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+---
+title: "Théorie des graphes et plus courts chemins"
+date: "2025-05-26"
+---
+
+# Les graphes
+
+\Huge
+
+Les graphes
+
+# Exercice
+
+* Établir la liste d'adjacence et appliquer l'algorithme de parcours en largeur au graphe
+
+```{.mermaid format=pdf width=400 loc=figs/}
+graph LR;
+ 1---2;
+ 1---3;
+ 1---4;
+ 2---3;
+ 2---6;
+ 3---6;
+ 3---4;
+ 3---5;
+ 4---5;
+```
+
+
+# Illustration: parcours en profondeur
+
+{width=80%}
+
+# Parcours en profondeur
+
+## Idée générale
+
+* Initialiser les sommets comme non-lus
+* Visiter un sommet
+* Pour chaque sommet visité, on visite un sommet adjacent s'il est pas encore visité récursivement.
+
+## Remarque
+
+* La récursivité est équivalent à l'utilisation d'une **pile**.
+
+# Parcours en profondeur
+
+## Pseudo-code (5min)
+
+. . .
+
+```C
+initialiser(graphe) // tous sommets sont non-visités
+pile = visiter(sommet, vide) // sommet est un sommet du graphe au hasard
+tant que !est_vide(pile)
+ v = dépiler(pile)
+ pile = visiter(v, pile)
+```
+
+## Que fait visiter?
+
+. . .
+
+```C
+pile visiter(sommet, pile)
+ sommet = visité
+ pour w = chaque arête de sommet
+ si w != visité
+ pile = empiler(pile, w)
+ retourne pile
+```
+
+
+# Exercice
+
+* Établir la liste d'adjacence et appliquer l'algorithme de parcours en profondeur au graphe
+
+```{.mermaid format=pdf width=400 loc=figs/}
+graph LR;
+ 1---2;
+ 1---3;
+ 1---4;
+ 2---3;
+ 2---6;
+ 3---6;
+ 3---4;
+ 3---5;
+ 4---5;
+```
+
+# Interprétation des parcours
+
+* Un graphe vu comme espace d'états (sommet: état, arête: action);
+ * Labyrinthe;
+ * Arbre des coups d'un jeu.
+
+. . .
+
+* BFS (Breadth-First) ou DFS (Depth-First) parcourent l'espace des états à la recherche du meilleur mouvement.
+ * Les deux parcourent *tout* l'espace;
+ * Mais si l'arbre est grand, l'espace est gigantesque!
+
+. . .
+
+* Quand on a un temps limité
+ * BFS explore beaucoup de coups dans un futur proche;
+ * DFS explore peu de coups dans un futur lointain.
+
+# Contexte: les réseaux (informatique, transport, etc.)
+
+* Graphe orienté;
+* Source: sommet `s`;
+* Destination: sommet `t`;
+* Les arêtes ont des poids (coût d'utilisation, distance, etc.);
+* Le coût d'un chemin est la somme des poids des arêtes d'un chemin.
+
+## Problème à résoudre
+
+* Quel est le plus court chemin entre `s` et `t`.
+
+# Exemples d'application de plus court chemin
+
+## Devenir riches!
+
+* On part d'un tableau de taux de change entre devises.
+* Quelle est la meilleure façon de convertir l'or en dollar?
+
+{width=80%}
+
+. . .
+
+* 1kg d'or => 327.25 dollars
+* 1kg d'or => 208.1 livres => 327 dollars
+* 1kg d'or => 455.2 francs => 304.39 euros => 327.28 dollars
+
+# Exemples d'application de plus court chemin
+
+## Formulation sous forme d'un graphe: Comment (3min)?
+
+{width=80%}
+
+
+# Exemples d'application de plus court chemin
+
+## Formulation sous forme d'un graphe: Comment (3min)?
+
+{width=60%}
+
+* Un sommet par devise;
+* Une arête orientée par transaction possible avec le poids égal au taux de change;
+* Trouver le chemin qui maximise le produit des poids.
