Applied Mathematics E-Notes, 7(2007), 16-22 c
Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/

ISSN 1607-2510

Orders With Ten Elements Are Circle Order∗
Roman Bayon†, Nik Lygeros‡, Jean-Sébastien Sereni§
Received 9 February 2006

Abstract
With a new method based on the notion of genetic algorithm and the explicit
enumeration of orders, we prove that all orders on at most 10 elements are circle
orders. This theorem represents the best partial result on Sidney-Sidney-Urrutia
Conjecture.

1

Order Dimension and Circle Orders

We are interested in orders of small size which are circle orders, in relation to theirdimension (see [6]). As is well known, the finite posets of dimension at most two
are just those which have inclusion representations using closed intervals of the realline R. Because a closed interval of R can also be considered as a sphere in R1 , itis natural to ask which posets have inclusion representations using circular disks in
R2 . For historical reasons, these posets are called circle orders. Schneinerman and
Wierman [11] showed in that Z3 is not a circle order, and then Hulbert [10] showedthat the same holds for N3 . Urrutia, after having proved that all finite orders withdimension at most three are regular n-gon orders for all n ≥ 3, conjectured that all
finite orders with dimension at most three are circle orders (see [13]). However, in 1999,
Felsner, Fishburn and Trotter [5] disproved this conjecture by using a new argumentfrom Ramsey theory. They think it would be possible to find an order which is nota circle order with dimension three and at most one hundred elements. The crossing
number of an order is the smallest integer m such that the order can be represented
by continuous real-valued functions in [0, 1], no two of whose graphs intersect in morethan m points. With this notion, Sidney, Sidney and Urrutia [12] showed the following
result.
THEOREM 1 [Sidney, Sidney and Urrutia, 1988]. For n ≥ 4, Ψn is not a circle
order.
To construct the family Ψn , start with the standard n-dimension order denoted byHn(Y, <), where Y = {u1 , . . ., un, v1, . . . , vn}, and ui < vj for i 6= j, the other elements∗ Mathematics Subject Classifications: 06A06, 05C85, 68-04.
† INPG, 46 Avenue Félix Viallet, 38031 Grenoble, France.
‡ Institut Girard Desargues – Département de Mathématiques, Université Lyon I, 43 Bld du 11 Nov.
1918, 69622 Villeurbanne, France.
§ MASCOTTE – I3S (CNRS/UNSA) - INRIA, BP 93, 2004 Route des lucioles, 06902 Sophia Antipolis Cedex, France.

16

Bayon et al.

17

being incomparable. Then for every subset Sk of {1, . . ., n} with exactly b n2 c or b n+1
2 celements (if n is even, both values are the same, if n is odd they are different), insert
in Hn(Y, <) a new element sk such that sk > uj for j ∈ Sk , sk < vi for i ∈/ Sk and
finally, sk > s0k if Sk0 ⊂ Sk .
To show that the dimension of Ψn is exactly n, it is sufficient to observe that Ψn is
contained in 2n and contains Hn.
Furthermore in 1989, thanks to Radon lemma, Brightwell and Winkler [1] obtained
the following theorem.
THEOREM 2 [Brightwell and Winkler, 1989]. For each n ≥ 1, the order with
2n+2 − 2 elements P2n+2 −2 is a n-sphere order, but not a (n − 1)-sphere order, whereP(2n+2 −2) = {A ⊆ {1, 2, . . ., n + 2} : 1 ≤ |A| ≤ n + 1}
with A ≤ B if and only if A ⊆ B and (|A| = 1 or |B| = n + 1).
Among those two families of orders which are not circle orders, there is an orderwith 14 elements belonging to both families: P2 = Ψ4 , we call it p14.
CONJECTURE 1 [Sidney, Sidney and Urrutia, 1988]. The smallest order which is
not a circle order is p14.
In 1991, Fraı̈ssé and Lygeros [7] showed that all orders with at most 7 elements
are circle orders, introducing a method based upon an exhaustive order computationand their automatic discrimination with the application of the theorem of Dushnik andMiller [4], and Hiraguchi [9]. However, the last part of the proof which is a fastidiouscase by case verification, cannot be applied to larger orders. So, it is necessary to
introduce a new method.

