# Is my proof of Asmuth-Bloom scheme's perfection correct?

I am trying to understand how Asmuth-Bloom's secret sharing is perfect. My "proof" so far (in a 3 out of 4 scheme):

$$y=s + a * m_0 \bmod m_1 * m_2 * m_3$$

Having $$k-1$$ shares, it will yield a valid solution for the CRT, but not a unique one. 2 shares yield $$x = y' \bmod m_1 * m_2$$
Therefore, we know that $$y = y' + z * m_1 * m_2$$ and also that $$y = [0;m_1*m_2*m_3]$$
But since we do not know the exact range of $$y$$, since $$m_3$$ is missing, we can not just simply guess $$z$$ and therefore guess $$y$$.
However, I see a sequence of following primes used a lot, so when we know $$m_1$$ and $$m_2$$ we could just assume $$m_3$$ is the next prime and guess values for $$z$$ and therefore $$y$$ right?

Example with numbers:
$$m_0 = 43,\space m_1=131, \space m_2=137, \space m_3 = 139, \space m_4=149, \space s=42, \space a = 476$$
$$y = 42 + 476 * 43$$
$$y = 20510 \bmod 2494633$$

$$s_1 = (74, 131), \space s_2 = (97, 137), \space s_3 = (77, 139), \space s_4 = (97, 149)$$

CRT with two shares:

$$x \equiv 74 \bmod 131$$
$$x \equiv 97 \bmod 137$$
$$x \equiv 2563 \bmod 17947$$

So now we could start guessing

$$y_0 = 2563 + 0*17947$$
$$y_1 = 2563 + 1*17947$$
$$y_2 = 2563 + 2*17947$$
$$y_n = 2563 + n*17947$$
and, with the knowledge (or assumption) that $$m_3$$ is the next prime after $$137$$, a max value of $$y_{138} = 2563 + 138 * 17947$$
since $$2563 + 139 * 17947$$ exceeds $$m_1*m_2*m_3$$

So we have $$138$$ values to guess the secret from.
$$s_n = y_n \bmod 43$$

From my conclusion, this would not be a perfect scheme, but everyone states it is. So where am i wrong?

Okay, so i was finally able to figure out a solution for this. The scheme does indeed yield 138 valid results to guess the secret from. However, since they already know the secret lies in $$\mathbb{Z}_{m_0}$$ there are only (in the example above) 42 possible secrets anyways. Since $$a$$ is uniformly distributed, the scheme does indeed leak some probabilistic information about the secret when you apply $$\bmod 43$$ to all 138 possible values, as seen in: