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RE: CBR – The Way to Split Coins (DRAFT)

in #gridcoin • 8 years ago (edited)

N = 10 and M = 100 then B = 64, and C = nB = 640

No, as sum(n) = N; usually we use n to count things, here I've used it (not a great choice) as a value of one utxo. To have a chance to stake 64 times (within 16h) you would have to split N = 10 into 64 chunks of n=0.15625 each (equal ns for simplicity).

@tomasbrod commented on splitting coins, but his message is not clear to me.

Also,

  • B – (time) maximum number of blocks that N coins can stake when M is a total number of staking coins

is not accurate enough

I meant

  • B – (time) maximum number of blocks that N coins can stake within 16 h when M is a total number of staking coins
    (now updated in the main text, thanks)
8 years ago in #gridcoin by
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     8 years ago  

    No, as sum(n) = N; usually we use n to count things, here I've used it (not a great choice) as a value of one utxo.

    I'm not sure this contradicts my original point, as the formula for B still only contains N, M, and t.

    To have a chance to stake 64 times (within 16h) you would have to split N = 10 into 64 chunks of n=0.15625 each (equal ns for simplicity).

    This is what I'm trying to get at, that the maximum number of blocks a person can stake is equal to the number of UTXOs that person has. Theoretically, if one makes more UTXOs than there are blocks in some time period, then a person could theoretically stake all of those blocks, though the probability of that will be extremely low.