+
+. . .
+
+## Problème
+
+* On aimerait plutôt avoir une somme...
+
+
+# Exemples d'application de plus court chemin
+
+## Conversion du problème en plus court chemin
+
+* Soit `taux(u, v)` le taux de change entre la devise `u` et `v`.
+* On pose `w(u,w)=-log(taux(u,v))`
+* Trouver le chemin poids minimal pour les poids `w`.
+
+{width=60%}
+
+* Cette conversion se base sur l'idée que
+
+$$
+\log(u\cdot v)=\log(u)+\log(v).
+$$
+
+# Applications de plus courts chemins
+
+## Quelles applications voyez-vous?
+
+. . .
+
+* Déplacement d'un robot;
+* Planificaiton de trajet / trafic urbain;
+* Routage de télécommunications;
+* Réseau électrique optimal;
+* ...
+
+# Algorithmes de plus courts chemins
+
+\Huge
+
+Algorithmes de plus courts chemins
+
+# Contexte: les réseaux (informatique, transport, etc.)
+
+* Graphe orienté;
+* Source: sommet `s`;
+* Destination: sommet `t`;
+* Les arêtes ont des poids (coût d'utilisation, distance, etc.);
+* Le coût d'un chemin est la somme des poids des arêtes d'un chemin.
+
+## Problème à résoudre
+
+* Quel est le plus court chemin entre `s` et `t`.
+
+# Plus courts chemins à source unique
+
+* Soit un graphe, $G=(V, E)$, une fonction de pondération $w:E\rightarrow\mathbb{R}$, et un sommet $s\in V$
+ * Trouver pour tout sommet $v\in V$, le chemin de poids minimal reliant $s$ à $v$.
+* Algorithmes standards:
+ * Dijkstra (arêtes de poids positif seulement);
+ * Bellman-Ford (arêtes de poids positifs ou négatifs, mais sans cycles).
+* Comment résoudre le problèmes si tous les poids sont les mêmes?
+
+. . .
+
+* Un parcours en largeur!
+
+# Algorithme de Dijkstra
+
+## Comment chercher pour un plus court chemin?
+
+. . .
+
+```
+si distance(u,v) > distance(u,w) + distance(w,v)
+ on passe par w plutôt qu'aller directement
+```
+
+# Algorithme de Dijkstra (1 à 5)
+
+* $D$ est le tableau des distances au sommet $1$: $D[7]$ est la distance de 1 à 7.
+* Le chemin est pas forcément direct.
+* $S$ est le tableau des sommets visités.
+
+::: columns
+
+:::: column
+
+
+
+::::
+
+:::: column
+
+. . .
+
+![1 visité, `D[2]=1`, `D[4]=3`.](figs/dijkstra_1.png)
+
+::::
+
+:::
+
+# Algorithme de Dijkstra (1 à 5)
+
+::: columns
+
+:::: column
+
+
+
+
+::::
+
+:::: column
+
+. . .
+
+![2 visité, `D[3]=2`, `D[7]=3`.](figs/dijkstra_2.png)
+
+::::
+
+:::
+
+# Algorithme de Dijkstra (1 à 5)
+
+::: columns
+
+:::: column
+
+
+
+
+::::
+
+:::: column
+
+. . .
+
+![3 visité, `D[7]=3` inchangé, `D[6]=6`.](figs/dijkstra_3.png)
+
+::::
+
+:::
+
+# Algorithme de Dijkstra (1 à 5)
+
+
+::: columns
+
+:::: column
+
+
+
+
+::::
+
+:::: column
+
+. . .
+
+![4 visité, `D[7]=3` inchangé, `D[5]=9`.](figs/dijkstra_4.png)
+
+::::
+
+:::
+
+# Algorithme de Dijkstra (1 à 5)
+
+::: columns
+
+:::: column
+
+
+
+
+::::
+
+:::: column
+
+. . .
