CSE5313 Coding and information theory for data science (Lecture 18)
Secret sharing
The president and the vice president must both consent to a nuclear missile launch.
We would like to share the nuclear code such that:
In other words:
- The two shares are everything.
- One share is nothing.
Solution
Scheme:
- The nuclear code is a field element , chosen at random (M arbitrary).
- Let .
- , where , i.e., for every .
- Fix (not random).
- .
- .
And then:
- One share reveals nothing about .
- I.e., (gradient could be anything)
- Two shares reveal .
- I.e., (two points determine a line).
Formalize the notion of secret sharing
Problem setting
A dealer is given a secret chosen from an arbitrary distribution .
The dealer creates shares and send to parties.
Two privacy parameters: .
Requirements:
For denote .
- Decodability: Any set of shares can reconstruct the secret.
- for all with .
- Security: Any set of shares reveals no information about the secret.
- for all with .
This is called -secret sharing scheme.
Interpretation
- , is a corrupted set of parties.
- An adversary which corrupts at most parties cannot infer anything about the secret.
Applications
- Secure distributed storage.
- Any hacked servers reveal nothing about the data.
- Secure distributed computing with a central server (e.g., federated learning).
- Any corrupted computation nodes know nothing about the data.
- Secure multiparty computing (decentralized).
- Any corrupted parties cannot know the inputs of other parties.
Scheme 1: Shamir secret sharing scheme
Parameters , and .
Fix , and distinct points . (public, known to all).
Given the dealer:
- Choose (uniformly random from ).
- Defines by .
- Send share to party .
Theorem valid encoding scheme
This is an -secret sharing scheme.
Decodability:
- , any shares can reconstruct by Lagrange interpolation.
Proof
Specifically, any parties can define the interpolation polynomial , where . (, for ).
, so for all .
Therefore, .
Privacy:
Need to show that for all with .
that is equivalent to show that and are independent for all with .
Proof
We will show that , for all and .
Let , and .
So,
So exactly one solution for is possible.
So for all .
Recall the law of total probability:
So .
Scheme 2: Ramp secret sharing scheme (McEliece-Sarwate scheme)
- Any know nothing
- Any knows everything
- Partial knowledge for
Parameters , and .
Fix , and distinct points . (public, known to all)
Given , the dealer:
- Choose (uniformly random from ).
- Defines .
- Send share to party .
Decodability
Similar to Shamir scheme, any shares can reconstruct by Lagrange interpolation.
Privacy
Similar to the proof of Shamir, exactly one value of is possible!
(‘s are uniform and independent).
Conclude similarly by the law of total probability.
.
Conditional mutual information
The dealer needs to communicate the shares to the parties.
Assumed: There exists a noiseless communication channel between the dealer and every party.
From previous lecture:
- The optimal number of bits for communicating (i’th share) to the i’th party is .
- Q: What is ?
Tools:
- Conditional mutual information.
- Chain rule for mutual information.
Definition of conditional mutual information
The conditional mutual information of and given is defined as:
where is the conditional entropy of given and .
The chain rule of mutual information
Conditioning reduces entropy.
Lower bound for communicating secret
Consider the Shamir scheme (, one message).
Q: What is with respect to ? A: Fix any of size , and let .
So the bits used for sharing the secret is at least the bits of actual secret.
In Shamir we saw: .
- If is uniform (standard assumption), then Shamir achieves this bound with equality.
- In ramp secret sharing we have (similar proof).
- Also optimal if is uniform.