Fig. 4

Given \(\pi = (+\,4 +5 +3 +1 -2 -6)\), we have in a the cp-graph \(G^{X^1}_{\pi}\) for \(X^1 = (3,3,0,-3,-3,0)\), in b the cp-graph \(G^{X^2}_{\pi}\) for \(X^2 = (3,-3,0,3,-3,0)\), in c the cp-graph \(G^{X^3}_{\pi}\) for \(X^3 = (3,-3,0,-3,3,0)\), in d the cp-graph \(G^{X^4}_{\pi}\) for \(X^4 = (- \, 3,3,0,3,-3,0)\), in e the cp-graph \(G^{X^5}_{\pi}\) for \(X^5 = (- \, 3,3,0,-3,3,0)\), and in f the cp-graph \(G^{X^6}_{\pi}\) for \(X^6 = (- \, 3,-3,0,3,3,0)\). \(X^1\) to \(X^6\) are the six possible VD-vectors for \(\pi\) with minimum crossing number (i.e., \(cn(X^i) = cn(\pi ) = 8)\) with \(i \in [1..6]\)). Note that, by definition, they are in S, but Algorithm 1 will not generate all of them. For instance, if the algorithm starts S (at line 9) with \(X^1\), it will not generate (in loop at lines 10–16) \(X^6\) since this VD-vector is reachable from \(X^1\) using two transformations but not using only one (note that they differ in exactly four displacement values). As proved in Theorem 1, since Algorithm 1 is not capable of generating all VD-vectors in S, then there is at least one VD-vector generated in the path between \(X^1\) and \(X^6\) with only one component (in this example, the four intermediate VD-vectors \(X^2\), \(X^3\), \(X^4\), and \(X^5\) that are generated by Algorithm 1 satisfy this condition)