Bounds and constructions for ${\overline{3}}$-separable codes with length 3

Abstract

Separable codes were introduced to provide protection against illegal redistribution of copyrighted multimedia material. Let ${\mathcal {C}}$ be a code of length n over an alphabet of q letters. The descendant code ${\mathsf{desc}}({\mathcal {C}}_0)$ of ${\mathcal {C}}_0 = \{\mathbf{c}_1, \mathbf{c}_2, \ldots , \mathbf{c}_t\} \subseteq {{\mathcal {C}}}$ is defined to be the set of words $\mathbf{x} = (x_1, x_2, \ldots ,x_n)^T$ such that $x_i \in \{c_{1,i}, c_{2,i}, \ldots , c_{t,i}\}$ for all $i=1, \ldots , n$, where $\mathbf{c}_j=(c_{j,1},c_{j,2},\ldots ,c_{j,n})^T$. ${\mathcal {C}}$ is a ${\overline{t}}$-separable code if for any two distinct ${\mathcal {C}}_1, {\mathcal {C}}_2 \subseteq {\mathcal {C}}$ with $|{\mathcal {C}}_1| \le t$, $|{\mathcal {C}}_2| \le t$, we always have ${\mathsf{desc}}({\mathcal {C}}_1) \ne {\mathsf{desc}}({\mathcal {C}}_2)$. Let $M({\overline{t}},n,q)$ denote the maximal possible size of such a separable code. In this paper, an upper bound on $M({\overline{3}},3,q)$ is derived by considering an optimization problem, and then two constructions for ${\overline{3}}\hbox {-SC}(3,M,q)$s are provided by means of perfect hash families and Steiner triple systems.

Appendix: Proof of Theorem 2.3

The necessity is clear from Theorem 1.3 and the discussion in Sect. 2. We consider its sufficiency. From the relationship between a separable code and a FPC in Theorem 1.3, it is sufficient to prove that for any distinct $A, B\subseteq {\mathcal {C}}$ with $|A|=|B|=3$, we have ${\mathsf{desc}}(A)\ne {\mathsf{desc}}(B)$ except that $A \bigcup B$ equals one of the forbidden configurations in (1) and (2).

Let $A=\{\mathbf{a}=(a_1, a_2, a_3)^T$, $\mathbf{b}=(b_1,b_2,b_3)^T$, $\mathbf{c}=(c_1,c_2,c_3)^T\}\subseteq {\mathcal {C}}$ and $B=\{\mathbf{d}=(d_1,d_2,d_3)^T$, $\mathbf{e}=(e_1,e_2,e_3)^T$, $\mathbf{f}=(f_1,f_2,f_3)^T\}$ $\subseteq {\mathcal {C}}$. Suppose that ${\mathsf{desc}}(A)={\mathsf{desc}}(B)$, $A\ne B$ and $|A|=|B|=3$. Then we have $\{a_1,b_1,c_1\}=\{d_1,e_1,f_1\}$, $\{a_2,b_2,c_2\}=\{d_2,e_2,f_2\}$ and $\{a_3,b_3,c_3\}=\{d_3,e_3,f_3\}$ by the definition of a separable code. There are three cases to be considered:

• (I) $|A\bigcap B|=2$. Without loss of generality, we may assume that $\mathbf{b}=\mathbf{e}$ and $\mathbf{c}=\mathbf{f}$, that is, $A=\{\mathbf{a},\mathbf{b},\mathbf{c}\}$, $B=\{\mathbf{b},\mathbf{c},\mathbf{d}\}$. We list them as follows.

  $$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} b_{1} &{} c_{1} &{} d_{1}\\ a_{2} &{} b_{2} &{} c_{2} &{} d_{2}\\ a_{3} &{} b_{3} &{} c_{3} &{} d_{3} \end{array}\right) \end{aligned}$$

  (6)

  Since ${\mathcal {C}}$ is a 2-FPC, the following statements hold.

