New secondary constructions of Bent functions

Abstract

In this paper, we first present a novel secondary construction of bent functions (building new bent functions from two already defined ones). Furthermore, the algebraic degree and algebraic immunity of the constructed functions are analysed. Finally, we apply the construction using as initial functions some specific bent functions and then specify sufficient conditions for the resulting bent functions not to be contained in the completed Maiorana–McFarland class. In the second part of the paper, we present a corrigendum of “Constructions of bent–negabent functions and their relation to the completed Maiorana–McFarland Class” (IEEE Trans Inf Theory 61(3):1496–1506, 2015).

Appendix

Proof of Lemma 3

Set $\textsc {f}_0(x)= {\mathop {\mathop {\bigoplus }\limits _{i=1}}\limits _{i\ne \mu }} ^{\frac{n}{2}}\phi _i(x^{(2)})\, x_{i}$ and $\textsc {g}_0(y)= {\mathop {\mathop {\bigoplus }\limits _{j=1}}\limits _{j\ne \rho }} ^{\frac{m}{2}}\psi _j(y^{(2)})\, y_{j}$. We have then

$$\begin{aligned} h(x,y)=\textsc {f}_0(x) \oplus \textsc {g}_0(y)\oplus \, \phi _{\mu }(x^{(2)})\psi _{\rho }(y^{(2)})\oplus \, \theta (x^{(2)})\oplus \varpi (y^{(2)}). \end{aligned}$$

• 1. If there exist $(a_1, 0_{\frac{n}{2}}, 0_{\frac{m}{2}-1}, a_4),(b_1, 0_{\frac{n}{2}}, 0_{\frac{m}{2}-1}, b_4) \in \varDelta$ such that $a_4\ne 0_{\frac{m}{2}}$, $b_4\ne 0_{\frac{m}{2}}$ and $a_4\ne b_4$. Then,

  $$\begin{aligned} D_{(a_1, 0_{\frac{n}{2}})}D_{(b_1, 0_{\frac{n}{2}})}\left( \textsc {f}_0(x)\oplus \theta (x^{(2)})\right) =0. \end{aligned}$$

  We have

  $$\begin{aligned}&D_{(a_1, 0_{\frac{n}{2}}, 0_{\frac{m}{2}-1}, a_4)}D_{(b_1, 0_{\frac{n}{2}}, 0_{\frac{m}{2}-1}, b_4)}h(x,y) \\&\quad = D_{( 0_{\frac{m}{2}-1},a_4)}D_{(0_{\frac{m}{2}-1},b_4)}\textsc {g}_0(y) \oplus D_{ a_4}D_{b_4}\varpi (y^{(2)}) \oplus \phi _{\mu }(x^{(2)}) D_{a_4}D_{b_4}\left( \psi _{\rho }(y^{(2)})\nonumber \right) . \end{aligned}$$

  (5)

  We know that $D_{( 0_{\frac{m}{2}-1},a_4)}D_{(0_{\frac{m}{2}-1},b_4)}\textsc {g}_0(y)$ equals ${\mathop {\mathop {\bigoplus }\limits _{j=1}}\limits _{j\ne \rho }} ^{\frac{m}{2}}\left( D_{a_4}D_{b_4}\psi _j(y^{(2)})\right) \, y_{j}$, which equals 0 or must depend on $y^{(1)}$. Clearly, $\phi _{\mu }(x^{(2)}) D_{a_4}D_{b_4}\left( \psi _{\rho }(y^{(2)})\right)$ equals 0 or must depend on $x^{(2)}$. However, $D_{ a_4}D_{b_4}\varpi (y^{(2)})$ does not equal a constant and only depends on $y^{(2)}$ since $D_{u_4}D_{v_4}\varpi (y^{(2)})$ is not a constant for any two different nonzero vectors $u_4,v_4\in {\mathbb F}_2^{\frac{m}{2}}\setminus \{0_{\frac{m}{2}}\}$. Hence,

  $$\begin{aligned} D_{(a_1, 0_{\frac{n}{2}}, 0_{\frac{m}{2}-1}, a_4)}D_{(b_1, 0_{\frac{n}{2}}, 0_{\frac{m}{2}-1}, b_4)}h(x,y)\ne 0. \end{aligned}$$