+
+![7 visité, `D[5]=7`, `D[6]=6` inchangé.](figs/dijkstra_5.png)
+
+::::
+
+:::
+
+# Algorithme de Dijkstra (1 à 5)
+
+::: columns
+
+:::: column
+
+
+
+
+::::
+
+:::: column
+
+. . .
+
+![`6` visité, `D[5]=7` inchangé.](figs/dijkstra_6.png)
+
+::::
+
+:::
+
+# Algorithme de Dijkstra (1 à 5)
+
+::: columns
+
+:::: column
+
+
+
+::::
+
+:::: column
+
+. . .
+
+
+
+::::
+
+:::
+
+# Algorithme de Dijkstra
+
+## Idée générale
+
+* On assigne à chaque noeud une distance $0$ pour $s$, $\infty$ pour les autres.
+* Tous les noeuds sont marqués non-visités.
+* Depuis du noeud courant, on suit chaque arête du noeud vers un sommet non visité et on calcule le poids du chemin à chaque voisin et on met à jour sa distance si elle est plus petite que la distance du noeud.
+* Quand tous les voisins du noeud courant ont été visités, le noeud est mis à visité (il ne sera plus jamais visité).
+* Continuer avec le noeud à la distance la plus faible.
+* L'algorithme est terminé losrque le noeud de destination est marqué comme visité, ou qu'on a plus de noeuds qu'on peut visiter et que leur distance est infinie.
+
+# Algorithme de Dijkstra
+
+## Pseudo-code (5min, matrix)
+
+\footnotesize
+
+. . .
+
+```C
+tab dijkstra(graph, s, t)
+ pour chaque v dans graphe
+ distance[v] = infini
+ q = ajouter(q, v)
+ distance[s] = 0
+ tant que non_vide(q)
+ // sélection de u t.q. la distance dans q est min
+ u = min(q, distance)
+ si u == t // on a atteint la cible
+ retourne distance
+ q = remove(q, u)
+ // voisin de u encore dans q
+ pour chaque v dans voisinage(u, q)
+ // on met à jour la distance du voisin en passant par u
+ n_distance = distance[u] + w(u, v)
+ si n_distance < distance[v]
+ distance[v] = n_distance
+ retourne distance
+```
+
+# Algorithme de Dijkstra
+
+* Cet algorithme, nous donne le plus court chemin mais...
+* ne nous donne pas le chemin!
+
+## Comment modifier l'algorithme pour avoir le chemin?
+
+. . .
+
+* Pour chaque nouveau noeud à visiter, il suffit d'enregistrer d'où on est venu!
+* On a besoin d'un tableau `precedent`.
+
+## Modifier le pseudo-code ci-dessus pour ce faire (3min matrix)
+
+# Algorithme de Dijkstra
+
+\footnotesize
+
+```C
+tab, tab dijkstra(graph, s, t)
+ pour chaque v dans graphe
+ distance[v] = infini
+ precedent[v] = indéfini
+ q = ajouter(q, v)
+ distance[s] = 0
+ tant que non_vide(q)
+ // sélection de u t.q. la distance dans q est min
+ u = min(q, distance)
+ si u == t
+ retourne distance
+ q = remove(q, u)
+ // voisin de u encore dans q
+ pour chaque v dans voisinage(u, q)
+ n_distance = distance[u] + w(u, v)
+ si n_distance < distance[v]
+ distance[v] = n_distance
+ precedent[v] = u
+ retourne distance, precedent
+```
+
+# Algorithme de Dijkstra
+
+## Comment reconstruire un chemin ?
+
+. . .
+
+```C
+pile parcours(precedent, s, t)
+ sommets = vide
+ u = t
+ // on a atteint t ou on ne connait pas de chemin
+ si u != s && precedent[u] != indéfini
+ tant que vrai
+ sommets = empiler(sommets, u)
+ u = precedent[u]
+ si u == s // la source est atteinte
+ retourne sommets
+ retourne sommets
+```
+
+# Algorithme de Dijkstra amélioré
+
+## On peut améliorer l'algorithme
+
+* Avec une file de priorité!
