On the Lov asz number of Certain Circulant Graphs
On the Lovasz number of Certain Circulant Graphs Valentin Brimkov
Bruno Codenottiy Mauro Leoncini x September 10, 1999
Valentino Crespi z
Abstract
The theta function of a graph, also known as the Lovasz number, has the remarkable property of being computable in polynomial time, despite being \sandwiched" between two hard to compute integers, i.e., clique and chromatic number. Very little is known about the explicit value of the theta function for special classes of graphs. In this paper we provide the explicit formula for the Lovasz number of the union of two cycles, in two special cases, and a practically ecient algorithm, for the general case.
1 Introduction The notion of capacity of a graph was introduced by Shannon in [14], and after that labeled as Shannon capacity. This concept arises in connection with a graph representation for the problem of communicating messages in a zero-error channel. One considers a graph G, whose vertices are letters from a given alphabet, and where adjacency indicates that two letters can be confused. In this setting, the maximum number of one-letter messages that can be sent without danger of confusion is given by the independence number of G, here denoted by (G). If (Gk ) denotes the maximum number of k-letter messages that can be safely communicated, we see that Department of Mathematics, Eastern Mediterranean University, Famagusta, TRNC. y Istituto di Matematica Computazionale del CNR, Via S. Maria 46, 56126-Pisa (Italy).
e-mail: [email protected]. z Department of Mathematics, Eastern Mediterranean University, Famagusta, TRNC. x Facolta di Economia, Universita di Foggia, Via IV Novembre 1, 71100 Foggia, Italy, and IMC-CNR, Via S. Maria 46, 56126 Pisa, Italy.
1
(Gk ) (G)k . Moreover one can readily show that equality does not hold in general (see,qe.g., [11]). The Shannon capacity of G is the number (G) = limk!1 k (Gk ) ; which, by the previous observations, satis es (G) (G), where equality does not need to occur.
It was very early recognized that the determination of the Shannon capacity is a very dicult problem, even for small and simple graphs (see [8, 13]). In a famous paper of 1979, Lovasz introduced the theta function #(G), with the explicit goal of estimating (G) [11]. Shannon capacity and Lovasz theta function attracted a lot of interest in the scienti c community, because of the applications to communication issues, but also due to the connections with some central combinatorial and computational questions in graph theory, like computing the largest clique and nding the chromatic number of a graph (see [2, 3, 4, 6] for a sample of the wealth of di erent results and applications of #(G) and (G)). Despite a lot of work in the eld, nding the explicit value of the theta function for interesting special classes of graphs is still an open problem . In this paper we present some results on the theta function of circulant graphs, i.e., graphs which admit a circulant adjacency matrix. We recall that a circulant matrix is fully determined by its rst row, each other row being a cyclic shift of the previous one. Such graphs span a wide spectrum, whose extremes are the single cycle and the complete graph. We either give a formula or an algorithm for computing the Lovasz number of circulant graphs given by the union of two cycles. The algorithm is based on the computation of the intersection of halfplanes and (although its running time is O(n log n) in the worst case, as compared with the linear time achievable through linear programming) is very ecient in practice, since it exploits the particular geometric structure of the intersection.
2 Preliminaries There are several equivalent de nitions for the Lovasz theta function (see, e.g., the survey by Knuth [10]). We give here the one that comes out of Theorem 6 in [11], because it requires only little technical machinery. De nition 1 Let G be a graph and let A be the family of matrices A = (aij ) such that aij = 0 if i and j are adjacent in G. Also, let 1 (A) 2 (A) : : : n (A) denote the eigenvalues of A. Then 1 (A) #(G) = max 1 ? (A) : A2A n 2
p Combining the fact that (G) #(G) with the easy lower bound (C5 ) 5, Lovasz has been able to determine exactly the capacity of C5 , the penp
tagon, which turns out to be 5. For several families of simple graphs, the value of #(G) is given by explicit formulas. For instance, in the case of odd cycles of length n we have =n) : #(Cn) = 1n+cos( cos(=n)
We now sketch the proof of correctness of the above formula (see [10] for more details), because it will be instrumental to the more general results obtained in this paper. With reference to the de nition of the Lovasz number which resorts to the minimum of the largest eigenvalue over all feasible matrices (Section 6 in [10]), in the case of n-cycles, we have that a feasible matrix has ones everywhere, except on the superdiagonal, subdiagonal and the upper-right and lower-left corners, i.e. it can be written as C = J + xP + xP ?1 , where J is a matrix whose entries are all equal to one, and P is the permutation matrix taking j into (j + 1) mod n. It is well known and easy to see that the eigenvalues of C are n + 2x, and x(!j + !?j ), for j = 1; : : : ; n ? 1, where ! = e2i=n . The minimum over x of the maximum of these values is obtained when n + 2x = ?2x cos =n, which immediately leads to the above formula.
