Weighted Triangle-free 2-matching Problem with Edge-disjoint Forbidden Triangles

Abstract

The weighted ${\mathcal {T}}$-free 2-matching problem is the following problem: given an undirected graph G, a weight function on its edge set, and a set ${\mathcal {T}}$ of triangles in G, find a maximum weight 2-matching containing no triangle in ${\mathcal {T}}$. When ${\mathcal {T}}$ is the set of all triangles in G, this problem is known as the weighted triangle-free 2-matching problem, which is a long-standing open problem. A main contribution of this paper is to give the first polynomial-time algorithm for the weighted ${\mathcal {T}}$-free 2-matching problem under the assumption that ${\mathcal {T}}$ is a set of edge-disjoint triangles. In our algorithm, a key ingredient is to give an extended formulation representing the solution set, that is, we introduce new variables and represent the convex hull of the feasible solutions as a projection of another polytope in a higher dimensional space. Although our extended formulation has exponentially many inequalities, we show that the separation problem can be solved in polynomial time, which leads to a polynomial-time algorithm for the weighted ${\mathcal {T}}$-free 2-matching problem.

Appendices

A Proof of Lemma 5

By symmetry, it suffices to consider $(G_1, b_1, {\mathcal {T}}_1)$. Since the tightness of (10) for $(S^*, F^*_0, F^*_1)$ implies that $x_1(\delta _{G_1}(r)) = 1$, we can easily see that $(x_1, y_1)$ satisfies (1), (2), (4)–(7). In what follows, we consider (10) for $(x_1, y_1)$ in $(G_1, b_1, {\mathcal {T}}_1)$. For edge sets $F'_0, F'_1 \subseteq E_1$, we denote $g(F'_0, F'_1) = \sum _{e \in F'_0} x_1(e) + \sum _{e \in F'_1} (1-x_1(e))$ to simplify the notation. For $(S', F'_0, F'_1) \in {\mathcal {F}}_1$, let $h(S', F'_0, F'_1)$ denote the left-hand side of (10). To derive a contradiction, let $(S', F'_0, F'_1) \in {\mathcal {F}}_1$ be a minimizer of $h(S', F'_0, F'_1)$ and assume that $h(S', F'_0, F'_1) < 1$. By changing the roles of $S'$ and $V' {\setminus } S'$ if necessary, we may assume that $r \not \in S'$.

For $T \in {\mathcal {T}}^+_{S^*}$, let $v_1, v_2, v_3, \alpha , \beta$, and $\gamma$ be as in Figs. 7–10. Let $G'_T = (V'_T, E'_T)$ be the subgraph of $G_1$ corresponding to T, that is, the subgraph induced by $\{r, p_1, p_2, v_2, v_3\}$ (Fig. 7), $\{r, p_3, v_1 \}$ (Fig. 8), $\{r, p_1, p_2, p_3, v_2, v_3\}$ (Fig. 9), or $\{r, p_4, v_1 \}$ (Fig. 10). Let ${\hat{S}} = S' \cap (V'_T {\setminus } \{v_1, v_2, v_3\})$, ${\hat{F}}_0 = F'_0 \cap E'_T$, and ${\hat{F}}_1 = F'_1 \cap E'_T$.

We show the following properties (P1)–(P9) in Sect. 1, and show that $(x_1, y_1)$ satisfies (10) by using these properties in Sect. 2.

• (P1) If T is of type (A) or (B) and $v_2, v_3 \not \in S'$, then $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even.

• (P2) If T is of type (A), $v_2, v_3 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even, then $g({\hat{F}}_0, {\hat{F}}_1) \ge \min \{x(\alpha )+x(\gamma ), 2-x(\alpha )-x(\gamma ) - 2 y_\beta \}$.

• (P3) If T is of type (B), $v_2, v_3 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even, then $g({\hat{F}}_0, {\hat{F}}_1) \ge y_{\alpha } + y_{\gamma } + y_{\alpha \beta } + y_{\beta \gamma }$.

• (P4) If T is of type (A) or (B), $v_2, v_3 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is odd, then $g({\hat{F}}_0, {\hat{F}}_1) \ge y_\emptyset + y_\beta + y_{\alpha \gamma }$.

• (P5) If T is of type (A) or (B), $v_2 \in S'$, $v_3 \not \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even, then $g({\hat{F}}_0, {\hat{F}}_1) \ge \min \{ x(\alpha )+x(\beta ), 2 - x(\alpha ) - x(\beta ) - 2 y_{\gamma }\}$.

• (P6) If T is of type (A) or (B), $v_2 \in S'$, $v_3 \not \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is odd, then $g({\hat{F}}_0, {\hat{F}}_1) \ge y_\emptyset + y_\gamma + y_{\alpha \beta }$.

• (P7) If T is of type (A$'$) or type (B$'$) and $v_1 \not \in S'$, then $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even.

• (P8) If T is of type (A$'$) or type (B$'$), $v_1 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even, then $g({\hat{F}}_0, {\hat{F}}_1) = \min \{x(\alpha )+x(\gamma ), 2-x(\alpha )-x(\gamma ) - 2 y_\beta \}$.

