Find an Orientation for a complete Bipartite graph $K_{r,s}$, where $r,s \geq 2$
I'm assuming i need to find a graph which is connected and that each edge belongs to a circuit, hence this would be a orientation of the graph?
Considering we have a complete bipartite graph I think the answer would be a circle circuit? however, how would i show this
As for the simple orientation, If you let the vertices be $\{u_1,u_2,...,u_r\}$ and $\{v_1,v_2,...,v_s\}$.
Then a simple orientation would just be "Let the direction of $u_iv_j$ start from $u_i$ and end with $v_j$ for all $1\leq i \leq r,1\leq j\leq s$".
For a strong orientation case, a valid solution would be "Let the direction of $u_iv_j$ start from $u_i$ and end with $v_j$ for all $1 \leq i \leq r, 1 \leq j \leq s$ where $i+j$ is even and start from $v_j$ and end with $u_i$ where $i+j$ is odd.".
To show this is a valid solution,
(1) Consider two arbitrary vertices from different components, $u_i,v_j$, if $i+j$ is even then $u_iv_j$ would be a path from $u_i$ to $v_j$, if $i+j$ is odd then $u_iv_{j+1}u_{i+1}v_j$ (If $i=r$ change $i+1$ to $i-1$ and if $j=s$ change $j+1$ to $j-1$) would be a path from $u_i$ to $v_j.$ Similarly, if $i+j$ is even then $v_ju_i$ is a path from $v_j$ to $u_i$ and if $i+j$ is odd then $v_ju_{i+1}v_{j+1}u_i$ (change the $+1$ to $-1$ for $i=r$ or $j=s$) would be a path from $v_j$ to $u_i$.
(2) Consider two arbitrary vertices from the same component, say $u_i,u_k$. Then from the previous case we know there exist a path $p_1$ from $u_i$ to $v_1$ and a path $p_2$ from $v_1$ to $u_k$ so $p_1p_2$ would be a path from $u_i$ to $u_k$.