Definition. Given a (v,k,λ)- BIBD, (X,B), a set of disjoint blocks of B which partition X is called a parallel class.

Transcription

1 Resolvable BIBDs

2 Definition Given a (v,k,λ)- BIBD, (X,B), a set of disjoint blocks of B which partition X is called a parallel class. A partition of B into parallel classes (there must be r of them) is called a resolution and a design which has at least one resolution is called a resolvable BIBD. Note that in order for a parallel class to exist, we must have that k v. [ v 0 mod k ] The corresponding restriction for a resolution is that b must be divisible by v/k. But since in a BIBD, bk = vr, we do not obtain another restriction.

3 Complete Graphs A (v,2,1)-bibd is a complete graph K v. (Note that K 2, while a complete graph, is not considered a BIBD since v = k and the design is not incomplete). In graph-theoretic terms, a parallel class is called a 1-factor (or complete matching) and the necessary condition for the existence of a 1-factor is that v is even. A resolution of such a design is called a 1-factorization.

4 K 6 A 1-factor of K 6 consists of 3 disjoint edges. In each of these two graphs, same colored edges form a 1- factor. A 1-factorization(resolution) of K 6.

5 1-factorizations of K 6 Theorem: There are six different 1-factorizations of K 6, any two isomorphic. Two disjoint 1-factors are contained in a unique 1-factorization. Proof: The union of two disjoint 1-factors is a graph, all of whose vertices have degree 2. Thus, it is the union of cycles which must be of even length since each cycle has the same number of edges from each 1-factor. However, the smallest possible such cycle has length 4 leaving the impossible cycle of length 2 as its complement, so this union must be a single 6-cycle (a hexagon). A 1-factor disjoint from these two either consists of the three long diagonals of the hexagon, or one long diagonal and the two perpendicular short ones. (in red in earlier example)

6 1-factorizations of K 6 Theorem: There are six different 1-factorizations of K 6, any two isomorphic. Two disjoint 1- factors are contained in a unique 1-factorization. Proof: (cont) Since there are three long and six short diagonals, the remaining 1-factors in a 1-factorization containing the first two must be of the second type and so are determined by the first two 1-factors. This proves the uniqueness of the 1-factorization (up to isomorphism). Now there are 15(8) = 120 choices of two disjoint 1- factors and any 1-factorization contains 5(4) = 20 such pairs, so there are 6 1-factorizations. Each of the 6 is isomorphic to the 1-factorization given earlier.

7 1-factorizations of K v Surprisingly, the necessary condition for the existence of a 1- factor in K v is also sufficient for the existence of a 1-factorization. Theorem: A (v,2,1)-bibd is resolvable iff v is even and v 4. Pf: By construction. If v is even, consider Z v-1 U { } where is a symbol not in Z v-1. With as the central vertex, construct: Symbolically, for each j we have edge j and then the edges joining j+i mod (v-1) and j i mod (v-1).

8 Back to K 6 While there are 6 vertices of K 6 and also 6 1-factorizations, there is no natural isomorphism between these two sets of objects. (I am suppressing the definition of a natural isomorphism.) However, if the 6 1-factorizations are considered as the vertices of a graph, and two vertices are joined if they share a common 1-factor (of K 6 ), then the graph so constructed is complete and there is a natural isomorphism between its 1-factorizations and the vertices of the original K 6. The abnormality here is due to the number 6 and this strangeness appears in other areas of mathematics.

9 Back to K 6-2 For instance, in group-theoretic language, the symmetric group of degree n, S n, has an outer automorphism if and only if n=6, and in category-theoretic language, the category whose objects are the n-element sets and whose morphisms are the bijections between them has a non-trivial functor to itself if and only if n=6. We shall use the 1-factors and 1-factorizations of K 6 to construct some other objects of combinatorial interest. The strangeness of 6 translates into the fact that the objects we construct will be richer (more complex) than they have any right to be.

