eric76 wrote:
It's obvious that you can't solve for N wizards.
We had N/2 for an even number of wizards earlier and N-3 above.
N-2 or N-1 is open for grabs.
For what it's worth, it is likely not even possible to guarantee N-2 promotions. A proof showing that N-2 is not possible would be quite interesting.
So can anyone show that N-2 can be guaranteed or prove that N-2 cannot be guaranteed?
18 Jul 2013, 10:13 am
