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Refactor the PowerCmp function

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Why we have this: 
This algorithm here directly follow the rule of ratio to win, and easier to understand. 
Just think about a smaller space, e.g. 100, instead of 2^256. 

When one miner has 10% power, it should have 10% ratio to win. that is, when the h is in {0..9}, it wins. Simply speaking, it is: 
     (h + 1) / 100 <= 10% ,  instead of     
     h / 99 < 10%     
The former is easier to understand, though both are equivalent when h is an integer and power_ratio < 1)
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Steven Li 2019-10-31 13:36:33 +08:00 committed by GitHub
parent 790ac7b510
commit 0aa56ba665
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1 changed files with 8 additions and 7 deletions
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15
chain/types/blockheader.go
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@ -166,24 +166,25 @@ func PowerCmp(eproof ElectionProof, mpow, totpow BigInt) bool {
/* /*
Need to check that Need to check that
h(vrfout) / max(h) < e * minerPower / totalPower (h(vrfout) + 1) / (max(h) + 1) <= e * minerPower / totalPower
max(h) == 2^256-1 max(h) == 2^256-1
which in terms of integer math means: which in terms of integer math means:
h(vrfout) * totalPower < e * minerPower * (2^256-1) (h(vrfout) + 1) * totalPower <= e * minerPower * 2^256
*/ */
h := sha256.Sum256(eproof) h := sha256.Sum256(eproof)
lhs := BigFromBytes(h[:]).Int lhs := BigFromBytes(h[:]).Int
lhs = lhs.Add(lhs, NewInt(1).Int)
lhs = lhs.Mul(lhs, totpow.Int) lhs = lhs.Mul(lhs, totpow.Int)
// rhs = minerPower * 2^256 - minerPower // rhs = minerPower * 2^256
// rhs = minerPower << 256 - minerPower // rhs = minerPower << 256
rhs := new(big.Int).Lsh(mpow.Int, 256) rhs := new(big.Int).Lsh(mpow.Int, 256)
rhs = rhs.Mul(rhs, blocksPerEpoch.Int) rhs = rhs.Mul(rhs, blocksPerEpoch.Int)
rhs = rhs.Sub(rhs, mpow.Int)
// return true if lhs is less than or equal to rhs
return lhs.Cmp(rhs) == -1 return lhs.Cmp(rhs) < 1
} }
func (t *Ticket) Equals(ot *Ticket) bool { func (t *Ticket) Equals(ot *Ticket) bool {