Talk:Secret sharing using the Chinese remainder theorem

Example is wrong[edit]

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)[reply]

if you'll find that, for example, $M'=155\mod 19*17$ $M'=155\mod 19*17$ , 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:
- $155\equiv 2\mod 3$
- $478\equiv 1\mod 3$
- $633\equiv 0\mod 3$

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)[reply]

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[reply]