1. **Problem Statement:** Find the Cartesian product of the complete graph $K_3$ and the cycle graph $C_4$. 2. **Definitions:** - The complete graph $K_3$ has 3 vertices, each connected to every other vertex. - The cycle graph $C_4$ has 4 vertices connected in a closed loop. - The Cartesian product $G \times H$ of graphs $G$ and $H$ has vertex set $V(G) \times V(H)$, and two vertices $(g,h)$ and $(g',h')$ are adjacent if either: - $g = g'$ and $h$ is adjacent to $h'$ in $H$, or - $h = h'$ and $g$ is adjacent to $g'$ in $G$. 3. **Step-by-step:** - $K_3$ has vertices $\{1,2,3\}$ with edges between every pair. - $C_4$ has vertices $\{a,b,c,d\}$ with edges $(a,b),(b,c),(c,d),(d,a)$. - The vertex set of $K_3 \times C_4$ is all pairs $(i,x)$ where $i \in \{1,2,3\}$ and $x \in \{a,b,c,d\}$, so total $3 \times 4 = 12$ vertices. - Edges connect vertices: - $(i,x)$ to $(i,y)$ if $x$ adjacent to $y$ in $C_4$. - $(i,x)$ to $(j,x)$ if $i$ adjacent to $j$ in $K_3$. 4. **Result:** The Cartesian product $K_3 \times C_4$ is a graph with 12 vertices arranged in 3 layers (for $K_3$ vertices), each layer a $C_4$ cycle, and edges connecting corresponding vertices between layers according to $K_3$ adjacency. 5. **Summary:** The Cartesian product combines the structures of $K_3$ and $C_4$ by connecting vertices in a grid-like pattern respecting adjacency in each graph. **Final answer:** The graph $K_3 \times C_4$ has 12 vertices and edges defined by adjacency in $K_3$ and $C_4$ as per the Cartesian product definition.