• 2. If there exist $(a_1,a_2,a_3,a_4),(b_1,b_2,$ $b_3,b_4)\in V$ such that $D_{a_2}D_{b_2}\phi _{\mu }(x^{(2)})\ne 0$, $a_3=b_3=0_{\frac{m}{2}-1}$. Then, according to $a_4$ and $b_4$, we have four cases: $a_4=b_4\ne 0_{\frac{m}{2}}$, $a_4=b_4= 0_{\frac{m}{2}}$, $a_4\ne b_4$ and $b_4= 0_{\frac{m}{2}}$ (or $a_4= 0_{\frac{m}{2}}$), $a_4\ne b_4$ and $b_4\ne 0_{\frac{m}{2}}$ and $a_4\ne 0_{\frac{m}{2}}$.

  • (a) $a_4=b_4\ne 0_{\frac{m}{2}}$: In this case $(c_1,c_2,0_{\frac{m}{2}-1},0_{\frac{m}{2}})\in V$, where $(c_1,c_2,0_{\frac{m}{2}-1},0_{\frac{m}{2}}) =(a_1,a_2,a_3,a_4)\oplus (b_1,b_2,b_3,b_4)$. We have

    $$\begin{aligned}&D_{(a_1, a_2, a_3, a_4)}D_{(c_1,c_2,0_{\frac{m}{2}-1},0_{\frac{m}{2}})}h(x,y) \nonumber \\&\quad = D_{(a_1,a_2)}D_{(c_1,c_2)}\textsc {f}_0(x)\oplus D_{a_2}D_{c_2}\theta (x^{(2)})\\&\qquad \oplus D_{c_2}\phi _{\mu }(x^{(2)})\psi _{\rho }(y^{(2)})\oplus D_{c_2}\phi _{\mu }(x^{(2)}\oplus a_2)\psi _{\rho }(y^{(2)}\oplus a_4) \nonumber . \end{aligned}$$

    (6)

    Note that, denoting $\psi _{\rho }(y^{(2)})=\bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}d_{I}\prod \limits _{i\in I}y_i,$ and $\psi _{\rho }(y^{(2)}\oplus a_4)=$ $\bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}d_{I}^{(a_4)}\prod \limits _{i\in I}y_i$, we have $D_{c_2}\phi _{\mu }(x^{(2)})\psi _{\rho }(y^{(2)})\oplus D_{c_2}\phi _{\mu }(x^{(2)}\oplus a_2)\psi _{\rho }(y^{(2)}\oplus a_4) =\bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}\left( d_{I}D_{c_2}\phi _{\mu }(x^{(2)})\oplus d_{I}^{(a_4)} D_{c_2}\phi _{\mu }(x^{(2)}\oplus a_2)\right) \prod \limits _{i\in I}y_i$. Since $\psi (y^{(2)})$ and $\phi (x^{(2)})$ have no nonzero linear structure, we have $\psi _{\rho }(y^{(2)})\ne \psi _{\rho }(y^{(2)}\oplus a_4)$ and $D_{a_2}D_{b_2}\phi _{\mu }(x^{(2)})= D_{c_2}\phi _{\mu }(x^{(2)})\oplus D_{c_2}\phi _{\mu }(x^{(2)}\oplus a_2) \ne 0$, then $D_{c_2}\phi _{\mu }(x^{(2)})\psi _{\rho }(y^{(2)})\oplus D_{c_2}\phi _{\mu }(x^{(2)}\oplus a_2)\psi _{\rho }(y^{(2)}\oplus a_4)$ must depend on $y^{(2)}$. However, $D_{(a_1,a_2)}D_{(c_1,c_2)}\textsc {f}_0(x)\oplus D_{a_2}D_{c_2}\theta (x^{(2)})$ can not depend on $y^{(2)}$, that is, it cannot contribute in the cancellation of the terms in $y^{(2)}$. Hence, we have

    $$\begin{aligned} D_{(a_1, a_2, a_3, a_4)}D_{\left( c_1,c_2,0_{\frac{m}{2}-1},0_{\frac{m}{2}}\right) }h(x,y)\ne 0. \end{aligned}$$