10 The (21,5,1)- design Let A = K 6. Define a structure whose points are the vertices and 1-factors of A, and whose lines (blocks) are the edges and 1-factorizations. The incidence relation holds between a point and an edge containing it, between an edge and a 1-factor containing it, and between a 1-factor and a 1- factorization containing it. Now notice that two edges either intersect in a unique point or are contained in a unique 1- factor. Two 1-factorizations intersect in a unique 1-factor and an edge and a 1-factorization are incident with a unique 1-factor. Thus, every pair of ``lines'' intersect in a unique ``point''. Similarly, two vertices determine a unique edge, two 1-factors either have a unique edge in common or determine a unique 1-factorization, and an element and a 1-factor are incident with a unique edge. Hence, every pair of ``points'' are incident with a unique ``line''.

11 The Projective Plane of Order 4 As there are 15 1-factors, we have 15+6 = 21 points and since there are 15 edges and 6 1-factorizations we get the same number of blocks. A vertex is in 5 edges and no 1- factorizations, and a 1-factor is contained in 2 1- factorizations and contains 3 edges. So, every point is contained in 5 blocks. This proves that we have constructed a (21,5,1) symmetric BIBD, a projective plane of order 4. We can prove that this design is unique after we introduce one geometric concept.

12 Hyperovals In a projective plane of order n, a set of k points with no three on the same line, is called a k-arc. The largest possible size k of a k-arc is n+2. (As there are n+1 lines through a point, given a point on a k-arc, there are at most n+1 other points on the k-arc.) An (n+2)-arc is called an hyperoval. A more careful counting argument shows that a necessary condition for the existence of hyperovals is that n is even. While they are not guaranteed to exist in general, they do always exist in the projective planes PG(2,q), when q = 2 e.

13 Hyperovals -2 From the definition of a k-arc, it follows that any line can contain only 2, 1 or 0 points of the k-arc. Such lines are called secant, tangent or exterior respectively. Since a hyperoval contains n+2 points, there can exist no tangent lines and so there are ½(n 2 + 3n + 2) secant lines and ½(n 2 n) exterior lines to a hyperoval. The restriction that n 2 in the definition of a projective plane can be shown to be equivalent to the existence of a 4-arc in the projective plane. We can now construct a hyperoval in a projective plane of order 4.

14 Hyperovals - 3 An hyperoval in a projective plane of order 4 consists of 6 points, no three on a line. There are 21 points & lines in a projective plane of order 4, and 5 points/line and lines/point. Start with a 4-arc. This determines 6 secant lines. Each pair of these secants meet, 3 at a time at a 4-arc point and 2 at a time on 3 other points (not in the 4-arc). These 7 points (4-arc + 3) form a subplane (a projective plane of order 2) missing one line, which shows that the extra three points are on a line. This line contains two other points. These two points are the only ones that can be added to the 4-arc to extend it to a 6-arc.

15 Uniqueness Theorem: There is, up to isomorphism, a unique projective plane of order 4. It has the property that any four points, no three collinear, are contained in a unique hyperoval. Proof: We have already seen that any quadrangle (4 points no three collinear, a 4-arc) in a projective plane of order 4 determines a unique hyperoval. To prove this result we have to show that the structure of a projective plane of order 4 containing a hyperoval Ω (a 6-set) is uniquely determined. We shall consider the points of O as being the vertices of K 6, and freely pass between the graph and the geometry.

16 Uniqueness Theorem: There is, up to isomorphism, a unique projective plane of order 4. It has the property that any four points, no three collinear, are contained in a unique hyperoval. Proof (cont): There are 15 secant lines meeting Ω in two points; they can be labeled with the edges of Ω. Now a point outside of Ω lies on three secant lines meeting Ω; the corresponding edges form a 1-factor, which can be used to label the point. Finally, let l be an exterior line of Ω. Each secant line (edge) meets the exterior line in a point (1-factor), so the 5 points of l form a 1-factorization, which can be used to label l. The 6 exterior lines thus correspond to the 6 1-factorizations. We have now recovered the description of the projective plane via 1-factors and 1-factorizations, showing that all projective planes of order 4 are isomorphic to this model.

17 Moore Graphs The diameter of a graph Γ is the maximum distance between two vertices of Γ. The girth of a graph Γ is the number of vertices in the shortest cycle in Γ. A Moore graph is a connected graph with diameter d and girth 2d+1. It can be shown that Moore graphs must be regular. The unique Moore graph with diameter 1 is K 3. If d > 2 then a Moore graph is just the cycle C 2d+1. Thus, the interesting case for Moore graphs occurs when d = 2. A Moore graph of diameter 2 and girth 5 must have degree 2, 3, 7 or 57. For k = 2 the unique Moore graph is the pentagon (C 5 ) and for k = 3 it is the Petersen graph. The existence of the Moore graph with k = 57 is unknown. We will denote a Moore graph of diameter 2 and degree k by M(k), and shall now investigate M(7).

18 Moore Graphs - 2 Moore graphs are a type of strongly regular graphs which means in this case: Given any two adjacent vertices the number of vertices adjacent to both is constant. Since our Moore graphs have girth 5, there are no 3-cycles and this constant is 0. Given any two non-adjacent vertices the number of vertices adjacent to both is another constant. Since our Moore graphs have diameter 2 for a pair of non-adjacent vertices there must be vertices adjacent to both, and you can't have two since there are no 4-cycles, so the constant is 1. Recall that the (open) neighborhood of a vertex x, N(x), consists of all vertices adjacent to x (but not x).

19 Uniqueness of Peterson Graph Let {, 0} be an edge in the Moore graph Γ of diameter 2 and valency k, and let A = N( ) \{0}, X = N(0) \{ }. Any further vertex has a unique neighbor in each of the sets A and X, and so, can be labeled with with the pair (a,x) in A x X. Every pair in A x X labels a unique vertex. So the vertex set is {,0} U A U X U (A x X). We know all the edges except for those within A x X. Furthermore, given a 1,a 2 ԑ A and x 1 ԑ X, there is a unique common neighbor of the non-adjacent points (a 1,x 1 ) and a 2, that is, a unique x 2 ԑ X such that (a 1,x 1 ) is joined to (a 2,x 2 ). From this we can see the uniqueness of the Petersen graph; for, if A = {a,b} and X = {x,y}, then we must have edges {(a,x), (b,y)} and {(a,y), (b,x)}.

20 Hoffman-Singleton Graph We can also see the non-existence of a Moore graph of degree 4. Let A = {a,b,c} and X = {x,y,z}. If (a,x) is joined to (b,y), then both of these vertices must be joined to (c,z), creating a triangle. We now give the construction of the Hoffman-Singleton graph, the unique Moore graph of degree 7, M(7). Let A and X be dual 6-sets. Note that the edges of A correspond to 1-factors of X and dually. Take the vertex set {,0} U A U X U (A x X), with the edges already specified. In addition, join (a,x) to (b,y) if and only if {a,b} is an edge of the 1-factor corresponding to {x,y} (or dually). We obtain a graph of degree 7 on 50 vertices.

21 Hoffman-Singleton Graph To show that it is a Moore graph, it suffices to show that the girth is 5. We can easily see that any cycle of length less than 5 must be contained in A x X. The non-existence of triangles follows from the properties of 1-factors and 1- factorizations, and the non-existence of quadrangles by a similar argument. We can again prove uniqueness using the following lemma. Lemma: M(7) does not contain an induced subgraph isomorphic to M(3) with an edge deleted.

22 Uniqueness Theorem: There is, up to isomorphism, a unique Moore graph of diameter 2 and degree 7. Proof: Let Γ be such a graph, and write the vertex set as {,0} U A U X U (A x X), where A = X = 6. Suppose that there is an edge {(a,x),(b,y)}. Then the set {, 0,a,b,x,y,(a,x),(a,y),(b,x),(b,y)} contains all but one of the edges of M(3). By the Lemma, the remaining edge {(a,y),(b,x)}, must also be present. We write {a,b} ~ {x,y} if this occurs. Now, for any x,y ԑ X, there are three disjoint pairs of elements a,b ԑ A for which {a,b} ~ {x,y}, forming a 1-factor on A. The five 1-factors corresponding to the pairs {x,y} with x fixed are easily seen to form a 1-factorization. So, A and X are dual sets, and we recover our previous construction.

23 Affine Planes Another important class of resolvable designs are the affine planes (in fact these form the model for the resolvable concept). Recall that an affine plane of order n is an (n 2, n, 1)-BIBD. Lemma: Let (P, L) be an affine plane of order n. For any line (block) L and any point x not on L, there is a unique line M containing x which does not intersect L. Pf: For each point y on L, there is a unique line containing the pair {x,y}. This accounts for n of the n+1 (= r) lines through x, leaving exactly one line through x which does not meet L. This property is known as Playfair's Axiom in geometry.