3 The function # of circulant graphs of degree 4
Let n be an odd integer and let j be such that 1 < j n?2 1 . Let C (n; j ) denote the circulant graph with vertex set f0; :::; n ? 1g and edge set ffi; i + 1 mod ng; fi; i + j mod ng; i = 0; :::; n ? 1g. By using the approach sketched in [10], we can easily obtain the following result.
Lemma 1 Let f0(x; y) = n2ij+ 2x + 2y and, for some xed value of j , fi (x; y) = 2x cos 2ni + 2y cos n , i = 1; :::; n ? 1. Then n ? 1 : (1) #(C (n; j )) = min max f ( x; y ) ; i = 0 ; 1 ; : : : ; i x;y i 2 Proof: Follows from the same arguments which lead to the known formula for the Lovasz number of odd cycles [10] (i.e., taking advantage of the fact that we can restrict the set of feasible matrices within the family of circulant matrices) and observing that, for i 1, fi (x; y) = fn?i(x; y). 3
3.1 A linear programming formulation
Throughout the rest of this paper we will consider the following linear programming formulation of (1).
minimize z (2) s.t. fi (x; y) ? z 0; i = 0; :::; n?2 1 z 0; where the fi(x; y)'s are de ned in Lemma 1. Consider the intersection C of the closed halfspaces de ned by z 0 and fi (x; y) ? z 0, i = 1; :::; n?2 1 (which is not empty, since any point (0; 0; k), k 0, satis es all the inequalities). C is a polyhedral cone with the apex at the origin. This follows from the two following facts, which can be easily veri ed: (1) the equations fi(x; y) ? z = 0, i 1, de ne hyperplanes through T the origin; (2) for any z0 > 0, the projection Qz0 of C fz = z0 g onto the xy plane is a polygon, i.e., Qz0 is bounded1 (see Figure 1, which corresponds to the graph C (13; 2)).
2 1.8 1.6 1.4 1.2
Z
1
0.8 0.6 1
0.4 0.2
0
Y
0 –1.4 –1.2 –1 –0.8 –0.6 –0.4 –0.2
0
X
0.2 0.4 0.6 0.8
1
–1 1.2 1.4
Figure 1: The polyedral cone for n = 13 and j = 2 cut at z = 2 Consider now the rst constraint of formulation (2). The region represented by such constraint is the halfspace above the plane A with equation 1
In the appendix we shall give a rigorous proof of this fact for the case j = 2.
4
n + 2x + 2y ? z = 0. It is then easy to see that the minimum z of (2) will correspond to the point P = (x; y; z) of C that is the last met by a sweeping line, parallel to the line y = ?x, which moves on the surface of A towards the negative ortant (we will simply refer to these as to the extremal vertices). In particular, x and y are the coordinates of the extremal vertex v of the convex polygon Qz in the third quadrant. The lines which de ne v have equations 2x cos + 2y cos( j ) = z and 2x cos + 2y cos( j ) = z, where = 2in 1 and = 2in 2 , for some indices i1 and i2 . The key property, which we will exploit both to determine a closed formula for the # (in the cases j = 2 and j = 3) and to implement an ecient algorithm for the general case of circulant graphs of degree 4, is that i1 and i2 can be computed using \any" projection polygon Qz0 , z0 > 0, and determining its extremal point in the third quadrant. Once i1 and i2 are known, z can be computed by solving the following linear system 8 > < > :
2x cos + 2y cos(j ) ? z = 0 2x cos + 2y cos(j ) ? z = 0 2x + 2y ? z = ?n
(3)
3.2 The special case j = 2
The detailed proof of the following theorem is deferred to the appendix.
Theorem 2
!
2 2 n (bn=3c + 1)) : (4) #(C (n; 2)) = n 1 ? ?2cos( n bn=3c) ? cos( 2 (cos( n bn=3c) ? 1)(cos( n (bn=3c + 1)) ? 1) 1 2
3.3 The special case j = 3
Consider again the projection polygon Qz0 , for some z0 > 0. We know from Section 3.1 that the value of # is the optimal value z of the objective function in the linear program (2), and that this value is achieved at the extremal vertex P of Qz in the third quadrant. Also, we know that any projection polygon Qz0 can be used to determine the two lines fi (x; y) ? z = 0, i 1, which form P . It turns out that nding such lines is easy when j = 3. In the following we will say that the line li has positive x-intercept (resp., y-intercept) if the intersection between li and the x-axis (resp., y-axis) has positive x-coordinate (resp., y-coordinate), otherwise we will say that the intercept is negative. The crucial observation is the following. Among the lines with negative x- and y-intercepts, lbn=2c is the one for which these 5
intersections are closest to the origin. It then follows that P must lay on this line and (after a moment of thought) that the second line forming P must be searched among those with positive slope. Let xi and yi denote the coordinates of the intersection between the line li and the line lbn=2c . Now, since lbn=2c is slightly steeper than the line with equation y = ?x, the line sought will be the one with positive x-intercept, negative y-intercept, and such that yi is maximum (see Figure 2). We shall prove that such line is the one with index n = d n?6 3 e.