• (P9) If T is of type (A$'$) or type (B$'$), $v_1 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is odd, then $g({\hat{F}}_0, {\hat{F}}_1) = y_\emptyset + y_\beta + y_{\alpha \gamma }$.

Note that each $T \in {\mathcal {T}}^+_{S^*}$ satisfies exactly one of (P1)–(P9) by changing the labels of $v_2$ and $v_3$ if necessary.

1.1 A.1 Proofs of (P1)–(P9)

1.1.1 A.1.1 When T is of type (A)

We first consider the case when T is of type (A).

Proof of (P1). Suppose that T is of type (A) and $v_2, v_3 \not \in S'$. If $b_1({{\hat{S}}}) + |{{\hat{F}}}_1|$ is odd, then either $p_1 \in {{\hat{S}}}$ and $|{{\hat{F}}}_1 \cap \delta _{G_1}(p_1)|$ is even or $p_2 \in {{\hat{S}}}$ and $|{{\hat{F}}}_1 \cap \delta _{G_1}(p_2)|$ is even. In the former case, $h(S', F'_0, F'_1) \ge \min \{x_1(e_1) + x_1(e_5), 2 - x_1 (e_1) - x_1(e_5) \} = 1$, which is a contradiction. The same argument can be applied to the latter case. Therefore, $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even.

Proof of (P2). Suppose that T is of type (A), $v_2, v_3 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even. If $p_1 \not \in S'$, then we define $(S'', F''_0, F''_1) \in {\mathcal {F}}_1$ as $(S'', F''_0, F''_1) = (S' \cup \{p_1\}, F'_0 {\setminus } \{e_5\}, F'_1\cup \{e_1\})$ if $e_5 \in F'_0$ and $(S'', F''_0, F''_1) = (S' \cup \{p_1\}, F'_0 \cup \{e_1\}, F'_1 {\setminus } \{e_5\})$ if $e_5 \in F'_1$. Since $h(S'', F''_0, F''_1) = h(S', F'_0, F'_1)$ holds, by replacing $(S', F'_0, F'_1)$ with $(S'', F''_0, F''_1)$, we may assume that $p_1 \in S'$. Similarly, we may assume that $p_2 \in S'$, which implies that ${{\hat{S}}} = \{p_1, p_2\}$, ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_1, e_2, e_3, e_4\}$, and $|{{\hat{F}}}_1|$ is even. Then, $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge \min \{ x(\alpha ) + x(\gamma ), 2-x(\alpha )-x(\gamma ) - 2 y_\beta \}$ by the following case analysis.

• If ${{\hat{F}}}_1 = \emptyset$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) = x_1(e_1) + x_1(e_2) + x_1(e_3) + x_1(e_4) = 2-x(\alpha )-x(\gamma ) - 2 y_\beta$.

• If $|{{\hat{F}}}_1| \ge 2$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge 2 - (x_1(e_1) + x_1(e_2) + x_1(e_3) + x_1(e_4)) = x(\alpha ) + x(\gamma ) + 2 y_\beta \ge x(\alpha ) + x(\gamma )$.

Proof of (P4). Suppose that T is of type (A), $v_2, v_3 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is odd. In the same way as (P2), we may assume that ${{\hat{S}}} = \{p_1, p_2\}$, ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_1, e_2, e_3, e_4\}$, and $|{{\hat{F}}}_1|$ is odd. Then, $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge y_\emptyset + y_\beta + y_{\alpha \gamma }$ by the following case analysis and by the symmetry of $v_2$ and $v_3$.

• If $|{{\hat{F}}}_1| = 3$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge 3 - (x_1(e_1) + x_1(e_2) + x_1(e_3) + x_1(e_4)) \ge 1 \ge y_\emptyset + y_\beta + y_{\alpha \gamma }$.

• If ${{\hat{F}}}_1 = \{e_1\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge (1 - x_1(e_1)) + x_1(e_2) \ge y_\emptyset + y_\beta + y_{\alpha \gamma }$.

• If ${{\hat{F}}}_1 = \{e_3\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge 1-x_1(e_3) \ge y_\emptyset + y_\beta + y_{\alpha \gamma }$.

Proof of (P5). Suppose that T is of type (A), $v_2 \in S'$, $v_3 \not \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even. In the same way as (P2), we may assume that $p_1 \in S'$. If $p_2 \in S'$, then $b_1(p_2) + |F'_1 \cap \delta _{G_1}(p_2)|$ is even by the same calculation as (P1). Therefore, we may assume that $p_2 \not \in S'$, since otherwise we can replace $(S', F'_0, F'_1)$ with $(S' {\setminus } \{p_2\}, F'_0 {\setminus } \delta _{G_1}(p_2), F'_1{\setminus } \delta _{G_1}(p_2))$ without increasing the value of $h(S', F'_0, F'_1)$. That is, we may assume that ${{\hat{S}}} = \{p_1\}$, ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_1, e_3\}$, and $|{{\hat{F}}}_1|$ is odd. Then, $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge \min \{(1 - x_1(e_1)) + x_1(e_3), x_1(e_1) + (1 - x_1(e_3)) \} = \min \{ x(\alpha )+x(\beta ), 2 - x(\alpha ) - x(\beta ) \} \ge \min \{ x(\alpha )+x(\beta ), 2 - x(\alpha ) - x(\beta ) - 2 y_{\gamma }\}$.