  • (b) $a_4=b_4= 0_{\frac{m}{2}}$: In this case we have

    $$\begin{aligned}&D_{\left( a_1, a_2,0_{\frac{m}{2}-1},0_{\frac{m}{2}}\right) } D_{\left( b_1,b_2,0_{\frac{m}{2}-1},0_{\frac{m}{2}}\right) }h(x,y) \nonumber \\&\quad = D_{(a_1,a_2)}D_{(b_1,b_2)}\textsc {f}_0(x)\oplus D_{a_2}D_{b_2}\theta (x^{(2)})\\&\qquad \oplus D_{a_2} D_{b_2}\phi _{\mu }(x^{(2)})\psi _{\rho }(y^{(2)})\nonumber . \end{aligned}$$

    (7)

    Since $D_{a_2} D_{b_2}\phi _{\mu }(x^{(2)})\ne 0$, we have

    $$\begin{aligned} D_{\left( a_1, a_2, ,0_{\frac{m}{2}-1},0_{\frac{m}{2}}\right) } D_{\left( b_1,b_2,0_{\frac{m}{2}-1},0_{\frac{m}{2}}\right) }h(x,y)\ne 0. \end{aligned}$$

  • (c) $a_4\ne b_4$ and $b_4= 0_{\frac{m}{2}}$ (or $a_4= 0_{\frac{m}{2}}$), without loss of generality, set $b_4= 0_{\frac{m}{2}}$: In this case we have

    $$\begin{aligned}&D_{(a_1, a_2, a_3, a_4)}D_{\left( b_1,b_2,0_{\frac{m}{2}-1},0_{\frac{m}{2}}\right) }h(x,y) \nonumber \\&\quad = D_{(a_1,a_2)}D_{(b_1,b_2)}\textsc {f}_0(x)\oplus D_{a_2}D_{b_2}\theta (x^{(2)})\\&\qquad \oplus \left( \phi _{\mu }(x^{(2)})\oplus \phi _{\mu }(x^{(2)}\oplus b_2) \right) \psi _{\rho }(y^{(2)})\nonumber \\&\qquad \oplus \left( \phi _{\mu }(x^{(2)}\oplus a_2)\oplus \phi _{\mu }(x^{(2)}\oplus b_2\oplus a_2) \right) \psi _{\rho }(y^{(2)}\oplus a_4) \nonumber . \end{aligned}$$

    (8)

    Thus, utilizing the method of item 2, we have $D_{(a_1, a_2, a_3, a_4)}D_{(b_1,b_2,0_{\frac{m}{2}-1},0_{\frac{m}{2}})}h(x,y)\ne 0$.

  • (d) $a_4\ne b_4$, $b_4\ne 0_{\frac{m}{2}}$ and $a_4\ne 0_{\frac{m}{2}}$: In this case we have

    

    (9)

