  We investigate the infinite version of $k$-switch problem of Greenwell and
Lov\'asz. Given infinite cardinals ${\kappa}$ and ${\lambda}$, for functions
$x,y\in {}^{\kappa}\lambda $ we say that they are totally different if $x(i)\ne
y(i)$ for each $i\in {\lambda}$. A function $F:{}^{\lambda}\kappa
\longrightarrow {\kappa} $ is a proper coloring if $F(x)\ne F(y)$ whenever $x$
and $y$ are totally different elements of ${}^\lambda{\kappa} $. We say that
  (i) $F$ is weakly uniform iff there are pairwise totally different functions
$\{r_{\alpha}:{\alpha}<{\kappa}\}\subset {}^{\lambda}{\kappa}$ such that
$F(r_{\alpha})={\alpha}$;
  (ii) $F$ is tight if there is no proper coloring $G:{}^{\lambda}\kappa
\longrightarrow {\kappa}$ such that there is exactly one $x\in
{}^{\lambda}{\kappa}$ with $G(x)\ne F(x)$.
  We show that given a proper coloring $F:{}^{\lambda}{\kappa}\to {\kappa}$,
the following statements are equivalent
  (i) $F$ is weakly uniform,
  (ii) there is a ${\kappa}$ complete ultrafilter $\mathscr{U}$ on ${\lambda}$
an there is a permutation ${\pi}\in Symm({\kappa})$ such that for each $x\in
^{\lambda}{\kappa}$ we have $$F(x)={\pi}({\alpha})\ \Longleftrightarrow \
\{i\in {\lambda}: x(i)={\alpha}\} \in \mathscr{U}.$$ We also show that there
are tight proper colorings which can not be obtain such a way.
