\begin{algorithm}
\DontPrintSemicolon
\KwData{$G=(X,U)$ such that $G^{tc}$ is an order.}
\KwResult{$G'=(X,V)$ with $V\subseteq U$ such that $G'^{tc}$ is an
interval order.}
\Begin{
  $V \longleftarrow U$\;
  $S \longleftarrow \emptyset$\; 
  \For{$x\in X$}{ 
    $NbSuccInS(x) \longleftarrow 0$\;
    $NbPredInMin(x) \longleftarrow 0$\;
    $NbPredNotInMin(x) \longleftarrow |ImPred(x)|$\;
    }
  \For{$x \in X$}{
    \If{$NbPredInMin(x) = 0$ {\bf and} $NbPredNotInMin(x) = 0$}{
      $AppendToMin(x)$}
    } 
    \nl\While{$S \neq \emptyset$}{\label{InRes1}
    \nlset{REM} remove $x$ from the list of $T$ of maximal index\;\label{InResR}
    \lnl{InRes2}\While{$|S \cap  ImSucc(x)| \neq |S|$}{ 
      \For{$ y \in  S-ImSucc(x)$}{
        \{ remove from $V$ all the arcs $zy$ : \}\;
        \For{$z \in  ImPred(y) \cap  Min$}{
          remove the arc $zy$ from $V$\;
          $NbSuccInS(z) \longleftarrow NbSuccInS(z) - 1$\;
          move $z$ in $T$ to the list preceding its present list\;
          \{i.e. If $z \in T[k]$, move $z$ from $T[k]$ to 
           $T[k-1]$\}\;
          }
        $NbPredInMin(y) \longleftarrow 0$\;
        $NbPredNotInMin(y) \longleftarrow 0$\;
        $S \longleftarrow S - \{y\}$\;
        $AppendToMin(y)$\;
        }
      }
    $RemoveFromMin(x)$\;
    }
  }  
\caption{IntervalRestriction\label{IR}}
\end{algorithm}