    Note that $\phi _{\mu }(x^{(2)})\psi _{\rho }(y^{(2)}) \oplus \phi _{\mu }(x^{(2)}\oplus a_2)\psi _{\rho }(y^{(2)}\oplus a_4)$ $\oplus \phi _{\mu }(x^{(2)}\oplus b_2)\psi _{\rho }(y^{(2)}\oplus b_4)\oplus$ $\phi _{\mu }(x^{(2)}\oplus b_2\oplus a_2)\psi _{\rho }(y^{(2)}\oplus b_4\oplus a_4)$ $=\bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}\left( d_{I}\phi _{\mu }(x^{(2)})\oplus d_{I}^{(a_4)}\phi _{\mu }(x^{(2)}\oplus a_2)\oplus d_{I}^{(b_4)}\phi _{\mu }(x^{(2)}\oplus b_2) \oplus d_{I}^{(b_4\oplus a_4)}\phi _{\mu }(x^{(2)}\oplus \right. \left. b_2\oplus a_2)\right) \prod \limits _{i\in I}y_i$, where $\psi _{\rho }(y^{(2)})=\bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}d_{I}\prod \limits _{i\in I}y_i,$ $\psi _{\rho }(y^{(2)}\oplus a_4)$ $= \bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}d_{I}^{(a_4)}\prod \limits _{i\in I}y_i$, $\psi _{\rho }(y^{(2)}\oplus b_4)=$ $\bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}d_{I}^{(b_4)}\prod \limits _{i\in I}y_i$, $\psi _{\rho }(y^{(2)}\oplus a_4\oplus b_4)=$ $\bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}d_{I}^{(a_4\oplus b_4)}\prod \limits _{i\in I}y_i$. Since $D_{a_2}D_{b_2}\phi _{\mu }(x^{(2)})\ne 0$ and $\phi _{\mu }(x^{(2)})$ has no nonzero linear structure, there must exist a set J such that $d_{J} \ne d_{J}^{(a_4)}$, that is, $d_{J}\phi _{\mu }(x^{(2)})\oplus d_{J}^{(a_4)}\phi _{\mu }(x^{(2)}\oplus a_2)\oplus d_{J}^{(b_4)}\phi _{\mu }(x^{(2)}\oplus b_2) \oplus d_{J}^{(b_4\oplus a_4)}\phi _{\mu }(x^{(2)}\oplus b_2\oplus a_2)$ depends on $x^{(2)}$. Hence, $\phi _{\mu }(x^{(2)})\psi _{\rho }(y^{(2)})\oplus \phi _{\mu }(x^{(2)}\oplus a_2)\psi _{\rho }(y^{(2)}\oplus a_4)$ $\oplus \phi _{\mu }(x^{(2)}\oplus b_2)\psi _{\rho }(y^{(2)}\oplus b_4)\oplus$ $\phi _{\mu }(x^{(2)}\oplus b_2\oplus a_2)\psi _{\rho }(y^{(2)}\oplus b_4\oplus a_4)$ depends on $x^{(2)}$ and $y^{(2)}$. Further, from Relation (9), we have $D_{(a_1, a_2, a_3, a_4)}D_{(b_1,b_2,b_3,b_4)}h(x,y)\ne 0$.

• 3. If there exist $(a_1, a_2, a_3, a_4)=(a_1,0_{\frac{n}{2}},0_{\frac{m}{2}-1},0_{\frac{m}{2}})\ne 0_{n+m-2}$, $(b_1,b_2,b_3,b_4)$ such that $b_2 \ne 0_{\frac{n}{2}}$. We have

  $$\begin{aligned}&D_{(a_1, a_2, a_3, a_4)}D_{(b_1,b_2,b_3,b_4)}h(x,y) = D_{\left( a_1,0_{\frac{n}{2}}\right) }D_{(b_1,b_2)}\textsc {f}_0(x)\ne 0 \end{aligned}$$

  (10)

  since $\phi (x^{(2)})$ has no nonzero linear structure and $b_2\ne 0_{\frac{n}{2}}$.

• 4. If there exist $(a_1, 0_{\frac{n}{2}}, 0_{\frac{m}{2}-1},a_4)\in V\cap \varDelta$ such that $a_4\ne 0_{\frac{m}{2}}$, $(b_1,b_2,b_3,b_4)\in V$ and $b_2\ne 0_{\frac{n}{2}}$. We have

  $$\begin{aligned}&D_{(a_1, a_2, a_3, a_4)}D_{(b_1,b_2,b_3,b_4)}h(x,y) \nonumber \\&\quad = D_{\left( a_1,0_{\frac{n}{2}}\right) }D_{(b_1,b_2)}\textsc {f}_0(x) \oplus D_{a_4}D_{b_4}\varpi (y^{(2)}) \oplus D_{\left( 0_{\frac{m}{2}-1},a_4\right) }D_{(b_3,b_4)}\textsc {g}_0(y) \nonumber \\&\qquad \oplus \phi _{\mu }(x^{(2)})D_{a_4}\psi _{\rho }(y^{(2)}) \oplus \phi _{\mu }(x^{(2)}\oplus b_2) D_{a_4}\psi _{\rho }(y^{(2)}\oplus b_4) . \end{aligned}$$

  (11)

  There are also two cases to be considered.

