Fig. 5

Given \(\pi = (-\, 1 -2 +12 -4 -5 -6 -7 +3 +9 +10 +11 +8 -13 -14)\), a and b show the two cp-graphs \(G^{X^1}_{\pi}\) and \(G^{X^2}_{\pi}\) for VD-vectors \(X^1 = (0,0,9,0,0,0,0,-\,5,0,0,0,-\,4,0,0)\) and \(X^2 = (0,0,-\,5,0,0,0,0,9,0,0,0,-\,4,0,0)\). \(X^1\) and \(X^2\) are the two VD-vectors with minimum crossing number (i.e., \(cn(X^1) = cn(X^2) = cn(\pi ) = 16\)), so \(\{X^1,X^2\} \in S\). Note that \(X^2\) (resp. \(X^1\)) can be obtained from \(X^1\) (resp. \(X^2\)) by \(T_{3,8}(X^1)\) (resp. \(T_{8,3}(X^2)\)), so Algorithm 1 will generate both VD-vectors, starting either with \(X^1\) or \(X^2\). Note that \(cc^-(G^{X^1}_{\pi }) = cc^-(G^{X^2}_{\pi }) = 4\), so \(d(\pi ,X^1) = d(\pi ,X^2) = 16+4 = 20\). In c we have the cp-graph \(G_{\pi}^{X^3}\) for VD-vector \(X^3 = (0,0,- \, 5,0,0,0,0,- \, 5,0,0,0,10,0,0)\) with \(cn(X^3) = 18 > cn(\pi )\), so \(X^3 \not \in S\), but \(X^3 \in S'\) since \(X^3 = T_{3,12}(X^1) = T_{8,12}(X^2)\). Note that \(cc^-(G^{X^3}_{\pi }) = 0\), so \(d(\pi ,X^3) = 18\). Among all VD-vectors in \(S \cup S'\), \(X^3\) is in fact the VD-vector that minimizes the sum and it follows that \(d(\pi ) = d(\pi, X^3) = 18\)