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improve performance to ~100 documents/sec by avoiding encrypted multiplications

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Arthur Lu
2024-11-21 22:26:38 +00:00
parent def623e6e1
commit 81a46f0cba
1 changed files with 3 additions and 3 deletions
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6
basic.py
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@@ -57,7 +57,7 @@ def hyperbolic_distance_parts(u, v): # returns only the numerator and denominato
#du = -(1 - u @ u) # for some reason we need to negate this #du = -(1 - u @ u) # for some reason we need to negate this
#dv = -(1 - v @ v) # for some reason we need to negate this #dv = -(1 - v @ v) # for some reason we need to negate this
#return diff @ diff, du * dv # returns the numerator and denominator #return diff @ diff, du * dv # returns the numerator and denominator
return diff @ diff return diff
# document matrix containing rows of document encoding vectors # document matrix containing rows of document encoding vectors
@@ -115,11 +115,11 @@ for i in range(len(s_res)):
#c_num = PyCtxt(pyfhel=HE_server, bytestring=s_res[i]) #c_num = PyCtxt(pyfhel=HE_server, bytestring=s_res[i])
c_num = PyCtxt(pyfhel=HE_server, bytestring=s_res[i]) c_num = PyCtxt(pyfhel=HE_server, bytestring=s_res[i])
#c_den = PyCtxt(pyfhel=HE_server, bytestring=s_res[i][1]) #c_den = PyCtxt(pyfhel=HE_server, bytestring=s_res[i][1])
p_num = HE_client.decrypt(c_num)[0] p_num = HE_client.decrypt(c_num)
#p_den = HE_client.decrypt(c_den)[0] #p_den = HE_client.decrypt(c_den)[0]
#dist = np.arccosh(1 + 2 * (p_num / p_den)) #dist = np.arccosh(1 + 2 * (p_num / p_den))
# compute final score # compute final score
dist = np.arccosh(1 + 2 * (p_num / (du * dv[i]))) dist = np.arccosh(1 + 2 * (p_num @ p_num / (du * dv[i])))
#print(dist) #print(dist)
c_res.append(dist) c_res.append(dist)
end = time.time() end = time.time()