+
+## Une file de priorité est
+
+* Une file dont chaque élément possède une priorité,
+* Elle existe en deux saveurs: `min` ou `max`:
+ * File `min`: les éléments les plus petits sont retirés en premier.
+ * File `max`: les éléments les plus grands sont retirés en premier.
+* On regarde l'implémentation de la `max`.
+
+## Comment on fait ça?
+
+. . .
+
+* On insère les éléments à haute priorité tout devant dans la file!
+
+# Les files de priorité
+
+## Trois fonction principales
+
+```C
+booléen est_vide(element) // triviale
+element enfiler(element, data, priorite)
+data defiler(element)
+rien changer_priorite(element, data, priorite)
+nombre priorite(element) // utilitaire
+```
+
+## Pseudo-implémentation: structure (1min)
+
+. . .
+
+```C
+struct element
+ data
+ priorite
+ element suivant
+```
+
+# Les files de priorité
+
+## Pseudo-implémentation: enfiler (2min)
+
+. . .
+
+```C
+element enfiler(element, data, priorite)
+ n_element = creer_element(data, priorite)
+ si est_vide(element)
+ retourne n_element
+ si priorite(n_element) > priorite(element)
+ n_element.suivant = element
+ retourne n_element
+ sinon
+ tmp = element
+ prec = element
+ tant que !est_vide(tmp) && priorite < priorite(tmp)
+ prec = tmp
+ tmp = tmp.suivant
+ prev.suivant = n_element
+ n_element.suivant = tmp
+ retourne element
+```
+
+# Les files de priorité
+
+## Pseudo-implémentation: defiler (2min)
+
+. . .
+
+```C
+data, element defiler(element)
+ si est_vide(element)
+ retourne AARGL!
+ sinon
+ tmp = element.data
+ n_element = element.suivant
+ liberer(element)
+ retourne tmp, n_element
+```
+
+# Algorithme de Dijkstra avec file de priorité min
+
+```C
+distance, precedent dijkstra(graphe, s, t):
+ distance[source] = 0
+ fp = file_p_vide()
+ pour v dans sommets(graphe)
+ si v != s
+ distance[v] = infini
+ precedent[v] = indéfini
+ fp = enfiler(fp, v, distance[v])
+ tant que !est_vide(fp)
+ u, fp = defiler(fp)
+ pour v dans voisinage de u
+ n_distance = distance[u] + w(u, v)
+ si n_distance < distance[v]
+ distance[v] = n_distance
+ precedent[v] = u
+ fp = changer_priorite(fp, v, n_distance)
+ retourne distance, precedent
+```
+
+# Algorithme de Dijkstra avec file
+
+\footnotesize
+
+```C
+distance dijkstra(graphe, s, t)
+---------------------------------------------------------
+ pour v dans sommets(graphe)
+O(V) si v != s
+ distance[v] = infini
+O(V) fp = enfiler(fp, v, distance[v]) // notre impl est nulle
+------------------O(V * V)-------------------------------
+ tant que !est_vide(fp)
+O(1) u, fp = defiler(fp)
+---------------------------------------------------------
+O(E) pour v dans voisinage de u
+ n_distance = distance[u] + w(u, v)
+ si n_distance < distance[v]
+ distance[v] = n_distance
+O(V) fp = changer_priorite(fp, v, n_distance)
+---------------------------------------------------------
+ retourne distance
+```
+
+* Total: $\mathcal{O}(|V|^2+|E|\cdot |V|)$:
+ * Graphe dense: $\mathcal{O}(|V|^3)$
+ * Graphe peu dense: $\mathcal{O}(|V|^2)$
+
+# Algorithme de Dijkstra avec file
+
+## On peut faire mieux
+
+* Avec une meilleure implémentation de la file de priorité:
+ * Tas binaire: $\mathcal{O}(|V|\log|V|+|E|\log|V|)$.
+ * Tas de Fibonnacci: $\mathcal{O}(|V|+|E|\log|V|)$
+* Graphe dense: $\mathcal{O}(|V|^2\log|V|)$.