2

Enumeration and Genetic Algorithm

We consider the list of non-isomorphic orders with n elements, obtained by the generation algorithm of Chaunier and Lygeros (see [2, 3]). Every order is represented in a
canonical way by a list of n(n−1)
bits. If P = (X, <) is a poset then for (x, y) ∈ X 2
2
we write x k y if and only if x and y are incomparable for <. A conjugate of a poset
P = (X, <) is a poset P 0 = (X, ≺) such that ∀(x, y) ∈ X 2 , x k y if and only if(x ≺ y or y ≺ x). As every order with dimension at most two is clearly a circle order,
we use the following theorem to detect them (see [4]).
THEOREM 3 [Dushnik and Miller, 1941]. A poset has dimension at most two if
and only if it has a conjugate order.
Thus, to know whether or not the dimension of a poset P is at most two, we only
need to know whether or not the graph G = (X, {(x, y) ∈ X 2 , x k y}) — called theincomparability graph of the poset P — admits a transitive orientation, which can bedone in linear time (see, for instance, [8]).
We obtain the following proportions.

18

Orders with Ten Elements
n
7
8
9
10

Proportion of 2-dimensional orders among orders with n elements
95.65% that is 1956/2045
87.03% that is 14794/16999
71.78% that is 131526/183231
51.88% that is 1331848/2567284

We explicit now some basic properties of a minimal counter-example P (X, <) tothe conjecture, regarding the number of elements. Let n be the number of elements of
P — therefore, all orders with n − 1 elements are circle orders. A sink is an elementx ∈ X such that ∀y ∈ X \ {x}, y < x. It is clear that P cannot have a sink. A sourceis an element x ∈ X such that ∀y ∈ X \ {x}, x < y. Then, P has no source. This isby virtue of the existence of Sidney-Sidney-Urrutia’s normal representation: be it byincreasing the circles radius, every circle order admits a part of the plane that is sharedby all the circles. Besides, we also know that P does not have two twin elements, where(x, y) ∈ X 2 are twin elements if and only if• x < y;• ∀z 6= y, x < z ⇒ y < z; and• ∀z 6= x, z < y ⇒ z < x.
There is a natural notion of duality for posets: let (X, <) be a poset, the binary
relation ≺ defined on X by∀(x, y) ∈ X 2 , x ≺ y if and only if y < x
is a partial order. The poset (X, ≺) is called the dual of the poset (X, <). As is wellknown (see [13]), the dual of a finite circle order is also a circle order.
Let n ∈ {8, 9, 10}, and consider the enumeration of all non-isomorphic partial orders
with n elements. We remove from the list every order that has a sink, a source or two
twin elements. Moreover, we remove an order in each couple of dual orders. It remains
783 orders with 8 elements, 16537 orders with 9 elements and 397603 orders with 10
elements, on which we run the representation search algorithm.
This program comes from genetic algorithms. To find a representation of a given
order P with n elements, we start by randomly generating a population of subjects:
a subject is a set of n circles, fulfilling some inclusion constraints. Every subject iscreated in a random way, the circles representing elements of big degree being more
likely to have smaller radii. Then, we define the fitness of every subject: the bigger
it is, the more the subject is close to a representation of P . A subject whose fitness
reaches the maximum value is a solution. To compute the fitness of a subject, we apply
a bonus of value n + 1 for each relation present in the subject that also is present in
P . In the same way, a bonus of one is given for any two incomparable elements ofthe subject that also are incomparable in P . On the contrary, if a relation is missingin the subject, then a penalty of value 2n is applied. For the relations present in thesubject but not in the original poset, we distinguish two cases: consider the height ofthe elements in the Hasse diagram of P , where minima of P have height zero while

Bayon et al.