Proof of (P6). Suppose that T is of type (A), $v_2 \in S'$, $v_3 \not \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is odd. In the same way as (P5), we may assume that ${{\hat{S}}} = \{p_1\}$, ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_1, e_3\}$, and $|{{\hat{F}}}_1|$ is even. Then, $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge \min \{ x_1(e_1) + x_1(e_3), 2 - x_1(e_1) - x_1(e_3) \} = \min \{ y_\emptyset + y_\gamma + y_{\alpha \beta }, 2 - (y_\emptyset + y_\gamma + y_{\alpha \beta })\} = y_\emptyset + y_\gamma + y_{\alpha \beta }$.

1.1.2 A.1.2 When T is of type (A$'$)

Second, we consider the case when T is of type (A$'$).

Proof of (P7). Suppose that T is of type (A$'$) and $v_1 \not \in S'$. If $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is odd, then ${{\hat{S}}} =\{p_3\}$ and $|{{\hat{F}}}_1|$ is odd. This shows that $h(S', F'_0, F'_1) \ge g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge 1$ by the following case analysis, which is a contradiction.

• If ${{\hat{F}}}_1 = \{e_1\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge (1-x_1(e_1)) + x_1(e_2) + x_1(e_9) \ge 1$. The same argument can be applied to the case of ${{\hat{F}}}_1 = \{e_2\}$ by the symmetry of $\alpha$ and $\gamma$.

• If ${{\hat{F}}}_1 = \{e_i\}$ for some $i \in \{3, 4, 8\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge (1 - x_1(e_i)) + x_1(e_9) \ge 1$.

• If ${{\hat{F}}}_1 = \{e_9\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) = 1 + 2 y_\emptyset \ge 1$.

• If $|{{\hat{F}}}_1| \ge 3$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge 3 - (x_1(e_1) + x_1(e_2) + x_1(e_3) + x_1(e_4) + x_1(e_8) + x_1(e_9)) \ge 1$.

Therefore, $b_1({{\hat{S}}}) + |{{\hat{F}}}_1|$ is even.

Proof of (P8). Suppose that T is of type (A$'$), $v_1 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even. Then, $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge \min \{ x(\alpha ) + x(\gamma ), 2-x(\alpha )-x(\gamma ) - 2 y_\beta \}$ by the following case analysis.

• If ${{\hat{F}}}_0 = \{e_8, e_9\}$ and ${{\hat{F}}}_1 = \emptyset$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) = x_1(e_8) + x_1(e_9) = x(\alpha ) + x(\gamma )$.

• If ${{\hat{F}}}_0 = \emptyset$ and ${{\hat{F}}}_1 = \{e_8, e_9\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) = (1 - x_1(e_8)) + (1-x_1(e_9)) = 2-x(\alpha )-x(\gamma ) \ge 2-x(\alpha )-x(\gamma ) - 2 y_\beta$.

• If ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_1, e_2, e_3, e_4\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge \min \{ x(\alpha ) + x(\gamma ), 2-x(\alpha )-x(\gamma ) - 2 y_\beta \}$ by the same calculation as (P2) in Sect. 1.

Proof of (P9). Suppose that T is of type (A$'$), $v_1 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is odd. Then, $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge y_\emptyset + y_\beta + y_{\alpha \gamma }$ by the following case analysis.

• If ${{\hat{F}}}_0 = \{e_8 \}$ and ${{\hat{F}}}_1 = \{e_9 \}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) = x_1(e_8) + (1-x_1(e_9)) = y_\emptyset + y_\beta + y_{\alpha \gamma }$.

• If ${{\hat{F}}}_0 = \{e_9 \}$ and ${{\hat{F}}}_1 = \{e_8 \}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) = (1 - x_1(e_8)) + x_1(e_9) \ge 1 \ge y_\emptyset + y_\beta + y_{\alpha \gamma }$.

• If ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_1, e_2, e_3, e_4\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge y_\emptyset + y_\beta + y_{\alpha \gamma }$ by the same calculation as (P4) in Sect. 1.

