1$ (if no set can be found then $m^{(0)}_n(\alpha|\beta)$ vanishes and if $r=n$ then $\beta = \alpha$ and $m^{(0)}_n(\alpha|\alpha)$ is given as above). If the set can be extended, e.g., by adding $i-1$ while still remaining $\beta$-consecutive then take its maximal extension. Assuming $\{i,i+1,\ldots, i+r\}$ is maximal already, redraw the graph by moving all points in the set along the boundary of the disk, until they are close to each other.\footnote{In order to make this precise one has to take the limit in which they become a single point. However, the more informal description is enough for most practical purposes.} The other external points must be kept fixed. If the lines coming out of $i$ and $i+r$ intersect then give a name to identify the intersection point, e.g.\ $R$ (If the lines do not intersect then go back to the original graph and move on to the next set of both $\alpha$- and $\beta$-consecutive external~points). Assuming the lines intersect, note that $\{ i,i+1,\ldots ,i+r, R\}$ form a convex polygon. Now remove the polygon from the graph, bring $R$ to the boundary of the disk and treat the new graph, which has $R$ as an external point, as a \emph{new} problem and repeat the procedure. If at any given point one fails to find a set which forms a polygon with at most one internal point after trying all of them, then $m^{(0)}_n(\alpha|\beta)=0$. This procedure comes to an end when one finds a graph where both orderings agree completely and therefore gives a single polygon with only `external' edges. Finally, after completing all iterations and obtaining a list of all polygons found in the process one computes for each polygon its corresponding sub-amplitude. These sub-amplitudes can be computed just as regular amplitudes, where both orderings coincide, because for scalar particles it is straightforward to go off-shell. The double partial amplitude, $m^{(0)}_n(\alpha|\beta)$, is then given by the product of all sub-amplitudes times a propagator for each internal point that was obtained in the process of removing polygons. In figure~\ref{fig:example1a} we present $m^{(0)}_8(I|54376218)$ as an illustrative example. In this particular case, by a simple rearrangement of the external points (from figure~\ref{fig:example1a}(b) to figure~\ref{fig:example1a}(c)) one finds a decomposition in terms of four convex polygons, namely, $\{ 2,1,8,B\}$, $\{ 3,4,5,A \}$, $\{ 6,7,C\}$ and $\{ A,B,C\}$. It is easy to obtain the corresponding sub-amplitudes (ignoring the signs) \be m^{(0)}(1,2,B,8|B,2,1,8) = \frac{1}{s_{21}}+\frac{1}{s_{18}},\quad m^{(0)}(3,4,5,A|5,4,3,A) = \frac{1}{s_{34}}+\frac{1}{s_{45}}, \nonumber \ee \be m^{(0)}(6,7,C|7,6,C) =1,\quad m^{(0)}(B,A,C|B,A,C) =1, \ee and propagators \be A \rightarrow \frac{1}{s_{345}},\quad B\rightarrow \frac{1}{s_{812}},\quad C\rightarrow \frac{1}{s_{67}}. \ee Putting all of them together gives, up to an overall sign, \be m^{(0)}_8(I|54376218) = \left(\frac{1}{s_{21}}+\frac{1}{s_{18}}\right)\left(\frac{1}{s_{34}}+\frac{1}{s_{45}}\right)\frac{1}{s_{345}s_{812}s_{67}}. \label{aso}\ee \begin{figure}[tb] \centering \includegraphics{example1a.png} ]]>