with both 
and 
in the class of circular codes is non-empty.
. As no two conjugated trinucleotides can belong to a circular code, we are in contradiction. So,
. 
Proposition 3. The class of self-complementary circular codes with both and in the class of circular codes is non-empty.
Proof. Consider, for example, the following set 
of 20 trinucleotides
It is enough to prove that 
is a self-complementary circular code and that its two conjugated classes 
and 
are also circular codes.
is a self-complementary circular code.
is self-complementary. Obvious by inspection.
is a circular code. We use Proposition 1 [17]. By way of contradiction, suppose that 
admits a 5LDCN. As 
can be
, 
, 
or
, it is enough to prove that each choice leads to a contradiction.
1) If 
then there is no possible 
as 
is not a suffix of any trinucleotide of
, contradiction.
2) If
, there are three possible
:
• if 
(a) or 
(b) then 
(c) but there is no possible 
as 
is not a prefix of any trinucleotide of
, contradiction
• if 
(d), there is a contradiction as no trinucleotide of 
has a prefix
.
3) If
, there are six possible
:
• if 
or
, contradiction (a) and (b)
• if 
then
, contradiction (c)
• if 
or 
then 
or
:
if
, there are three possible
: if 
or 
then
, similarly to (c), contradiction, and if
, similarly to (d), contradiction
if
, contradiction (c)
• if
, contradiction (d).
4) If
, similarly to (c), contradiction.
As, for each letter, we cannot complete the assumed 5LDCN for
, we are in contradiction. Hence, 
is a circular code.
is a circular code. We have to prove that
is a circular code. By way of contradiction, assume that 
admits a 5LDCN.
1) If
, there are four possible
:
, 
, 
and
, but no possible
, contradiction.
2) If
, there are three possible
:
, 
and
, but no possible
, contradiction.
3) If
, there are six possible
:
, 
and
, and the cases
, 
and 
already seen, but no possible
, contradiction.
4) If
, there is no possible
, contradiction.
Hence, 
is also a circular code.
is a circular code. Finally, we have to prove that
is a circular code. By way of contradiction, assume that 
admits a 5LDCN.
1) If
, there is no possible
, contradiction.
2) If
, there are six possible
:
, 
, 
, 
, 
and
, but no possible
, contradiction.
3) If
, there are three possible
:
, 
and 
which are cases already seen, contradiction.
4) If
, there are four possible
:
, 
, 
and
, but no possible
, contradiction.
Hence, as 
and
, 
is also a circular code. 
Proposition 4. The class of self-complementary circular codes 
having 20 elements with neither 
nor 
in the class of circular codes is non-empty.
Proof. Consider, for example, the following set 
of 20 trinucleotides
It is enough to prove that 
is a self-complementary circular code and that neither its conjugated class 
nor its conjugated class 
are circular codes.
is a self-complementary circular code.
is self-complementary. Obvious by inspection.
is a circular code. We use Proposition 1 [17]. By way of contradiction, assume that 
admits a 5LDCN.
1) If 
then there is one possible 
but no possible
, contradiction.
2) If
, there are two possible
:
• if 
then 
(a) and 
(b) but there is no possible
, contradiction
• if 
(c) then there is no possible
, contradiction.
3) If 
we have seven possible
:
• if 
then 
or
:
if 
(d) then 
or
:
- if 
then 
and 
but there is no possible
, contradiction
- if 
then there is no possible
, contradiction
if
, contradiction (a)
• if
, similarly to (b), contradiction
• if
, 
or 
then
, contradiction (a)
• if 
then 
or
, contradiction (a) and (d)
• if
, contradiction (c).
4) If
, similarly to (a), contradiction.
Hence, 
is a circular code.
is not a circular code. We have
We use a technique developed in [23]. Observe that 
contains 
So,
is a 5LDCN for this 4-element subset of 
and, a fortiori, for 
itself which, consequently, is not a circular code.
is not a circular code. We have
We again use a technique developed in [23]. Remark that 
contains
. So,
is a 5LDCN for this 4-element subset of 
and, a fortiori, for 
itself which, consequently, is not a circular code. 
We need the propositions hereafter and, in particular the following one which states a general property of the involutional antiisomorphisms such as the complementary map
.
Proposition 5. A subset 
of 
is a circular code if and only if 
is a circular code.
Proof. Suppose, first, that 
is not a circular code and that 
is a circular code. So 
has a 5LDCN. This means that there are 13 nucleotides, say
such that the trinucleotides
and
Now, consider the sequence
All the following trinucleotides belong to
:
and
as they are the complement of trinucleotides in
. So, 
admits a 5LDCN and it cannot be a circular code. Contradiction.
The case 
is a circular code and 
is not a circular code is similar. 
Proposition 6. Let 
be a self-complementary subset of
. If 
is partitioned into three classes such that two of them are the complement of each other then necessarily the third one is self-complementary.
Proof. Let
, 
and 
be the three classes of an arbitrary partition of 
and suppose that 
and 
are complementary, i.e. 
and 
satisfy
. Let 
be a trinucleotide of
. We claim that
. Indeed, in the opposite case, 
should not be the complement of 
because
. We also claim that
. Indeed, in the opposite case, 
should not be the complement of 
because
. It remains the case
. So, 
is self-complementary. 
Remark 1. Clearly, if
, 
and 
constitute an arbitrary partition of 
then the self-complementarity of 
is not enough to ensure that 
and 
are complementary of each other. This remark is again true if, in addition, 
is a self-complementary circular code having 20 elements. Indeed in this case, it is easy to make a partition 
in two classes 
and 
that are not complementary of each other. Any case, if we consider the partition of 
in the three classes given by a self-complementary trinucleotide circular code 
having 20 elements and by its two conjugated classes 
and 
then the necessary and sufficient condition holds (Proposition 7 below).
Proposition 7. A trinucleotide circular code 
having 20 elements is self-complementary if and only if 
and 
are complement of each other.
Proof if part. It is a trivial consequence of Proposition 6.
Only if part. Suppose that 
is self-complementary and consider the partition
, 
and 
of
. Suppose that the trinucleotide, say
, belongs to
. Then, also
.
We have
and
.
As 
is a generic trinucleotide of 
and as
and
then 
is the complement of
. 
As a consequence, we have the following proposition.
Proposition 8. If a trinucleotide circular code 
having 20 elements is self-complementary then either 1) 
and 
are both circular codes or 2) 
and 
are not circular codes (both have a necklace).
Proof. We have four possibilities:
is a circular code and 
is a circular code;
is a circular code and 
is not a circular code;
is not a circular code and 
is a circular code;
is not a circular code and 
is not a circular code.
Now, by applying Propositions 3 and 4, we have that the first and the last possibilities can be effectively realized.
Suppose that, by way of contradiction, the second possibility is realized. So, 
is a circular code. By Proposition 7, we have
. So, by Proposition 5, 
must also be a circular code. Contradiction.
Suppose that, by way of contradiction, the third possibility is realized. So, 
is a circular code. By Proposition 7, we have
. So, by Proposition 5, 
must also be a circular code. Contradiction.
So, only the first and the last possibilities can occur. 
Hence, our proposition holds.
Proposition 9. The 528 self-complementary circular codes having 20 elements are partitioned into two classes: one class contains codes with the two permuted sets 
and 
which are both circular codes while the other class contains codes with the two permuted sets 
and 
which both are not circular codes.
Proof. It is enough to apply Proposition 8 to each of the 528 trinucleotide circular codes having 20 elements. 
4. Acknowledgements
We thank Jacques Justin for his advices. The second author thanks the Dipartimento di matematica U. Dini for giving him a friendly hospitality.
REFERENCES