1.1.3 A.1.3 When T is of type (B)

Third, we consider the case when T is of type (B). Let $G^+=(V^+, E^+)$ be the graph obtained from $G'_T = (V'_T, E'_T)$ in Fig. 9 by adding a new vertex $r^*$, edges $e_{11}= r r^*$, $e_{12}= v_2 r^*$, $e_{13}= v_3 r^*$, and self-loops $e_{14}, e_{15}, e_{16}$ that are incident to $v_2$, $v_3$, and $r^*$, respectively (Fig. 12). We define $b_T: V^+ \rightarrow {\mathbf {Z}}_{\ge 0}$ as $b_T(v) = 1$ for $v \in \{r, p_1, p_2, p_3\}$ and $b_T(v) = 2$ for $v \in \{r^*, v_2, v_3\}$. We also define $x_T: E^+ \rightarrow {\mathbf {Z}}_{\ge 0}$ as $x_T(e) = x_1(e)$ for $e \in E'_T$ and $x_T(e_{11}) = y_{\alpha } + y_{\gamma } + y_{\alpha \beta } + y_{\beta \gamma }$, $x_T(e_{12}) = y_{\alpha } + y_{\beta } + y_{\alpha \gamma } + y_{\beta \gamma }$, $x_T(e_{13}) = y_{\beta } + y_{\gamma } + y_{\alpha \beta } + y_{\alpha \gamma }$, $x_T(e_{14}) = y_{\emptyset } + y_{\gamma }$, $x_T(e_{15}) = y_{\emptyset } + y_{\alpha }$, and $x_T(e_{16}) = y_{\emptyset }$. For $J \in {\mathcal {E}}_T$, define $b_T$-factors $M_J$ in $G^+$ as follows:

$$\begin{aligned}&M_{\emptyset } = \{e_1, e_7, e_{14}, e_{15}, e_{16} \},&M_{\alpha } = \{e_4, e_8, e_{11}, e_{12}, e_{15} \},&M_{\beta } = \{e_1, e_8, e_9, e_{12}, e_{13} \}, \\&M_{\gamma } = \{e_3, e_9, e_{11}, e_{13}, e_{14} \},&M_{\alpha \beta } = \{e_5, e_8, e_9, e_{11}, e_{13} \},&M_{\alpha \gamma } = \{e_2, e_8, e_9, e_{12}, e_{13} \}, \\&M_{\beta \gamma } = \{e_6, e_8, e_9, e_{11}, e_{12}\}.&&\end{aligned}$$

Then, we obtain $\sum _{J \in {\mathcal {E}}_T} y_1(J) =1$ and $\sum _{J \in {\mathcal {E}}_T} y_1(J) x_{M_J} = x_T$, where $x_{M_J} \in {\mathbf {R}}^{E^+}$ is the characteristic vector of $M_J$. This shows that $x_T$ is in the $b_T$-factor polytope in $G^+$. Therefore, $x_T$ satisfies (3) with respect to $G^+$ and $b_T$. We now show (P1), (P3), (P4), (P5), and (P6).

Fig. 12

Construction of $G^+$

Proof of (P1). Suppose that T is of type (B) and $v_2, v_3 \not \in S'$. If $b_1({{\hat{S}}}) + |{{\hat{F}}}_1|$ is odd, then $b_T({{\hat{S}}}) + |{{\hat{F}}}_1|$ is also odd. Since $x_T$ satisfies (3) with respect to $G^+$ and $b_T$, we obtain $g({\hat{F}}_0, {\hat{F}}_1) \ge 1$. This shows that $h(S', F'_0, F'_1) \ge 1$, which is a contradiction. Therefore, $b_1({{\hat{S}}}) + |{{\hat{F}}}_1|$ is even.

Proof of (P3). Suppose that T is of type (B), $v_2, v_3 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even. Since $b_T({{\hat{S}}} \cup \{r^*, v_2, v_3\}) + |{{\hat{F}}}_1 \cup \{e_{11}\}|$ is odd and $x_T$ satisfies (3), we obtain $g({{\hat{F}}}_0, {{\hat{F}}}_1) + (1-x_T(e_{11})) \ge 1$. Therefore, $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge x_T(e_{11}) = y_\alpha + y_\gamma + y_{\alpha \beta } + y_{\beta \gamma }$.

Proof of (P4). Suppose that T is of type (B), $v_2, v_3 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is odd. Since $b_T({{\hat{S}}} \cup \{r^*, v_2, v_3\}) + |{{\hat{F}}}_1|$ is odd and $x_T$ satisfies (3), we obtain $g({{\hat{F}}}_0, {{\hat{F}}}_1) + x_T(e_{11}) \ge 1$. Therefore, $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge 1 - x_T(e_{11}) = y_\emptyset + y_\beta + y_{\alpha \gamma }$.

Proof of (P5). Suppose that T is of type (B), $v_2 \in S'$, $v_3 \not \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even. If ${{\hat{S}}} \cap \{p_1, p_3\} \not = \emptyset$ and $p_2 \not \in {{\hat{S}}}$, then we can add $p_2$ to $S'$ without decreasing the value of $h(S', F'_0, F'_1)$. Therefore, we can show $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge \{x(\alpha ) + x(\beta ), 2 - x(\alpha ) - x(\beta ) - 2 y_\gamma \}$ by the following case analysis.

