Finished relations.md

docs/en/mathematics/set-theory/relations.md

@@ -79,8 +79,6 @@ Some relations have special properties
 
     Let $\Pi_R$ be the set of $R$-equivalence classes. Then by reflexivity of $R$ we find that each element $a \in S$ is inside the class $[a]_R$ of $\Pi_R$. If an element $a \in S$ is in the classes $[b]_R$ and $[c]_R$ of $\Pi_R$, then by the previous lemma we find $[b]_R = [a]_R$ and $[c]_R = [a]_R$. Then $[b]_R = [c]_R$, therefore each element $a \in S$ is inside a unique member of $\Pi_R$, which therefore is a partition of $S$.
 
-<br>
-
 ## Composition of relations
 
 If $R_1$ and $R_2$ are two relations between a set $S$ and $T$, new relations can be formed between $S$ and $T$ by taking the intersection $R_1 \cap R_2$, the union $R_1 \cup R_2$ or the complement $R_1 \backslash R_2$. Furthermore a relation $R^\top$ from $T$ to $S$ can be considered as the relation $\{(t,s) \in T \times S \;|\; (s,t) \in R\}$ and the identity relation from $T$ to $S$ is given by $I = \{(s, t) \in S \times T \;|\; s = t\}$
@@ -99,8 +97,6 @@ Another way of making new relations out of existing ones is by taking the compos
 
     Similarly, if $s \in S$ and $v \in V$ with $s(R_1;R_2);R_3v$. Then a $u \in U$ with $s(R_1;R_2)u$ and $uR_3v$ can be found. Then there is also a $t \in T$ with $sR_1t$ and $tR_2u$. For this $t$ there is $tR_2;R_3u$ and $sR_1t$ and hence $sR_1;(R_2;R_3)v$.
 
-<br>
-
 ## Transitive closure
 
 > *Lemma*: let $\ell$ be a collection of relations $R$ on a set $S$. If all relations $R$ in $\ell$ are transitive, reflexive or symmetric then the relation $\bigcap_{R \in \ell} R$ is also transitive, reflexive or symmetric respectively.
@@ -157,4 +153,31 @@ It may be observed that the reflexive closure of $R$ equals the relation $I \cup
 
     :   Now suppose that for some $k \in \mathbb{N}$ we have $R^k \subseteq T$. Then by assumption $R^{k+1} \subseteq T$. Let $(s_1, s_3) \in R^{k+1} = R^k ; R$, then $(s_1, s_2) \in R$ and $(s_2, s_3) \in R^k$ for some $s_2$. Hence $(s_1, s_2), (s_2, s_3) \in T$ and by transitivity of $T$, $(s_1, s_3) \in T$. 
 
-    Hence if the claim holds for some $k \in \mathbb{N}$ then it also holds for $k+1$. The principle of natural induction implies now that $\forall n \in \mathbb{N}$ we have $R^n \subseteq \bar R \subseteq T$. 
+    Hence if the claim holds for some $k \in \mathbb{N}$ then it also holds for $k+1$. The principle of natural induction implies now that $\forall n \in \mathbb{N}$ we have $R^n \subseteq \bar R \subseteq T$. 
+
+Suppose a relation $R$ on a finite set $S$ of size $n$ is given by its adjacency matrix $A_R$. Then Warshall's algorithm is an method for finding the adjacency matrix of the transitive closure of the relation $R$. 
+
+> *Algorithm* **- Warshall's algoritm**: for an adjacency matrix $A_R = M_0$ of relation $R$ on $n$ elements there will be $n$ steps taken to obtain the adjacency matrix of the transitive closure of the relation $R$. Let $R_i$ and $C_i$ be the $i$th row and column of $A_R$. In each step a new matrix $M_i$ is obtained with $C_i \times R_i$ added to $M_{i-1}$. After $n$ steps $A_{\bar R}$ is obtained.
+
+For example let $R$ be an relation on $S = \{1,2,3,4\}$ with $R = \{(2,1), (2,3), (3,1), (3,4), (4,1), (4,3)\}$, determining the transitive closure $\bar R$ of $R$ with Warshall's algorithm.
+
+:   The adjacency matrix of the relation $R$ is given by
+
+    $$
+        A_R = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0\end{pmatrix}.
+    $$
+
+    We have $C_1 = \{2,3,4\}$ and $R_1 = \varnothing$, therefore $C_1 \times R_1 = \varnothing$ and no additions will be made, $M_1 = A_R$.
+
+    We have $C_2 = \varnothing$ and $R_2 = \{1,3\}, therefore $C_2 \times R_2 = \varnothing$ and no additions will be made, $M_2 = M_1$.
+
+    We have $C_3 = \{2,4\}$ and $R_3 = \{1,4\}$, therefore $C_3 \times R_3 = \{(2,1), (2,4), (4,1), (4,4)\}$ obtaining the matrix
+
+    $$
+        M_3 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1\end{pmatrix}.
+    $$
+    We have $C_4 = \{2,3,4\} and $R_4 = \{1,3,4\}, therefore $C_4 \times R_4 = \{(2,1), (2,3), (2,4), (3,1), (3,3), (3,4), (4,1), (4,3), (4,4)\}$ obtaining the final matrix
+
+    $$
+    M_4 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1\end{pmatrix} = A_{\bar R}.
+    $$