IBM / fhe-toolkit-linux

IBM Fully Homomorphic Encryption Toolkit For Linux. This toolkit is a Linux based Docker container that demonstrates computing on encrypted data without decrypting it! The toolkit ships with two demos including a fully encrypted Machine Learning inference with a Neural Network and a Privacy-Preserving key-value search.
MIT License
1.43k stars 158 forks source link

Question #162

Closed Hang-shao closed 3 years ago

Hang-shao commented 3 years ago

Where

For the next step, one should be aware of a mathematical result known as Fermat's little theorem (FLT). FLT states that given any non-zero integer a and a prime number p then a raised to the power p-1 always results in a number pn + 1 where n is an integer.

What

This is a problem. What is n? FLT:a^p-1=p^(n+1)?Is it real FLT?

How

It should "any non-zero integer a and a prime number p then a raised to the power p-1 always results in anumber 1 mod p "

github-actions[bot] commented 3 years ago

Thank you very much for helping us to improve the quality of our software. The FHE Toolkit develiopment team want to sincerely thank you for submitting your first issue and joining our technical community! We will get started on your issue immediately.

boland25 commented 3 years ago

@Hang-shao thanks for pointing this out. Can you show us where you came across this in the repo? thanks!

boland25 commented 3 years ago

@Hang-shao ah I see it, its in the Readme of BGV_country_db_lookup, thanks!

boland25 commented 3 years ago

@Hang-shao We are correcting the text to read like so....

For the next step, one should be aware of a mathematical result known as Fermat's little theorem (FLT). FLT implies that given any non-zero integer a and a prime number p, if a is not divisible by p then a^(p-1)-1 is an integer multiple of p. Or to put it another way, a^(p-1) modulo p equals 1. This is formally captured below. Using this result we can modify it slightly to produce a function f(x) as stated below.

Hang-shao commented 3 years ago

@boland25 My goodness, thank you I understand,imageimage thanks again