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.
HenrySeptember 24, 2023
Hi,
How can you be sure one of the last two is fertilized? Where is the proof?
Emil-KunkinOctober 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.
