I think I may have an entry, although it looks ugly.
This room is a Hanoi-style room with Force Arrows, and has 145 orbs. It requires 2^145 - 1 orb hits to complete, or...
...44,601,490,397,061,246,283,071,436,545,296,723,011,960,831...
... which is 4.46 * 10^43 to 3 s.f.
However, the total number of moves needed is quite a bit larger.
Moves required...
...To get to the first orb = 315.
...To move to and hit all the orbs (ignoring all the side squares - imagine the orbs are in a very long row) = 2^147 + 2^146 - 298 = 2.68 * 10^44 moves to 3 s.f.
...For all rotations required (I don't know how to reduce this sequence) = ((2^126 - 1) + (2^108 - 1) + (2^90 - 1) + (2^72 - 1) + (2^54 - 1) + (2^36 - 1) + (2^18 - 1) * 8 = 6.81 * 10^38 moves to 3 s.f..
So, including assumptions, this room requires 2.67609623 * 10^44 moves to 9 s.f (This is what my calculator display can only display up to).
Including the bits that have been assumed non-existent, I think the total value comes up at around 268-270
quattuordecillion in American notation, or 268-270 septillion in British notation.
Of course, it or my calculations may be buggy, and it might not work.
[Edited by agaricus5 at
Local Time:09-26-2004 at 05:11 PM]
[Edited by agaricus5 at
Local Time:09-26-2004 at 05:11 PM]
[Edited by agaricus5 at
Local Time:09-26-2004 at 10:32 PM]
Edit - This hold has a larger solution than the original Eternity by Watcher, but there is a bug in it that removes about 2^144 of the orb hits required to finish the room. Get the fixed version below.
[Edited by agaricus5 at
Local Time:09-26-2004 at 10:49 PM]
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Resident Medic/Mycologist