methode yu lin esquissée

[ancetre.git] / classEquiv.tex

diff --git a/classEquiv.tex b/classEquiv.tex
index cdc1280d2bf86f1628f3160b8853d08da090e0b7..b77c51ac3bddcd20373d15cbd79facfd35360925 100644 (file)
--- a/classEquiv.tex
+++ b/classEquiv.tex
@@ -6,10 +6,10 @@ 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
 $g_{i}$ and $g_{j}$.
 
 The first step of this stage consists in building the following non-oriented

-graph furthere denoted as to \emph{similarity graphe}.
+graph further denoted as to \emph{similarity graph}.

 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 
 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 $t$.
+threshold $t$.

 
 We then define the relation $\sim$  such that
 $ x \sim y$ if $x$ and $y$ belong in the same connected component.
 
 We then define the relation $\sim$  such that
 $ x \sim y$ if $x$ and $y$ belong in the same connected component.


@@ -21,21 +21,21 @@ 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 the set of genes 
 by $\sim$, denoted $X/\sim = \{\dot{x} | x \textrm{ is a gene}\}$. 
 are also elements of the same equivalence class.
 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 each gene  into it respective equivalence classe by $\sim$.
+defined by $\pi(x) = \dot{x}$
+which maps each gene  into it respective equivalence class by $\sim$.

 
 
 
 
 
 
 
 

-For each genome $[g_l,\ldot,g{l+m}]$, the second step computes 
+For each genome $[g_l,\ldots,g{l+m}]$, the second step computes 

 the projection of each gene according to $\pi$. 
 The resulting genome  which is 
 $$
 the projection of each gene according to $\pi$. 
 The resulting genome  which is 
 $$

-[\pi(g_l),\ldot,\pi(g{l+m})]
+[\pi(g_l),\ldots,\pi(g{l+m})]

 $$ 
 is again of size $m$.
 
 $$ 
 is again of size $m$.
 

-Intuitivelly speaking, for two genes $g_i$ and $g_j$ 
+Intuitively 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) 
 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) 


@@ -48,8 +48,8 @@ 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.
 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  $\dot{x}$.
+This results as the set of all the class $\dot{x}$
+such that each genome has an gene $x$ in  $\dot{x}$.

 The pan genome is computed similarly: the union of all the 
 projected genomes in computed here.
 
 The pan genome is computed similarly: the union of all the 
 projected genomes in computed here.