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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