3 there is a strong 3-subrelation. (1,2,5,6), 
(1,3,7,8), (1,4,9,10), (1,2,11,12), (1,3,13,14), 
(1,4,15,16), (1,2,17,18), (1,3,19,20), (1,4,21,22). The 
BP is an RBP (equal spacing for all the sets), for 
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