19

maxima have maximum height. Let x < y be a relation present in the subject but not
in P . If the height of x is less than the height of y, then the value of the penalty appliedis n. But if the height of x is at least the height of y then the value of the penalty
applied is 2n. Note that this test is elementary thanks to the canonical representationof the poset P given by the enumeration algorithm. Thus the fitness of any subject isn(n−1)+n×(number of relations of P ), with equality if and only if the subject
at most2
is a representation of P .
The following three steps form a so-called generation. A maximum number ofgenerations is fixed at the beginning. If the algorithm has not found a representation
after this number of generations, it stops.
1. Each subject, except the best two, undergoes local disturbances so as to improve
its fitness;
2. the fitness of each subject is re-evaluated; and3. worst subjects are replaced by new ones without any crossing process.
We have implemented two kinds of local disturbances: the algorithm can remove
an inclusion between two circles, or force an inclusion.
To remove the inclusion of C1 in C2, we move the circle that scores less points in thefitness evaluation process of the subject. The centre is translated in a random direction,with the minimum ratio that cancels the inclusion.
To force the inclusion of C1 in C2, again only the circle that scores less points inthe fitness evaluation process of the subject is changed, say C1. The forcing aims at
minimising the changes of position. The radius of C1 becomes r × (radius of C2 ) wherer is randomly chosen between 12 and 34 . Then the centre of C1 is translated in the
direction defined by the two centres, with a ratio that minimises the movement andachieves the inclusion.
Between two generations, some subjects of the population might undergo somemutations. The subjects are chosen randomly, and a mutation consists of forcing someneeded inclusion not achieved by the subject.
All the solutions returned by the representation search algorithm were validated bya formal certificate. This certificate computes and compares only integers, using the
GMP multiprecision library.
With this new method, we extended the previous result on orders with at most 7
elements to orders with up to 10 elements. We thus obtained the following theorem.
THEOREM 4. All orders with at most 10 elements are circle orders.
We underline here the importance of the local aspect of the disturbances applied
to the subjects. For instance, when moving a circle C so as to force or to remove an
inclusion, if we move accordingly all the circles that C contains, or that are contained
by C, then the algorithm is much less powerful.
As the Urrutia conjecture on 3-dimensional orders has been disproved by the resultof [5], it does not exist for the moment a global method able to directly prove SidneySidney-Urrutia conjecture. Thus our method, even if it is exhaustive, remains the only
one which can theoretically prove it. Furthermore, the representation search algorithm

20

Orders with Ten Elements

has always found a representation until now. A failure of the algorithm could exhibit
a potentially good candidate to disprove the conjecture. Hence our method can also
help in finding a counter-example, and by the way increase our understanding of circle
orders.

3

Numerical Tables

These tables show the number of orders that remain after the discrimination process.

Orders with 8 elements

relations
7
8
9
10
11
12
13

orders
3
13
28
59
82
115
121

relations
14
15
16
17
18
19
20

orders
123
96
78
38
21
4
2

Orders with 9 elements

relations
8
9
10
11
12
13
14
15
16
17
18

orders
11
43
136
303
593
932
1374
1721
2022
2111
2027

relations
19
20
21
22
23
24
25
26
27
28

Orders with 10 elements

orders
1735
1381
959
612
334
153
62
22
4
2

Bayon et al.

21
relations
9
10
11
12
13
14
15
16
17
18
19
20
21
22

orders
37
166
554
1483
3231
6157
10351
15795
22047
28517
34189
38273
39947
39111

relations
23
24
25
26
27
28
29
30
31
32
33
34
35
36

orders
35719
30697
24492
18349
12678
8212
4811
2642
1257
571
214
76
23
7

Acknowledgements. The authors would like to thank all the people who offeredcomputer cycles, especially Patrice Deloche and Pierre Hyvernat.

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[2] C. Chaunier and N. Lygerōs, The number of orders with thirteen elements, Order,
9(3)(1992), 203–204.
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[9] T. Hiraguchi, On the dimension of partially ordered sets, Sci. Rep. Kanazawa
Univ., 1(1951), 77–94.

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Orders with Ten Elements[10] G. Hulbert, A short proof that N3 is not a circle containment order, Order, 5(1988),
235–237.
[11] E. R. Scheinerman and J. C. Wierman, On circle containment orders, Order,
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[13] J. Urrutia, Partial orders and Euclidean geometry, In Algorithms and order (Ottawa, ON, 1987), pages 387–434. Kluwer Acad. Publ., Dordrecht, 1989.