# Asmuth-Bloom Secret Sharing $m_0$ public?

i am trying to figure out how Asmuth-Bloom secret sharing works and followed the example on wikipedia and replaced it with own numbers.

Namely, i used:
$$S=3$$
$$k=3$$
$$n=4$$
$$m_0 = 13, m_1=45, m_2=46, m_3=47, m_4=49$$

I've been able to construct the shares

$$s_1 = (17, 45), s_2 = (2, 46), s_3 = (34, 47), s_4=(6,49)$$

I then result in $$x \equiv 692 \mod 46*47*48$$ which is correct, since $$S = 692 \equiv 3 \mod 13$$ which is my secret.

However, i wonder where $$m_0 = 13$$ is coming from for the final secret reconstruction, it seems like it is not explained anywhere. The only thing i could find is this stating it would be "kept secret by the dealer unless stated otherwise".

Is $$m_0$$ just either a public parameter? Or do i need the dealer to store it "secretly" (which would require the dealer to reconstruct the secret)?

## 1 Answer

According to Dragan&Tiplea,On the Asymptotic Idealness of the Asmuth-Bloom Threshold Secret Sharing Scheme $$m_0$$ is a public parameter.
The scheme itself does not require $$m_0$$ to be kept secret. However, by keeping it secret you could eliminiate the possibility to get a probabilistic distribution of secrets as explained here but if $$m_0$$ for some reason is lost, the secret is not recoverable, which kind of defeats the purpose (or parts of it) of a secret sharing scheme.