Constructing t-designs with prescribed automorphism groups

Constructing t-designs with prescribed automorphism groups 2 Mario Osvin Pavčević University of Zagreb Croatia e-mail: mario.pavcevic@fer.hr /37

Dear organizers, thank you very much for the invitation! 2 2/37

Dear organizers, thank you very much for the invitation! 2 This talk emerged unter the influence of my colleagues to whom I owe big thanks and want to mention in particular my close collaborators from the University of Zagreb, Croatia: Vedran Krčadinac Ivica Martinjak Anamari Nakić as well as the colleagues from the University of, Germany: Reinhard Laue Axel Kohnert Alfred Wassermann 2/37

Definition. A t-(v, k, λ) design D is a pair (P, B), consisting of a v-element set of points P and a collection B of its k-element subsets called blocks, such that each t-element subset of P is contained in exactly λ blocks. Every t-design is also an s-design, for all s t, with parameters ) v s = v, k s = k, λ s = λ In particular, each point is contained in r = λ = λ ( v t ( k t ( v s t s ( k s t s blocks and since the empty set is contained in every block, the number of blocks equals to ( v t) ) ) b = λ = λ ( k. t) ) 2 3/37

Examples. 2-(7, 3, ) design P = {,, 2, 3, 4, 5, 6} B = {{, 2, 3}, {,, 4}, {, 2, 5}, {, 3, 6}, {, 5, 6}, {2, 4, 6}, {3, 4, 5}} 2 4/37

Examples. 2-(7, 3, ) design 2 P = {,, 2, 3, 4, 5, 6} B = {{, 2, 3}, {,, 4}, {, 2, 5}, {, 3, 6}, {, 5, 6}, {2, 4, 6}, {3, 4, 5}} 4/37 J I

3-(8, 4, ) design P = {,, 2, 3, 4, 5, 6, 7} B = {{, 2, 3, 7}, {,, 4, 7}, {, 2, 5, 7}, {, 3, 6, 7}, {, 5, 6, 7}, {2, 4, 6, 7}, {3, 4, 5, 7}, {, 4, 5, 6}, {2, 3, 5, 6}, {, 3, 4, 6}, {, 2, 4, 5}, {, 2, 3, 4}, {,, 3, 5}, {,, 2, 6}} 2 5/37

3-(8, 4, ) design P = {,, 2, 3, 4, 5, 6, 7} B = {{, 2, 3, 7}, {,, 4, 7}, {, 5, 6, 7}, {, 4, 5, 6}, {2, 3, 5, 6}, {, 2, 3, 4}, 2 {, 2, 5, 7}, {2, 4, 6, 7}, {, 3, 4, 6}, {,, 3, 5}, {, 3, 6, 7}, {3, 4, 5, 7}, {, 2, 4, 5}, {,, 2, 6}} 5/37 J I

Besides as a list of blocks, a t-design can be represented by a - matrix M = [m ij ] called incidence matrix. If we denote the points of D by P = {p, p 2,..., p v } and the blocks of D by B = {B, B 2,..., B v }, then the entries of the incidence matrix are defined by {, if pi B m ij = j, otherwise 2 6/37

Besides as a list of blocks, a t-design can be represented by a - matrix M = [mij ] called incidence matrix. If we denote the points of D by P = {p, p2,..., pv } and the blocks of D by B = {B, B2,..., Bv }, then the entries of the incidence matrix are defined by, if pi Bj mij =, otherwise 2 M = 6/37 J I

2 7/37 J I

For the sake of simplicity, rename the vertices of the cube: 2 8/37 J I

Goal: construct t-(v, k, λ) designs effectively 2 9/37

Goal: construct t-(v, k, λ) designs effectively 2 Reasons: existence question (hard) enumeration problem (hard) get an example with given parameters quickly (not so hard) 9/37

Goal: construct t-(v, k, λ) designs effectively 2 Reasons: existence question (hard) enumeration problem (hard) get an example with given parameters quickly (not so hard) We use the deterministic approach (generating all, classification), adding some constraints. In this way, the constructed structures will be even more regular. 9/37

