Question about Dynkin Lehmann Scheffe Theorem I'm self-studying for an examination, and I would like to understand how to use the Dynkin Lehmann Scheffe theorem for an applied question. 
I am using Bickel and Doksum's "Mathematical Statistics" (2007 edition), and there is only a sentence describing the process:
Let P = {$P_{\theta}$ : $\theta$ $\in$  $\Theta$} where $P_{\theta}$ is discrete concentrated on X = {$x_1$,$x_2$,....}. Let $p(x,\theta)$ $\equiv$ $P_{\theta}$$[X = x]$ $\equiv$ $L_x(\theta)$ > 0 on X.
It can be shown that $\frac{L_x(.)}{L_x(\theta_0)}$ is minimally sufficient.
I found a question with the solution, but I would like to see how the process works.
Here is the question:

Here is the solution:

I apologize about not having shown any work, but there is a dearth of information available on this theorem, so I just wanted to see how the process for this type of problem works.
 A: The following claim makes no sense:

$\frac{L_x(.)}{L_x(\theta_0)}$ is minimal sufficient

Lehmann and Scheffé proved:

If $T(x)=T(y)$ $\iff$ $\theta \mapsto \dfrac{p(x,\theta)}{p(y,\theta)}$ is a constant function, then $T$ is minimal sufficient.

Call $(*)$ = "$\theta \mapsto \frac{p(x,\theta)}{p(y,\theta)}$ is a constant function". 
Taking an arbitrary $\theta_0 \in \Theta$ ($\theta_0=3$ in your example) then $(*)$ means that $\dfrac{p(x,\theta)}{p(y,\theta)} =  \dfrac{p(x,\theta_0)}{p(y,\theta_0)}$ or in other words $\dfrac{p(x,\theta)}{p(x,\theta_0)} =  \dfrac{p(y,\theta)}{p(y,\theta_0)}$ for every $\theta$. The annotations in the table of the solution you posted are the values of $\frac{p(x,\theta)}{p(x,\theta_0)}$ for every $x$ and $\theta$. Call $r(x)$ the vector ${\left(\frac{p(x,\theta)}{p(x,\theta_0)}\right)}_{\theta \in \Theta}$  Then the procedure consists in assigning a value to $T(x)$ shared by all $x$ having the same $r(x)$. For example $r(1)=r(2)$ in your problem, then we assign a common value to $T(1)$ and $T(2)$. Next, $r(3)$ is "alone" in your table ($r(x)\neq r(3)$ for $x\neq 3$), then assign a value to $T(3)$ different from the previously assigned values, and so on... Constructing $T$ by this way, then the condition of Lehmann-Scheffé's theorem is fulfilled, and then the theorem applies.
Update
Sorry I have understand that now:

$\frac{L_x(.)}{L_x(\theta_0)}$ is minimal sufficient

this is nothing but my $r(x)$ !
