New answers tagged secret-sharing
-2
There is a problem with this explanation.
At stage 3 you write:
" 3. All parties locally compute [z]=[c]+[x]⋅e+[y]⋅d−e⋅d "
You mean that:
z_1 = c_1 + x_1 * e + y_1 * d - e*d
...
z_n = c_n +x_n * e + y_n * d - e*d
When you compute the sum of all shares you want z=xy. However,
z_1+...+z_n = c + x * e + y * d - n * ed = c + x(y-b) + y*(x-a) -n*(x-a)...
0
Also, to answer the question you asked:
Is an encryption mechanism like this exists ?
Yes, blinded decryption techniques exist. There are techniques where party A has a key, party B has a ciphertext, and party A decrypts the ciphertext for party B, and A does not learn anything about the plaintext.
One way to do this is with RSA encryption; the normal ...
0
If the goal is to store the secret on two servers and Alice so that, if Alice had access to both servers, you can recover the secret, but neither server by itself could (even if you could recover Alice's stored value).
If so, this problem is essentially secret sharing. In this case, it can be done fairly easily:
Alice generates two random strings $P, Q$ ...
1
GMW protocol says that A and B can locally compute $c_i = a_i \oplus b_i$ and the result should be same for both.
I highlighted the part that is incorrect. They started with secret shares of $a=0$ and secret shares of $b=1$. Now they have secret shares of the XOR $c= a \oplus b =1$.
Only if the secret value $c$ is zero will their shares be the same (and ...
2
Aren't they delievering by the network the same as the encypted messages?
No. At least not in the plain.
What doing it end-to-end encrypted?
Try to search for the key agreement protocol (e. g. Diffie-Hellman ) and asymmetric encryption (e.g. RSA ).
The "asymmetric" encryption allows sending encrypted and signed messages without sharing the private (...
Top 50 recent answers are included