  • (a) $b_4= 0_{\frac{m}{2}}$: In this case $\phi _{\mu }(x^{(2)})D_{a_4}\psi _{\rho }(y^{(2)}) \oplus \phi _{\mu }(x^{(2)}\oplus b_2) D_{a_4}\psi _{\rho }(y^{(2)}\oplus b_4)=D_{b_2}\phi _{\mu }(x^{(2)})D_{a_4}\psi _{\rho }(y^{(2)})$ depends on $x^{(2)}, y^{(2)}$ since both $\phi (x^{(2)})$ and $\psi (y^{(2)} )$ have no nonzero linear structure and $a_4\ne 0_{\frac{m}{2}}, b_2\ne 0_{\frac{n}{2}}$. Further, from Relation (11), we have

    $$\begin{aligned} D_{(a_1, a_2, a_3, a_4)}D_{(b_1,b_2,b_3,b_4)}h(x,y) \ne 0. \end{aligned}$$

  • (b) $b_4\ne 0_{\frac{m}{2}}$: In this case we have $\phi _{\mu }(x^{(2)})D_{a_4}\psi _{\rho }(y^{(2)}) \oplus \phi _{\mu }(x^{(2)}\oplus b_2) D_{a_4}\psi _{\rho }(y^{(2)}\oplus b_4)$ $=\bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}\left( (d_{I}\oplus d_{I}^{(a_4)})\phi _{\mu }(x^{(2)})\oplus \ (d_{I}^{(b_4)}\oplus d_{I}^{(b_4\oplus a_4)}) \phi _{\mu }(x^{(2)}\oplus b_2) \right) \prod \limits _{i\in I}y_i$, where $\psi _{\rho }(y^{(2)})=\bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}d_{I}\prod \limits _{i\in I}y_i,$ $\psi _{\rho }(y^{(2)}\oplus a_4)=$ $\bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}d_{I}^{(a_4)}\prod \limits _{i\in I}y_i$, $\psi _{\rho }(y^{(2)}\oplus b_4)=$ $\bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}d_{I}^{(b_4)}\prod \limits _{i\in I}y_i$, $\psi _{\rho }(y^{(2)}\oplus a_4\oplus b_4)=$ $\bigoplus \limits _{I\subset \{\frac{m}{2}+1,\ldots ,m\}}d_{I}^{(a_4\oplus b_4)}\prod \limits _{i\in I}y_i$. Since $a_4\ne 0_{\frac{m}{2}}, b_4\ne 0_{\frac{m}{2}}, b_2\ne 0_{\frac{n}{2}}$ and both $\phi (x^{(2)})$ and $\psi (y^{(2)} )$ have no nonzero linear structure, we know $\phi _{\mu }(x^{(2)})D_{a_4}\psi _{\rho }(y^{(2)}) \oplus \phi _{\mu }(x^{(2)}\oplus b_2) D_{a_4}\psi _{\rho }(y^{(2)}\oplus b_4)$ depends on $x^{(2)}, y^{(2)}$. Further, from Relation (11), we have

    $$\begin{aligned} D_{(a_1, a_2, a_3, a_4)}D_{(b_1,b_2,b_3,b_4)}h(x,y) \ne 0. \end{aligned}$$

• 5. If there exist $(a_1, a_2, a_3, a_4)=(a_1,0_{\frac{n}{2}},0_{\frac{m}{2}-1},a_4), (b_1,b_2,$ $b_3,b_4)=(b_1,0_{\frac{n}{2}},b_3,b_4)$ such that $a_4=b_4\ne 0_{\frac{m}{2}}$ and $b_3\ne 0_{\frac{m}{2}-1}$. We have

  $$\begin{aligned}&D_{(a_1, a_2, a_3, a_4)}D_{(b_1,b_2,b_3,b_4)}h(x,y) \nonumber \\&\quad = D_{\left( 0_{\frac{m}{2}-1},a_4\right) }D_{(b_3,a_4)}\textsc {g}_0(y)\nonumber \\&\quad = b_3\cdot \left( D_{a_4}\psi _1(y^{(2)}), \ldots , D_{a_4}\psi _{\rho -1}(y^{(2)}),\right. \left. D_{a_4}\psi _{\rho +1}(y^{(2)}),\ldots ,D_{a_4}\psi _{\frac{m}{2}}(y^{(2)})\right) \ne 0\nonumber \\ \end{aligned}$$

  (12)

  since $\psi (y^{(2)})$ has no nonzero linear structure and $b_3 \ne 0_{\frac{m}{2}-1}$.

