Quote from: gmaxwell on March 11, 2015, 08:04:25 PM

Quote from: Crowex on March 11, 2015, 12:04:20 PM

To extend this to a t of n case they must make sure that each and all of the possible subsets of t signers hold the same multiplicative secret. I can see how this could possibly be done with a dealer but I cant understand how it is done without a dealer.
Of course this could just be due to my lack of understanding but I reserve the right to keep asking stupid questions in order to try and improve my understanding.

I don't see it either, except by repeating the process for each of the the m choose n satisfactions; which you can't do for a single secret without a dealer. Thats why I was asking in the other thread where this was implemented when someone said it was.

Glad to see you detected there might be a problem too, I had a bit of self doubt after I first posted a comment and blog question (hence the edit) but then the more I look at it the more I am convinced that their scheme does not extend to the t-of-n case in the way that they described.
I don't think it will work ~~with or~~ without a dealer for t-of-n. ~~The methods I imagined might be used with a dealer scheme seem to have security problems.~~

I'll see if I get a response to my last blog question that will provide any clarification.