24 Affine Planes - 2 In any affine plane define a binary relation ~ on the lines of the plane by: L ~ M if either L = M or L and M are disjoint. Lemma: In an affine plane ~ is an equivalence relation. Pf: This is straightforward. Lemma: The equivalence classes of ~ are parallel classes. Pf: Consider an equivalence class of ~. Any pair of distinct lines in this class are disjoint. Consider any line L of this class. For any point off L, there is a unique element of the class containing the point by Playfair's axiom. We get a partition of the points by the lines of the class.

25 Affine Planes - 3 Theorem: Any affine plane is a resolvable design. Pf: Every line of the affine plane is contained in an equivalence class of ~ which is a parallel class. Thus, these equivalence classes form a resolution. This property of affine planes makes it fairly simple to prove that every affine plane can be extended to a projective plane which has the original affine plane as a residual design.

26 Affine Planes - 4 Theorem: Every affine plane of order n can be extended to a projective plane of order n. This projective plane has the affine plane as a residual design. Pf: As the affine plane has n 2 points and n 2 +n lines, we must add n+1 new points and 1 new line to get the n 2 +n+1 points and lines of a projective plane. Note that a parallel class contains n lines and a resolution contains n+1 parallel classes. The n+1 new points will be symbols each identified with one of the parallel classes. These new points are added to each line of the parallel class with which they are identified. The one new line consists entirely of the new points. Each line now has n+1 points.

27 Affine Planes - 5 Theorem: Every affine plane of order n can be extended to a projective plane of order n. This projective plane has the affine plane as a residual design. Pf: (cont) We need only verify that each pair of points is contained in exactly one line. Two affine plane points are contained in a unique line of the affine plane. Two new points are contained in the unique new line. A new point and an affine plane point are both contained on the line of the parallel class associated to the new point that contains the affine point in the extended design. By removing the new line and all its points we obtain a residual design of the projective plane which is clearly the original affine plane.

28 AG(2,q) As we have seen, there is a special class of projective planes which are constructed from finite fields (PG(2,q)). The corresponding affine planes are denoted by AG(2,q), and we can give a direct construction of these from the finite fields. Construction: Let F = GF(q). A point is an ordered pair (x,y) ԑ F x F. The lines come in two types. Type 1 lines are of the form L c for each c in F where L c = {(c,y) y ԑ F}. The type 2 lines are of the form L a,b for each pair (a,b) ԑ F x F, defined by L a,b = {(x,y) y = ax + b}. We have constructed q 2 points and q 2 + q blocks (lines). Each line has q points.

29 AG(2,q) - 2 To show that we have constructed a (q 2,q,1) -BIBD, we need to show that each pair of points is contained in a unique line. Let (r,s) and (u,v) be two points. Consider the equations s = ar + b v = au + b Subtracting these we get: s v = a(r u). If r = u, then there is a solution for a only if s = v, which means the two points are the same. If r = u and the points are distinct, then the two points are only on the line L r. Now if r u, a and b are uniquely determined and the two points are on the line L a,b.

30 Isomorphism We now have two constructions of AG(2,q) and need to show that they produce isomorphic designs. Since the Singer automorphism of PG(2,q) is transitive on the lines, all the possible residual designs (obtained by choosing different lines to remove) are isomorphic. We may therefore take our favorite line (2-dim subspace) to remove. This will be <(1,0,0), (0,1,0)>. Note that the points on this line are the 1-dim subspaces with representatives (x,y,z) which have z = 0. The points of the residual design will have representatives of the form (x,y,z) with z 0. We may always take the representative of the form (x/z. y/z, 1) for these points.

31 Isomorphism We define the map between the two representations by (x,y) (x,y,1). This map is a bijection, so we need only show that it maps blocks to blocks to show that it is an isomorphism. <(0,1,0), (c,0,1)>\(0,1,0) = {(c,0,1) + y(0,1,0) = (c,y,1) y ԑ F} L c. <(1,a,0), (0,b,1)>\(1,a,0) = {(0,b,1) + x(1,a,0) = (x, ax+b,1)} L a,b. Therefore, blocks get mapped to blocks and the two representations are isomorphic.