Proof of Theorem 3

Functions f and g being bent, h is bent, according to Theorem 2. We prove that h does not belong to ${M}^{\#}$, by using Lemma 2. We need to show that there does not exist an $(\frac{n+m-2}{2})$-dimensional subspace V of ${\mathbb F}_2^{\frac{n}{2}-1} \times {\mathbb F}_2^{\frac{n}{2}}\times {\mathbb F}_2^{\frac{m}{2}-1}\times {\mathbb F}_2^{\frac{m}{2}}$ such that, for any $(a_1, a_2, a_3, a_4),$ $(b_1, b_2, b_3, b_4)\in V$:

$$\begin{aligned} D_{(a_1, a_2, a_3, a_4)}D_{(b_1, b_2, b_3, b_4)}h(x,y)=0. \end{aligned}$$

(13)

We denote the elements of V by $(v_1^{(1)},v_2^{(1)},v_3^{(1)},v_4^{(1)}),(v_1^{(2)},v_2^{(2)},v_3^{(2)},v_4^{(2)}),\ldots ,$

$(v_1^{(2^{\frac{n+m-2}{2}})},v_2^{(2^{\frac{n+m-2}{2}})},v_3^{(2^{\frac{n+m-2}{2}})},v_4^{(2^{\frac{n+m-2}{2}})})$. Let the subspace of ${\mathbb F}_2^{n+m-2}$ given by $\{(x^{(1)},0_{\frac{n}{2}},0_{\frac{m}{2}-1}, y^{(2)})\mid x^{(1)}\in {\mathbb F}_2^{\frac{n}{2}-1}, y^{(2)}\in {\mathbb F}_2^{\frac{m}{2}}\}$ be denoted by $\varDelta$. We consider two cases: $V= \varDelta$ and $V \ne \varDelta$.

• 1. For $V=\varDelta$, we can find two vectors $(a_1, 0_{\frac{n}{2}}, 0_{\frac{m}{2}-1}, a_4),(b_1, 0_{\frac{n}{2}}, 0_{\frac{m}{2}-1}, b_4) \in \varDelta$ such that $a_4\ne 0_{\frac{m}{2}}$, $b_4\ne 0_{\frac{m}{2}}$ and $a_4\ne b_4$. Thus, a, b satisfy the item 1 of Lemma 3.

• 2. For $V\ne \varDelta$, we split the proof into three cases depending on the cardinality of $V\cap \varDelta$.

  • (a) For $|V\cap \varDelta |= 1$, that is, $V\cap \varDelta =\{0_{n+m-2}\}$, we have $(v_2^{(i)},v_3^{(i)})\ne (v_2^{(j)},v_3^{(j)})$ for every $i\ne j$. Indeed, if there exist indices $i_1,j_1$ such that $(v_2^{(i_1)},v_3^{(i_1)})= (v_2^{(j_1)},v_3^{(j_1)})$, then we have $(v_1^{(i_1)},v_2^{(i_1)},v_3^{(i_1)},v_4^{(i_1)})\oplus (v_1^{(j_1)},v_2^{(j_1)},v_3^{(j_1)},v_4^{(j_1)})\in V\cap \varDelta$, that is, $(v_1^{(i_1)},v_2^{(i_1)},v_3^{(i_1)},v_4^{(i_1)})= (v_1^{(j_1)},v_2^{(j_1)},v_3^{(j_1)},v_4^{(j_1)})$. This implies then that $|\{(v_2^{(1)},v_3^{(1)}),(v_2^{(2)},v_3^{(2)}),\ldots , (v_2^{(2^{\frac{n+m-2}{2}})}, v_3^{(2^{\frac{n+m-2}{2}})})\}|=|V|= 2^{\frac{n+m-2}{2}}$, that is, $\{(v_2^{(1)},v_3^{(1)}),(v_2^{(2)},v_3^{(2)}),\ldots , (v_2^{(2^{\frac{n+m-2}{2}})}, v_3^{(2^{\frac{n+m-2}{2}})})\}= {\mathbb F}_2^{\frac{n}{2}}\times {\mathbb F}_2^{\frac{m}{2}-1}$. Hence, according to the hypothesis that $\phi (x^{(2)})$ has no nonzero linear structure, we have $\deg (\phi _{\mu }(x^{(2)}))\ge 2$, and there must exist two vectors $(a_1,a_2,a_3,a_4),(b_1,b_2,$ $b_3,b_4)\in V$ such that $D_{a_2}D_{b_2}\phi _{\mu }(x^{(2)})\ne 0$, and $a_3=b_3=0_{\frac{m}{2}-1}$. Thus a, b satisfy the item 2 of Lemma 3.

