SNPs data.
\end{enumerate}
-In~\cite{10.1371/journal.pone.0052841} the authors analyse 435 Mycobacterium Tuberculosis complex isolates of the same clade. By focusing on the H37Rv genome,
+In~\cite{10.1371/journal.pone.0052841} the authors analyze 435 Mycobacterium Tuberculosis complex isolates of the same clade. By focusing on the H37Rv genome,
they produce 13382 SNPs. Later, they compare 44 genomes to this one
regarding these SNP. The way they extract this phylogenetic tree is
not detailed. They focus then on Percy256 and Percy556 since both these
$g_{i}$ 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
-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.
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
$$
-[\pi(g_l),\ldot,\pi(g{l+m})]
+[\pi(g_l),\ldots,\pi(g{l+m})]
$$
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)
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 approache is further based on the ability to decide how far is each
+The approach is further based on the ability to decide how far is each
genome from each others. To achieve this, we combine XXX metrics which are
detailed in this part.
%1/ On SNPs of the core genome strict
All the $y$ are thus aligned
thanks to a global alignment tool. The SNPs may thus be extracted.
-For each genome, one can thus compute the vector of boolean values
-memorizing at index $i$ wether the SNP $i$ is present in one of its gene
-(postive value) or not (null value).
+For each genome, one can thus compute the vector of Boolean values
+memorizing at index $i$ whether the SNP $i$ is present in one of its gene
+(positive value) or not (null value).
A Hamming distance between two vectors allows to build the distance
between two genes.
-This metric is further refered as to $m_S$.
+This metric is further referred as to $m_S$.
% plus il y a de diff, plus le nombre est élevé
The $m_S$ method does not consider genes to have the same incidence in the
metric value. A gene with many SNPs has a larger influence in
the metric computation than a gene with fewer ones.
-The metric further refered as to $m_{|S|}$ gives the same weight to each gene
+The metric further referred as to $m_{|S|}$ gives the same weight to each gene
without considering the number of SNP it contains.
% plus il y a de diff, plus le nombre est élevé
-
-%3/ On gene content (symmetric difference)
-The third metric consider the symetric difference $\Delta$
-between the two sets $G_1$ and $G_2$ of genes.
+\subsection{Symmetric Difference based metric}
+The third metric consider the symmetric difference $\Delta$
+between the two sets $G_1$ and $G_2$ of genes recalled hereafter
$$
G_1\Delta G2 =
-(G1\cup G_2)\setminus (G1\cap G_2) = (G1\setminus G_2)\cup(G_2\setminus G1)
+(G1\cup G_2)\setminus (G1\cap G_2) = (G1\setminus G_2)\cup(G_2\setminus G1).
$$
+The cardinality of $G_1\Delta G2$, give the metric.
+This metric is furthered referred as to $m_{\Delta}$.
+
+Practically, let $k$ be the number of all the equivalence classes. Due to the definition of the pan genome, this number is equal to the cardinality of this set.
+For each genome, if we only consider which gene belongs into it \textit{i.e.}, if we abstract away all the position this gene appears, this genome may be
+memorized as a vector of $k$ Boolean values. The element at index $i, 0 \le i \le k-1$ is true if and only if the $i$-th gene of the pan genome belongs to this
+one.
+This metric is equal to the Hamming distance between the two corresponding
+vectors of Boolean values.
+
\end{document}
% 4/ Using EPFL method
