Timeline for Algorithm Design for only Mutual Information Sharing
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 16, 2012 at 1:26 | comment | added | hackartist | I know but Bob has to play by the rules still because he could gain advantage by just looking at T[1] all the time... sure this weeds out the possibility of Eve listening in. But for the bits Alice sends a table for, Bob has advantage and for the bits Bob sends a table for Alice has the advantage. It needs to be that neither one can have any idea for any bit that is 0 in their string and this would allow them to cheat if they wanted to on some of their 0 bits. | |
| May 16, 2012 at 1:17 | comment | added | Charles Fu | Well to prevent eavesdropping you can always wrap this with a PKS right? Alice should let $T = \{E_b(a_0\text{ AND }0), E_b(a_0\text{ AND }1)\}$ where $E_b$ is the encryption function using Bob's public key. Bob then can first obtain the value then decrypt it to see the real value of $T[b_0]$. | |
| May 16, 2012 at 1:11 | comment | added | hackartist | What I mean is someone who is listening to the messages in between will see the table Alice sends for a0 and can just look at the value of a0 AND 1 if they wanted to find out that bit. Sure from Bob's point of view if he plays by the rules and only looks at T[b0] and not T[1] all the time this works but that assumes Bob and any eavesdroppers play by those rules. | |
| May 16, 2012 at 1:00 | comment | added | hackartist | ok, but Alice has revealed a0 to Bob... so they would have to decide randomly which ones each reveals to the other like this... unless I am missing something I want neither to have any idea of the value of the others' bits unless that party has a 1 in that position. | |
| May 16, 2012 at 0:50 | history | answered | Charles Fu | CC BY-SA 3.0 |