Algorithm Correctness

Here we will give an outline of the proof. For the details of the proof please refer to [1].

Below we will show in some detail a result, which proves the last statement of the above outline.

  

Figure: Schematic proof of the key observation. This figure depicts a graph G (blank), original clique graph H (green) and some other clique graph C (green with pattern). Consider $E(H) \Delta E(G)$ and $E(C) \Delta E(G)$. Each edge (u,v) which belongs to neither C nor to H, but belongs to G increments both $E(H) \Delta E(G)$ and $E(C) \Delta E(G)$. Each edge (u,v) which belongs to both C and H, but does not belong to G also increments both $E(H) \Delta E(G)$ and $E(C) \Delta E(G)$. Thus, the difference between $E(H) \Delta E(G)$ and $E(C) \Delta E(G)$ is due to patches that correspond to $E(H) \Delta E(C)$. Let us denote by $\Delta ^{*}$ the symmetric difference constrained to edges in $E(C) \Delta E(H)$. But $\vert E(C) \Delta ^{*} E(G)\vert + \vert E(H) \Delta ^{*} E(G)\vert = \vert E(H) \Delta E(C)\vert$. Thus $\vert E(C) \Delta ^{*} E(G)\vert < \vert E(H) \Delta ^{*} E(G)\vert$ iff $\vert E(H) \Delta ^{*} E(G)\vert > \frac {\vert E(H) \Delta E(C)\vert}{2}$. Hence, $\vert E(C) \Delta E(G)\vert < \vert E(H) \Delta E(G)\vert$ iff $\vert E(H) \Delta ^{*} E(G)\vert > \frac {\vert E(H) \Delta E(C)\vert}{2}$.