  • $\mathbf{d}\not \in {\mathsf{desc}}(\{\mathbf{a},\mathbf{b}\})$ holds because $\mathbf{d}\not \in \{\mathbf{a},\mathbf{b}\}$. Without loss of generality, we may assume that $d_{1}\not \in \{a_{1},b_{1}\}$, then $c_1=d_1$ and $a_1=b_1$ always hold since $\{a_1,b_1,c_1\} =\{b_1,c_1,d_1\}$. Then (6) can be written as follows.

    $$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} a_{1} &{} c_{1} &{} c_{1}\\ a_{2} &{} b_{2} &{} c_{2} &{} d_{2}\\ a_{3} &{} b_{3} &{} c_{3} &{} d_{3} \end{array}\right) \end{aligned}$$

    (7)

  • $\mathbf{d}\not \in {\mathsf{desc}}(\{\mathbf{b},\mathbf{c}\})$ holds because $\mathbf{d}\not \in \{\mathbf{b},\mathbf{c}\}$. Without loss of generality, we may assume that $d_{2}\not \in \{b_{2},c_{2}\}$, then $a_2=d_2$ always holds since $\{a_2,b_2,c_2\} =\{b_2,c_2,d_2\}$. Then (7) can be written as follows.

    $$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} a_{1} &{} c_{1} &{} c_{1}\\ a_{2} &{} b_{2} &{} c_{2} &{} a_{2}\\ a_{3} &{} b_{3} &{} c_{3} &{} d_{3} \end{array}\right) \end{aligned}$$

    (8)

  • $\mathbf{d}\not \in {\mathsf{desc}}(\{\mathbf{a},\mathbf{c}\})$ holds because $\mathbf{d}\not \in \{\mathbf{a},\mathbf{c}\}$. Then we must have $d_{3}\not \in \{a_{3},c_{3}\}$, which implies $b_3=d_3$ and $a_3=c_3$ since $\{a_3,b_3,c_3\}=\{b_3,c_3,d_3\}$. Then (8) can be written as follows.

    $$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} a_{1} &{} c_{1} &{} c_{1}\\ a_{2} &{} b_{2} &{} c_{2} &{} a_{2}\\ a_{3} &{} b_{3} &{} a_{3} &{} b_{3} \end{array}\right) \end{aligned}$$

    (9)

    Obviously, (9) equals $\triangle _1$ in (1), which is a forbidden configuration in ${\mathcal {C}}$.

• (II) $|A\bigcap B|=1$. Without loss of generality, we may assume that $\mathbf{c}=\mathbf{f}$, that is, $A=\{\mathbf{a},\mathbf{b},\mathbf{c}\}$, $B=\{\mathbf{c},\mathbf{d},\mathbf{e}\}$. We list them as follows.

  $$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} b_{1} &{} c_{1} &{} d_{1} &{} e_{1}\\ a_{2} &{} b_{2} &{} c_{2} &{} d_{2} &{} e_{2}\\ a_{3} &{} b_{3} &{} c_{3} &{} d_{3} &{} e_{3} \end{array}\right) \end{aligned}$$

  (10)

  Since ${\mathcal {C}}$ is a 2-FPC, the following statements hold.

  • $\mathbf{d}\not \in {\mathsf{desc}}(\{\mathbf{a},\mathbf{b}\})$ holds because $\mathbf{d}\not \in \{\mathbf{a},\mathbf{b}\}$. Without loss of generality, we may assume that $d_{1}\not \in \{a_{1},b_{1}\}$, then $c_1=d_1$ and $a_1=b_1=e_1$ always hold since $\{a_1,b_1,c_1\}=\{c_1,d_1,e_1\}$. Then (10) can be written as follows.

    $$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} a_{1} &{} c_{1} &{} c_{1} &{} a_{1}\\ a_{2} &{} b_{2} &{} c_{2} &{} d_{2} &{} e_{2}\\ a_{3} &{} b_{3} &{} c_{3} &{} d_{3} &{} e_{3} \end{array}\right) \end{aligned}$$

    (11)

  • $\mathbf{e}\not \in {\mathsf{desc}}(\{\mathbf{a},\mathbf{b}\})$ holds because $\mathbf{e}\not \in \{\mathbf{a},\mathbf{b}\}$. Without loss of generality, we may assume that $e_{2}\not \in \{a_{2},b_{2}\}$, then $c_2=e_2$ and $a_2=b_2=d_2$ always hold since $\{a_2,b_2,c_2\}=\{c_2,d_2,e_2\}$. Then (11) can be written as follows.

