Fig. 14

Example for the reduction from an instance \(({\mathfrak {C}},S)\) of Set Splitting to an instance \((\vec {G},\sigma )\) with \(k=0\) of BPURC, as specified in the proof of Thm. 30. In this example, we have \(S=\{a,b,c,d\}\) and \({\mathfrak {C}}=\{C_1,C_2,C_3\}\). By construction, all arcs are bidirectional and thus, arrow heads are omitted in the drawing of \((\vec {G},\sigma )\). A solution for \(({\mathfrak {C}},S)\) is \(S_1=\{a,d\}\) and \(S_2=\{b,c\}\). The latter is equivalent to a solution of BPURC by “separating” the a- and d-gadget from the b- and c-gadget as indicated by the dashed line. The latter yields a bipartition \({\mathcal {V}}=\{V_1,V_2\}\) of \(V(\vec {G})\) that solves BPURC with input \((\vec {G},\sigma ,k=0)\). Note, slight changes of the input \(({\mathfrak {C}},S)\) to \(S'=S\setminus \{d\}\) and \({\mathfrak {C}}'=\{C_1,C_2,C_3\setminus \{d\}\}\) would yield an instance of Set Splitting that has no yes-answer. In this case, the d-gadget would disappear from \((\vec {G},\sigma )\) resulting in the digraph \((\vec {G}',\sigma ')\). It is easy to see that there is no bipartition \({\mathcal {V}}=\{V_1,V_2\}\) of \(V(\vec {G}')\) such that \(\sigma (V_1)=\sigma (V_2)=\sigma (V(\vec {G}'))\) and no gadget gets split up between \(V_1\) and \(V_2\); two necessary properties to obtain a solution for BPURC with input \((\vec {G}',\sigma ')\) and \(k=0\) (cf. proof of Thm. 30)