Since r,e1,s,e2∈[β]m,e3∈[β], so rTe1∈[mβ2],e2Ts∈[mβ2],e3∈[β], then
v−uTs∈[2mβ2+β]+⌈q/2⌋μ
If the choice of parameters m,β satisfies 2mβ2+β<q/4, then the decryptor can just determine:
μ={10v−uTs is more close to q/2v−uTs is more close to 0
The security of scheme is based on the difficulties of the following problem:
Definition(Learning With Error, LWE): For positive integers m,n,q, and β≪q, the LWEn,m,q,β problem asks to distinguish between the following two distributions:
(A,As+e), where A←Zqn×m,s←[β]m,e←[β]n.
(A,u), where A←Zqn×m,u←Zqn.
We assume the that the problem is hard, i.e. the two distribution is indistinguishable. The LWE problem has been proven to have a reduction to the difficult problem in lattice problem.
The public key (A,t) is indistinguishable from uniform distribution due to the LWE assumption,