+* Graphe peu dense: $\mathcal{O}(|V|\log|V|)$.
+
+# Algorithme de Dijkstra (exercice, 5min)
+
+{width=60%}
+
+* Donner la liste de priorité, puis...
+
+## A chaque étape donner:
+
+* Le tableau des distances à `a`;
+* Le tableau des prédécesseurs;
+* L'état de la file de priorité.
+
+# Algorithme de Dijkstra (corrigé)
+
+
+
+# Algorithme de Dijkstra (corrigé)
+
+
+
+# Algorithme de Dijkstra (corrigé)
+
+
+
+# Algorithme de Dijkstra (corrigé)
+
+
+
+# Algorithme de Dijkstra (corrigé)
+
+
+
+# Algorithme de Dijkstra (corrigé)
+
+
+
+# Limitation de l'algorithme de Dijkstra
+
+## Que se passe-t-il pour?
+
+{width=50%}
+
+## Quel est le problème?
+
+. . .
+
+* L'algorithme n'essaiera jamais le chemin `s->x->y->v` et prendra direct `s->v`.
+* Ce problème n'apparaît que s'il y a des poids négatifs.
+
+
+# Plus cours chemin pour toute paire de sommets
+
+## Comment faire pour avoir toutes les paires?
+
+. . .
+
+* Appliquer Dijkstra sur tous les sommets d'origine.
+* Complexité:
+ * Graphe dense: $\mathcal{O}(|V|)\mathcal{O}(|V|^2\log|V|)=\mathcal{O}(|V|^3\log|V|)$.
+ * Graphe peu dense: $\mathcal{O}(|V|)\mathcal{O}(|V|\log|V|)=\mathcal{O}(|V|^2\log|V|)$.
+
+. . .
+
+## Solution alternative: Floyd--Warshall
+
+* Pour toutes paires de sommets $u,v\in V$, trouver le chemin de poids minimal reliant $u$ à $v$.
+* Complexité $\mathcal{O}(|V|^3)$, indiqué pour graphes denses.
+* Fonctionne avec la matrice d'adjacence.
+
+# Algorithme de Floyd--Warshall
+
+## Idée générale
+
+* Soit l'ensemble de sommets $V=\{1, 2, 3, 4, ..., n\}$.
+* Pour toute paire de sommets, $i,j$, on considère tous les chemins passant par les sommets intermédiaires $\in\{1, 2, ..., k\}$ avec $k\leq n$.
+* On garde pour chaque $k$ la plus petite valeur.
+
+## Principe
+
+* A chaque étape, $k$, on vérifie s'il est plus court d'aller de $i$ à $j$ en passant par le sommet $k$.
+* Si à l'étape $k-1$, le coût du parcours est $p$, on vérifie si $p$ est plus petit que $p_1+p_2$, le chemin de $i$ à $k$, et $k$ à $j$ respectivement.
+
+# Algorithme de Floyd--Warshall
+
+## The algorithme
+
+Soit $d_{ij}(k)$ le plus court chemin de $i$ à $j$ passant par les sommets $\in\{1,2,...,k\}$
+
+$$
+d_{ij}(k)=\left\{
+\begin{array}{ll}
+ w(i,j), & \mbox{si } k=0,\\
+ \min(d_{ij}(k-1),d_{ik}(k-1)+d_{kj}(k-1)), & \mbox{sinon}.
+\end{array}
+\right.
+$$
+
+# Algorithme de Floyd--Warshall (exemple)
+
+
+::: columns
+
+:::: column
+
+
+
+
+::::
+
+:::: column
+
+## Que vaut $D^{(0)}$ (3min)?
+
+. . .
