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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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