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SubjectRe: [ANNOUNCE] udev 0.1 release
Hello, world!\n

> 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 <>
Faculty of Math and Physics, Charles University, Prague, Czech Rep., Earth
God is real, unless declared integer.
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