Meaning of the bit precision Bis

[mmazz]

Hello,

I have a complete beginner’s question about the meaning of bit precision on the CKKS scheme. I already read post related (like meaning of the bit precision) but I keep having doubts.

When decrypting and having a statement like “Estimated precision: xx bits.”

What I think that I understand when I encrypt with CKKS a real number, is that if I had a number (a double) with 10 bits of interger part and 20 bits of decimal part and use a scaling factor of 20, I already I’m loosing 10 bits of precision. Getting as best a number with 10 bits of integer part and 10 bits of decimal part, Right?

In this case, the answer of the Estimated precision will be 20 bits?

So ideally, if I have only one element encrypted, the estimated precision are the total correct bits of the mantissa of the representation of the decrypted result?

If all this is correct, can someone expand more of what Yuriy said at the post that I mentioned:

Roughly it corresponds to the [scaling factor] − [log2 of the average L1 norm of the difference between approximate FHE result and floating-point result in the clear]

Thank you for your help and sorry if is a really beginner question.

[ypolyakov]

@mmazz

What I think that I understand when I encrypt with CKKS a real number, is that if I had a number (a double) with 10 bits of interger part and 20 bits of decimal part and use a scaling factor of 20, I already I’m loosing 10 bits of precision. Getting as best a number with 10 bits of integer part and 10 bits of decimal part, Right?

No, you are not losing 10 bits. You have a 10-bit integer. After scaling up by 2^{20}, you will get a 30-bit integer. Of course, as soon as you encrypt it (i.e., add some noise), the effective precision will get reduced by 10-15 bits or so.

@mmazz

In this case, the answer of the Estimated precision will be 20 bits?

The estimated precision also accounts for LWE noise (see above). The noise cannot be ignored!

@mmazz

If all this is correct, can someone expand more of what Yuriy said at the post that I mentioned:

Roughly it corresponds to the [scaling factor] − [log2 of the average L1 norm of the difference between approximate FHE result and floating-point result in the clear]

At a high level, the estimated precision prints outs the decimal precision, i.e., the number of bits after the decimal point. If the magnitude of encrypted numbers is much larger than 1, then the decimal precision will significanly drop after a multiplication (the approximation error in one ciphertext gets multiplied by the encrypted message in the other ciphertext).