Figure 2: Lines forming the extremal vertex P . To this end, observe rst that the requirement of positive x-intercept and negative y-intercept implies 12n < i < n4 (recall that li has equation cos 2ni z0 y = ? cos maximizes yi we show that, for 6i x + cos 6i ). To prove that ln n n n any integer i 6= n in the interval 12 < i < n4 , the three points vi = (xi ; yi ), vn = (xn ; yn ) and O = (0; 0) form a clockwise circuit. We already know (see the Appendix) that this amounts to proving that d(vi ; vn ; O) = xi yn ?yixn < 0. This is easy; the only formal diculty is working with the integer part of n?3 . Clearly this might be circumvent by dealing with the three di erent 6 cases, namely n = 6k +1, n = 6k +3, and n = 6k +5, for some positive integer 6
k. For simplicity we shall prove the rst case only. Now, for n = 6k + 1 we have d n?6 3 e = n?6 1 and, using z0 = 2, ( 3 ? 3n ) yn = cos?3coscosn(?cos ? )+cos2
cos n ?cos n xn = cos 3 cos ( ? )+cos2
n
3
3n
n
3
6i
n
n
n
3
3n
n
2i
n
n
n yi = cos coscos6in +cos ?cos 3 cos 2i
n +cos n xi = cos 3 cos cos 2i ?cos 6i cos n
3
n
n
n
After some relatively simple algebra we obtain + + ; d(vi ; vn ; O) = 3 ? cos n cos 3 ? 3n + cos2 n cos 3n cos 2ni ? cos n cos 6ni
, = cos( 2i ) cos ? cos 3 , and = 6i where = cos n cos n + cos n n n n ? 6 i 3 cos 3 ? 3n cos n + cos n , and it is easy to check that ; ; > 0 while the denominator of d(vi ; vn ; O) is negative for the admissible values of i . We are now able to determine the value of the # function of C (n; 3).
Theorem 3
#(C (n; 3)) = n 1 ?
! n?3 n?3 cos 2d 6 e +cos2 2d 6 e ?1 ? cos cos2 n n n n 3 n?3 2 d n? 6 e ?1)(1?cos +cos 2 d 6 e ) (cos +1)(cos n n n n
:
(5)
Proof: #(C (n; 3)) is the value of z2 in the solution2 to thelinear system (3) +cos cos +cos ?1 2i1 where j = 3, i.e., z = n 1 ? (1?cos cos )(cos ?1)(cos +cos +1) ; where = n and = 2in 2 . By the previous results we know that i1 = b n2 c and i2 = d n?6 3 e.
Plugging these values into the expression for z we get the desired result.
4 An ecient algorithm and computational results Although the Lovasz number can be computed in polynomial time, the available algorithms are far from simple and ecient (see, e.g., [1]). It is thus desirable to devise ecient algorithms tailored to the computation of # for special classes of graphs. By reduction to linear programming, the theta function of circulant graphs can be computed in linear time, provided that the number of cycles is independent of n. The corresponding algorithms are not necessarily ecient in practice, though. We brie y describe a practically ecient algorithm for computing #(C (n; j )), i.e., in case of two cycles. The algorithm rst determines the 2 lines forming the extremal vertex of Q1 in the third quadrant, then solves the resulting 3 3 linear system (i.e., 7
the system (3)). More precisely, the algorithm incrementally builds the intersection of the halfplanes which de ne Q1 (considering only the third quadrant) and keeps track of the extremal point. The running time is O(n log n) in the worst case (i.e., it does not improve upon the optimal algorithms for computing the intersection of n arbitrary halfplanes or, equivalently, the convex hull of n points in the plane). However, it does make use of the properties of the lines bounding the halfplanes to keep the number of vertices of the incremental intersection close to the minimum possible. In some cases (such as C (n; 2)) this is still (n), but for most values of n and j it turns out to be substantially smaller. Using the above algorithm we have performed some preliminary experiments to get insights about the behavior of the theta function for the special class of circulant graphs considered in this abstract. Actually, since the value sandwiched by the clique and the chromatic number of C (n; j ) is the theta function of C (n; j ) (i.e., the complementary graph of C (n; j )), the results refer to #(C (n; j )) = #(C (nn;j )) . Table 1 shows #(C (n; j )) approximated to the four decimal place, for a number of values of n and j . It is immediate to note that, for a xed value of j , the values of the theta function seem to slowly approach, as n grows, the lower bound (given by the clique number), which happens to be 2 almost always (obvious exceptions occur when 3 divides n and j = n3 . 51 101 201 301 401 501 1001 2001 3001 4001 5001 10001
