I'm trying to understand the following RLWE encryption scheme from this Chris Peikert's paper

enter image description here Say i choose $q = 97$, $n=8$ and the polynomial $a = 96 x^8+30 x^7+76 x^6+12 x^5+57 x^4+77 x^3+70 x^2+49 x+83$

secret key $s = 8 x^8+7 x^7+6 x^6+5 x^5+4 x^4+3 x^3+2 x^2+x$

and error $ e = x^6+x^4+x^2+1 $

then $b = a \cdot s +e \mod 97 = 19 x^7+84 x^6+35 x^5+40 x^4+94 x^3+69 x^2+59 x+63$

Suppose i want to encrypt the message $z =1010101 = x^6+x^4+x^2+1$

I choose $r = x^6+x^4+x^2+1$

$e1 = x^6+x^4+x^2+1$

$e2 = x^6+x^4+x^2+1$

then we have $u = a \cdot r + e1 = 71 x^7+94 x^6+11 x^5+39 x^4+84 x^3+22 x^2+27 x+76$

and $v = b \cdot r + e2 + q/2 \cdot z = (13 x^7+63 x^6+72 x^5+89 x^4+2 x^3+9 x^2+8 x+65) + (49 x^6+49 x^4+49 x^2+49) = 13 x^7+15 x^6+72 x^5+41 x^4+2 x^3+58 x^2+8 x+17$

to decrypt i compute $v-u \cdot s = (13 x^7+14 x^6+72 x^5+40 x^4+2 x^3+57 x^2+8 x+16) - (29 x^7+62 x^6+74 x^5+78 x^4+91 x^3+89 x^2+91 x+46) = -16 x^7-47 x^6-2 x^5-37 x^4-89 x^3-31 x^2-83 x-29 $

rounding each coefficient with respect to $(-q/4, q/4)$ gives 11111010

what am I doing wrong?

  • $\begingroup$ do you know what is $\lfloor q/2 \rceil $ $\endgroup$
    – kelalaka
    Feb 26 '20 at 9:09
  • $\begingroup$ yes, rounding to nearest integer; however rounding to 49 instead of 48 does not change the result. $\endgroup$
    – Deus Ex
    Feb 27 '20 at 4:11
  • $\begingroup$ Are you performing the reductions modulo $q$ using the centered representation, i.e., reducing everything into $-q/2, ..., q/2$? Moreover, are you really computing $u - s\cdot v$ instead of $v - s\cdot u$ in the decryption step or it is just a typo? $\endgroup$ Feb 27 '20 at 8:19
  • $\begingroup$ yes, it's a typo. I just reduced every coefficient mod q. I only rounded to $(-q/2...q/2)$ in the last step. Do i need to mod $(-q/2...q/2)$ for every calculation? The paper only says $mod q$. $\endgroup$
    – Deus Ex
    Feb 28 '20 at 0:26

convert $v-u \cdot s$ to centered representation as described here seems to work

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.