 Solution

Read each Scala trait linearization from the most specific implementation toward the base. For a declaration with parents $P_1,\dots ,P_r$, concatenate their linearizations from right to left, deleting a duplicate from a left operand when it already occurs in the right operand. This can move a repeated ancestor later than a superficial “rightmost trait first” scan suggests. Omitting the common AnyRef, Any suffix, the answers areFor example, T contributes T,L,D,I,E, while O contributes O,Y,S,H,E. In A, the separately mentioned Y also appears in the later merge with O, and N's ancestors also occur in T; resolving these duplicates yields the final order displayed above. The inherited trait orders are the displayed lists after removing their leading class names. ^[ref]