+
+$$
+D^{(0)}=\begin{bmatrix}
+0 & 2 & 4 & \infty & 3 \\
+2 & 0 & 8 & \infty & 1 \\
+6 & 2 & 0 & 4 & 3 \\
+1 & \infty & \infty & 0 & 5 \\
+\infty & \infty & \infty & 1 & 0 \\
+\end{bmatrix}
+$$
+
+::::
+
+:::
+
+# Algorithme de Floyd--Warshall (exemple)
+
+
+::: columns
+
+:::: column
+
+## On part de $D^{(0)}$?
+
+$$
+D^{(0)}=\begin{bmatrix}
+0 & 2 & 4 & \infty & 3 \\
+2 & 0 & 8 & \infty & 1 \\
+6 & 2 & 0 & 4 & 3 \\
+1 & \infty & \infty & 0 & 5 \\
+\infty & \infty & \infty & 1 & 0 \\
+\end{bmatrix}
+$$
+
+
+::::
+
+:::: column
+
+## Que vaut $D^{(1)}$ (3min)?
+
+. . .
+
+$$
+D^{(0)}=\begin{bmatrix}
+0 & 2 & 4 & \infty & 3 \\
+2 & 0 & \mathbf{6} & \infty & 1 \\
+6 & 2 & 0 & 4 & 3 \\
+1 & \mathbf{3} & \mathbf{5} & 0 & \mathbf{4} \\
+\infty & \infty & \infty & 1 & 0 \\
+\end{bmatrix}
+$$
+
+::::
+
+:::
+
+# Algorithme de Floyd--Warshall (exemple)
+
+
+::: columns
+
+:::: column
+
+## On part de $D^{(0)}$
+
+$$
+D^{(0)}=\begin{bmatrix}
+0 & 2 & 4 & \infty & 3 \\
+2 & 0 & 8 & \infty & 1 \\
+6 & 2 & 0 & 4 & 3 \\
+1 & \infty & \infty & 0 & 5 \\
+\infty & \infty & \infty & 1 & 0 \\
+\end{bmatrix}
+$$
+
+
+::::
+
+:::: column
+
+## Que vaut $D^{(1)}$ (3min)?
+
+. . .
+
+$$
+D^{(1)}=\begin{bmatrix}
+0 & 2 & 4 & \infty & 3 \\
+2 & 0 & \mathbf{6} & \infty & 1 \\
+6 & 2 & 0 & 4 & 3 \\
+1 & \mathbf{3} & \mathbf{5} & 0 & \mathbf{4} \\
+\infty & \infty & \infty & 1 & 0 \\
+\end{bmatrix}
+$$
+
+## Exemple
+
+$$
+D_{42}^{(1)}=D_{41}^{(0)}+D_{12}^{(0)}=1+2<\infty.
+$$
+
+::::
+
+:::
+
+# Algorithme de Floyd--Warshall (exemple)
+
+::: columns
+
+:::: column
+
+## On part de $D^{(1)}$
+
+$$
+D^{(1)}=\begin{bmatrix}
+0 & 2 & 4 & \infty & 3 \\
+2 & 0 & 6 & \infty & 1 \\
+6 & 2 & 0 & 4 & 3 \\
+1 & 3 & 5 & 0 & 4 \\
+\infty & \infty & \infty & 1 & 0 \\
+\end{bmatrix}
+$$
+
+
+::::
+
+:::: column
+
+## Que vaut $D^{(2)}$ (3min)?
+
+. . .
+
+$$
+D^{(2)}=\begin{bmatrix}
+0 & 2 & 4 & \infty & 3 \\
+2 & 0 & 6 & \infty & 1 \\
+\mathbf{4} & 2 & 0 & 4 & 3 \\
+1 & 3 & 5 & 0 & 4 \\
+\infty & \infty & \infty & 1 & 0 \\
+\end{bmatrix}
+$$
+
+::::
+
+:::
+
+# Algorithme de Floyd--Warshall (exemple)
+
+::: columns
+
+:::: column
+
+## On part de $D^{(2)}$
+
+$$
+D^{(2)}=\begin{bmatrix}
+0 & 2 & 4 & \infty & 3 \\
+2 & 0 & 6 & \infty & 1 \\
+4 & 2 & 0 & 4 & 3 \\
+1 & 3 & 5 & 0 & 4 \\
+\infty & \infty & \infty & 1 & 0 \\
+\end{bmatrix}
+$$
+
+
+::::
+
+:::: column
+
+## Que vaut $D^{(3)}$ (3min)?