  • (b) For $|V\cap \varDelta |= 2$, there exists one vector $(a_1, 0_{\frac{n}{2}}, 0_{\frac{m}{2}-1},a_4)\in V$ such that $(a_1,a_4)\ne 0_{ \frac{n}{2}-1+ \frac{m}{2}}$. Additionally, there must exist one vector $(v_1^{(l)},v_2^{(l)},v_3^{(l)},v_4^{(l)})$ such that $v_2^{(l)}\ne 0_{\frac{n}{2}}$. Indeed, since $|V|= 2^{\frac{n+m-2}{2}}$ and $n>2$, then there are at least four vectors $v^{(i_1)},v^{(i_2)},v^{(i_3)},v^{(i_4)}$ such that $v_3^{(i_1)}=v_3^{(i_2)}=v_3^{(i_3)}=v_3^{(i_4)}$. Suppose $v_2^{(i)}= 0_{\frac{n}{2}}$ for $i=1,2,\ldots ,2^{\frac{n+m-2}{2}}$, then there are at least three vectors $v^{(i_2)}\oplus v^{(i_1)},v^{(i_3)}\oplus v^{(i_1)}, v^{(i_4)}\oplus v^{(i_1)}\in V\cap \varDelta$, which is in contradiction with $|V\cap \varDelta |= 2$. Hence, there must exist one vector $(v_1^{(l)},v_2^{(l)},v_3^{(l)},v_4^{(l)})$ such that $v_2^{(l)}\ne 0_{\frac{n}{2}}$. We set $(b_1,b_2,b_3,b_4)=(v_1^{(l)},v_2^{(l)},v_3^{(l)},v_4^{(l)})$. There are two cases to be considered.

    • i. If $a_4=0_{\frac{m}{2}}$, then $a_1\ne 0_{\frac{n}{2}-1}$. Thus a, b satisfy the item 3 of Lemma 3.

    • ii. If $a_4\ne 0_{\frac{m}{2}}$, then a, b satisfy the item 4 of Lemma 3.

  • (c) For $|V\cap \varDelta |=t> 2$ (i.e., $|V\cap \varDelta |=t\ge 4$ ), without loss of generality, set $V\cap \varDelta =\{v^{(1)},v^{(2)},\ldots , v^{(t)}\}$, we consider three cases: $|\{v^{(1)}_4,\ldots , v^{(t)}_4\}|>2$, $|\{v^{(1)}_4,\ldots ,$ $v^{(t)}_4\}|=2$ and $|\{v^{(1)}_4,\ldots , v^{(t)}_4\}|=0$.

    • i. If there exist at least two vectors $(a_1,0_{\frac{n}{2}},0_{\frac{m}{2}-1},a_4), (b_1,0_{\frac{n}{2}},0_{\frac{m}{2}-1},b_4)\in V\cap \varDelta$ such that $a_4\ne 0_{\frac{m}{2}},b_4\ne 0_{\frac{m}{2}}$ and $a_4\ne b_4$, then, a, b satisfy the item 1 of Lemma 3.

    • ii. If $\{v^{(1)}_4,\ldots ,v^{(t)}_4 \}=\{0_{\frac{m}{2}}, v^{(\ell )}_4\}$, where $(v^{(\ell )}_1,0_{\frac{n}{2}},0_{\frac{m}{2}-1},v^{(\ell )}_4)\in V\cap \varDelta$ such that $v^{(\ell )}_4\ne 0_{\frac{m}{2}}$, $\ell \le t$, then $|V\cap \varDelta |=t\le 2^{\frac{n}{2}}$ since $x^{(1)}\in {\mathbb F}_2^{\frac{n}{2}-1}$. Further, there are two cases to be considered.

