X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/ancetre.git/blobdiff_plain/5e17bc6675db329838a7a6902e3956f16cfbbcf1..f5256e331a2301f2774366bfed13512b3373e1db:/classEquiv.tex?ds=inline

diff --git a/classEquiv.tex b/classEquiv.tex
index 5d9f19d..cdc1280 100644
--- a/classEquiv.tex
+++ b/classEquiv.tex
@@ -2,8 +2,8 @@ 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 $G$.
+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}.
@@ -11,30 +11,29 @@ 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
+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 $g_j$.
@@ -44,3 +43,13 @@ 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.
 
+
+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.
+