# Monochromatic Edges in Complete Multipartite Hypergraphs
Similarly to Case 4A, \(F_{4}\) and \(F_{6}\) have fewer monochromatic edges than \(F_{5}\) and \(F_{7}\), respectively, and \(F_{6}\) has fewer monochromatic edges than \(F_{4}\). Next, we have to compare between \(F_{3}\) and \(F_{6}\) to determine which form \(c_{X}^{*}\) is. Due to the complexity of the comparison, we will conclude that \(c_{X}^{*}\) is in the form \(F_{3}\) or \(F_{6}\). Case 5B: \(n_{3}\leqslant X\leqslant\lfloor\frac{n_{1}+n_{2}+n_{3}}{2}\rfloor\) and \(H\) is a type B hypergraph (See Table 8). Similarly to Case 4A, \(F_{4}\) and \(F_{6}\) have fewer monochromatic edges than \(F_{5}\) and \(F_{7}\), respectively, and \(F_{6}\) has fewer monochromatic edges than \(F_{4}\). Next, we will compare between colorings in the forms \(F_{8}\) and \(F_{9}\). If \(n_{1}=n_{2}\), then both forms are the same. Thus,  
\[S(0,X-n_{3},n_{3})=S(0,X-n_{3},0)=S(X-n_{3},0,0)==S(X-n_{3},0,n_{3}).\]  
The equality holds only when \(n_{1}=n_{2}\) or \(n_{2}-1<2\). Since \(3\leqslant n_{2}\), we have that \(F_{8}\) gives fewer monochromatic edges than \(F_{9}\). Finally, we will compare between colorings in the forms \(F_{6}\) and \(F_{8}\). If \(n_{2}=n_{3}\), then both forms are the same. We may suppose that \(n_{2}<n_{3}\). Thus, since \(X\leqslant\lfloor\frac{N}{2}\rfloor\), \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Possible Forms** & \(\mathbf{x_{1}}\) & \(\mathbf{x_{2}}\) & \(\mathbf{x_{3}}\) & \(\mathbf{S(x_{1},x_{2},x_{3})}\) \\ \hline \(F_{3}\) & \(0\) & \(0\) & \(X\) & \(S(0,0,X)\) \\ \(F_{4}\) & \(n_{1}\) & \(X-n_{1}\) & \(0\) & \(S(n_{1},X-n_{1},0)\) \\ \(F_{5}\) & \(n_{1}\) & \(0\) & \(X-n_{1}\) & \(S(n_{1},0,X-n_{1})\) \\ \(F_{6}\) & \(X-n_{2}\) & \(n_{2}\) & \(0\) & \(S(X-n_{2},n_{2},0)\) \\ \(F_{7}\) & \(0\) & \(n_{2}\) & \(X-n_{2}\) & \(S(0,n_{2},X-n_{2})\) \\ \hline \hline \end{tabular}
\end{table} 
Table 7: Possible canonical forms for Case 4B. \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Possible Forms** & \(\mathbf{x_{1}}\) & \(\mathbf{x_{2}}\) & \(\mathbf{x_{3}}\) & \(\mathbf{S(x_{1},x_{2},x_{3})}\) \\ \hline \(F_{4}\) & \(n_{1}\) & \(X-n_{1}\) & \(0\) & \(S(n_{1},X-n_{1},0)\) \\ \(F_{5}\) & \(n_{1}\) & \(0\) & \(X-n_{1}\) & \(S(n_{1},0,X-n_{1})\) \\ \(F_{6}\) & \(X-n_{2}\) & \(n_{2}\) & \(0\) & \(S(X-n_{2},n_{2},0)\) \\ \(F_{7}\) & \(0\) & \(n_{2}\) & \(X-n_{2}\) & \(S(0,n_{2},X-n_{2})\) \\ \(F_{8}\) & \(X-n_{3}\) & \(0\) & \(n_{3}\) & \(\[S(X-n_{2},n_{2},0) = S(X-n_{2},0,0)\] \[= \left[ \begin{pmatrix}0\\ 2\end{pmatrix} \ +\ \begin{pmatrix}1\\ 2\end{pmatrix} +\cdots+ \begin{pmatrix}X-n_{3}-1\\ 2\end{pmatrix} \right]\] \[+ \left[ \begin{pmatrix}X-n_{3}\\ 2\end{pmatrix} \ +\ \begin{pmatrix}X-n_{3}+1\\ 2\end{pmatrix} +\cdots+ \begin{pmatrix}X-n_{2}-1\\ 2\end{pmatrix} \right]\] \[- \left[ \begin{pmatrix}n_{1}+n_{2}-X\\ 2\end{pmatrix} \ +\ \begin{pmatrix}n_{1}+n_{2}-X+1\\ 2\end{pmatrix} \ +\ \cdots\ +\ \begin{pmatrix}n_{1}+n_{3}-X\ -\ 1\\ 2\end{pmatrix} \right]\] \[- \left[ \begin{pmatrix}n_{1}+n_{3}-X\\ 2\end{pmatrix} \ +\ \begin{pmatrix}n_{1}+n_{3}-X+1\\ 2\end{pmatrix} \ +\ \cdots\ +\ \begin{pmatrix}n_{1}-1\\ 2\end{pmatrix} \right]\] \[= S(X-n_{3},0,0)=S(X-n_{3},0,n_{3}).\]  
The equality holds only when \(X=\frac{N}{2}\) or \(n_{1}+n_{3}-X\ -\ 1<2\). First, if \(X=\frac{N}{2}\), then both forms have the same number of monochromatic edges. However, if we relabel the names of the colors, then both forms are the same. We will focus on the condition of \(n_{1}+n_{3}-X<3\), where \(3\leqslant n_{1}\leqslant n_{2}<n_{3}\leqslant X<\frac{N}{2}\) and \(n_{1}+n_{2}>n_{3}\). Suppose that \(n_{1}+n_{3}-X<3\). If \(N\) is even, then \(3>n_{1}+n_{2}-X\geqslant n_{1}+n_{3}-\left(\frac{N}{2}-1\right)=\frac{n_{1}+n_ {3}-n_{2}}{2}+1\). Thus, \(n_{1}+n_{3}-n_{2}<4\). Since \(1\leqslant n_{3}-n_{2}\), we have that \(n_{1}<3\), which contradicts \(3\leqslant n_{1}\). If \(N\) is odd, then \(3>n_{1}+n_{2}-X\geqslant n_{1}+n_{3}-\left(\frac{N-1}{2}\right)=\frac{n_{1}+n _{3}-n_{2}+1}{2}\). Thus, \(n_{1}+n_{3}-n_{2}<5\). Since \(1\leqslant n_{3}-n_{2}\), we have that \(n_{1}<4\).