    $$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} a_{1} &{} c_{1} &{} c_{1} &{} a_{1}\\ a_{2} &{} a_{2} &{} c_{2} &{} a_{2} &{} c_{2}\\ a_{3} &{} b_{3} &{} c_{3} &{} d_{3} &{} e_{3} \end{array}\right) \end{aligned}$$

    (12)

  • $\mathbf{d}\not \in {\mathsf{desc}}(\{\mathbf{b},\mathbf{c}\})$ holds because $\mathbf{d}\not \in \{\mathbf{b},\mathbf{c}\}$. Then we must have $d_{3}\not \in \{b_{3},c_{3}\}$, which implies $a_3=d_3$ since $\{a_3,b_3,c_3\}=\{c_3,d_3,e_3\}$. Then (12) can be written as follows.

    $$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} a_{1} &{} c_{1} &{} c_{1} &{} a_{1}\\ a_{2} &{} a_{2} &{} c_{2} &{} a_{2} &{} c_{2}\\ a_{3} &{} b_{3} &{} c_{3} &{} a_{3} &{} e_{3} \end{array}\right) \end{aligned}$$

    Obviously, $\mathbf{d}\in {\mathsf{desc}}(\{\mathbf{a},\mathbf{c}\})$ holds. This contradicts the definition of a 2-FPC since $\mathbf{d}\not \in \{\mathbf{a},\mathbf{c}\}$.

  So the case of $|A\bigcap B|=1$ can never happen.

• (III) $|A\bigcap B|=0$. We list A and B as follows.

  $$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} b_{1} &{} c_{1} &{} d_{1} &{} e_{1} &{} f_{1}\\ a_{2} &{} b_{2} &{} c_{2} &{} d_{2} &{} e_{2} &{} f_{2}\\ a_{3} &{} b_{3} &{} c_{3} &{} d_{3} &{} e_{3} &{} f_{3} \end{array}\right) \end{aligned}$$

  (13)

  Since ${\mathcal {C}}$ is a 2-FPC, the following statements hold.

  • $\mathbf{d}\not \in {\mathsf{desc}}(\{\mathbf{a},\mathbf{b}\})$ holds because $\mathbf{d}\not \in \{\mathbf{a},\mathbf{b}\}$. Without loss of generality, we may assume that $d_{1}\not \in \{a_{1},b_{1}\}$, then $c_1=d_1$ always holds since $\{a_1,b_1,c_1\} =\{d_1,e_1,f_1\}$. Then (13) can be written as follows.

    $$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} b_{1} &{} c_{1} &{} c_{1} &{} e_{1} &{} f_{1}\\ a_{2} &{} b_{2} &{} c_{2} &{} d_{2} &{} e_{2} &{} f_{2}\\ a_{3} &{} b_{3} &{} c_{3} &{} d_{3} &{} e_{3} &{} f_{3} \end{array}\right) \end{aligned}$$

    (14)

  • $\mathbf{d}\not \in {\mathsf{desc}}(\{\mathbf{a},\mathbf{c}\})$ holds because $\mathbf{d}\not \in \{\mathbf{a},\mathbf{c}\}$. Without loss of generality, we may assume that $d_{2}\not \in \{a_{2},c_{2}\}$, then $b_2=d_2$ always holds since $\{a_2,b_2,c_2\} =\{d_2,e_2,f_2\}$. Then (14) can be written as follows.

    $$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} b_{1} &{} c_{1} &{} c_{1} &{} e_{1} &{} f_{1}\\ a_{2} &{} b_{2} &{} c_{2} &{} b_{2} &{} e_{2} &{} f_{2}\\ a_{3} &{} b_{3} &{} c_{3} &{} d_{3} &{} e_{3} &{} f_{3} \end{array}\right) \end{aligned}$$

    (15)