• Suppose that ${{\hat{S}}} = \{p_1, p_2, p_3\}$, which implies that ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_1, e_2, e_6, e_9\}$ and $|{{\hat{F}}}_1|$ is odd.

  • If ${{\hat{F}}}_1 = \{e_i\}$ for $i \in \{1, 2, 6\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge (1- x_1(e_i)) + x_1(e_9) \ge x(\alpha ) + x(\beta )$.

  • If ${{\hat{F}}}_1 = \{e_9\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) = y_\alpha + y_\beta + y_{\alpha \gamma } + y_{\beta \gamma } + 2 y_\emptyset = 2 - x(\alpha ) - x(\beta ) - 2 y_\gamma$.

  • If $|{{\hat{F}}}_1| = 3$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge 3 - (x_1(e_1) + x_1(e_2) + x_1(e_6) + x_1(e_9)) \ge x(\alpha ) + x(\beta )$.

• Suppose that ${{\hat{S}}} = \{p_1, p_2\}$, which implies that ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_1, e_2, e_4, e_6, e_7\}$ and $|{{\hat{F}}}_1|$ is even.

  • If ${{\hat{F}}}_1 = \emptyset$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) = x_1(e_1) + x_1(e_2) + x_1(e_4) + x_1(e_6) + x_1(e_7) = 2 - x(\alpha ) - x(\beta ) - 2 y_\gamma$.

  • If $|{{\hat{F}}}_1| \ge 2$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge 2 - (x_1(e_1) + x_1(e_2) + x_1(e_4) + x_1(e_6) + x_1(e_7)) \ge x(\alpha ) + x(\beta )$.

• Suppose that ${{\hat{S}}} = \{p_2\}$, which implies that ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_3, e_5, e_7\}$ and $|{{\hat{F}}}_1|$ is odd.

  • If ${{\hat{F}}}_1 = \{e_i\}$ for $i \in \{3, 7\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge (1- x_1(e_i)) + x_1(e_5) \ge x(\alpha ) + x(\beta )$.

  • If ${{\hat{F}}}_1 = \{e_5\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge (1-x_1(e_5)) + x_1(e_7) \ge 2 - x(\alpha ) - x(\beta ) - 2 y_\gamma$.

  • If ${{\hat{F}}}_1 = \{e_3, e_5, e_7\}$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) = 3 - (x_1(e_3) + x_1(e_5) + x_1(e_7)) \ge x(\alpha ) + x(\beta )$.

• Suppose that ${{\hat{S}}} = \{p_2, p_3\}$, which implies that ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_3, e_4, e_5, e_9\}$ and $|{{\hat{F}}}_1|$ is even.

  • If ${{\hat{F}}}_1 = \emptyset$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) = x_1(e_3) + x_1(e_4) + x_1(e_5) + x_1(e_9) = x(\alpha ) + x(\beta ) + 2 y_{\gamma } \ge x(\alpha ) + x(\beta )$.

  • If $|{{\hat{F}}}_1| \ge 2$, then $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge 2 - (x_1(e_3) + x_1(e_4) + x_1(e_5) + x_1(e_9)) = 2- x(\alpha ) - x(\beta ) - 2 y_{\gamma }$.

• If ${{\hat{S}}} = \emptyset$, then ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_5, e_8\}$ and $|{{\hat{F}}}_1|$ is even. Therefore, $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge \min \{x_1(e_5) + x_1(e_8), 2-x_1(e_5)-x_1(e_8) \} \ge \min \{x(\alpha ) + x(\beta ), 2 - x(\alpha ) - x(\beta ) - 2 y_\gamma \}$.

Proof of (P6). Suppose that T is of type (B), $v_2 \in S'$, $v_3 \not \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is odd. Since $b_T({{\hat{S}}} \cup \{v_2 \}) + |{{\hat{F}}}_1|$ is odd and $x_T$ satisfies (3), we obtain $g({{\hat{F}}}_0, {{\hat{F}}}_1) + x_T(e_{12}) \ge 1$. Therefore, $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge 1 - x_T(e_{12}) = y_\emptyset + y_\gamma + y_{\alpha \beta }$.

1.1.4 A.1.4 When T is of type (B$'$)

Finally, we consider the case when T is of type (B$'$).

Proof of (P7). Suppose that T is of type (B$'$) and $v_1 \not \in S'$. If $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is odd, then ${{\hat{S}}} =\{p_4\}$ and $h(S', F'_0, F'_1) \ge \min \{x_1(e_1) + x_1(e_{10}), 2 - x_1(e_1) - x_1(e_{10})\} = 1$, which is a contradiction. Therefore, $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even.