4 2.2446 2.2383 2.2366 2.2363 2.2362 2.2362 2.2361 2.2361 2.2361 2.2361 2.2361 2.2361
5 2.0474 2.0121 2.0031 2.0014 2.0008 2.0005 2.0001 2 2 2 2 2
6 2.1227 2.1122 2.1103 2.1099 2.11 2.11 2.1099 2.1099 2.1099 2.1099 2.1099 2.1099
7 2.0838 2.0228 2.0059 2.0027 2.0015 2.001 2.0002 2.0001 2 2 2 2
b n4 c
2.1297 2.2383 2.2366 2.2363 2.2362 2.2362 2.2361 2.2361 2.2361 2.2361 2.2361 2.2361
d n4 e
2.2446 2.1162 2.1113 2.1099 2.1102 2.1102 2.1099 2.1099 2.1099 2.1099 2.1099 2.1099
b n3 c
3 2.2383 3 2.0005 2.2362 3 2.2361 3 2 2.2361 3 2.2361
b n3 c + 1 2.0173 2.0044 2.0011 2.2363 2.0003 2.0002 2 2 2.2361 2 2 2
n?3 2
2.2446 2.2383 2.2366 2.2363 2.2362 2.2362 2.2361 2.2361 2.2361 2.2361 2.2361 2.2361
Table 1: Some computed values of #(C (n; j )). This is con rmed by the results in Table 2, which depicts the behavior of the relative distance dnj of #(C (n; j )) from the clique number. We only consider odd values of j (so that the clique number is always 2); we also rule out the cases where j = n3 , for which we know there is a (relatively) large gap between clique number and theta function. More precisely, Table 2 shows: 8
(1) the maximum relative distance M = maxj;n dnj , where n rangesPover all odd integers from 9 to n ; (2) the average relative distance = N1n j;n dnj , where Nn is the number of admissible pairs (n; j ); (3) the average quadratic P distance = N1n j;n (dnj ? )2 . n
101 201 301 401 501 1001
M
0.372402 0.372402 0.372402 0.372402 0.372402 0.372402
0.056077 0.033712 0.024840 0.019897 0.016734 0.009657
0.004343 0.002600 0.001876 0.001471 0.001214 0.000653
Table 2: Relative distances of #(C (n; j )) from the clique number. The regularities presented by the value of the theta function and by the geometric structure of the optimal lines suggest the possibilities of further analytic investigations. Forpinstance, we have observed that, for j = 4, the formula i = b 2n arccos ?1?4 5 c seems to correctly predict the index of the rst optimal line, in perfect agreement with the experimental results. In general, for j even and j max sin (2x ?n 1) ; sin (2x +n 3)
16
n
= sin (2x ?n 1) ;
which implies that x^ cannot be larger than n3 . This fact allows us to conclude that the integer value which minimizes gn (x) (and hence z ) is one among b n3 c ? 1, b n3 c and d n3 e. We now prove that the value sought is b n3 c by showing that gn (b n3 c? 1) ? gn (b n3 c) and gn (d n3 e) ? gn (b n3 c) are both positive. For simplicity, we shall use the following notation: fb c?1 = cos 2(bn=n3c?1) , fb c = cos 2(bnn=3c) , fde = cos 2(dnn=3e) , and fd e+1 = cos 2(dn=n3e+1) . We have
f + f ? 1=2 f + f ? 1=2 gn(d n3 e) ? gn (b n3 c) = (f d e ? 1)(d ef+1 ? 1) ? (f b c ? 1)(def ? 1) de d e+1 bc de f f ? f + f f ? f ? 1f + 1 = d e b c (f de? 1)(dfe+1 ?b c1)(f d e+1 ? 1)2 b c 2 ? bc de d e+1 fb c fd e+1 ? fbc + fd e fd e+1 ? fde ? 21 fd e+1 + 12 (fb c ? 1)(fd e ? 1)(fd e+1 ? 1) (f ? f )( 21 + fde ) : = (f ?b c1)(f d e+1 bc d e ? 1)(fd e+1 ? 1) The last expression is positive since the denominator is negative, fb c ? fd e+1 > 0, and fde < ? 21 . Similarly, f + f ? 1=2 f + f ? 1=2 gn (b n3 c ? 1) ? gn (b n3 c) = (f bc?1 ? 1)(b cf ? 1) ? (f bc ? 1)(def ? 1) b c?1 bc bc de f f ? f 1 + fbc fde ? fb c ? 12 fd e + 12 ? = bc?1 de(f b c? b c?1 ? 1)(fb c ? 1)(fd e ? 1) fbc fbc?1 ? fb c + fd e fbc?1 ? fd e ? 12 fb c?1 + 12 (fb c?1 ? 1)(fb c ? 1)(fd e ? 1) (f e ? fb c?1 )( 21 + fbc ) = (f d ? ; b c?1 1)(fb c ? 1)(fd e ? 1)) and again the numerator is negative since fd e ? fb c?1 < 0 and fbc > ? 21 .