What is a tactical decomposition? 2 /37

What is a tactical decomposition? 2 M = /37 J I

What is a tactical decomposition? 2 M = Suppose that there is a partition /37 P = P t P2 t t Pm and B = B t B2 t t Bn, so that every submatrix Mij of M, consisting of rows of Pi and columns of Bj (i =,..., m; j =,..., n) has a constant number of s in each row and column. We shall call such a decomposition of the incidence matrix a tactical decomposition. J I

Two tactical decompositions are trivial in case of a t-(v, k, λ) design: the whole M itself (m = n = ) m = v and n = b (each entry alone) So, we look for non-trivial tactical decompositions! 2 /37

Two tactical decompositions are trivial in case of a t-(v, k, λ) design: the whole M itself (m = n = ) m = v and n = b (each entry alone) So, we look for non-trivial tactical decompositions! 2 Denote by Further denote by ρ ij = the number of s in each row of M ij κ ij = the number of s in each column of M ij. p = {B B p B}, for any p P and B = {p P p B}, for any B B, then we can formulate the coefficients ρ ij and κ ij as ρ ij = p B j, p P i κ ij = B P i, B B j. /37

In one of our previous examples 2 2/37 J I

In one of our previous examples 2 these coefficients are: 2 2 [ρij ] = 2 2 [κij ] = 4 2 2 2 2 4 2 2 2 2 2/37 J I

Find necessary conditions which a TD of a t-(v, k, λ) has to fulfil! Count the total number of s in M ij in two ways. 2 3/37

Find necessary conditions which a TD of a t-(v, k, λ) has to fulfil! Count the total number of s in M ij in two ways. You get: P i ρ ij = B j κ ij. () 2 3/37

Find necessary conditions which a TD of a t-(v, k, λ) has to fulfil! Count the total number of s in M ij in two ways. You get: P i ρ ij = B j κ ij. () 2 3/37 Interprete it as a double counting of {(p, B) p P i, B B j, p B}.

Take any point p P i. Look at the following set of triples: {(p, q, B) q P l, p B, q B} 2 4/37

Take any point p P i. Look at the following set of triples: {(p, q, B) q P l, p B, q B} 2 A double counting gives here the following equation: n ρ ij κ lj = p q = I know that! q P l j= 4/37

The generalized version of this formula is n ρ ij κ l jκ l2 j κ ls j = q P l j= p q q 2 q s, (2) q 2 P l2 q s P ls 2 which one gets by taking a fixed point p P i and counting the set {(p, q, q 2,..., q s, B) q P l,..., q s P ls, p B, q B,..., q s B} in two different ways, for any appropriate integer s. 5/37

Are these equations all that I know about ρ ij and κ ij? 2 6/37

Are these equations all that I know about ρ ij and κ ij? Add the trivial conditions n ρ ij = r, i, and j= m κ ij = k, j. (3) i= 2 6/37

Are these equations all that I know about ρ ij and κ ij? Add the trivial conditions n ρ ij = r, i, and j= m κ ij = k, j. (3) i= 2 Together with (2), which can be rewritten via () only in terms of ρ ij, that s it! q P l P i ρ ij = B j κ ij n ρ ij κ l jκ l2 j κ ls j = j= p q q 2 q s q 2 P l2 q s P ls 6/37

Another example of a tactical decomposition of a 3-(8, 4, ) design: 2 7/37 J I

What is an automorphism of a design? 2 8/37

What is an automorphism of a design? A well-known notion: a permutation of points which preserves the blocks. 2 ϕ = (,, 2, 3)(4, 5, 6, 7) is an automorphism. hϕi is an automorphism group, a subgroup of the full automorphism group AutD. Point orbits P = {,, 2, 3} t {4, 5, 6, 7} partition P Block orbits B = {{,, 2, 3}} t {{4, 5, 6, 7}} t other 4 sides t {{, 2, 4, 6}, {, 3, 5, 7}} t other 4 diag. par. edges t 2 indep. sets partition B 8/37 J I