Proof of (P8). Suppose that T is of type (B$'$), $v_1 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even. If $p_4 \not \in S'$, then we define $(S'', F''_0, F''_1) \in {\mathcal {F}}_1$ as $(S'', F''_0, F''_1) = (S' \cup \{p_4\}, F'_0 {\setminus } \{e_{10}\}, F'_1\cup \{e_1\})$ if $e_{10} \in F'_0$ and $(S'', F''_0, F''_1) = (S' \cup \{p_4\}, F'_0 \cup \{e_1\}, F'_1 {\setminus } \{e_{10}\})$ if $e_{10} \in F'_1$. Since $h(S'', F''_0, F''_1) = h(S', F'_0, F'_1)$, by replacing $(S', F'_0, F'_1)$ with $(S'', F''_0, F''_1)$, we may assume that $p_4 \in S'$. Then, since ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_1, e_2\}$ and $|{{\hat{F}}}_1|$ is odd, we obtain $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge \min \{ (1-x_1(e_1)) + x_1(e_2), x_1(e_1) + (1-x_1(e_2)) \} \ge \min \{ x(\alpha ) + x(\gamma ), 2 - x(\alpha ) - x(\gamma ) - 2y_\beta \}$.

Proof of (P9). Suppose that T is of type (B$'$), $v_1 \in S'$, and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is odd. In the same way as (P8), we may assume that ${{\hat{S}}} = \{p_4\}$, ${{\hat{F}}}_0 \cup {{\hat{F}}}_1 = \{e_1, e_2\}$, and $|{{\hat{F}}}_1|$ is even. Then, $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge \min \{ x_1(e_1) + x_1(e_2), (1-x_1(e_1)) + (1-x_1(e_2)) \} = \min \{ y_\emptyset + y_\beta + y_{\alpha \gamma }, 2 - (y_\emptyset + y_\beta + y_{\alpha \gamma })\} = y_\emptyset + y_\beta + y_{\alpha \gamma }$.

1.2 A.2 Condition (10)

Recall that $r \not \in S'$ is assumed and note that $x_1(\delta _{G_1}(r)) = 1$. Let ${\mathcal {T}}_{(P3)} \subseteq {\mathcal {T}}^+_{S^*}$ be the set of triangles satisfying the conditions in (P3), i.e., the set of triangles of type (B) such that $v_2, v_3 \in S'$ and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is even. Since $y_{\alpha } + y_{\gamma } + y_{\alpha \beta } + y_{\beta \gamma } = 1 - x_1(e^{T}_1) - x_1(e^{T}_2)$ holds for each triangle $T \in {\mathcal {T}}^+_{S^*}$ of type (B), if there exist two triangles $T, T' \in {\mathcal {T}}_{(P3)}$, then $h(S', F'_0, F'_1) \ge (1- x_1(e^{T}_1) - x_1(e^{T}_2)) + (1- x_1(e^{T'}_1) - x_1(e^{T'}_2)) \ge 2 - x_1(\delta _{G_1}(r)) = 1$, which is a contradiction. Similarly, if there exists a triangle $T \in {\mathcal {T}}_{(P3)}$ and an edge $e \in (\delta _{G_1}(r) {\setminus } E'_{T}) \cap F'_1$, then $h(S', F'_0, F'_1) \ge (1- x_1(e^{T}_1) - x_1(e^{T}_2)) + (1- x_1(e)) \ge 2 - x_1(\delta _{G_1}(r)) = 1$, which is a contradiction. Therefore, either ${\mathcal {T}}_{(P3)} = \emptyset$ holds or ${\mathcal {T}}_{(P3)}$ consists of exactly one triangle, say T, and $(\delta _{G_1}(r) {\setminus } E'_{T}) \cap F'_1 = \emptyset$.

Assume that ${\mathcal {T}}_{(P3)} = \{T\}$ and $(\delta _{G_1}(r) {\setminus } E'_{T}) \cap F'_1 = \emptyset$. Define $(S'', F''_0, F''_1) \in {\mathcal {F}}_1$ as $S'' = S' \cup V'_{T}$, $F''_0 = (F'_0 \triangle \delta _{G_1}(r)) {\setminus } E'_{T}$, and $F''_1 = F'_1 {\setminus } E'_{T}$, where $\triangle$ denotes the symmetric difference. Note that $(F''_0, F''_1)$ is a partition of $\delta _{G_1} (S'')$, $b_1(S'') + |F''_1| = (b_1 (S') + b_1({\hat{S}}) ) + (|F'_1| - |{{\hat{F}}}_1|) \equiv 1 \pmod {2}$, and $h(S', F'_0, F'_1) - h(S'', F''_0, F''_1) \ge (1 - x_1(e^{T}_1) - x_1(e^{T}_2)) - x_1( \delta _{G_1}(r) {\setminus } \{ x_1(e^{T}_1), x_1(e^{T}_2)\}) = 0$. By these observations, $(S'', F''_0, F''_1) \in {\mathcal {F}}_1$ is also a minimizer of h. This shows that $(V'' {\setminus } S'', F''_0, F''_1) \in {\mathcal {F}}_1$ is a minimizer of h such that $r \in V'' {\setminus } S''$. Furthermore, if a triangle $T' \in {\mathcal {T}}^+_{S^*}$ satisfies the conditions in (P3) with respect to $(V'' {\setminus } S'', F''_0, F''_1)$, then $T'$ is a triangle of type (B) such that $v_2, v_3 \not \in S'$ and $b_1({\hat{S}}) + |{{\hat{F}}}_1|$ is odd with respect to $(S', F'_0, F'_1)$, which contradicts (P1). Therefore, by replacing $(S', F'_0, F'_1)$ with $(V'' {\setminus } S'', F''_0, F''_1)$, we may assume that ${\mathcal {T}}_{(P3)} = \emptyset$.

