Now attempting to work on the cases where the dimension D obeys D mod 4=2 or D mod 4=1. If D mod 4=2 and 2D-2 is the sum of two squares, e.g. D=2: 2D-2=2=1^1+1^1, D=6: 2D-2=10=3^2+1^2, D=10: 2D-2=18=3^2+3^2, D=14: 2D-2=26=5^2+1^2, D=18: 2D-2=34=5^2+3^2, but not D=22: 2D-2=42=you lose! it has been conjectured that there always exists a DxD sign matrix of the form W X Y Z where each block is (D/2)x(D/2) and each block has constant row sums (and constant column sums) and any two different rows of the whole matrix have either dot product = 0 or dot product = 2, and always 0 if the two rows being dotted lie in different halves. Any such configuration achieves maximum possible determinant ("Ehlich-Wojtas bound for Hadamard maximum determinant problem"). Now, any Ehlich-Wojtas configuration in D+1 dimensions (after "normalizing" so the first column is all -1s then omitting it) corresponds to a slightly-non-regular simplex inscribed in a D-cube, having only two edge lengths. If the larger edge lengths are shrunk to equal the smaller, then I presume the resulting shrunk regular simplex still is embedabble in the cube because it is embeddable in its unshrunk version? If so, the result would be CONJECTURE: If D mod 4=1 and 2D is a sum of two squares, then a regular simplex with edge length = sqrt(D/2) will fit inside a unit-side D-cube. EXAMPLE: D=5: edge length = sqrt(5/2) = 1.5811388 which would improve versus Lunnon's 1.511173070871355. And again this would be optimal to within a constant factor for all applicable D. The construction I gave earlier for D a multiple of 4, (using constant weight codes) also can be done when D mod 4 = 2 and shows that when D=6,10,14,18,22,26,30 edge length >= sqrt((D-2)/2) is achieved using a constant weight code with weight= 1, 3, 4, 6, 8, 9, 11 respectively [and Hamming distance = (D-2)/2]. It is not optimal when D=6, because Lunnon achieved the longer edge length sqrt(5/2). But anyway this all seems enough to make us believe the CONJECTURE: In D dimensions, it is always possible to inscribe a regular D-simplex in a D-cube of unit side, with D-simplex volume exceeding a positive constant times the "Hadamard" volume arising from simplex edge length = sqrt((D+1)/2). Another construction could be based on "bi-weight codes" which are N-bit words with A 1-bits in the first M bits, and B 1-bits in the last N-M bits, and then add an extra vertex or two of the form (x,x,x,x,...x,y,y,y,...,y). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)