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I think @Murch has been the only other one who has engaged

It's actually not hard, but just too notation heavy for most people I think. But I can't really think of a way to do it with less notation.

The main difference seems to be that the chain doesn't trivially reduce if Bob finds the next block after Alice has chosen not to publish a previous block, so you have to include that possibility as well.

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This is almost correct, except Alice still has to decide whether to publish what she has when Bob finds the next block, or keep mining privately.

So the correct equation is:

I'll give you the bounty since it was close enough.

The trick is now figuring out when , i.e. when it would be preferable to mine in secret versus publishing.

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That makes sense. I was implicitly assuming that if she didn't publish at b, then she still wouldn't at b+1, but maybe that's wrong. She probably cuts her losses at some point.

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yup and i haven't solved it all the way through. I imagine there will be some results along the lines of:

if it's optimal to publish at (a,b) then it's also optimal to publish at (a+1,b)

if it's optimal to secretly mine at (a,b), then it's also optimal to secretly mine at (a,b-1)

or something like that. Haven't proved either result though.

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71 sats \ 1 reply \ @Murch 7 May

I’m attending a workshop this week, I’ll have to skip this one, sorry.

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no worries, feel free to try it whenever

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Other people aren't doing graduate macro for fun?

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I started off thinking the math would be fairly simple, then realized that the right way to do it would have to be continuous time dynamic programming

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It is neat that this turned out to be a less trivial problem than it seemed at first, even with the simplifying assumptions.

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