Talk:Secret sharing using the Chinese remainder theorem

Example is wrong
The example seems to be wrong. You can take two the the modular equations and still find the key easily. 71.190.80.195 (talk) 13:43, 8 May 2009 (UTC) To find the secret you need to define M' in the ring of 11*13*17 (or bigger), not lesser one. Vlsergey (talk) 07:32, 11 August 2010 (UTC)
 * if you'll find that, for example, $$M' = 155 \mod 19*17$$, it means 478 is also a solution, and 633 as well. Since you don't know third equation you don't know which to choose, and all three of them gives different secret:
 * $$155 \equiv 2 \mod 3$$
 * $$478 \equiv 1 \mod 3$$
 * $$633 \equiv 0 \mod 3$$

This condition is crucial because the scheme is built on choosing the secret as an integer between the two products, and this condition ensures that at least k shares are needed to fully recover the secret, no matter how they are chosen. The following statement is wrong. You are able to solve Mignotte's Secret Sharing Scheme with k-1 shares when using the right program. Click here to see the pdf with a broader explenation. --2001:985:A9A9:1:1DD4:72AC:A2EE:7D85 (talk) 21:50, 30 April 2019 (UTC)Elise