17
Bruno Codenottiy Mauro Leoncini x September 10, 1999
Valentino Crespi z
Abstract
The theta function of a graph, also known as the Lovasz number, has the remarkable property of being computable in polynomial time, despite being \sandwiched" between two hard to compute integers, i.e., clique and chromatic number. Very little is known about the explicit value of the theta function for special classes of graphs. In this paper we provide the explicit formula for the Lovasz number of the union of two cycles, in two special cases, and a practically ecient algorithm, for the general case.
1 Introduction The notion of capacity of a graph was introduced by Shannon in [14], and after that labeled as Shannon capacity. This concept arises in connection with a graph representation for the problem of communicating messages in a zero-error channel. One considers a graph G, whose vertices are letters from a given alphabet, and where adjacency indicates that two letters can be confused. In this setting, the maximum number of one-letter messages that can be sent without danger of confusion is given by the independence number of G, here denoted by (G). If (Gk ) denotes the maximum number of k-letter messages that can be safely communicated, we see that Department of Mathematics, Eastern Mediterranean University, Famagusta, TRNC. y Istituto di Matematica Computazionale del CNR, Via S. Maria 46, 56126-Pisa (Italy).
e-mail: [email protected]. z Department of Mathematics, Eastern Mediterranean University, Famagusta, TRNC. x Facolta di Economia, Universita di Foggia, Via IV Novembre 1, 71100 Foggia, Italy, and IMC-CNR, Via S. Maria 46, 56126 Pisa, Italy.
1
(Gk ) (G)k . Moreover one can readily show that equality does not hold in general (see,qe.g., [11]). The Shannon capacity of G is the number (G) = limk!1 k (Gk ) ; which, by the previous observations, satis es (G) (G), where equality does not need to occur.
It was very early recognized that the determination of the Shannon capacity is a very dicult problem, even for small and simple graphs (see [8, 13]). In a famous paper of 1979, Lovasz introduced the theta function #(G), with the explicit goal of estimating (G) [11]. Shannon capacity and Lovasz theta function attracted a lot of interest in the scienti c community, because of the applications to communication issues, but also due to the connections with some central combinatorial and computational questions in graph theory, like computing the largest clique and nding the chromatic number of a graph (see [2, 3, 4, 6] for a sample of the wealth of di erent results and applications of #(G) and (G)). Despite a lot of work in the eld, nding the explicit value of the theta function for interesting special classes of graphs is still an open problem . In this paper we present some results on the theta function of circulant graphs, i.e., graphs which admit a circulant adjacency matrix. We recall that a circulant matrix is fully determined by its rst row, each other row being a cyclic shift of the previous one. Such graphs span a wide spectrum, whose extremes are the single cycle and the complete graph. We either give a formula or an algorithm for computing the Lovasz number of circulant graphs given by the union of two cycles. The algorithm is based on the computation of the intersection of halfplanes and (although its running time is O(n log n) in the worst case, as compared with the linear time achievable through linear programming) is very ecient in practice, since it exploits the particular geometric structure of the intersection.
2 Preliminaries There are several equivalent de nitions for the Lovasz theta function (see, e.g., the survey by Knuth [10]). We give here the one that comes out of Theorem 6 in [11], because it requires only little technical machinery. De nition 1 Let G be a graph and let A be the family of matrices A = (aij ) such that aij = 0 if i and j are adjacent in G. Also, let 1 (A) 2 (A) : : : n (A) denote the eigenvalues of A. Then 1 (A) #(G) = max 1 ? (A) : A2A n 2
p Combining the fact that (G) #(G) with the easy lower bound (C5 ) 5, Lovasz has been able to determine exactly the capacity of C5 , the penp
tagon, which turns out to be 5. For several families of simple graphs, the value of #(G) is given by explicit formulas. For instance, in the case of odd cycles of length n we have =n) : #(Cn) = 1n+cos( cos(=n)
We now sketch the proof of correctness of the above formula (see [10] for more details), because it will be instrumental to the more general results obtained in this paper. With reference to the de nition of the Lovasz number which resorts to the minimum of the largest eigenvalue over all feasible matrices (Section 6 in [10]), in the case of n-cycles, we have that a feasible matrix has ones everywhere, except on the superdiagonal, subdiagonal and the upper-right and lower-left corners, i.e. it can be written as C = J + xP + xP ?1 , where J is a matrix whose entries are all equal to one, and P is the permutation matrix taking j into (j + 1) mod n. It is well known and easy to see that the eigenvalues of C are n + 2x, and x(!j + !?j ), for j = 1; : : : ; n ? 1, where ! = e2i=n . The minimum over x of the maximum of these values is obtained when n + 2x = ?2x cos =n, which immediately leads to the above formula.
3 The function # of circulant graphs of degree 4
Let n be an odd integer and let j be such that 1 < j n?2 1 . Let C (n; j ) denote the circulant graph with vertex set f0; :::; n ? 1g and edge set ffi; i + 1 mod ng; fi; i + j mod ng; i = 0; :::; n ? 1g. By using the approach sketched in [10], we can easily obtain the following result.
Lemma 1 Let f0(x; y) = n2ij+ 2x + 2y and, for some xed value of j , fi (x; y) = 2x cos 2ni + 2y cos n , i = 1; :::; n ? 1. Then n ? 1 : (1) #(C (n; j )) = min max f ( x; y ) ; i = 0 ; 1 ; : : : ; i x;y i 2 Proof: Follows from the same arguments which lead to the known formula for the Lovasz number of odd cycles [10] (i.e., taking advantage of the fact that we can restrict the set of feasible matrices within the family of circulant matrices) and observing that, for i 1, fi (x; y) = fn?i(x; y). 3
3.1 A linear programming formulation
Throughout the rest of this paper we will consider the following linear programming formulation of (1).
minimize z (2) s.t. fi (x; y) ? z 0; i = 0; :::; n?2 1 z 0; where the fi(x; y)'s are de ned in Lemma 1. Consider the intersection C of the closed halfspaces de ned by z 0 and fi (x; y) ? z 0, i = 1; :::; n?2 1 (which is not empty, since any point (0; 0; k), k 0, satis es all the inequalities). C is a polyhedral cone with the apex at the origin. This follows from the two following facts, which can be easily veri ed: (1) the equations fi(x; y) ? z = 0, i 1, de ne hyperplanes through T the origin; (2) for any z0 > 0, the projection Qz0 of C fz = z0 g onto the xy plane is a polygon, i.e., Qz0 is bounded1 (see Figure 1, which corresponds to the graph C (13; 2)).
2 1.8 1.6 1.4 1.2
Z
1
0.8 0.6 1
0.4 0.2
0
Y
0 –1.4 –1.2 –1 –0.8 –0.6 –0.4 –0.2
0
X
0.2 0.4 0.6 0.8
1
–1 1.2 1.4
Figure 1: The polyedral cone for n = 13 and j = 2 cut at z = 2 Consider now the rst constraint of formulation (2). The region represented by such constraint is the halfspace above the plane A with equation 1
In the appendix we shall give a rigorous proof of this fact for the case j = 2.
4
n + 2x + 2y ? z = 0. It is then easy to see that the minimum z of (2) will correspond to the point P = (x; y; z) of C that is the last met by a sweeping line, parallel to the line y = ?x, which moves on the surface of A towards the negative ortant (we will simply refer to these as to the extremal vertices). In particular, x and y are the coordinates of the extremal vertex v of the convex polygon Qz in the third quadrant. The lines which de ne v have equations 2x cos + 2y cos( j ) = z and 2x cos + 2y cos( j ) = z, where = 2in 1 and = 2in 2 , for some indices i1 and i2 . The key property, which we will exploit both to determine a closed formula for the # (in the cases j = 2 and j = 3) and to implement an ecient algorithm for the general case of circulant graphs of degree 4, is that i1 and i2 can be computed using \any" projection polygon Qz0 , z0 > 0, and determining its extremal point in the third quadrant. Once i1 and i2 are known, z can be computed by solving the following linear system 8 > < > :
2x cos + 2y cos(j ) ? z = 0 2x cos + 2y cos(j ) ? z = 0 2x + 2y ? z = ?n
(3)
3.2 The special case j = 2
The detailed proof of the following theorem is deferred to the appendix.
Theorem 2
!
2 2 n (bn=3c + 1)) : (4) #(C (n; 2)) = n 1 ? ?2cos( n bn=3c) ? cos( 2 (cos( n bn=3c) ? 1)(cos( n (bn=3c + 1)) ? 1) 1 2
3.3 The special case j = 3
Consider again the projection polygon Qz0 , for some z0 > 0. We know from Section 3.1 that the value of # is the optimal value z of the objective function in the linear program (2), and that this value is achieved at the extremal vertex P of Qz in the third quadrant. Also, we know that any projection polygon Qz0 can be used to determine the two lines fi (x; y) ? z = 0, i 1, which form P . It turns out that nding such lines is easy when j = 3. In the following we will say that the line li has positive x-intercept (resp., y-intercept) if the intersection between li and the x-axis (resp., y-axis) has positive x-coordinate (resp., y-coordinate), otherwise we will say that the intercept is negative. The crucial observation is the following. Among the lines with negative x- and y-intercepts, lbn=2c is the one for which these 5
intersections are closest to the origin. It then follows that P must lay on this line and (after a moment of thought) that the second line forming P must be searched among those with positive slope. Let xi and yi denote the coordinates of the intersection between the line li and the line lbn=2c . Now, since lbn=2c is slightly steeper than the line with equation y = ?x, the line sought will be the one with positive x-intercept, negative y-intercept, and such that yi is maximum (see Figure 2). We shall prove that such line is the one with index n = d n?6 3 e.