  • $\mathbf{d}\not \in {\mathsf{desc}}(\{\mathbf{b},\mathbf{c}\})$ holds because $\mathbf{d}\not \in \{\mathbf{b},\mathbf{c}\}$. Then we must have $d_{3}\not \in \{b_{3},c_{3}\}$, which implies $a_3=d_3$ since $\{a_3,b_3,c_3\}=\{d_3,e_3,f_3\}$. Then (15) can be written as follows.

    $$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} b_{1} &{} c_{1} &{} c_{1} &{} e_{1} &{} f_{1}\\ a_{2} &{} b_{2} &{} c_{2} &{} b_{2} &{} e_{2} &{} f_{2}\\ a_{3} &{} b_{3} &{} c_{3} &{} a_{3} &{} e_{3} &{} f_{3} \end{array}\right) \end{aligned}$$

  • $\mathbf{e}\not \in {\mathsf{desc}}(\{\mathbf{a},\mathbf{b}\})$ holds because $\mathbf{e}\not \in \{\mathbf{a},\mathbf{b}\}$. There must exist an index $1\le i\le 3$ such that $e_{i}\not \in \{a_{i},b_{i}\}$ holds.

    • (1.1) If $e_{1}\not \in \{a_{1},b_{1}\}$ holds, then $c_1=e_1$ always holds since $\{a_1,b_1,c_1\}=\{c_1,e_1,f_1\}$;

    • (1.2) If $e_{2}\not \in \{a_{2},b_{2}\}$ holds, then $c_2=e_2$ always holds since $\{a_2,b_2,c_2\}=\{b_2,e_2,f_2\}$;

    • (1.3) If $e_{3}\not \in \{a_{3},b_{3}\}$ holds, then $c_3=e_3$ always holds since $\{a_3,b_3,c_3\}=\{a_3,e_3,f_3\}$.

  • $\mathbf{e}\not \in {\mathsf{desc}}(\{\mathbf{a},\mathbf{c}\})$ holds because $\mathbf{e}\not \in \{\mathbf{a},\mathbf{c}\}$. There must exist an integer $1\le i\le 3$ such that $e_{i}\not \in \{a_{i},c_{i}\}$ holds.

    • (2.1) If $e_{1}\not \in \{a_{1},c_{1}\}$ holds, then $b_1=e_1$ always holds since $\{a_1,b_1,c_1\}=\{c_1,e_1,f_1\}$;

    • (2.2) If $e_{2}\not \in \{a_{2},c_{2}\}$ holds, then $b_2=e_2$ always holds since $\{a_2,b_2,c_2\}=\{b_2,e_2,f_2\}$;

    • (2.3) If $e_{3}\not \in \{a_{3},c_{3}\}$ holds, then $b_3=e_3$ always holds since $\{a_3,b_3,c_3\}=\{a_3,e_3,f_3\}$.

  • $\mathbf{e}\not \in {\mathsf{desc}}(\{\mathbf{b},\mathbf{c}\})$ holds because $\mathbf{e}\not \in \{\mathbf{b},\mathbf{c}\}$. There must exist an index $1\le i\le 3$ such that $e_{i}\not \in \{b_{i},c_{i}\}$ holds.

    • (3.1) If $e_{1}\not \in \{b_{1},c_{1}\}$ holds, then $a_1=e_1$ always holds since $\{a_1,b_1,c_1\}=\{c_1,e_1,f_1\}$;

    • (3.2) If $e_{2}\not \in \{b_{2},c_{2}\}$ holds, then $a_2=e_2$ always holds since $\{a_2,b_2,c_2\}=\{b_2,e_2,f_2\}$;

    • (3.3) If $e_{3}\not \in \{b_{3},c_{3}\}$ holds, then $a_3=e_3$ always holds since $\{a_3,b_3,c_3\}=\{a_3,e_3,f_3\}$.

  • $\mathbf{f}\not \in {\mathsf{desc}}(\{\mathbf{a},\mathbf{b}\})$ holds because $\mathbf{f}\not \in \{\mathbf{a},\mathbf{b}\}$. There must exist an index $1\le i\le 3$ such that $f_{i}\not \in \{a_{i},b_{i}\}$ holds.