32 Quasi-residual Designs A natural question to ask is: if a BIBD has the parameters of a residual design must it be a residual design? We have just seen an example where this is true, affine planes are residual projective planes, but are there other cases? A BIBD with r = k + λ is called a quasiresidual design. Note that this is sufficient for a design to have the right parameters to be a residual design. The parameters of a residual design would be (v-k, v-1, k, k-λ,λ) and if we let these numbers be (v', b', r', k', λ') then k' + λ' = k λ + λ = k = r'.

33 Quasi-residual Designs Prop: Every residual design is quasiresidual. Pf: Clear. Theorem: Any quasiresidual design with λ = 1 is residual. Pf: A quasiresidual design with λ = 1 is a (v,k,1) design with r = k + 1. But this implies that v-1 = (k+1)(k-1) so v = k 2 and the design is an affine plane, which is residual. Theorem: Any quasiresidual design with λ = 2 is residual. This is the Hall-Connor Theorem which is considerably more difficult to prove than the theorem above. Proof omitted.

34 * blocks have 4 elements in common, ** blocks disjoint if this were residual, after adding a new block and 3 new elements to each block, any pair of blocks would need to meet in 3 common elements. Quasi-residual Designs The following example due to Bhattacharya (1944) shows that the corresponding result for λ > 2 is much more complex. A (16, 24, 9, 6, 3) BIBD (quasi-residual): 0167DE 2467AC 1278CF 2478BD 0568BC* 1469CE 2369BF 235CDE 3468BE 1389AC 2569AD ** 0367AF 1379BD 4579EF 0579BC* 012ABE 1568DF 034CDF 145ABF 0289EF 3578AE 0489AD ABCDEF**

35 Bose's Inequality Theorem: If there exists a resolvable (v,b,r,k,λ)-bibd, then b v + r 1. Pf: As in the alternate proof of Fisher's Inequality, we set up the following notation. Let A be the incidence matrix of the BIBD. Let s j be the j th row of A T. Note that the s j 's are all vectors in R v and there are b of them. Let S be the span <s j 1 j b>, that is, all the linear combinations (over R) of the s j 's. Recall that it was shown that the standard basis vectors e i were linear combinations of the s j 's. Now we are given a resolvable BIBD. For each i, with 1 i r, we define:

36 Bose's Inequality - 2 Theorem: If there exists a resolvable (v,b,r,k,λ)-bibd, then b v + r 1. Pf: (cont) m i = i 1 v k 1 and n i = iv k. Label the blocks so that the r parallel classes are: i = {A j m i j n i }. Since these are parallel classes we have: n i j=m i s j = 1,1,,1 for 1 i r. So n s = 1 mi j=m 1 n s i s for 2 i r. j j j=m i 1

37 Bose's Inequality - 3 Theorem: If there exists a resolvable (v,b,r,k,λ)-bibd, then b v + r 1. Pf: (cont) That is to say, the r 1 vectors in the set S '={s m2,, s mr } can be expressed as linear combinations of the b - r + 1 vectors in S\S'. Now, since the b vectors in S span R v, it follows that the b-r+1 vectors of S\S' span R v. Since R v has dimension v we obtain b r + 1 v.

38 An alternative The following result gives an alternate version of Bose's inequality. Lemma: For a BIBD, b v + r 1 iff r k + λ. Pf: If b v + r 1 then b > v and so r > k. Then vr k v r 1 v r k v k r 1 r k k r k 1 r 1 k r 1 r k r k 1 r k r k k r 1 r k 1 r k r k 1 r k.

39 Affine Resolvable BIBD's A resolvable BIBD with b = v + r -1 (or equivalently, if r = k+ λ) is said to be an affine resolvable BIBD. Affine planes (and their relatives the affine spaces) are affine resolvable designs. This observation gives us: Corollary: Let q be a prime power and m 2. There exists an affine resolvable (q m, q m-1,λ)-bibd where λ = (q m-1-1)/(q-1). Notice that affine resolvable designs are quasiresidual.

