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CSE5313 Coding and information theory for data science (Lecture 18)

The president and the vice president must both consent to a nuclear missile launch.

We would like to share the nuclear code such that:

Share1, Share2 \mapsto Nuclear Code
Share1 \not\mapsto Nuclear Code
Share2 \not\mapsto Nuclear Code
Share1 \not\mapsto Share2
Share2 \not\mapsto Share1

In other words:

The two shares are everything.
One share is nothing.

Solution

Scheme:

The nuclear code is a field element m \in \mathbb{F}_q, chosen at random m \sim M (M arbitrary).
Let p(x) = m + rx \in \mathbb{F}_q[x].
- r \sim U, where U = Uniform \mathbb{F}_q, i.e., Pr(\alpha = 1/q) for every \alpha \in \mathbb{F}_q.
Fix \alpha_1, \alpha_2 \in \mathbb{F}_q (not random).
s_1 = p(\alpha_1) = m + r\alpha_1, s_1 \sim S_1.
s_2 = p(\alpha_2) = m + r\alpha_2, s_2 \sim S_2.

And then:

One share reveals nothing about m.
I.e., I(S_i; M) = 0 (gradient could be anything)
Two shares reveal p \Rightarrow reveal p(0) = m.
I.e., H(M|S_1, S_2) = 0 (two points determine a line).

Problem setting

A dealer is given a secret m chosen from an arbitrary distribution M.

The dealer creates n shares s_1, s_2, \cdots, s_n and send to n parties.

Two privacy parameters: t,z\in \mathbb{N} z<t.

Requirements:

For \mathcal{A}\subseteq[n] denote S_\mathcal{A} = \{s_i:i\in \mathcal{A}\}.

Decodability: Any set of t shares can reconstruct the secret.
- H(M|S_\mathcal{T}) = 0 for all \mathcal{T}\subseteq[n] with |\mathcal{T}|\geq t.
Security: Any set of z shares reveals no information about the secret.
- I(M;S_\mathcal{Z}) = 0 for all \mathcal{Z}\subseteq[n] with |\mathcal{Z}|\leq z.

This is called $(n,z,t)$-secret sharing scheme.

Interpretation

\mathcal{Z} \subseteq [n], |\mathcal{Z}| \leq z is a corrupted set of parties.
An adversary which corrupts at most z parties cannot infer anything about the secret.

Applications

Secure distributed storage.
- Any \leq z hacked servers reveal nothing about the data.
Secure distributed computing with a central server (e.g., federated learning).
- Any \leq z corrupted computation nodes know nothing about the data.
Secure multiparty computing (decentralized).
- Any \leq z corrupted parties cannot know the inputs of other parties.

Parameters n,t, and z=t-1.

Fix \mathbb{F}_q, q>n and distinct points \alpha_1, \alpha_2, \cdots, \alpha_n \in \mathbb{F}_q\setminus \{0\}. (public, known to all).

Given m\sim M the dealer:

Choose r_1, r_2, \cdots, r_z \sim U_1, U_2, \cdots, U_z (uniformly random from \mathbb{F}_q).
Defines p\in \mathbb{F}_q[x] by p(x) = m + r_1x + r_2x^2 + \cdots + r_zx^z.
Send share s_i = p(\alpha_i) to party i.

Theorem valid encoding scheme

This is an $(n,t-1,t)$-secret sharing scheme.

Decodability:

\deg p=t-1, any t shares can reconstruct p by Lagrange interpolation.

Proof

Specifically, any t parties \mathcal{T}\subseteq[n] can define the interpolation polynomial h(x)=\sum_{i\in \mathcal{T}} s_i \delta_{i}(x), where \delta_{i}(x)=\prod_{j\in \mathcal{T}\setminus \{i\}} \frac{x-\alpha_j}{\alpha_i-\alpha_j}. (\delta_{i}(\alpha_i)=1, \delta_{i}(\alpha_j)=0 for j\neq i).

\deg h=\deg p=t-1, so h(x)=p(x) for all x\in \mathcal{T}.

Therefore, h(0)=p(0)=m.

Privacy:

Need to show that I(M;S_\mathcal{Z})=0 for all \mathcal{Z}\subseteq[n] with |\mathcal{Z}|=z.

that is equivalent to show that M and s_\mathcal{Z} are independent for all \mathcal{Z}\subseteq[n] with |\mathcal{Z}|=z.

Proof

We will show that \operatorname{Pr}(s_\mathcal{Z}|M=m)=\operatorname{Pr}(M=m), for all s_\mathcal{Z}\in S_\mathcal{Z} and m\in M.

Let m,\mathcal{Z}=(i_1,i_2,\cdots,i_z), and s_\mathcal{Z}.


\begin{bmatrix}
m & U_1 & U_2 & \cdots & U_z
\end{bmatrix} = \begin{bmatrix}
1 & 1 & 1 & \cdots & 1 \\
\alpha_{i_1} & \alpha_{i_2} & \alpha_{i_3} & \cdots & \alpha_{i_n} \\
\alpha_{i_1}^2 & \alpha_{i_2}^2 & \alpha_{i_3}^2 & \cdots & \alpha_{i_n}^2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\alpha_{i_1}^{z} & \alpha_{i_2}^{z} & \alpha_{i_3}^{z} & \cdots & \alpha_{i_n}^{z}
\end{bmatrix}=s_\mathcal{Z}=\begin{bmatrix}
s_{i_1} \\ s_{i_2} \\ \vdots \\ s_{i_z}
\end{bmatrix}

So,


\begin{bmatrix}
U_1 & U_2 & \cdots & U_z
\end{bmatrix} = (s_\mathcal{Z}-\begin{bmatrix}
m & m & m & \cdots & m
\end{bmatrix})
\begin{bmatrix}
\alpha_{i_1}^{-1} & \alpha_{i_2}^{-1} & \alpha_{i_3}^{-1} & \cdots & \alpha_{i_n}^{-1} \\
\end{bmatrix}

\begin{bmatrix}
1 & 1 & 1 & \cdots & 1 \\
\alpha_1 & \alpha_2 & \alpha_3 & \cdots & \alpha_n \\
\alpha_1^2 & \alpha_2^2 & \alpha_3^2 & \cdots & \alpha_n^2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\alpha_1^{z-1} & \alpha_2^{z-1} & \alpha_3^{z-1} & \cdots & \alpha_n^{z-1}
\end{bmatrix}^{-1}

So exactly one solution for U_1, U_2, \cdots, U_z is possible.

