## ‘Two-at-a-Time’ approach with >3 treat­ments [Design Issues]

Dear Helmut,

I vaguelly recall that for 4x4 William design, each row (sequence) and each column (period) should contain A, B, C, and D each exactly once. I think that there's a typo in the 3rd sequence of your first William design. Your 3rd and 4th column do not seem right. So 3rd period would have 2 Bs and 4th had 2 Ds. CABD should be CADB. So It would be ABCD/BDAC/CADB/DCBA. In such case, it would be balanced for pair-wise comparisons.

$$\small{\begin{array}{c|cccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV}\\\hline 1 & \text{A} & \text{B} & \text{C} & \text{D}\\ 2 & \text{B} & \text{D} & \text{A} & \text{C}\\ 3 & \text{C} & \text{A} & \text{D} & \text{B}\\ 4 & \text{D} & \text{C} & \text{B} & \text{A}\\ \end{array}{\color{Blue}\mapsto} \begin{array}{c|cccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV} \\\hline 1 & \text{A} & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{D}\\ 2 & {\color{Red}\bullet} & \text{D} & \text{A} & {\color{Red}\bullet}\\ 3 & {\color{Red}\bullet} & \text{A} & \text{D} & {\color{Red}\bullet}\\ 4 & \text{D} & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{A}\\ \end{array}{\color{Blue}\wedge} \begin{array}{c|cccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV}\\\hline 1 & {\color{Red}\bullet} & \text{B} & {\color{Red}\bullet} & \text{D}\\ 2 & \text{B} & \text{D} & {\color{Red}\bullet} & {\color{Red}\bullet}\\ 3 & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{D} & \text{B}\\ 4 & \text{D} & {\color{Red}\bullet} & \text{B} & {\color{Red}\bullet}\\ \end{array}{\color{Blue}\wedge} \begin{array}{c|cccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV}\\\hline 1 & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{C} & \text{D}\\ 2 & {\color{Red}\bullet} & \text{D} & {\color{Red}\bullet} & \text{C}\\ 3 & \text{C} & {\color{Red}\bullet} & \text{D} & {\color{Red}\bullet}\\ 4 & \text{D} & \text{C} & {\color{Red}\bullet} & {\color{Red}\bullet}\\ \end{array}}$$

All the best,
Shuanghe

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