Proposition Rows and columns of an incidence matrix M of a t- (v, k, λ) design D corresponding to the point and block orbits obtained under an action of an automorphism group G AutD form a tactical decomposition. Remark Not every tactical decomposition of a t-(v, k, λ) design comes from an action of an automorphism group of it! Our next constructions shall prove it. There are (at least) two known different approaches how to construct designs with the additional constraint that an automorphism group acts on it: Finding all candidates for tactical decomposition matrices [ρ ij ] and blowing them up to incidence matrices. (in addition: using this method, find designs with tactical decompositions which are not orbits) Finding the Kramer-Mesner matrix and solving the linear system of equations. (in addition: improve the method by implementing the knowledge coming from the tactical decomposition matrices) 2 9/37

Blowing up TDM s and getting designs without any automorphisms 2 2/37

Blowing up TDM s and getting designs without any automorphisms General construction procedure. Prescribe a group G. 2. Prescribe its action on points and blocks. 3. Classify all tactical decomposition matrix candidates, e.g. matrices [ρ ij ] fulfilling the equations () (3). 4. Try to construct incidence matrices of the designs, consistent with achieved TD matrices, by forgetting the group action at this stage, expanding each entry of the TD-matrix to its full size in an incidence matrix. Since there is a computer program, written by V. Krčadinac, which solves the step 3 above quite general (for any t-design) we shall concentrate on step 4 - changing the coefficient ρ ij by a matrix with ρ ij s in each row and κ ij s in each column. 2 2/37

Assume now that a cyclic group of order p, p prime, acts on a design. Then P i, B j {, p} The only interesting (non-unique) case for step 4 is 2 P i = B j = p. 2/37

Assume now that a cyclic group of order p, p prime, acts on a design. Then P i, B j {, p} The only interesting (non-unique) case for step 4 is 2 P i = B j = p. Note that Alltop s lemma gives immediately ρ ij = κ ij. Hence, the problem in to replace the coefficient ρ ij by a p p matrix M ij posessing ρ ij s in each row and each column, taking care of the design properties. 2/37

If you want to forget the cyclic action, the number of possibilities becomes quite large very soon. 2 22/37

If you want to forget the cyclic action, the number of possibilities becomes quite large very soon. Throw a look at the table: 2 p ρ ij Cyc ij All ij 2 2 2 3 3 6 2 3 6 5 5 2 2 24 3 24 4 5 2 7 7 54 2 2 394 3 35 689388 4 35 689388 5 2 394 6 7 54 22/37

In case p = 2: can t avoid the cyclic action! 2 23/37

In case p = 2: can t avoid the cyclic action! In case p 5: the number of possibilities is quite large for an exhaustive search and comparison with the cyclic case. 2 23/37

In case p = 2: can t avoid the cyclic action! In case p 5: the number of possibilities is quite large for an exhaustive search and comparison with the cyclic case. So, the convenient and interesting case is p = 3. Here, the number of possibilites in non-unique cases have doubled, and we can speak of the cyclic and anti-cyclic matrices of order 3 as replacements for ρ ij, if ρ ij {, 2}, e.g. if ρ ij =, then cyclic possibilities:, anti-cyclic possibilities:,,, 2 23/37

Results for symmetric (36, 5, 6) designs The number of fixed points (and blocks) F {, 3, 6, 9}. 2 F TDM s 9 2 6 4 3 334 84 24/37

Results for symmetric (36, 5, 6) designs The number of fixed points (and blocks) F {, 3, 6, 9}. 2 F TDM s 9 2 6 4 3 334 84 An exhaustive search of step 4 in the cyclic case gives F TDM s niso 9 2 99 6 4 2368 3 334 79662 84 5872 24/37

Theorem There are exactly 46 symmetric designs with parameters (36, 5, 6) admitting an action of an automorphism of order 3. 2 25/37

Theorem There are exactly 46 symmetric designs with parameters (36, 5, 6) admitting an action of an automorphism of order 3. 2 Here is a complete list of the automorphism group orders and their frequencies among all constructed designs. Aut(D) 3 6 9 2 8 2 24 27 3 niso 36733 364 629 275 73 2 45 25 2 Aut(D) 36 42 48 54 72 8 96 8 44 niso 33 2 8 2 5 5 Aut(D) 62 26 24 243 324 36 384 432 486 niso 6 2 3 2 2 2 Aut(D) 648 52 944 3888 296 584 niso 2 25/37

