$\{((g_1,g_2),r_{12}), (g_1,g_3),r_{13}), (g_{n-1},g{n}),r_{n-1.n})\}$ 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 $G$.
+is a pair that gives the similarity rate $r_{ij}$ between the two genes
+$g_{i}$ and $g_{j}$.
The first step of this stage consists in building the following non-oriented
graph furthere denoted as to \emph{similarity graphe}.
$g_{i}$ 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
+We then define the relation $\sim$ 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 $\dot{x}= \{y \in G | x \sim y\}$
+Let $\dot{x}= \{y | x \sim y\}$
denotes the equivalence class to which $x$ belongs.
-All the elements of $G$ which are equivalent to each other
+All the genes 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 $\pi: G \to G/\mathord{\sim}$
+Let us then consider the set of all equivalence classes of the set of genes
+by $\sim$, denoted $X/\sim = \{\dot{x} | x \textrm{ is a gene}\}$.
defined by \pi(x) = \dot{x}
-which maps elements of $G$ into their respective equivalence classes by $\sim$.
+which maps each gene into it respective equivalence classe by $\sim$.
-For each genome $G=[g_l,\ldot,g{l+m}]$, the second step computes
+For each genome $[g_l,\ldot,g{l+m}]$, the second step computes
the projection of each gene according to $\pi$.
-Let $G'$ the resulting genome which is
+The resulting genome which is
$$
-G'=[\pi(g_l),\ldot,\pi(g{l+m})]
+[\pi(g_l),\ldot,\pi(g{l+m})]
$$
-is of size $m$.
+is again 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 $g_j$.
is greater than $t$.
Genes $g_i$ and $g_j$ may thus have a common ancestor.
+
+We compute the core genome as follow.
+Each genome is projected according to $\pi$. We then consider the
+intersection of all the projected genomes which are considered as sets of genes
+and not as sequences of genes.
+This results as the set of all the class representents $\dot{x}$
+such that each geneome has an gene $x$ in $\dot{x}$.
+The pan genome is computed similarly: the union of all the
+projected genomes in computed here.
+