+
+. . .
+
+$$
+D^{(3)}=\begin{bmatrix}
+0 & 2 & 4 & \mathbf{8} & 3 \\
+2 & 0 & 6 & \mathbf{10} & 1 \\
+4 & 2 & 0 & 4 & 3 \\
+1 & 3 & 5 & 0 & 4 \\
+\infty & \infty & \infty & 1 & 0 \\
+\end{bmatrix}
+$$
+
+::::
+
+:::
+
+# Algorithme de Floyd--Warshall (exemple)
+
+::: columns
+
+:::: column
+
+## On part de $D^{(3)}$
+
+$$
+D^{(3)}=\begin{bmatrix}
+0 & 2 & 4 & 8 & 3 \\
+2 & 0 & 6 & 10 & 1 \\
+4 & 2 & 0 & 4 & 3 \\
+1 & 3 & 5 & 0 & 4 \\
+\infty & \infty & \infty & 1 & 0 \\
+\end{bmatrix}
+$$
+
+::::
+
+:::: column
+
+## Que vaut $D^{(4)}$ (3min)?
+
+. . .
+
+$$
+D^{(4)}=\begin{bmatrix}
+0 & 2 & 4 & 8 & 3 \\
+2 & 0 & 6 & 10 & 1 \\
+4 & 2 & 0 & 4 & 3 \\
+1 & 3 & 5 & 0 & 4 \\
+\mathbf{2} & \mathbf{4} & \mathbf{6} & 1 & 0 \\
+\end{bmatrix}
+$$
+
+::::
+
+:::
+
+# Algorithme de Floyd--Warshall (exemple)
+
+::: columns
+
+:::: column
+
+## On part de $D^{(4)}$
+
+$$
+D^{(4)}=\begin{bmatrix}
+0 & 2 & 4 & 8 & 3 \\
+2 & 0 & 6 & 10 & 1 \\
+4 & 2 & 0 & 4 & 3 \\
+1 & 3 & 5 & 0 & 4 \\
+2 & 4 & 6 & 1 & 0 \\
+\end{bmatrix}
+$$
+
+::::
+
+:::: column
+
+## Que vaut $D^{(5)}$ (3min)?
+
+. . .
+
+$$
+D^{(5)}=\begin{bmatrix}
+0 & 2 & 4 & \mathbf{4} & 3 \\
+2 & 0 & 6 & \mathbf{2} & 1 \\
+4 & 2 & 0 & 4 & 3 \\
+1 & 3 & 5 & 0 & 4 \\
+2 & 4 & 6 & 1 & 0 \\
+\end{bmatrix}
+$$
+
+::::
+
+:::
+
+# Algorithme de Floyd--Warshall
+
+## The pseudo-code (10min)
+
+* Quelle structure de données?
+* Quelle initialisation?
+* Quel est le code pour le calcul de la matrice $D$?
+
+# Algorithme de Floyd--Warshall
+
+## The pseudo-code
+
+* Quelle structure de données?
+
+```C
+int distance[n][n];
+```
+
+. . .
+
+* Quelle initialisation?
+
+```C
+matrice ini_floyd_warshall(distance, n, w)
+ pour i de 1 à n
+ pour j de 1 à n
+ distance[i][j] = w(i,j)
+ retourne distance
+```
+
+# Algorithme de Floyd--Warshall
+
+## The pseudo-code
+
+* Quel est le code pour le calcul de la matrice $D$?
+
+```C
+matrice floyd_warshall(distance, n, w)
+ pour k de 1 à n
+ pour i de 1 à n
+ pour j de 1 à n
+ distance[i][j] = min(distance[i][j],
+ distance[i][k] + distance[k][j])
+ retourne distance
+```
+
+# Algorithme de Floyd--Warshall
+
+## La matrice de précédence
+
+* On a pas encore vu comment reconstruire le plus court chemin!