A (not quite exhaustive) search of step 4 in the cyclic + anti-cyclic case gives F niso Cyc niso All 9 99 876 6 2368 885 3 79662 3849 5872 59836 all 46 66587 This search was not exhaustive in case F = ; the number of structures is quite large and we had to stop it. Altogether, the following statement concludes this search: Proposition There are at least 675363 symmetric designs with parameters (36, 5, 6) and 53692 of them don t admit any non-trivial automorphisms. 2 26/37

Results for symmetric (4, 6, 6) designs The number of fixed points (and blocks) F {5, }. 2 27/37

Results for symmetric (4, 6, 6) designs The number of fixed points (and blocks) F {5, }. 2 Outcome of step 3: F TDM s 5 834 27/37

Results for symmetric (4, 6, 6) designs The number of fixed points (and blocks) F {5, }. Outcome of step 3: F TDM s 5 834 An exhaustive search of step 4 in the cyclic case gives F TDM s niso 376 5 834 34258 Theorem There are exactly 345584 symmetric designs with parameters (4, 6, 6) admitting an action of an automorphism of order 3. 2 27/37

The following table gives a concise overview of the described results for the cyclic case. Aut(D) F = F = 5 all 3 2976 34224 34527 6 94 225 39 9 42 42 5 4 4 3 2 2 376 34258 345584 2 28/37

The following table gives a concise overview of the described results for the cyclic case. Aut(D) F = F = 5 all 3 2976 34224 34527 6 94 225 39 9 42 42 5 4 4 3 2 2 376 34258 345584 2 An exhaustive search in step 4 in the cyclic + anti-cyclic case gives F niso Cyc niso All 376 988 5 34258 43276 all 345584 4448 28/37

Altogether, the following statement concludes this search: Proposition There are at least 4448 symmetric designs with parameters (4, 6, 6) and 959 of them don t admit any non-trivial automorphisms. We conclude this search with an informative table on the number of non-isomorphic (4, 6, 6) designs admitting only a tactical decomposition with partition sizes and 3 and their automorphism group orders. Aut(D) F = F = 5 all 674 88423 959 2 345 345 3 2994 34224 34527 6 94 225 39 9 42 42 5 4 4 3 2 2 988 43276 4448 2 29/37

Kramer-Mesner approach improved by TDM knowledge 2 3/37

Kramer-Mesner approach improved by TDM knowledge 2 G = group of permutations of {, 2,..., v} 3/37

Kramer-Mesner approach improved by TDM knowledge 2 G = group of permutations of {, 2,..., v} T, T 2,..., T m orbits on t-subsets of {,..., v} K, K 2,..., K n orbits on k-subsets of {,..., v} 3/37

Kramer-Mesner approach improved by TDM knowledge 2 G = group of permutations of {, 2,..., v} T, T 2,..., T m orbits on t-subsets of {,..., v} K, K 2,..., K n orbits on k-subsets of {,..., v} a ij = { K K j T K }, T T i (does not depend on choice!) A G tk = [a ij ] Kramer-Mesner matrix Theorem. A simple t-(v, k, λ) design with G as a group of automorphisms exists if and only if the system of linear equations A G tk x = λj has a {, }-solution. The main problem when applying: the size of this linear system! 3/37

Main idea: Incorporate the knowledge that gives you the tactical decomposition matrix when building the Kramer-Mesner matrix, to reduce the number of columns! 2 3/37

Main idea: Incorporate the knowledge that gives you the tactical decomposition matrix when building the Kramer-Mesner matrix, to reduce the number of columns! Example Have a look at the parameters 2-(7, 3, ) 2 Define the point set to be P = {,, 2, 3, 4, 5, 6} If we assume G to be trivial, its all orbits are of cardinality. Therefore, there are ( ( 7 2) orbits on the 2-element subsets of P and 7 ) 3 orbits on the 3-element subsets of P. We need to solve the system A 23 x = λj. KM-attempt: solve this 2 35 system! 3/37

