Use CKKS homomorphism to find approximate reciprocal (approximate multiplicative inverse)

algorithm for reciprocal

Two numbers are reciprocals of each other, which means that the multiplication of these two numbers equals 1. For example, a and b are reciprocals of each other, then ab=1.

The reciprocal of 5 is 0.2, we can find it very simply, but how to find the reciprocal of a number in the ciphertext field?

An algorithm is given in the article "An investigation of complex operations with word-size homomorphic encryption".

We assume that y=1-x, and the modulus of y is less than (for real numbers, the absolute value) 0.5, then the following formula holds

Substitute x=1-y, and use the square difference formula to prove it quickly.

When r is large enough, obviously, the above formula is 1, and the reciprocal of x we ​​require can be found through the formula on the left

In this way, we have an iterative method to calculate the reciprocal of a number using only addition and multiplication.

However, the above method may have a problem, that is, the x at this time can only be in the range of [0.5,1], and the y involved in the calculation is a decimal.

We multiply the left and right sides of the above formula by one at the same time , let z=xp, y=pz then have

In this way, it can be calculated when the modulus of y is less than p/2.

However, obviously, when p=1, the amount of calculation will be much smaller

TenSEAL uses CKKS code example

import tenseal as ts
ctx=ts.context(ts.SCHEME_TYPE.CKKS, poly_modulus_degree=8192, coeff_mod_bit_sizes=[44,30,30,30,30,44])
ctx.global_scale=2**30
import numpy as np 
m1=np.array([1.2,0.6,0.4])
c0=ts.ckks_vector(ctx,m1)
p1=1
p=p1
r=4
t=p-c0
c=t
v=p+c
for i in range(r-1):
    c=c*c
    p=p*p
    v=v*(p+c)
m=v.decrypt()
m=np.array(m)/(p1**(2**(r)))

print(m)

print(m*m1)

When r=2 or 3, the result is more accurate