So \operatorname{Pr}(U_1, U_2, \cdots, U_z|M=m)=\frac{1}{q^z} for all m\in M.

Recall the law of total probability:


\operatorname{Pr}(s_\mathcal{Z})=\sum_{m'\in M} \operatorname{Pr}(s_\mathcal{Z}|M=m') \operatorname{Pr}(M=m')=\frac{1}{q^z}\sum_{m'\in M} \operatorname{Pr}(M=m')=\frac{1}{q^z}

So \operatorname{Pr}(s_\mathcal{Z}|M=m)=\operatorname{Pr}(M=m)\implies I(M;S_\mathcal{Z})=0.

Any z know nothing
Any t knows everything
Partial knowledge for z<s<t

Parameters n,t, and z<t.

Fix \mathbb{F}_q, q>n and distinct points \alpha_1, \alpha_2, \cdots, \alpha_n \in \mathbb{F}_q\setminus \{0\}. (public, known to all)

Given m_1, m_2, \cdots, m_n \sim M, the dealer:

Choose r_1, r_2, \cdots, r_z \sim U_1, U_2, \cdots, U_z (uniformly random from \mathbb{F}_q).
Defines p(x) = m_1+m_2x + \cdots + m_{t-z}x^{t-z-1} + r_1x^{t-z} + r_2x^{t-z+1} + \cdots + r_zx^{t-1}.
Send share s_i = p(\alpha_i) to party i.

Decodability

Similar to Shamir scheme, any t shares can reconstruct p by Lagrange interpolation.

Privacy

Similar to the proof of Shamir, exactly one value of U_1, \cdots, U_z is possible!

\operatorname{Pr}(s_\mathcal{Z}|m_1, \cdots, m_{t-z}) = \operatorname{Pr}(U_1, \cdots, U_z) = the above = 1/q^z

($U_i$'s are uniform and independent).

Conclude similarly by the law of total probability.

$\operatorname{Pr}(s_\mathcal{Z}|m_1, \cdots, m_{t-z}) = \operatorname{Pr}(s_\mathcal{Z}) \implies I(S_\mathcal{Z}; M_1, \cdots, M_{t-z}) = 0.

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 s_i (i'th share) to the i'th party is H(s_i).
Q: What is H(s_i|M)?

Tools:

Conditional mutual information.
Chain rule for mutual information.

Definition of conditional mutual information

The conditional mutual information I(X;Y|Z) of X and Y given Z is defined as:


\begin{aligned}
I(X;Y|Z)&=H(X|Z)-H(X|Y,Z)\\
&=H(X|Z)+H(X)-H(X)-H(X|Y,Z)\\
&=(H(X)-H(X|Y,Z))-(H(X)-H(X|Z))\\
&=I(X; Y,Z)- I(X; Z)
\end{aligned}

where H(X|Y,Z) is the conditional entropy of X given Y and Z.

The chain rule of mutual information


I(X;Y,Z)=I(X;Y|Z)+I(X;Z)

Conditioning reduces entropy.

Lower bound for communicating secret

Consider the Shamir scheme (z = t - 1, one message).

Q: What is H(s_i) with respect to H(M) ? A: Fix any \mathcal{T} = \{i_1, \cdots, i_t\} \subseteq [n] of size t, and let \mathcal{Z} = \{i_1, \cdots, i_{t-1}\}.


\begin{aligned}
H(M) &= I(M; S_\mathcal{T}) + H(M|S_\mathcal{T}) \text{(by def. of mutual information)}\\
&= I(M; S_\mathcal{T}) \text{(since S_\mathcal{T} suffice to decode M)}\\
&= I(M; S_{i_t}, S_\mathcal{Z}) \text{(since S_\mathcal{T} = S_\mathcal{Z} ∪ S_{i_t})}\\
&= I(M; S_{i_t}|S_\mathcal{Z}) + I(M; S_\mathcal{Z}) \text{(chain rule)}\\
&= I(M; S_{i_t}|S_\mathcal{Z}) \text{(since \mathcal{Z} ≤ z, it reveals nothing about M)}\\
&= I(S_{i_t}; M|S_\mathcal{Z}) \text{(symmetry of mutual information)}\\
&= H(S_{i_t}|S_\mathcal{Z}) - H(S_{i_t}|M,S_\mathcal{Z}) \text{(def. of conditional mutual information)}\\
\leq H(S_{i_t}|S_\mathcal{Z}) \text{(entropy is non-negative)}\\
\leq H(S_{i_t}|S_\mathcal{Z}) \text{(conditioning reduces entropy). \\
\end{aligned}

So the bits used for sharing the secret is at least the bits of actual secret.

In Shamir we saw: H(s_i) \geq H(M).

If M is uniform (standard assumption), then Shamir achieves this bound with equality.
In ramp secret sharing we have H(s_i) \geq \frac{1}{t-z}H(M_1, \cdots, M_{t-z}) (similar proof).
Also optimal if M is uniform.

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