AP plays a Chancellor of the Spires and chooses to cast Fated Infatuation from NAP's graveyard. AP intends to make ‘infinite Chancellor tokens’ and stack his entire deck using the scry 2 part of Fated Infatuation. NAP doesn't believe that the AP can actually stack the entire deck as it would be incredibly complicated and take way too long to actually do and calls over a judge. (It is possible, just really hard to do without first knowing exactly what you want in what order, and it would take a very long time.) What do we tell the players?

It is possible to do it within known finite number of iterations (less than
3600 iterations for sixty-cards deck), which is the requierement for loop.
I would ask the AP to explain the scrying method to NAP, so he can prove
that he knows how to do it. Then I would ask the NAP whether he wants to
respond at some point and then I would let AP rearrange his library however
he wants.

I agree with David that, assuming NAP has no responses and AP can explain how the process works, I would allow AP to shortcut the scry 2 triggers by just looking through his deck to sort it in the order he wants. I expect this to be done at a quick pace, though. The shortcut, in my mind, should not take much longer then the time that would be required for a basic search and thorough shuffle.

If AP makes an arbitrarily large number (infinite not allowed, he has to choose a number) of those tokens, then it's likely he only needs to stack a handful of cards. After all, his next attack phase should be rather difficult for NAP to survive. :)

That's good, because - as others have said - he needs to perform that stacking in a reasonable amount of time. While we don't have a set limit, I think anyone watching would get a bit impatient after just a few minutes, and soon it falls under General Unwanted Behaviors. (Intentionally wasting time in this manner is probably a Serious Problem.)

d:^D

… and miracle Terminus.

As for the actual mechanism though, I suspect I'm not the only one who doesn't get it (or I'm just having a case of the Mondays). Can anyone describe the “trick”?

Most simple Version:
Say you have 60 Cards left in your library. We can number those 1-60, where 1 is the one on top, and 60 the one on the bottom.
Now you make an order in which you want the cards, i1-i60, where each i specifies one card of those previously numbered.
You can now ‘hand down’ the card behind i60 by scrying through your library to the 60th slot, and leave the rest in the previous order - if i60 is among them, put the other card on the bottom and leave it top, otherwise, put both on bottom. When you reach card 1 again (now together with i60), but i60 on the bottom. Your deck now is order as before, but has i60 on the bottom instead of 60, and some pushed up 1 number.
We repeat this process now for i59, and so on, and so forth, until the library is stacked.

>
> As for the actual mechanism though, I suspect I'm not the only one who
> doesn't get it (or I'm just having a case of the Mondays). Can anyone
> describe the “trick”?


Assume 50 cards in the library at this point. We know it will probably be
less, but if we can do it for 50, it's easier to do it for less. For the
first 25 scry 2 uses, the player records what the cards are and puts them
on the bottom, not changing the order. Then the player, knowing the exact
contents of the library, can decide what order he wants it in. He does more
scry 2 actions, keeping a card on top only if it's going to be one of the
top 2 cards at the end.

When both of the top 2 cards are in the scry, he then puts both on the
bottom and searches for the next card. When he finds it, he then keeps it,
cycling through the remainder of the deck to get the initial two cards to
the top again, then throws them back on the bottom and adds the third card
to the chain. Repeat the process until all cards are sorted, and then do up
to 24 more to reset the deck to the desired order.

We know the maximum number of repetitions it takes to make this happen for
an N card library: N x N. And if you get it earlier, you can simply leave
the cards on top for the remaining iterations. So we have a definite number
of iterations and a definite end state.

Dominick Riesland, aka Rabbitball
Creator of the Cosmversal Grimoire
“As soon as men decide that all means are permitted to fight an evil, then
their good becomes indistinguishable from the evil that they set out to
destroy.”
- Christopher Dawson

Bubble Sort.

Basicaly, you scry through the library until you find card you want at the bottom of your library and you put it on the bottom. Then you scry again until you find a card you want to have second from bottom. Once you have it, you keep it on the top and scry away other cards util you have your 2 last cards on top. You put both of them on the bottom and then you scry for the card which shoud be third from the bottom. You keep it on the top until you get to the two other cards. Now you put all 3 of them on the bottom. And so on, until the library is sorted as you wish.

To simplify it, let's assume you have 6 cards in library named A,B,C,D,E,F in random order, which you want to sort alphabeticaly (A on top).

Here are the steps:

Top -> A C F B D E <- Bottom - Scrying until ‘F’ is found.
Top -> F B D E A C <- Bottom - ‘F’ is found -> let's put it on the bottom
Top -> B D E A C F <- Bottom - Searching for ‘E’
Top -> E A C F B D <- Bottom - ‘E’ is found -> let's keep it on top and scry away other cards until we found ‘F’ again
Top -> E C F B D A <- Bottom - keep scrying
Top -> E F B D A C <- Bottom - now we have last 2 cards. Let's put them both at the bottom
Top -> B D A C E F <- Bottom - ‘D’ is found -> let's keep it on top and scry away other cards until we found ‘E’ again
Top -> D A C E F B <- Bottom - keep scrying
Top -> D C E F B A <- Bottom - keep scrying
Top -> D E F B A C <- Bottom - ‘E’ is found -> let's put all three cards at the bottom
Top -> F B A C D E <- Bottom - keep putting
Top -> B A C D E F <- Bottom - searching for C
Top -> C D E F B A <- Bottom - ‘C’ is found -> let's keep it on top and scry away other cards until we found ‘D’ again. Oh, ‘D’ is found as well, so scry all four of them to the bottom.
Top -> E F B A C D <- Bottom - keep putting
Top -> B A C D E F <- Bottom - let's seach for ‘B’. Oh ‘B’ is found. -> let's keep it on top and scry away other cards until we found ‘C’ again
Top -> B C D E F A <- Bottom - ‘C’ is found -> scry all five cards to the bottom.
Top -> D E F A B C <- Bottom
Top -> F A B C D E <- Bottom
Top -> A B C D E F <- Bottom - Since 5 of 6 last cards are sorted on the bottom, the last cards has to be on the top.

OK, four (!) explanations are probably plenty… and we've more than answered the original question, too.

Time to shut this thread down, I think.