# Table 1 Presentation of examples of diagnostic assessment on Criterion A

Symptom number 1 2 3 4 5 6 7 8 9   Episode
row 1 1 1 1 0 1 1 0 1 1 (mod 2) 1
row 2 0 1 0 0 1 0 0 1 1 (mod 2) 0
row 3 0 0 0 1 0 0 0 1 0 (mod 2) 0
row 4 1 1 0 1 1 0 1 0 1 (mod 2) 1
row 5 0 1 1 1 0 0 0 0 1 (mod 2) 0
row 6 1 1 1 0 1 1 0 0 1 (mod 2) 1
row 7 (= row 4) 1 1 0 1 1 0 1 0 1 (mod 2) 1
row 8 (= row 2) 0 1 0 0 1 0 0 1 1 (mod 2) 1
row 9 (= row 5) 0 1 1 1 0 0 0 0 1 (mod 2) 1
row 10 (= row 4) 1 1 0 1 1 0 1 0 1 (mod 2) 1
Total sum 5 9 4 6 7 2 3 4 9   5
Average 0.5 0.9 0.4 0.6 0.7 0.2 0.3 0.4 0.9   0.5
1. Examples for symptoms 1–9 in Criterion A having values of 0 or 1 are shown. Each row is an assessment during a session. Rows 3 and 5 are equivalent to row 1; i.e., row = row 3 = row 5. Additionally, row 4 = row 6 = row 7 = row 9 = row 10, and row 2 = row 8. The expression of these examples can be simplified as in Table 2. In this case, the order of ‘which items should be effective on the scale’, is Aall(1–9) = [11|12|13|14|15|16|17|18|19||0 10 |0 11 |0 12 |…] (mod 2); all symptoms 1–9 in Criterion A should be effective, and this could be reinterpreted as the result of the operation (selection for effectiveness) Aall(1–9) (= A(0→all(1–9))) acting on the identity order A0 = [01|02|03|04|05|06|07|08|09||0 10 |0 11 |0 12 |…] (mod 2); i.e., A0 * A(0→all(1–9)) = Aall(1–9). A0 could be also regarded as an undiagnosed state. The rows whose components are equivalent to each other are compressed in the earliest rows of Table 2 and are highlighted silver in Table 1. Additionally, the diagnosis is given in the extreme right column; rows 1, 4, 6, 7 and 10 meet Criterion A of a ‘major depressive episode’ and have a diagnosis value of 1 (whereas rows not meeting Criterion A have a value of 0)