假設壞礦工的算力為4.3%,請問確認次數z幾次,可達雙重支付攻擊成功機率p< 0.0001
Hint:
#include <iostream>
#include <cmath>
using namespace std;
double AttackerSuccessProbability(double q, int z)
{
double p = 1.0 - q;
double lambda = z * (q / p);
double sum = 1.0;
int i, k;
for (k = 0; k <= z; k++)
{
double poisson = exp(-lambda);
for (i = 1; i <= k; i++)
poisson *= lambda / i;
sum -= poisson * (1 - pow(q / p, z - k));
}
return sum;
}
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25 則留言:
z=5 p=1.35207e-005
1.35207e-005
--------------------------------
Process exited after 0.003546 seconds with return value 0
請按任意鍵繼續 . . .
總共要計算5次,才能算出機率。
Z=5次
P=1.35207e-5
Z=5 , P=1.35207e-5
Z=5次,P=35207e-5
Z=5 , p = 1.35207e-005
Z=5次
P=1.35207e-005
z=5,P=1.35207e-005
z= 5
P= 1.35207e-005
P=1.35207e-005
Z=5
P=1.35207-005
z = 5
p = 1.35207e-005
1.35207e-005
z=5
p=1.35207e-005
1.35207e-005
1.35207e-005
Z=5次
P=1.35207e-5
Z=5次
P=1.35207e-005
z = 5
p = 1.35207e-005
z = 5
p = 1.35207e-005
1.35207e-005
P=1.35207e-005
Z=5
P=1.35207e-005
Z=5
P=1.35207e-005
Z=5
Z=5
p = 1.35207e-005
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