Figure 2: Lines forming the extremal vertex P . To this end, observe rst that the requirement of positive x-intercept and negative y-intercept implies 12n < i < n4 (recall that li has equation cos 2ni z0 y = ? cos maximizes yi we show that, for 6i x + cos 6i ). To prove that ln n n n any integer i 6= n in the interval 12 < i < n4 , the three points vi = (xi ; yi ), vn = (xn ; yn ) and O = (0; 0) form a clockwise circuit. We already know (see the Appendix) that this amounts to proving that d(vi ; vn ; O) = xi yn ?yixn < 0. This is easy; the only formal diculty is working with the integer part of n?3 . Clearly this might be circumvent by dealing with the three di erent 6 cases, namely n = 6k +1, n = 6k +3, and n = 6k +5, for some positive integer 6
k. For simplicity we shall prove the rst case only. Now, for n = 6k + 1 we have d n?6 3 e = n?6 1 and, using z0 = 2, ( 3 ? 3n ) yn = cos?3coscosn(?cos ? )+cos2
cos n ?cos n xn = cos 3 cos ( ? )+cos2
n
3
3n
n
3
6i
n
n
n
3
3n
n
2i
n
n
n yi = cos coscos6in +cos ?cos 3 cos 2i
n +cos n xi = cos 3 cos cos 2i ?cos 6i cos n
3
n
n
n
After some relatively simple algebra we obtain + + ; d(vi ; vn ; O) = 3 ? cos n cos 3 ? 3n + cos2 n cos 3n cos 2ni ? cos n cos 6ni
, = cos( 2i ) cos ? cos 3 , and = 6i where = cos n cos n + cos n n n n ? 6 i 3 cos 3 ? 3n cos n + cos n , and it is easy to check that ; ; > 0 while the denominator of d(vi ; vn ; O) is negative for the admissible values of i . We are now able to determine the value of the # function of C (n; 3).
Theorem 3
#(C (n; 3)) = n 1 ?
! n?3 n?3 cos 2d 6 e +cos2 2d 6 e ?1 ? cos cos2 n n n n 3 n?3 2 d n? 6 e ?1)(1?cos +cos 2 d 6 e ) (cos +1)(cos n n n n
:
(5)
Proof: #(C (n; 3)) is the value of z2 in the solution2 to thelinear system (3) +cos cos +cos ?1 2i1 where j = 3, i.e., z = n 1 ? (1?cos cos )(cos ?1)(cos +cos +1) ; where = n and = 2in 2 . By the previous results we know that i1 = b n2 c and i2 = d n?6 3 e.
Plugging these values into the expression for z we get the desired result.
4 An ecient algorithm and computational results Although the Lovasz number can be computed in polynomial time, the available algorithms are far from simple and ecient (see, e.g., [1]). It is thus desirable to devise ecient algorithms tailored to the computation of # for special classes of graphs. By reduction to linear programming, the theta function of circulant graphs can be computed in linear time, provided that the number of cycles is independent of n. The corresponding algorithms are not necessarily ecient in practice, though. We brie y describe a practically ecient algorithm for computing #(C (n; j )), i.e., in case of two cycles. The algorithm rst determines the 2 lines forming the extremal vertex of Q1 in the third quadrant, then solves the resulting 3 3 linear system (i.e., 7
the system (3)). More precisely, the algorithm incrementally builds the intersection of the halfplanes which de ne Q1 (considering only the third quadrant) and keeps track of the extremal point. The running time is O(n log n) in the worst case (i.e., it does not improve upon the optimal algorithms for computing the intersection of n arbitrary halfplanes or, equivalently, the convex hull of n points in the plane). However, it does make use of the properties of the lines bounding the halfplanes to keep the number of vertices of the incremental intersection close to the minimum possible. In some cases (such as C (n; 2)) this is still (n), but for most values of n and j it turns out to be substantially smaller. Using the above algorithm we have performed some preliminary experiments to get insights about the behavior of the theta function for the special class of circulant graphs considered in this abstract. Actually, since the value sandwiched by the clique and the chromatic number of C (n; j ) is the theta function of C (n; j ) (i.e., the complementary graph of C (n; j )), the results refer to #(C (n; j )) = #(C (nn;j )) . Table 1 shows #(C (n; j )) approximated to the four decimal place, for a number of values of n and j . It is immediate to note that, for a xed value of j , the values of the theta function seem to slowly approach, as n grows, the lower bound (given by the clique number), which happens to be 2 almost always (obvious exceptions occur when 3 divides n and j = n3 . 51 101 201 301 401 501 1001 2001 3001 4001 5001 10001
