Difference between revisions of "Open Problems:79"
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In this problem, there are $k$ players and an eavesdropper. Input is a random bit $X \in \{0,1\}$, also called "the secret." The secret owner $J$ is chosen uniformly at random from $[k]$. Players have private randomness and they can communicate publicly on a shared blackboard visible to everyone. Players are said to win if | In this problem, there are $k$ players and an eavesdropper. Input is a random bit $X \in \{0,1\}$, also called "the secret." The secret owner $J$ is chosen uniformly at random from $[k]$. Players have private randomness and they can communicate publicly on a shared blackboard visible to everyone. Players are said to win if | ||
| − | |||
* everyone learns the secret, and | * everyone learns the secret, and | ||
* eavesdropper does not guess the secret owner correctly. | * eavesdropper does not guess the secret owner correctly. | ||
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We are interested in maximizing the probability that players win (maximum success probability). | We are interested in maximizing the probability that players win (maximum success probability). | ||
| − | For $k=2$, the following trivial protocol has success probability $0.25$. Just output $1$. It can be shown easily that maximum success probability is at most $0.5$. The bounds can be improved to $0.3384 \le $ maximum success probability $\le 0.361$. | + | For $k=2$, the following trivial protocol has success probability $0.25$. Just output $1$. It can be shown easily that maximum success probability is at most $0.5$. The bounds can be improved to $0.3384 \le $ maximum success probability $\le 0.361$. Here is a protocol that achieves success probability $1/3$. |
| + | |||
| + | '''First round:''' | ||
| + | If Alice has the secret, | ||
| + | |||
| + | * with probability $2/3$ she asks Bob to speak second | ||
| + | * with probability $1/3$ she speaks second. | ||
| + | |||
| + | Else (vice-versa) | ||
| + | * with probability $2/3$ she speaks second | ||
| + | * with probability $1/3$ she asks Bob to speak second. | ||
| + | |||
| + | '''Second round:''' | ||
| + | If the speaker has the secret, announce it, else announce a random bit. | ||
| + | |||
| + | |||
For large $k$, the following bounds can be shown: $0.5644 \le $ maximum success probability $\le 0.75$. | For large $k$, the following bounds can be shown: $0.5644 \le $ maximum success probability $\le 0.75$. | ||
| + | |||
Can we improve these bounds? | Can we improve these bounds? | ||
Revision as of 21:45, 31 March 2017
| Suggested by | Joshua Brody |
|---|---|
| Source | Banff 2017 |
| Short link | https://sublinear.info/79 |
Cryptogenography is concerned with the following question. "How to share a secret without revealing the secret owner?" Please see Brody et al. [BJSW-14] and Jakobsen [J-16].
In this problem, there are $k$ players and an eavesdropper. Input is a random bit $X \in \{0,1\}$, also called "the secret." The secret owner $J$ is chosen uniformly at random from $[k]$. Players have private randomness and they can communicate publicly on a shared blackboard visible to everyone. Players are said to win if
- everyone learns the secret, and
- eavesdropper does not guess the secret owner correctly.
We are interested in maximizing the probability that players win (maximum success probability).
For $k=2$, the following trivial protocol has success probability $0.25$. Just output $1$. It can be shown easily that maximum success probability is at most $0.5$. The bounds can be improved to $0.3384 \le $ maximum success probability $\le 0.361$. Here is a protocol that achieves success probability $1/3$.
First round: If Alice has the secret,
- with probability $2/3$ she asks Bob to speak second
- with probability $1/3$ she speaks second.
Else (vice-versa)
- with probability $2/3$ she speaks second
- with probability $1/3$ she asks Bob to speak second.
Second round: If the speaker has the secret, announce it, else announce a random bit.
For large $k$, the following bounds can be shown: $0.5644 \le $ maximum success probability $\le 0.75$.
Can we improve these bounds?
