How to Construct Strongly Secure Network Coding Scheme

Abstract

We say that a network coding scheme is strongly $1$-secure if a source node $s$ can multicast $n$ field elements $\{m_1, \cdots , m_n\}$ to a set of sink nodes $\{t_1, \cdots , t_q\}$ in such a way that any single edge leaks no information on any $S \subset \{m_1, \cdots , m_n\}$ with $|S|=n-1$, where $n=\min _{t_i}$max-flow$(s,t_i)$ is the maximum transmission capacity. We also say that a strongly $h$-secure network coding scheme is strongly $(h+1)$-secure if any $h+1$ edges leak no information on any $S \subset \{m_1, \cdots , m_n\}$ with $|S|=n-(h+1)$.

In this paper, we show the first explicit algorithm which can construct strongly $k$-secure network coding schemes. In particular, it runs in polynomial time for fixed $k$.

Appendices

A Example of Secure Linear Network Coding Schemes

Fig. 3.

$1$-Secure linear network coding scheme (mod3)

Fig. 4.

Strongly $1$-secure linear network coding scheme (mod5)

Fig. 5.

Strongly $2$-secure linear network coding scheme (mod11)

B Example of $D$-Zero Projection

Consider

$$ U=\left( \begin{array}{ccc} 1 &{} 1 &{} 2 \\ 0 &{} 1 &{} 1 \end{array} \right) $$

over $\mathsf{F}_5$. Then

$$ U_{\{1,2\}} = \left( \begin{array}{cc} 1 &{} 1 \\ 0 &{} 1 \end{array} \right) , U_{\{1,3\}} = \left( \begin{array}{cc} 1 &{} 2 \\ 0 &{} 1 \end{array} \right) , U_{\{2,3\}} = \left( \begin{array}{cc} 1 &{} 2 \\ 1 &{} 1 \end{array} \right) . $$

Therefore

$$\begin{aligned} \mathsf{Rank}_2=\{\{1,2\}, \{1,3\}, \{2,3\}\} \end{aligned}$$

because

$$\begin{aligned} rank(U_{\{1,2\}}) = rank(U_{\{1,3\}})=rank(U_{\{2,3\}})=2. \end{aligned}$$

Next

$$\begin{aligned} U_{\{1,2\}, \{1\}}&= (1,1) \\ U_{\{1,3\}, \{1\}}&= (1,2) \\ U_{\{2,3\}, \{1\}}&= (1,2) \end{aligned}$$

Therefore from Lemma 3, there exists a $\{1\}$-zero projection of $W$ because

$$\begin{aligned} rank(U_{\{1,2\}, \{1\}})= rank(U_{\{1,3\}, \{1\}})=rank(U_{\{2,3\}, \{1\}})=1 \end{aligned}$$

Let

$$\begin{aligned} \mathbf{b}_{\{1,2\}}&= -(1,0)^T+(1,1)^T=(0, 1)^T \\ \mathbf{b}_{\{1,3\}}&= -2\cdot (1,0)^T+(2,1)^T=(0, 1)^T \\ \mathbf{b}_{\{2,3\}}&= -2\cdot (1,1)^T+(2,1)^T=(0, -1)^T. \end{aligned}$$

Then $\mathbf{b}_A$ is a $\{1\}$-zero projection of $U_A$ for $A=\{1,2\}, \{1,3\}$ and $\{2,3\}$. Therefore $W$ such that

$$ W=(\mathbf{b}_{\{1,2\}}, \mathbf{b}_{\{1,3\}}, \mathbf{b}_{\{2,3\}}) =\left( \begin{array}{ccc} 0 &{} 0 &{} 0 \\ 1 &{} 1 &{} -1 \end{array} \right) $$

is a $\{1\}$-zero projection of $U$. Finally the second row of $W$ consists of nonzero elements. Therefore

$$\begin{aligned} rank(U_{\{1,2\},\{1,2\}})=rank(U_{\{1,3\},\{1,2\}})=rank(U_{\{2,3\},\{1,2\}})=2 \end{aligned}$$

from Lemma 4.

C Proof of Theorem 3

At line 5 and line 6, we can show that there exists such a $D$-zero projection $W_D$ of $U^*$ by induction on $i$ based on Lemma 3. (See the footnotes of Sec. 5.3.) At line 8, the $(i+1)$th row of $Y_i$ consists nonzero elements, and the other rows are the same as those of $X_i$. Therefore the final $Y_{n-1}$ looks as follows, where $U^*=T \cdot U$. It is also easy to see that $W_D^*$ is a $D$-zero projection of $U^*$ for all $D \subset \{1, \cdots , n\}$ such that $|D|<k$ (Fig. 6).

Fig. 6.

The final $Y_{n-1}$

In the above figure, since all the elements of $U^*$ are nonzeros, it is clear that

$$\begin{aligned} rank(U^*_A)=rank(U^*_{A, \{1\}})= \cdots = rank(U^*_{A, \{n\}})=1 \end{aligned}$$

for any $A \in \mathsf{Rank}_1$. Next from Lemma 4 and from the above figure, we have

$$\begin{aligned} rank(U^*_A)=rank(U^*_{A, \{1,2\}})= \cdots = rank(U^*_{A, \{n-1,n\}})=2 \end{aligned}$$

for any $A \in \mathsf{Rank}_2$. Similarly, we can see that

$$\begin{aligned} rank(U^*_A)=rank(U^*_{A,B}) \end{aligned}$$

for any $A \in \mathsf{Rank}_j$ and any $B \subset \{1, \cdots , n\}$ such that $|B|= j$ for $j=1, \cdots , k$. Therefore $U^*$ is strongly $k$-secure from Proposition 3.

Lemma 5

Let $L_k$ be the number of columns of the final $X$. Then

$$\begin{aligned} L_k \le |\mathcal{E}| + \sum _{i=1}^{k-1} {n-1 \atopwithdelims ()i} {|\mathcal{E}| \atopwithdelims ()i+1}. \end{aligned}$$

Proof

Let $\#A$ denote the number of columns of a matrix $A$. Then

$$\begin{aligned} L_k=\#U + \sum _{h=1}^{k-1} {n-1 \atopwithdelims ()h} \sum _{|D|=h} \#W_D \end{aligned}$$

If $|D|=h$, then we have

$$\begin{aligned} \#W_D = |\mathsf{Rank}_{h+1}| \le {|\mathcal{E}| \atopwithdelims ()h+1} \end{aligned}$$

from Eq. (1), Therefore we have this lemma. $\square$

Therefore at line 7, we can compute each $T_i$ if $|\mathsf{F}| \ge L \ge L_k$ in time $O(nL)$ from Theorem 1. To compute all $T_i$, it takes time $O(n^2L)$.

At line 5, it takes $O(n|D|^2)$ time to compute each $W_D$. To compute all $W_D$, it takes time $O(\sum _{i=1}^k ni^2 {n-1 \atopwithdelims ()i})$ which is bounded by $O(n^2L)$.

Finally the time complexity of line 2 and line 9 is bounded by $O(n^2L)$. Therefore our algorithm runs in time $O(n^2L)$.