A Construction of Perfect Secret Sharing

In this section we introduce our construction of perfect secret sharing scheme on G⁰_k,n whose information ratio is equal to 2.

Our constructions follow the ideal outlined in Section 1.3. In order to construct a perfect secret sharing scheme with ratio max_v∈Gdim(L_v)/dim(L_s), we start with a high-dimensional vector space F, and assign linear subspaces to the vertices and the secret so that

• if vu is an edge of the graph, then the linear span of the subspaces L_v and L_u contains the subspace L_s which is assigned to the secret, and

• if {v₁, v₂, ..., v_k} is an independent set, then Span({L_v1, L_v2, ..., L_vk}) ∩ L_s = {0}.

In our construction, F is a d(kn + 1)-dimensional vector space and subspaces will be given as the linear span of certain vectors. We split these coordinates into kn + 1 groups of d coordinates each. Now, we need some more definition and notation to help us describe our construction. If x and y are two `-dimensional vectors, then x<sup>k</sup> is defined as the k`-dimensional vector obtained by repeating the coordinates of x k times. The vector x ⊕ y is 2`-dimensional vector obtained by concatenating vector y after x. For example , if x = (010) and y = (101), then x³ = (010010010) and y ⊕ x² = (101010010).

Construction 3.2. Let λ₁, λ₂, ..., λ_km be km distinct integers, and let λ_x− λ_y be denoted as λ_x,y.

The subspace L_s assigned to the secret is spanned by the following d vectors:

(1000 · · · 0)^kn+1, (0100 · · · 0)^kn+1, (0010 · · · 0)^kn+1, ..., (000 · · · 01)^kn+1.

The subspace L_wj assigned to w_j is spanned by the following d vectors:

(0 · · · 0)^k(j−1)⊕ (100 · · · 0)^k⊕ (0 · · · 0)^k(n−j)+1, (0 · · · 0)^k(j−1)⊕ (010 · · · 0)^k⊕ (0 · · · 0)^k(n−j)+1,

...

(0 · · · 0)^k(j−1)⊕ (00 · · · 01)^k⊕ (0 · · · 0)^k(n−j)+1.

Furthermore, the subspace L_vi,j assigned to v_i,j is spanned by the following 2d vectors:

(100 · · · 0)^k(j−1)⊕ (00 · · · 0)^k⊕ (100 · · · 0)^k(n−j)+1,

Figure 3.2 shows the graphs G⁰_2,2 and we give our construction of secret sharing scheme on it in Example 3.3.

Example 3.3.

L_s = Span{(100100100100100), (010010010010010), (001001001001001)}

L_w1 = Span{(100100000000000), (010010000000000), (000000000000000)}

Figure 3.2: G⁰_2,2

L_w2 = Span{(000000100100000), (000000010010000), (000000001001000)}

L_v1,1 = Span{(000000100100100), (000000010010010), (000000001001001), (λ_1,100λ_1,200λ_1,300λ_1,400λ₁00), (0λ_1,100λ_1,200λ_1,300λ_1,400λ₁0), (00λ_1,100λ_1,200λ_1,300λ_1,400λ₁)}

L_v2,1 = Span{(000000100100100), (000000010010010), (000000001001001), (λ_2,100λ_2,200λ_2,300λ_2,400λ₂00), (0λ_2,100λ_2,200λ_2,300λ_2,400λ₂0), (00λ_2,100λ_2,200λ_2,300λ_2,400λ₂)}

L_v1,2 = Span{(100100000000100), (010010000000010), (001001000000001), (λ_3,100λ_3,200λ_3,300λ_3,400λ₃00), (0λ_3,100λ_3,200λ_3,300λ_3,400λ₃0), (00λ3,100λ3,200λ3,300λ3,400λ3)}

L_v2,2 = Span{(100100000000100), (010010000000010), (001001000000001), (λ4,100λ4,200λ4,300λ4,400λ400), (0λ4,100λ4,200λ4,300λ4,400λ40), (00λ_4,100λ_4,200λ_4,300λ_4,400λ₄)}

Theorem 3.4. Constriction 3.2. defines a perfect secret sharing scheme on G⁰_k,n with information 2.