In what follows, we construct $(S, F_0, F_1) \in {\mathcal {F}}$ for which (x, y) violates (10) to derive a contradiction. We initialize $(S, F_0, F_1)$ as $S = S' \cap V$, $F_0 = F'_0 \cap E$, and $F_1 = F'_1 \cap E$, and apply the following procedures for each triangle $T \in {\mathcal {T}}^+_{S^*}$.

• Suppose that T satisfies the condition in (P1) or (P7). In this case, we do nothing.

• Suppose that T satisfies the condition in (P2) or (P8). If $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge x(\alpha ) + x(\gamma )$, then add $\alpha$ and $\gamma$ to $F_0$. Otherwise, since $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge 2-x(\alpha )-x(\gamma ) - 2 y_\beta$, add $\alpha$ and $\gamma$ to $F_1$.

• Suppose that T satisfies the condition in (P4) or (P9). In this case, add $\alpha$ to $F_0$ and add $\gamma$ to $F_1$.

• Suppose that T satisfies the condition in (P5). If $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge x(\alpha ) + x(\beta )$, then add $\alpha$ and $\beta$ to $F_0$. Otherwise, since $g({{\hat{F}}}_0, {{\hat{F}}}_1) \ge 2-x(\alpha )-x(\beta ) - 2 y_\gamma$, add $\alpha$ and $\beta$ to $F_1$.

• Suppose that T satisfies the condition in (P6). In this case, add $\alpha$ to $F_0$ and add $\beta$ to $F_1$.

Note that exactly one of the above procedures is applied for each $T \in {\mathcal {T}}^+_{S^*}$, because ${\mathcal {T}}_{(P3)} = \emptyset$.

Then, we see that $(S, F_0, F_1) \in {\mathcal {F}}$ holds and the left-hand side of (10) with respect to $(S, F_0, F_1)$ is at most $h(S', F'_0, F'_1)$ by (P1)–(P9). Since $h(S', F'_0, F'_1) < 1$ is assumed, (x, y) violates (10) for $(S, F_0, F_1) \in {\mathcal {F}}$, which is a contradiction. $\square$

B Proof of Lemma 6

We first show that $M_1 \oplus M_2$ forms a ${\mathcal {T}}$-free b-factor. We can easily see that replacing $(M_1 \cup M_2) \cap \{e^f \mid f \in {\tilde{F}}^*_0\}$ with $\{ f \in {\tilde{F}}^*_0 \mid {e}^f \in M_1 \cap M_2 \}$ does not affect the degrees of vertices in V. Since $M_1 \cup M_2$ contains exactly one of $\{e^f_u, e^f_v\}$ or ${e}^f_r (= {e}^f_{r'})$ for $f = uv \in {\tilde{F}}^*_1$, replacing $(M_1 \cup M_2) \cap \{e^f_u, e^f_r, e^f_v \mid f = uv \in {\tilde{F}}^*_1\}$ with $\{ f \in {\tilde{F}}^*_1 \mid e^f_r \not \in M_1 \cap M_2 \}$ does not affect the degrees of vertices in V.

For every $T \in {\mathcal {T}}^+_{S^*}$ of type (A) or (A$'$), since $|\varphi (M_1, M_2, T) \cap \{\alpha , \gamma \}| = |M_T \cap \{e_8, e_9\}|$, $|\varphi (M_1, M_2, T) \cap \{\alpha , \beta \}| = |M_T \cap \{e_3, e_5\}|$, and $|\varphi (M_1, M_2, T) \cap \{\beta , \gamma \}| = |M_T \cap \{e_4, e_6\}|$ hold by the definition of $\varphi (M_1, M_2, T)$, replacing $M_T$ with $\varphi (M_1, M_2, T)$ does not affect the degrees of vertices in V.

Furthermore, for every $T \in {\mathcal {T}}^+_{S^*}$ of type (B) or (B$'$), since $|\varphi (M_1, M_2, T) \cap \{\alpha , \gamma \}| = |M_T \cap \{e_2, e_{10}\}|$, $|\varphi (M_1, M_2, T) \cap \{\alpha , \beta \}| = |M_T \cap \{e_5, e_8\}|$, and $|\varphi (M_1, M_2, T) \cap \{\beta , \gamma \}| = |M_T \cap \{e_6, e_9\}|$ hold by the definition of $\varphi (M_1, M_2, T)$, replacing $M_T$ with $\varphi (M_1, M_2, T)$ does not affect the degrees of vertices in V.