      • A. If there exists one vector $v^{(p)}$ such that $v_2^{(p)}\ne 0_{\frac{n}{2}}$, we set $(a_1, a_2, a_3, a_4)=(v^{(\ell )}_1,0_{\frac{n}{2}},$ $0_{\frac{m}{2}-1},v^{(\ell )}_4),$ $(b_1,b_2,b_3,b_4)=v^{(p)}$. then, a, b satisfy the item 4 of Lemma 3.

      • B. If $v_2^{(i)}= 0_{\frac{n}{2}}$ for $i=1,2,\ldots , 2^{\frac{n+m}{2}-1}$, then there must exist a vector $v^{(i_1)}$ such that $v_3^{(i_1)}\ne 0_{\frac{m}{2}-1}$ and $v_4^{(i_1)}= v^{(\ell )}_4(\ne 0_{\frac{m}{2}})$. Indeed, since $n>2,m>4$, we have $2^{\frac{n+m-2}{2}}-2^{\frac{n}{2}}=2^{\frac{n}{2}-1}\cdot (2^{\frac{m}{2}}-2)>2^{\frac{m}{2}}$, there exist two vectors $v^{(j_1)},v^{(j_2)}$ such that $v_4^{(j_1)}= v_4^{(j_2)}$ (i.e., $v_3^{(j_1)}\ne v_3^{(j_2)}$ since $v_2^{(j_1)}=v_2^{(j_2)}= 0_{\frac{n}{2}}$), where $j_1,j_2>t$. Thus, we can set $v^{(i_1)}=v^{(j_1)}\oplus v^{(j_2)}\oplus (v^{(\ell )}_1,0_{\frac{n}{2}},0_{\frac{m}{2}-1},v^{(\ell )}_4)$. Hence, we set $(a_1, a_2, a_3, a_4)=(v^{(\ell )}_1,0_{\frac{n}{2}},0_{\frac{m}{2}-1},v^{(\ell )}_4),$ $(b_1,b_2,b_3,b_4)=(v^{(i_1)}_1,0_{\frac{n}{2}},v_3^{(i_1)},v^{(\ell )}_4)$. Thus, a, b satisfy the item 5 of Lemma 3.

    • iii. If $v^{(i)}_4= 0_{\frac{m}{2}}$ for $i=1,2,\ldots ,t$, then there must exist one vector $v^{(l)}\in V\backslash \varDelta$ such that $v^{(l)}_2\ne 0_{\frac{n}{2}}$, where $l> t$. To show this, if $v^{(j)}_2= 0_{\frac{m}{2}}$ for $j=1,2,\ldots , 2^{\frac{n+m-2}{2}}$, and if $v^{(j_1)}_3=v^{(j_2)}_3$, we have $v^{(j_1)}_4=v^{(j_2)}_4$ since $v^{(i)}_4= 0_{\frac{m}{2}}$ for $i=1,2,\ldots ,t$, where $j_1,j_2 >t$. Moreover,

      $$\begin{aligned} 2^{\frac{n+m-2}{2}}>2^{\frac{n-2}{2}}\cdot 2^{\frac{m-2}{2}}=|{\mathbb F}_2^{\frac{n}{2}-1}|\cdot |{\mathbb F}_2^{\frac{m}{2}-1}|, \end{aligned}$$

      so there is at least one vector $v^{(l)}\in V\backslash \varDelta$ such that $v^{(l)}_2\ne 0_{\frac{n}{2}}$. We set $(a_1, a_2, a_3, a_4)=(a_1,0_{\frac{n}{2}},0_{\frac{m}{2}-1},0_{\frac{m}{2}})\in (V\cap \varDelta )\backslash \{0_{n+m-2}\}, (b_1,b_2,b_3,b_4)=v^{(l)}$. Thus, a, b satisfy the item 2 of Lemma 3.

Combining items 1 and 2, we deduce that h does not belong to ${M}^{\#}$.