There's an off by one error :) Needs to be (amt+1) as f32
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Ahhh, yes you're right! It thought this was right: ceil(log16(17)) == 2 ceil(log16(16)) == 1 But turns out: hex(17) == 11 (len 2) hex(16) == 10 (len 2) hex(15) == F (len 1) Big oops; indeed needs + 1. Thanks for pointing out!
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Nice one!
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Now that I see it it seems really obvious that you can use a logarithm to do that
but before I saw it I would have never thought of that
nice one! - 1 more reply
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