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Let S' be a Steiner triple system on v' points with incidence matrix Ms. 9 = ( 0 0 0 0 0 0 0 0 0 . o 0 0 0 0 0 000 In this way, we get the incidence matrix of a group divisible design V. It can be completed to the incidence matrix of an ST S(v) by adjoining the v' columns ei = (000. : . ooq 111 pOO . : . 000/, i = 1, 2, ... , v'. 3(i -I) 3(v' -i) We denote the new Steiner triple system on v = 3v ' points by S and its incidence matrix by Ms. Denote the rows of X by gl, g2, g3, the rows of Y by g4, gs, g6, and the rows of Z by g7, gg, g9· The binary code generated by gl, g2, ...

4 Assume thatC is a binary code oflength N = v(v-1)/6andminimumdistance r = (v - 1)/2 (r odd) such that Ar = v and the next nonzero weight is 2r - 2. Then the minimum weight codewords form the (point by block) incidence matrix of an ST S( v). Proof Since the minimum weight is odd, the Hamming distance between any two codewords of minimum weight is at least 2r - 2, hence the scalar product (over the reals) of any two minimum weight codewords is at most 1. Consequently, the v codewords of minimum weight form a constant weight code C* of distance d = 28 ~ 2r - 2.

G, U} is in Figure 3. n o It is often useful to put numbers on the Hasse diagram to provide more information. One method is to put TIF beside each point F. Although there is the slight possibility of confusion caused by having two labels for each point, such a sized Hasse diagram conveys 49 ORTHOGONAL PARTITIONS IN DESIGNED EXPERIMENTS U A C AI\B E BI\C E Figure 3. Hasse diagram for a rectangle (Example 2) Figure 4. Indexed Hasse diagram for a cuboid (Example 3) all the essential information to the statistician.

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