4 2.2446 2.2383 2.2366 2.2363 2.2362 2.2362 2.2361 2.2361 2.2361 2.2361 2.2361 2.2361
5 2.0474 2.0121 2.0031 2.0014 2.0008 2.0005 2.0001 2 2 2 2 2
6 2.1227 2.1122 2.1103 2.1099 2.11 2.11 2.1099 2.1099 2.1099 2.1099 2.1099 2.1099
7 2.0838 2.0228 2.0059 2.0027 2.0015 2.001 2.0002 2.0001 2 2 2 2
b n4 c
2.1297 2.2383 2.2366 2.2363 2.2362 2.2362 2.2361 2.2361 2.2361 2.2361 2.2361 2.2361
d n4 e
2.2446 2.1162 2.1113 2.1099 2.1102 2.1102 2.1099 2.1099 2.1099 2.1099 2.1099 2.1099
b n3 c
3 2.2383 3 2.0005 2.2362 3 2.2361 3 2 2.2361 3 2.2361
b n3 c + 1 2.0173 2.0044 2.0011 2.2363 2.0003 2.0002 2 2 2.2361 2 2 2
n?3 2
2.2446 2.2383 2.2366 2.2363 2.2362 2.2362 2.2361 2.2361 2.2361 2.2361 2.2361 2.2361
Table 1: Some computed values of #(C (n; j )). This is con rmed by the results in Table 2, which depicts the behavior of the relative distance dnj of #(C (n; j )) from the clique number. We only consider odd values of j (so that the clique number is always 2); we also rule out the cases where j = n3 , for which we know there is a (relatively) large gap between clique number and theta function. More precisely, Table 2 shows: 8
(1) the maximum relative distance M = maxj;n dnj , where n rangesPover all odd integers from 9 to n ; (2) the average relative distance = N1n j;n dnj , where Nn is the number of admissible pairs (n; j ); (3) the average quadratic P distance = N1n j;n (dnj ? )2 . n
101 201 301 401 501 1001
M
0.372402 0.372402 0.372402 0.372402 0.372402 0.372402
0.056077 0.033712 0.024840 0.019897 0.016734 0.009657
0.004343 0.002600 0.001876 0.001471 0.001214 0.000653
Table 2: Relative distances of #(C (n; j )) from the clique number. The regularities presented by the value of the theta function and by the geometric structure of the optimal lines suggest the possibilities of further analytic investigations. Forpinstance, we have observed that, for j = 4, the formula i = b 2n arccos ?1?4 5 c seems to correctly predict the index of the rst optimal line, in perfect agreement with the experimental results. In general, for j even and j max sin (2x ?n 1) ; sin (2x +n 3)
16
n
= sin (2x ?n 1) ;
which implies that x^ cannot be larger than n3 . This fact allows us to conclude that the integer value which minimizes gn (x) (and hence z ) is one among b n3 c ? 1, b n3 c and d n3 e. We now prove that the value sought is b n3 c by showing that gn (b n3 c? 1) ? gn (b n3 c) and gn (d n3 e) ? gn (b n3 c) are both positive. For simplicity, we shall use the following notation: fb c?1 = cos 2(bn=n3c?1) , fb c = cos 2(bnn=3c) , fde = cos 2(dnn=3e) , and fd e+1 = cos 2(dn=n3e+1) . We have
f + f ? 1=2 f + f ? 1=2 gn(d n3 e) ? gn (b n3 c) = (f d e ? 1)(d ef+1 ? 1) ? (f b c ? 1)(def ? 1) de d e+1 bc de f f ? f + f f ? f ? 1f + 1 = d e b c (f de? 1)(dfe+1 ?b c1)(f d e+1 ? 1)2 b c 2 ? bc de d e+1 fb c fd e+1 ? fbc + fd e fd e+1 ? fde ? 21 fd e+1 + 12 (fb c ? 1)(fd e ? 1)(fd e+1 ? 1) (f ? f )( 21 + fde ) : = (f ?b c1)(f d e+1 bc d e ? 1)(fd e+1 ? 1) The last expression is positive since the denominator is negative, fb c ? fd e+1 > 0, and fde < ? 21 . Similarly, f + f ? 1=2 f + f ? 1=2 gn (b n3 c ? 1) ? gn (b n3 c) = (f bc?1 ? 1)(b cf ? 1) ? (f bc ? 1)(def ? 1) b c?1 bc bc de f f ? f 1 + fbc fde ? fb c ? 12 fd e + 12 ? = bc?1 de(f b c? b c?1 ? 1)(fb c ? 1)(fd e ? 1) fbc fbc?1 ? fb c + fd e fbc?1 ? fd e ? 12 fb c?1 + 12 (fb c?1 ? 1)(fb c ? 1)(fd e ? 1) (f e ? fb c?1 )( 21 + fbc ) = (f d ? ; b c?1 1)(fb c ? 1)(fd e ? 1)) and again the numerator is negative since fd e ? fb c?1 < 0 and fbc > ? 21 .
17