Proof. To show that Construction 3.2. is a perfect secret sharing scheme on G⁰_k,n, we need to check the following conditions.

1. the span of Lw1, Lw2, ..., Lwn must be trivial, 2. the span of L_vi,j and L_wj must contain L_s,

3. the span of L_vi,j and {L_wm : m 6= j} intersects L_s in the trivial space {0}, and

4. the span of two different Lvi,j and Lvm,n should contain Ls.

Since the linear span of all subspaces L_wj’s contains those vectors where all coor-dinates in the (kn + 1)-th group are zero and any non-trivial linear combination of L_s has non-zero coordinates in each group, we have

Span{L_w1, L_w2, ..., L_wn} ∩ L_s = {0}, The first requirement for the independent set W is satisfied.

To verify the second condition, for each ` ∈ {1, 2, ..., d}, the sum of the `-th gen-erating vector of L_vi,j and L_wj gives the `-th generating element of L<sub>s</sub>. For example, when ` = 1

This implies that the linear span of L_vi,j and L_wj contains L_s as required.

Observe that the first d generating vectors in L_vi,j have all 0 in the (k(j −1)+1)-th to the (kj)-th groups, and the other d generating vectors in L_vi,j have all 0 in the (k(j − 1) + i)-th group. Hence the linear span of L_vi,j and all other L_wm’s with j 6= m has all zero coordinates in this group and therefore contains only the zero element from Ls.

In order to have the last condition satisfied, subtracting the d+1 generating vector of L_vs,r from the d + 1 generating vector of L_vi,j with (i, j) 6= (s, r) gives

The linear span of this vector contains the first generating vector of L_s. Since each generating vector of L_s can be obtained in the same way, the last condition holds as well.

With dim(L_vs) = d, dim(L_wj) = d and dim(L_vi,j) = 2d, we also know that the per-fect secret sharing scheme we have constructed has information ratio 2. 

By Theorem 3.1. and Theorem 3.4. we have the following corollary.

Corollary 3.5.

2 − 2⁻ⁿ⁺¹ 6 R(G⁰k,n) 6 2.

3.3 A Construction of Perfect Secret Sharing Scheme on G

Recall that in the graph G⁰⁰_k,n defined at the beginning of this chapter, there is no edge between the vertices from different Vi’s.

Construction 3.6. Let λ₁, λ₂, ..., λ_km be km distinct integers. For convenience, let λ_x− λ_y be denoted by λ_x,y and

a_i,j,m =

( λ_k(j−1)+i , where m = k(t − 1) + i for t = 1 · · · n λ_k(j−1)+i,m , otherwise.

Assign to L_s the subspace spanned by the following d vectors:

(1000 · · · 0)^kn+1, (0100 · · · 0)^kn+1, (0010 · · · 0)^kn+1, ..., (000 · · · 01)^kn+1.

Assign to L_wj the subspace spanned by the following d vectors:

(0 · · · 0)^k(j−1)⊕ (100 · · · 0)^k⊕ (0 · · · 0)^k(n−j)+1, (0 · · · 0)^k(j−1)⊕ (010 · · · 0)^k⊕ (0 · · · 0)^k(n−j)+1,

...

(0 · · · 0)^k(j−1)⊕ (00 · · · 01)^k⊕ (0 · · · 0)^k(n−j)+1.

In addition, L_vi,j is assigned the subspace spanned by the following 2d vectors:

(100 · · · 0)^k(j−1)⊕ (00 · · · 0)^k⊕ (100 · · · 0)^k(n−j)+1,

(010 · · · 0)^k(j−1)⊕ (00 · · · 0)^k⊕ (010 · · · 0)^k(n−j)+1, ...