40 Hadamard Matrices Theorem: If there exists an Hadamard matrix of order 4m, then there exists an affine resolvable (4m, 2m, 2m-1)- BIBD. Pf: From an Hadamard matrix of order 4m we can construct a (symmetric) (4m-1, 2m-1, m-1) BIBD, (X,A ) and it's block complement, a (4m-1, 2m, m)-bibd say (X, B). Let be a symbol not in X. Add this symbol to each block in A and form the design (X',C) with X' = X U and C = B U A ', where A ' = {A U A in A}. We now have v = 4m, b = 8m-2, r = 4m-1 and k = 2m. Given a pair of elements of X, they will be in m-1 blocks of A ' and m blocks of B. The symbol appears with any element z in 2m-1 blocks of A '.

41 Example Starting with an order 8 Hadamard matrix, we have constructed the (7,3,1) BIBD and its block complement a (7,4,2) BIBD. The (8,4,3) affine resolvable BIBD constructed by this technique would be: {, 1,4,5} {2,3,6,7} {, 2,5,6} {1,3,4,7} {, 3,6,7} {1,2,4,5} {, 1,4,7} {2,3,5,6} {, 1,2,5} {3,4,6,7} {, 2,3,6} {1,4,5,7} {, 3,4,7} {1,2,5,6}

42 Block Intersection Theorem: Any two blocks from different parallel classes of an affine resolvable (v,k,λ)-bibd intersect in exactly k 2 /v points. Proof omitted. Example: A resolvable (28,7,2)- BIBD would have r = 9 and b = 63. Since 9 = 7 + 2, a resolvable (28,7,2)-BIBD would be affine resolvable. But then, blocks from different parallel classes would meet in 49/28 points. However 49/28 = 7/4 is not an integer, so there can not exist a resolvable (28,7,2)- BIBD. Note: (28,7,2)-designs exist, but none are resolvable.

43 Symmetric BIBD's Theorem: If there exists an affine resolvable (v,b,r,k,λ) BIBD then there exists a (symmetric) ((r+1)v, kr, kλ)- BIBD. Pf: Let (X,A) be an affine resolvable (v,b,r,k,λ) BIBD with parallel classes Π 1,..., Π r and X = {x 1,...,x v }. Define the vxv matrices M h = (m h ), where ij mh = 1 if there is a block A in ij Π h such that x i and x j are in A, and 0 otherwise. Letting M 0 be the all 0 vxv matrix, the circulant matrix 0 M 1 M r M M = M 1 M 2 M 0 M r M 0 M r 1 is the incidence matrix of a symmetric BIBD.

44 Symmetric BIBD's Theorem: If there exists an affine resolvable (v,b,r,k,λ) BIBD then there exists a (symmetric) ((r+1)v, kr, kλ)- BIBD. Pf: (cont) We can see this as follows (let μ = k 2 /v): T M h M l = μj for 0 h l 0 and T M h M h = km h for 1 h r. Let N be the incidence matrix of (X,A), so we have: r h=1 r NN T = M h M h T =k h=1 r h=1 M h and therefore M h =k NN T =k r I J. Let L i be the i th row of M, 0 i r. We then have, L h L j T = (r-1)μj for h j and L h L h T = k(r-λ)i + kλj.

45 Symmetric BIBD's Theorem: If there exists an affine resolvable (v,b,r,k,λ) BIBD then there exists a (symmetric) ((r+1)v, kr, kλ)- BIBD. Pf: (cont) However, (r-1)μ = (r-1)k 2 /v = k[(r-1)k/v] = k[(r-1)k + r r]/v = k[r-k + rk -r]/v = k[λ + rk r]/v = k[λv]/v = kλ. Thus, MM T = k(r-λ)i + kλj. Constant row and column sums follow from the construction of M. So M is the incidence matrix of a symmetric ((r+1)v, kr, kλ)- BIBD.

46 Example Corollary. Let q be a prime power. There exists a (symmetric) (q 2 (q+2), q(q+1), q) BIBD. Pf: Use an affine plane of order q in the previous construction. Example: Starting with the affine plane of order 2 we obtain the parallel classes Π 1 = {12, 34}, Π 2 = {13, 24} and Π 3 = {14, 23}. We then have: M M M 1 = = =

47 Example (16, 6, 2) symmetric BIBD