# Monochromatic Edges in Complete Multipartite Hypergraphs
### Hypergraphs with \(n_{1}<3\) or \(n_{2}<3\)  
In this subsection, we will prove the remaining cases, which are unbalance complete tripartite hypergraphs with some classes smaller than 3. These cases are easy and straightforward but contain fuzzy details. First, we will consider an unbalanced complete tripartite 3-uniform hypergraphs with \(n_{1}\leqslant n_{2}<n_{3}\geqslant 3\). There are three possibilities for these hypergraphs:  
Case i: \(n_{1}=n_{2}=1\) and \(n_{3}\geqslant 3\). Since we have that the first two classes are smaller than 3, no edge can be contained in the first two classes. Thus,  
\[M(H,c)=\binom{X}{3}+\binom{N-X}{3}-\binom{x_{3}}{3}-\binom{n_{3}-x_{3}}{3}.\]  
We have that \(M(H,c)\geqslant 0\). Suppose that \(M(H,c)=0\). Since \(X\geqslant x_{3}\) and \(N-X\geqslant n_{3}-x_{3}\), we have that \(\binom{X}{3}=\binom{x_{3}}{3}\) and \(\binom{N-X}{3}=\binom{n_{3}-x_{3}}{3}\). Since \(N=n_{1}+n_{2}+n_{3}=1+1+n_{3}\geqslant 5\), at least three vertices are colored the same, such as red, i.e., \(X\geqslant 3\). Then, \(X=x_{3}\), which implies that \(N-X=n_{3}+2-X>n_{3}-x_{3}\). Consequently, \(3>N-X=n_{3}+2-x_{3}\), i.e., \(n_{3}=x_{3}\). This means that \(c\) is a coloring such that the third class contains all red vertices and no blue vertex, and the first two classes are all blue. Hence, we have already determined the minimum coloring and showed that it is unique up to a permutation of colors and classes. Case ii: \(n_{1}=1,n_{2}=2\) and \(n_{3}\geqslant 3\). Again, no edge can be contained in the first two classes. Thus,  
\[M(H,c)=\binom{X}{3}+\binom{N-X}{3}-\binom{x_{3}}{3}-\binom{n_{3}-x_{3}}{3}.\]  
We will show that \(M(H,c)\geqslant 1\). We have that \(X\geqslant x_{3}\) and \(N-X\geqslant n_{3}-x_{3}\). Since \(N=n_{1}+n_{2}+n_{3}=1+1+n_{3}\geqslant 6\), at least three vertices are colored the same, such as red, i.e., \(X\geqslant 3\). If \(X>x_{3}\), then \(M(H,c)>0\). Suppose that \(X=x_{3}\). Then, \(N-X=n_{3}+3-x_{3}>n_{3}-x_{3}\geqslant 3\). Hence, \(M(H,c)>0\). Next, suppose that \(M(H,c)=1\). If \(X>x_{3}\), then  
\[1\leqslant\binom{X-~{}1}{2}\leqslant\binom{X}{3}-\binom{x_{3}}{3}\leqslant \binom{X}{3}+\binom{N-X}{3}-\binom{x_{3}}{3}-\binom{n_{3}-x_{3}}{3}=M(H,c).\]  
This implies that \(X=3\), and so, \(N-X\geqslant 3\); moreover, \(\binom{N-X}{3}=\binom{n_{3}-x_{3}}{3}\). Hence, \(N-X=n_{3}-x_{3}\). Now, we have \(N-3=N-X=n_{3}-x_{3}=N-3-x_{3}\), i.e., \(x_{3}=0\). This means \(c\) is a coloring such that the third class contains no red vertices, and the first two classes are all red. Suppose that \(X=x_{3}\). Then,  
\[1=M(H,c)=\binom{X}{3}+\binom{N-X}{3}-\binom{x_{3}}{3}-\binom{n_{3}-x_{3}}{3}= \binom{n_{3}-x_{3}+3}{3}-\binom{n_{3}-x_{3}}{3}.\]It is only possible when \(n_{3}-x_{3}=0\) or when \(c\) is a coloring such that the third class contains all red vertices and no blue vertex and the first two classes are all blue. Note that if we relabel the names of colors, then both colorings are the same. Hence, we have already determined the minimum coloring and showed that it is unique up to a permutation of colors and classes. Case iii: \(n_{1}=2,n_{2}=2\) and \(n_{3}\geqslant 3\). Similarly, no edges can be contained in the first two classes. Thus,  
\[M(H,c)=\binom{X}{3}+\binom{N-X}{3}-\binom{x_{3}}{3}-\binom{n_{3}-x_{3}}{3}.\]  
We will show that \(M(H,c)\geqslant 4\). We have that \(X\geqslant x_{3}\) and \(N-X\geqslant n_{3}-x_{3}\). If there is a color, say red, such that all vertices of that color are only in the third class, then, we have \(X=x_{3}\) and \(N-X=n_{3}+4-x_{3}\geqslant 4\). Thus,  
\[M(H,c) =\binom{N-X}{3}-\binom{N-X-4}{3}\] \[=\binom{N-X-1}{2}+\binom{N-X-2}{2}+\binom{N-X-3}{2}+\binom{N-X-4} {2}\] \[\geqslant\binom{3}{2}+\binom{2}{2}=4.\]  
The equality holds only when \(N-X=4\), i.e., \(n_{3}=x_{3}\). Hence, if \(M(H,c)=4\), then \(c\) is a coloring such that the third class contains all red vertices but no blue vertex and the first two classes are all blue.