(00 · · · 01)^k(j−1)⊕ (00 · · · 0)^k⊕ (00 · · · 01)^k(n−j)+1,

" _kn

Figure 3.3 shows the graphs G⁰⁰_2,2 and we give our construction of secret sharing scheme on it in Example 3.7.

Figure 3.3: G⁰⁰_2,2

Example 3.7.

L_s = Span{(100100100100100), (010010010010010), (001001001001001)}

L_w1 = Span{(100100000000000), (010010000000000), (000000000000000)}

Lw2 = Span{(000000100100000), (000000010010000), (000000001001000)}

L_v1,1 = Span{(000000100100100), (000000010010010), (000000001001001), (a1,1,100a1,1,200a1,1,300a1,1,400λ100),

(0a_1,1,100a_1,1,200a_1,1,300a_1,1,400λ₁0), (00a_1,1,100a_1,1,200a_1,1,300a_1,1,400λ₁)}

Lv2,1 = Span{(000000100100100), (000000010010010), (000000001001001), (a_2,1,100a_2,1,200a_2,1,300a_2,1,400λ₂00),

(0a_2,1,100a_2,1,200a_2,1,300a_2,1,400λ₂0), (00a2,1,100a2,1,200a2,1,300a2,1,400λ2)}

L_v1,2 = Span{(100100000000100), (010010000000010), (001001000000001), (a_1,2,100a_1,2,200a_1,2,300a_1,2,400λ₃00),

(0a1,2,100a1,2,200a1,2,300a1,2,400λ30), (00a_1,2,100a_1,2,200a_1,2,300a_1,2,400λ₃)}

L_v2,2 = Span{(100100000000100), (010010000000010), (001001000000001), (a2,2,100a2,2,200a2,2,300a2,2,400λ400),

(0a_2,2,100a_2,2,200a_2,2,300a_2,2,400λ₄0), (00a_2,2,100a_2,2,200a_2,2,300a_2,2,400λ₄)}

Theorem 3.8. Constriction 3.6. defines a perfect secret sharing scheme on G⁰⁰_k,n with information 2.

Proof. To show that Construction 3.6. is a perfect secret sharing scheme on G⁰⁰_k,n, we need to check the following condition.

1. the span of L_w1, L_w2, ..., L_wn is trivial,

2. the span of L_vi,j and L_wj must contain L_s,

3. the span of L_vi,j and {L_wm : m 6= j} intersects L_s in {0},

4. the span of two different Lv and Lu, where v, u ∈ Vi, should contains Ls, and 5. the span of two different L_v and L_u, where v ∈ V_i and u ∈ V_j with i 6= j, should

be the trivial space {0}.

Note that Construction 3.6. is very similar to Construction 3.2, the only difference lies in the last d generating vectors of each Lvi,j for 1 6 i 6 k and 1 6 j 6 n. Hence the first, second, and third conditions hold by the proof of Theorem 3.4.

To verify that the forth condition holds as well, we observe that ( " _kn

The first generating vector of Ls can be obtained from the (d + 1)-th generating vectors of L_vi,j and L_vi,r with j 6= r. The linear span of L_vi,j and L_vi,r contains the generating vectors of L_s, hence the forth condition is also satisfied. To check the fifth condition, one can easily verify that any generating vector of L_s cannot be generated by the vectors in any two different vector subspaces L_vi,j and L_vs,r with s 6= i.

In this construction, L_vi,j is generated by 2d linearly independent vectors, L_wj and L_s are both generated by d linearly independent vectors, thus dim(L_vi,j) = 2d and dim(Lwj) = dim(Ls) = d. This shows that the information ratio of Construction 3.6. is also 2.



By Theorem 3.1. and Theorem 3.8. we have the following corollary.

Corollary 3.9.

2 − 2⁻ⁿ⁺¹ 6 R(G⁰⁰k,n) 6 2.