What improvement gives our idea? 2 32/37

What improvement gives our idea? 3 3 3 3 3 2 A G 3 : 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 3 2 2 2 2 2 32/37

What improvement gives our idea? 3 3 3 3 3 2 A G 3 : 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 3 2 2 2 2 2 Note: only those k-orbits with columns in A G k identical to a column of a TDM can be taken into further consideration - we have an additional necessary condition! Hence, eliminate the other k-orbits. 32/37 Plain KM A G 23 7 3 KM + TD A G 23 7 7 + 3 extra equations

Example Parameters 3-(8, 4, ) P = {,, 2, 3, 4, 5, 6, 7} G = h(,, 2, 3)(4, 5, 6, 7)} 2 33/37 J I

Example Parameters 3-(8, 4, ) P = {,, 2, 3, 4, 5, 6, 7} G = h(,, 2, 3)(4, 5, 6, 7)} Plain KM: a 4 2 linear system. 2 33/37 J I

Example Parameters 3-(8, 4, ) P = {,, 2, 3, 4, 5, 6, 7} G = h(,, 2, 3)(4, 5, 6, 7)} 2 Plain KM: a 4 2 linear system. KM + TD: the action of G on blocks is not given, hence you have to check all possibilities. Only two lead to TD matrices: Block orbits [2, 4, 4, 4] The system of size 4 8 + 4 additional equations Solution exists! (Which one is that?) Block orbits [,, 2, 2, 4, 4] The system of size 4 2 + 4 additional equations Solution exists! 33/37 J I

Example Parameters 2-(28, 4, ) P = {,, 2,..., 27} G = ρ = (,, 2, 3, 4, 5, 6)(7, 8, 9,,, 2, 3) (4, 5, 6, 7, 8, 9, 2)(2, 22, 23, 24, 25, 26, 27), σ = (, 6)(2, 5)(3, 4)(8, 3)(9, 2)(, ) (5, 2)(6, 9)(7, 8)(22, 27)(23, 26)(24, 25) So, I have chosen the dihedral group of order 4 to act on points in [7, 7, 7, 7]. Plain KM: a 36 533 linear system. KM + TD: the action of G on blocks can be different, but only in orbits of sizes 7 and 4 TD matrices exist only in 3 cases: Block orbits: [7, 7, 7, 7, 7, 4, 4] Number of TDM s: The system contradictory 2 34/37

Block orbits: [7, 7, 7, 7, 7, 7, 7, 4] Number of TDM s: 6 System sizes:,,,,, 44 97 Solution exists! Block orbits: [7, 7, 7, 7, 7, 7, 7, 7, 7] Number of TDM s: 7 System sizes:,,, 45 36, 44 34, 45 36, 43 37 Solution exists for TDM s no. 5 and 7! Comparison: Plain KM: linear system of size 36 533. KM + TD: 5 linear systems of sizes: 44 97, 45 36, 44 34, 45 36, 43 37. 2 35/37

Example Parameters 2-(65, 5, ) Choice of an automorphism group: the non-abelian group of order 39, acting on points in orbits of length [3, 3, 3, 3, 3]. 2 36/37

Example Parameters 2-(65, 5, ) Choice of an automorphism group: the non-abelian group of order 39, acting on points in orbits of length [3, 3, 3, 3, 3]. 2 Plain KM: I couldn t compute the KM-matrix in reasonable time! (Probably my fault, as it usually is the case...) KM + TD: the action of G on blocks can be different, but only in orbits of sizes 3 and 39. Here we shall take the usually hardest case - as many long orbits as possible. Take the block orbits to be [3, 39, 39, 39, 39, 39]. Number of TDM s: System size: 66 35. (How doable!?!) 36/37

Final comments 2 We use GAP as the background for all the computations. Hence, we have to add the limitations of GAP to our own limitations. 37/37

Final comments 2 We use GAP as the background for all the computations. Hence, we have to add the limitations of GAP to our own limitations. Even with limitations, I hope to be able to run something for you, if you give me a limited problem, at any time of this conference, or after it. 37/37