    • (4.1) If $f_{1}\not \in \{a_{1},b_{1}\}$ holds, then $c_1=f_1$ always holds since $\{a_1,b_1,c_1\}=\{c_1,e_1,f_1\}$;

    • (4.2) If $f_{2}\not \in \{a_{2},b_{2}\}$ holds, then $c_2=f_2$ always holds since $\{a_2,b_2,c_2\}=\{b_2,e_2,f_2\}$;

    • (4.3) If $f_{3}\not \in \{a_{3},b_{3}\}$ holds, then $c_3=f_3$ always holds since $\{a_3,b_3,c_3\}=\{a_3,e_3,f_3\}$.

  • $\mathbf{f}\not \in {\mathsf{desc}}(\{\mathbf{a},\mathbf{c}\})$ holds because $\mathbf{f}\not \in \{\mathbf{a},\mathbf{c}\}$. There must exist an index $1\le i\le 3$ such that $f_{i}\not \in \{a_{i},c_{i}\}$ holds.

    • (5.1) If $f_{1}\not \in \{a_{1},c_{1}\}$ holds, then $b_1=f_1$ always holds since $\{a_1,b_1,c_1\}=\{c_1,e_1,f_1\}$;

    • (5.2) If $f_{2}\not \in \{a_{2},c_{2}\}$ holds, then $b_2=f_2$ always holds since $\{a_2,b_2,c_2\}=\{b_2,e_2,f_2\}$;

    • (5.3) If $f_{3}\not \in \{a_{3},c_{3}\}$ holds, then $b_3=f_3$ always holds since $\{a_3,b_3,c_3\}=\{a_3,e_3,f_3\}$.

  • $\mathbf{f}\not \in {\mathsf{desc}}(\{\mathbf{b},\mathbf{c}\})$ holds because $\mathbf{f}\not \in \{\mathbf{b},\mathbf{c}\}$. There must exist an index $1\le i\le 3$ such that $f_{i}\not \in \{b_{i},c_{i}\}$ holds.

    • (6.1) If $f_{1}\not \in \{b_{1},c_{1}\}$ holds, then $a_1=f_1$ always holds since $\{a_1,b_1,c_1\}=\{c_1,e_1,f_1\}$;

    • (6.2) If $f_{2}\not \in \{b_{2},c_{2}\}$ holds, then $a_2=f_2$ always holds since $\{a_2,b_2,c_2\}=\{b_2,e_2,f_2\}$;

    • (6.3) If $f_{3}\not \in \{b_{3},c_{3}\}$ holds, then $a_3=f_3$ always holds since $\{a_3,b_3,c_3\}=\{a_3,e_3,f_3\}$.

It is easy to check that for any $i\in \{1,2,3\}$ or any $i\in \{4,5,6\}$, and for any $j\in \{1,2,3\}$, once the case (i.j) above occurs, none of $(i'.j)$ cases occurs for any $i'\in \{1,2,3\}\setminus \{i\}$ or any $i'\in \{4,5,6\}\setminus \{i\}$, respectively. So there are $3\times 2\times 1\times 3\times 2\times 1=36$ cases to be considered.