+* On définit, $P_{ij}^{(k)}$, qui est le prédécesseur du sommet $j$ depuis $i$ avec les sommets intermédiaires $\in\{1, 2, ..., k\}$.
+$$
+P^{(0)}_{ij}=\left\{
+\begin{array}{ll}
+ \mbox{vide}, & \mbox{si } i=j\mbox{, ou }w(i,j)=\infty\\
+ i, & \mbox{sinon}.
+\end{array}
+\right.
+$$
+
+* Mise à jour
+$$
+P^{(k)}_{ij}=\left\{
+\begin{array}{ll}
+ P^{(k-1)}_{\mathbf{i}j}, & \mbox{si } d_{ij}^{(k)}\leq d_{ik}^{(k-1)}+d_{kj}^{(k-1)}\\
+ P^{(k-1)}_{\mathbf{k}j}, & \mbox{sinon}.
+\end{array}
+\right.
+$$
+
+. . .
+
+* Moralité: si le chemin est plus court en passant par $k$, alors il faut utiliser son prédécesseur!
+
+# Algorithme de Floyd--Warshall
+
+## La matrice de précédence (pseudo-code, 3min)
+
+. . .
+
+```C
+matrice, matrice floyd_warshall(distance, n, w)
+ pour k de 1 à n
+ pour i de 1 à n
+ pour j de 1 à n
+ n_distance = distance[i][k] + distance[k][j]
+ if n_distance < distance[i][j]
+ distance[i][j] = n_distance
+ précédence[i][j] = précédence[k][j]
+ retourne distance, précédence
+```
+
+# Algorithme de Floyd--Warshall (exercice)
+
+
+::: columns
+
+:::: column
+
+
+
+
+::::
+
+:::: column
+
+## Que vaut $P^{(0)}$ (3min)?
+
+. . .
+
+$$
+P^{(0)}=\begin{bmatrix}
+- & 1 & 1 & - & 1 \\
+2 & - & 2 & - & 2 \\
+3 & 3 & - & 3 & 3 \\
+4 & - & - & - & 4 \\
+- & - & - & 5 & - \\
+\end{bmatrix}
+$$
+
+::::
+
+:::
+
+# Algorithme de Floyd--Warshall (exercice)
+
+
+::: columns
+
+:::: column
+
+
+
+
+::::
+
+:::: column
+
+## Que vaut $P^{(5)}$ (10min)?
+
+. . .
+
+$$
+P^{(5)}=\begin{bmatrix}
+- & 1 & 1 & 5 & 1 \\
+2 & - & 1 & 5 & 2 \\
+2 & 3 & - & 3 & 3 \\
+4 & 1 & 1 & - & 1 \\
+4 & 1 & 1 & 5 & - \\
+\end{bmatrix}
+$$
+
+::::
+
+:::
+
+# Exercice: retrouver le chemin entre 1 et 4 (5min)
+
+$$
+P=\begin{bmatrix}
+- & 1 & 1 & 5 & 1 \\
+2 & - & 1 & 5 & 2 \\
+2 & 3 & - & 3 & 3 \\
+4 & 1 & 1 & - & 4 \\
+4 & 1 & 1 & 5 & - \\
+\end{bmatrix}
+$$
+
+. . .
+
+## Solution
+
+* Le sommet $5=P_{14}$, on a donc, $5\rightarrow 4$, on veut connaître le prédécesseur de 5.
+* Le sommet $1=P_{15}$, on a donc, $1\rightarrow 5\rightarrow 4$. The end.
+
+# Exercice complet
+
+## Appliquer l'algorithme de Floyd--Warshall au graphe suivant
+
+{width=50%}
+
+* Bien indiquer l'état de $D$ et $P$ à chaque étape!
+* Ne pas oublier de faire la matrice d'adjacence évidemment...
+
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