I would probably set this up as a multilinear game, with the crops on the bottom, and the F/I on the top.
So know S > O > R, not FF, if P not 3 then P is F (or is P is I, then P is three), and if V > R then O is F.
Lets look at this question, where V is 2
_ _ _ _ _
_ V _ _ _
Since V must be before R, we know that O is fertilized. We also know there are now only twi things that could go first: P and S, so lets try out two scenarios with P and S.
F _ _ F _
P V S O R
We know that P is F, and that S O R are in order.
_ _ _ _ _
S V _ _ _
We know much less about this second scenario, but we know that if P is three, O is 4, and if O is three P is 5 or 4, so either way, one of the last two is fertilized.
Henryon September 24, 2023
Hi,
How can you be sure one of the last two is fertilized? Where is the proof?
Emil-Kunkinon October 4, 2023
Apologies if I wasn't clear earlier. There are two possibilities here, either s is third or s is first. The scenario where s is third is shown above, and in that case we know that O is fertilized in spot 4.
In the case that S is first all we know is as follows
_ _ _ _ _
S V _ _ _
We know in this case that O is fertilized, since V must be before R. So, if O isn't third, then O must be fertilized in one of the last two slots. However, what happens if we put O third?
_ _ F _ _
S V O_ _
Well we actually know that one of the last two will be P, and since that p will not be in third, that p must be fertilized.
So, no matter what we do, we must have a fertilized option in one of the two last spots.
