Well if the counter doesn’t reset (because the genie exists outside of time and therefore grants all the wishes “simultaneously” from its own perspective) we definitely get a problem, because granting 3 makes it impossible for 3 to be granted, and we get the paradox implied by the comic
Do the opposite of next
Do not grant 3
Ignore 1
If you grant 1 and 2, then you cannot grant 3 (since 3 implies not granting 1). If you grant 3, then 2 cannot be granted (since it implies not granting 3). This is the simple form of the paradox.
But you said the counter didn’t reset? If it grants 2 in the “second loop”, that implies 1 was granted (since we didn’t invert 2), but you can’t grant 2 (uninverted) and also grant 1.
If you’re operating with a time-loop recursion, you run into the problem of my initial comment. If you try to grant all three wishes simultaneously, you run into the obvious contradiction. The only way you get out is if you allow a time-loop recursion, but for some reason count the ignored guess as a granted guess in the inner loop(s).
Well if the counter doesn’t reset (because the genie exists outside of time and therefore grants all the wishes “simultaneously” from its own perspective) we definitely get a problem, because granting 3 makes it impossible for 3 to be granted, and we get the paradox implied by the comic
If you grant 1 and 2, then you cannot grant 3 (since 3 implies not granting 1). If you grant 3, then 2 cannot be granted (since it implies not granting 3). This is the simple form of the paradox.
Correct. The genie grants 1, 2, and 2 in the second loop.
But you said the counter didn’t reset? If it grants 2 in the “second loop”, that implies 1 was granted (since we didn’t invert 2), but you can’t grant 2 (uninverted) and also grant 1.
If you’re operating with a time-loop recursion, you run into the problem of my initial comment. If you try to grant all three wishes simultaneously, you run into the obvious contradiction. The only way you get out is if you allow a time-loop recursion, but for some reason count the ignored guess as a granted guess in the inner loop(s).