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#!/usr/bin/env python | |
#-*- encoding: utf-8 -*- | |
# This code is a part of stackexchange answer | |
# http://bitcoin.stackexchange.com/a/20060/2343 | |
# License: cc-wiki with attribution required | |
import numpy as np | |
HR = 550 # GHash/s, own hashrate | |
THR0 = 1.2e7 # GHash/s, network hashrate | |
DA = 1.2 # Expected difficulty adjustment | |
N = 26*2016 # For how much block make prediction? ~ 1 year | |
NB = 2016 # Number of block between difficulty adjustments | |
T0 = 10 # minutes, desired block time | |
def THR(n): | |
return THR0*np.exp((n/NB)*np.log(DA)) | |
p = HR/THR(np.arange(N)) | |
# To caclulate distribution we use algorithm from | |
# Yili Hong "On Computing the Distribution Function for the Sum of | |
# Independent and Non-identical Random Indicators", 2011 | |
# http://www.stat.vt.edu/research/Technical_Reports/TechReport11-2.pdf | |
x = np.zeros(N+1, dtype=np.complex128) | |
w = 2*np.pi / (N+1) | |
x[0] = 1.0 + 0.j | |
for l in xrange(1, int(np.ceil(N/2))+1): | |
z = 1.0 - p[:] + p[:]*np.cos(w*l) + 1.j * p[:]*np.sin(w*l) | |
dl = np.exp(np.sum(np.log(np.absolute(z[:])))) | |
ang = np.sum(np.angle(z[:])) | |
al = dl*np.cos(ang) | |
bl = dl*np.sin(ang) | |
x[l] = al + 1.j*bl | |
if l%1000 == 0: print l, '/', N/2 | |
for l in xrange(int(np.ceil(N/2))+1, N+1): | |
x[l] = np.conj(x[N+1-l]) | |
x = x[:] / (N+1) | |
y = np.fft.fft(x) | |
print DA | |
for i in enumerate(np.real(y[:10])): | |
print i[0], i[1] |
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