This step considers as input the set 
$\{((g_1,g_2),r_{12}), (g_1,g_3),r_{13}), (g_{n-1},g{n}),r_{n-1.n})\}$ of 
$\frac{n(n-1)}{2}$ elements. 
Each one $(g_i,g_j),r_{ij})$ where $i < j$, 
is a pair that gives the similarity rate $r_{ij}$ between the genes  
$g_{i}$ and $g_{j}$ in $G$.

The first step of this stage consists in building the following non-oriented
graph furthere denoted as to \emph{similarity graphe}.
In this one, the vertices are the genes. There is an edge between 
$g_{i}$ and $g_{j}$ if the rate $r_{ij}$ is greater than a given similarity 
treeshold $t$.

We then define the relation $\sim \in G \times G $ such that
$ x \sim y$ if $x$ and $y$ belong in the same connected component.
Mathematically speaking, it is obvious that this 
defines an equivalence relation. 
Let $\dot{x}= \{y \in G | x \sim y\}$
denotes the equivalence class to which $x$ belongs.
All the elements of $G$ which are  equivalent to each other
are also elements of the same equivalence class.
Let us then consider the set of all equivalence classes of $G$ 
by $\sim$, denoted $X/\sim = \{\dot{x} | x \in G\}$. 
We then consider the projection $\pi: G \to G/\mathord{\sim}$
defined by \pi(x) = \dot{x} 
which maps elements of $G$ into their respective equivalence classes by $\sim$.




For each genome $G=[g_l,\ldot,g{l+m}]$, the second step computes 
the projection of each gene according to $\pi$. 
Let $G'$ the resulting genome  which is 
$$
G'=[\pi(g_l),\ldot,\pi(g{l+m})]
$$ 
is of size $m$.

Intuitivelly speaking, for two genes $g_i$ and $g_j$ 
in the same equivalence class, there is path from  $g_i$ and $g_j$.
It signifies that  each evolution step 
(represented by an edge in the similarity graph) 
has produced a gene s.t. the similarity with the previous one 
is greater than $t$. 
Genes $g_i$ and $g_j$ may thus have a common ancestor.