Since $b(v) = b_1(v)$ for $v \in S^*$ and $b(v) = b_2(v)$ for $v \in V^* {\setminus } S^*$, this shows that $M_1 \oplus M_2$ forms a b-factor. Since $M_j$ is ${\mathcal {T}}_j$-free for $j \in \{1, 2\}$, $M_1 \oplus M_2$ is a ${\mathcal {T}}$-free b-factor.

We next show that $x = \sum _{(M_1, M_2) \in {\mathcal {M}}} \lambda _{(M_1, M_2)} x_{M_1 \oplus M_2}$. By the definitions of $x_1, x_2, M_1 \oplus M_2$, and $\lambda _{(M_1, M_2)}$, it holds that

$$\begin{aligned} x(e) = \sum _{(M_1, M_2) \in {\mathcal {M}}} \lambda _{(M_1, M_2)} x_{M_1 \oplus M_2}(e) \end{aligned}$$

(11)

for $e \in E {\setminus } \bigcup _{T \in {\mathcal {T}}^+_{S^*}} E(T)$.

Let $T \in {\mathcal {T}}^+_{S^*}$ be a triangle of type (A) for $(G_1, b_1, {\mathcal {T}}_1)$ and let $\alpha , \beta$, and $\gamma$ be as in Figs. 7 and 8. By the definition of $\varphi (M_1, M_2, T)$, we obtain

$$\begin{aligned}&\sum _{(M_1, M_2) \in {\mathcal {M}}} \lambda _{(M_1, M_2)} x_{M_1 \oplus M_2}(\beta )\\&\qquad = \sum \{ \lambda _{(M_1, M_2)} \mid \varphi (M_1, M_2, T) = \{\alpha , \beta \}, \{\beta , \gamma \},\hbox { or }\{\beta \} \} \\&\qquad = x_1(e_3) + x_1(e_4) + x_2(e_7) = y_{\alpha \beta } + y_{\beta \gamma } + y_\beta = x(\beta ). \end{aligned}$$

We also obtain

$$\begin{aligned}&\sum _{(M_1, M_2) \in {\mathcal {M}}} \lambda _{(M_1, M_2)} x_{M_1 \oplus M_2}(\alpha )\\&\qquad = \sum \{ \lambda _{(M_1, M_2)} \mid \varphi (M_1, M_2, T) \not = \{\gamma \}, \{\beta , \gamma \}, \{\beta \} \} \\&\qquad = 1 - x_1(e_1) - x_1(e_4) - x_2(e_7) = 1- y_{\emptyset } - y_{\gamma } - y_{\beta \gamma } - y_\beta = x(\alpha ). \end{aligned}$$

Since a similar equality holds for $\gamma$ by symmetry, (11) holds for $e \in \{\alpha , \beta , \gamma \}$. Since T is a triangle of type (A$'$’) for $(G_1, b_1, {\mathcal {T}}_1)$ if and only if it is of type (A) for $(G_2, b_2, {\mathcal {T}}_2)$, the same argument can be applied when T is a triangle of type (A$'$) for $(G_1, b_1, {\mathcal {T}}_1)$.

Let $T \in {\mathcal {T}}^+_{S^*}$ be a triangle of type (B) for $(G_1, b_1, {\mathcal {T}}_1)$ and let $\alpha , \beta$, and $\gamma$ be as in Figs. 9 and 10. By the definition of $\varphi (M_1, M_2, T)$, we obtain

$$\begin{aligned}&\sum _{(M_1, M_2) \in {\mathcal {M}}} \lambda _{(M_1, M_2)} x_{M_1 \oplus M_2}(\beta )\\&\quad = \sum \{ \lambda _{(M_1, M_2)} \mid \varphi (M_1, M_2, T) \not = \emptyset , \{\alpha \}, \{\gamma \}, \{\alpha , \gamma \} \} \\&\quad = 1 - x_1(e_2) - x_1(e_3) - x_1(e_4) - x_1(e_7) = 1 - y_{\alpha \gamma } - y_{\gamma } - y_{\alpha } - y_{\emptyset } = x(\beta ). \end{aligned}$$

We also obtain

$$\begin{aligned}&\sum _{(M_1, M_2) \in {\mathcal {M}}} \lambda _{(M_1, M_2)} x_{M_1 \oplus M_2}(\alpha )\\&\quad = \sum \{ \lambda _{(M_1, M_2)} \mid \varphi (M_1, M_2, T) = \{\alpha \}, \{\alpha , \beta \},\hbox { or }\{\alpha , \gamma \} \} \\&\quad = x_1(e_2) + x_1(e_4) + x_1(e_5) = y_{\alpha \gamma } + y_{\alpha } + y_{\alpha \beta } = x(\alpha ). \end{aligned}$$

Since a similar equality holds for $\gamma$ by symmetry, (11) holds for $e \in \{\alpha , \beta , \gamma \}$. The same argument can be applied when T is a triangle of type (B$'$) for $(G_1, b_1, {\mathcal {T}}_1)$.

Therefore, (11) holds for every $e \in E$, which complete the proof. $\square$