Start with 3 against 3 (known shown with[]):
Scale even:
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1: H L L : H L L [L]
2: take left 3 coins; 1 against 1:
if even, then 2 L's, else heavy side = H and other side = L
3: repeat with right 3 coins.
So 3 weighs required
Scale not even:
==========
1: HLL : [LLL] or HHL : [LLL] (we know light side = 3 L's)
2: take the 3 coins from heavy side; weigh 1 against 1:
(if HLL) H : L or L : L
(if HHL) H : L or H : H
3:
(if even after 2):
weigh one against one of the known L's:
if even then L, other H; if not even, then H, other L
OR
(if not even after 2):
we know the ones weighed, since heavy side = H and other side = L;
we don't know if the coin not weighed is H or L:
weigh it against one of the known L's.
So 3 weighs does it again.
Answer: 3 operations.