$$\begin{aligned}&\{(1.3),(2.2),(3.1),(4.2),(5.1),(6.3)\},\ \ \ \{(1.1),(2.3),(3.2),(4.2),(5.1),(6.3)\},\\&\{(1.1),(2.3),(3.2),(4.2),(5.3),(6.1)\},\ \ \ \{(1.1),(2.3),(3.2),(4.3),(5.1),(6.2)\},\\&\{(1.1),(2.3),(3.2),(4.3),(5.2),(6.1)\},\ \ \ \{(1.2),(2.1),(3.3),(4.1),(5.3),(6.2)\},\\&\{(1.2),(2.1),(3.3),(4.2),(5.3),(6.1)\},\ \ \ \{(1.2),(2.1),(3.3),(4.3),(5.1),(6.2)\},\\&\{(1.2),(2.1),(3.3),(4.3),(5.2),(6.1)\},\ \ \ \{(1.2),(2.3),(3.1),(4.1),(5.3),(6.2)\},\\&\{(1.2),(2.3),(3.1),(4.2),(5.1),(6.3)\},\ \ \ \{(1.2),(2.3),(3.1),(4.3),(5.2),(6.1)\},\\&\{(1.3),(2.1),(3.2),(4.1),(5.3),(6.2)\},\ \ \ \{(1.3),(2.1),(3.2),(4.2),(5.1),(6.3)\},\\&\{(1.3),(2.1),(3.2),(4.3),(5.2),(6.1)\},\ \ \ \{(1.3),(2.2),(3.1),(4.1),(5.3),(6.2)\},\\&\{(1.3),(2.2),(3.1),(4.2),(5.3),(6.1)\},\ \ \ \{(1.3),(2.2),(3.1),(4.3),(5.1),(6.2)\},\\&\{(1.1),(2.2),(3.3),(4.1),(5.2),(6.3)\},\ \ \ \{(1.1),(2.2),(3.3),(4.1),(5.3),(6.2)\},\\&\{(1.1),(2.2),(3.3),(4.2),(5.1),(6.3)\},\ \ \ \{(1.1),(2.2),(3.3),(4.2),(5.3),(6.1)\},\\&\{(1.1),(2.2),(3.3),(4.3),(5.1),(6.2)\},\ \ \ \{(1.1),(2.2),(3.3),(4.3),(5.2),(6.1)\},\\&\{(1.1),(2.3),(3.2),(4.1),(5.2),(6.3)\},\ \ \ \{(1.1),(2.3),(3.2),(4.1),(5.3),(6.2)\},\\&\{(1.2),(2.1),(3.3),(4.1),(5.2),(6.3)\},\ \ \ \{(1.2),(2.1),(3.3),(4.2),(5.1),(6.3)\},\\&\{(1.2),(2.3),(3.1),(4.1),(5.2),(6.3)\},\ \ \ \{(1.2),(2.3),(3.1),(4.2),(5.3),(6.1)\},\\&\{(1.3),(2.1),(3.2),(4.1),(5.2),(6.3)\},\ \ \ \{(1.3),(2.1),(3.2),(4.3),(5.1),(6.2)\},\\&\{(1.3),(2.2),(3.1),(4.1),(5.2),(6.3)\},\ \ \ \{(1.3),(2.2),(3.1),(4.3),(5.2),(6.1)\},\\&\{(1.2),(2.3),(3.1),(4.3),(5.1),(6.2)\},\ \ \ \{(1.3),(2.1),(3.2),(4.2),(5.3),(6.1)\}. \end{aligned}$$

It is readily checked that none of the first 18 subcases satisfies the condition (i) in this theorem. For example, consider the subcase $\{(1.3),(2.2),(3.1),(4.2),(5.1),(6.3)\}$, that is,

$$\begin{aligned} (1.3) \ c_3=e_3; \ (2.2) \ b_2 = e_2; \ (3.1) \ a_1 = e_1;\ (4.2) \ c_2=f_2; \ (5.1) \ b_1 = f_1; \ (6.3) \ a_3 = f_3. \end{aligned}$$

Then the corresponding subcode can be written as follows.

$$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} b_{1} &{} c_{1} &{} c_{1} &{} a_{1} &{} b_{1}\\ a_{2} &{} b_{2} &{} c_{2} &{} b_{2} &{} b_{2} &{} c_{2}\\ a_{3} &{} b_{3} &{} c_{3} &{} a_{3} &{} c_{3} &{} a_{3} \end{array}\right) \end{aligned}$$

where $\mathbf{e}=(a_{1},b_{2},c_{3})^T$ and $\mathbf{f}=(b_{1},c_{2},a_{3})^T$. Obviously, $\mathbf{b}\in {\mathsf{desc}}(\{\mathbf{e},\mathbf{f}\})$ holds since $b_{3}\in \{a_3,c_3\}$. This is a contradiction to the definition of a 2-FPC because of the assumption $\mathbf{b}\not \in \{\mathbf{e},\mathbf{f}\}$. It is also easy to check that none of the next 16 subcases satisfies the condition that $|B|=3$. Finally the remaining 2 subcases correspond to the forbidden configuration in (2).

The proof is complete.$\square$