I too am interested in Hadamard matrices especially since they can be constructed from symmetric balanced incomplete block designs. For instance the sbibd (31, 6,1) which is basically a finite geometry (n^2+n+1,n+1,1) where n is prime (5 in this case) can easily be constructed from the set of four mutual orthogonal latin squares of order 5. Once we have this design then we ltake the incidence matrix of this design and modify it a bit by replacing the zeros with negative ones and adding a new row 1 and column 1 consisting of all 1's. Now the other interesting fact is that the symmetric designs that I build have incidence matrices that are equal to their own transpose. That is to say they are self dual. Further to this Hadamards in standard form can be reverse engineered into sbibd's as you are likely aware. I have tried in vain to produce a Hadamard of order 668 by attempting to co construct a sbibd with parameters (667,333,166) without any success.This is what I call a "times two plus one design".Perhaps the Australian lady can assist you with a more appropriate answer. (Jennifer Salsberry I believe is her name). She has been at this for many years now.

Hi Rob: In my first answer I talked about constructing a symmetric block design with parameters (31,6,1) from a saet of 43 mols of order 5. What I should have said is to use the design (7,3,1) and then use the times two plus algoritm to get (15,7,3). Repaet the algoritm to get (31,15,7) . Then take the incidence matrix of (31,15,7) and c onvert the zeros to -1's and add a fist row and a first col of 1's only. When squaring thsi 32 by 32 matrix you will see 32's down the main diagonal and zero's elsewhere. Now I can use the squaring methaod since all of my designs are self dual. Sorry about the confusion in my first answer.

Jennifer Seberry is the name you're looking for in your first comment, Donald; she and I have published together; we go back a long way, to my Grad School days. I'll suggest that, unless you're using some extraordinary algebraic tools and intermediate objects, you are unlikely to find the SBIBD corresponding to the H668 directly. It is quite a bit more likely that you'll find the H668 first and derive the design from it. But the methods being applied to that problem today are all inadequate for it, with the possible exception of an approach that uses Turyn sequences -- but the needed ones in this case are still a bit big for our current computational power. I suspect there may be a "back door" approach that gets this one in a non-traditional way. I may try to guide one of my research students along this path this summer.

The 2x2 block structure is interesting because, if it is true, it is an elementary fact about substructure that does not appear to be directly deduced from the definition of Hadamard matrices in any obvious way. Yet it (the remaining part of the conjecture) seems so likely (in my view) that it would be a shock to find out it is false.

While the designs you describe satisfy both parts of the 2x2 conjecture, they do so for trivial reasons. Note that the (remaining part of the) conjecture does not say merely that you can always "get" a H4n with a decomposition into 2x2 blocks of rank 2. It says that EVERY H4n has such a decomposition -- or rather can be permuted into one such. This is not at all trivial. Consider that there are 60 equivalence classes of H24. So you might verify, for example, that the Paley class matrix satisfies the conjecture -- no surprise given it's elegant internal structure. But what about the other 59 classes, including many that are "sporadic" and some whose automorphism groups are essentially trivial? Yet they all satisfy the conjecture. What about the over 400 classes in order 28? You mention the (31,15,7) design obtained from two doublings of the (7,3,1) design corresponding to the Fano plane. But there are literally millions of classes of H32s. This construction clearly produces a matrix satisfying the conjecture -- but the vast majority of the classes do not arise in any analogous fashion. If they satisfy the conjecture, for what reason?

I always smile when I see design theorists use what you call the "times two plus algorithm" it is possibly the most awkward way to obtain what is an extremely simple construction for these designs. If H is the corresponding Hadamard matrix, the matrix corresponding to the "doubled" design is simply

H H

H -H

---that's it. The proof that it works is one short line of algebra. Design theory has its place, but the tendency of design theorists to do half a page of tedious combinatorial counting (as in the standard proof of Fisher's Inequality) that can be done by one or two lines of matrix arithmetic accomplishes the same goal completely transparently demonstrates an intrinsic weakness in the conventional approach to the subject. In my view the business of math is not only arriving at rigorous demonstrations but doing so in ways that provide clarity and insight. I am a combinatorist by trade, but I always prefer the proof that shows why so that a neophyte can grasp it rather than obtaining the result via a dark, uninformative path.

Rob: Have developed an algorithm to prodduce large designs from symmetric designs.Let the symmetric design D have the set of parameters (v,k,llamda)The the new set that I shown exists is (v,C(v,2)C(k,2),lamda,lamda--1). I can produce these frfom known symmetric designs. Also I can produce these from Steiner triple systems and room squares for unresolved deigns. Am hoping to publish once I have resolved the exustance of say (121,16,2) or (81,16,3).

Hi Donald. Do you mean (v, C(v,2),C(k,2),lambda, lambda-1)?

Good luck finding the biplane of order 14, though I agree that this is an interesting case. Computationally it may be as challenging as the resolution of the existence/nonexistence of a projective plane of order 10. I don't know the status of the other design you mention. Is it the first unresolved case with lambda=3?

Hi Rob: There are several cases with llamda equal to three that we are interested in. They are (81,16,3) ,(115,19,3),(155,22,3),(20128,3). The existance of these designs along with several dozen others still remain in question. Various other small oines have llamda values (4,5,6,9,10,12,15). I would like to communicate via e-mail. My address is [email protected] . Then I could reveal my approaches to resolving some of these mysteries. Talk to you again Rob.