> Someone PLEASE tell me the simpler way to do this.
If you have an arbitrary system of splitters and you hook another 3-way
splitter to any of its outputs, you lose one output and gain 3 new
outputs, so the total number of outputs increases by 2. Hence if you
take K inputs and N 3-way splitters, the network has K+2N outputs,
no matter how the splitters are connected (of course unless you create
a cycle :-) ).
So in our case, we are searching for the smallest possible N such that
5+2N >= 4000, which equals ceil((4000-5)/2) = 1998.
Have a nice fortnight
-- Martin `MJ' Mares <mj@ucw.cz> http://atrey.karlin.mff.cuni.cz/~mj/ Faculty of Math and Physics, Charles University, Prague, Czech Rep., Earth God is real, unless declared integer. - To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/