An extended approach for lifting clique tree inequalities

Abstract

We present a new lifting approach for strengthening arbitrary clique tree inequalities that are known to be facet defining for the symmetric traveling salesman problem in order to get stronger valid inequalities for the symmetric quadratic traveling salesman problem (SQTSP). Applying this new approach to the subtour elimination constraints (SEC) leads to two new classes of facet defining inequalities of SQTSP. For the special case of the SEC with two nodes we derive all known conflicting edges inequalities for SQTSP. Furthermore we extend the presented approach to the asymmetric quadratic traveling salesman problem (AQTSP).

Appendix

Theorem 18

Inequalities (14) define facets of $P_{\mathbf{SQTSP }_n}$ for $I, S_1, S_2 \subset V, V=I \dot{\cup }S_1 \dot{\cup }S_2,$ $I \cap S_1 = \emptyset$, $I\cap S_2 = \emptyset ,\, S_1\cap S_2 = \emptyset ,$ $S_1 \ne \emptyset ,$ $|I|\ge 3$, $2 \le |S_2| \le |I|$ and ($|S_1\cup S_2|\ge 5$ or $|S_2|\ge 3$).

Proof

We used the proof-framework of the proof of the dimension of $P_{\mathbf{SQTSP }_n}$ in Fischer and Helmberg (2012). To keep the proof self-contained we will repeat the notations used there. We prove this result by constructing $f(n):=3\cdot \left( {\begin{array}{c}n\\ 3\end{array}}\right) + \left( {\begin{array}{c}n\\ 2\end{array}}\right) - n^2$ affinely (linearly) independent tours in three main steps. In the first step we determine the rank of some specially structured tours ${\bar{C}}_{dim}^{{\bar{n}},1}$ by means of a computer algebra system and take the largest affinely independent subset $C_{dim}^{{\bar{n}},1} \subset {\bar{C}}_{dim}^{{\bar{n}},1}$. In the second and the third step we build tours so that each tour contains at least one 2-edge that is not contained in any tour constructed before. So, considering a matrix formed by the incidence vectors of these tours, we get a block with full row rank and a lower triangular matrix with ones on the main diagonal and zeros in the block of the first step for those variables that form the main diagonal in the second and third step. It is easy to see that the constructed matrix has full row rank.

We set, w. l. o. g., $I=\{i_1=n-|I|+1, \ldots , i_{|I|}=n\}$, $S_1=\{1, \ldots , |S_1|\}$ and denote by ${\bar{I}}$ all nodes of $I$ that are not explicitly mentioned in the tours, in arbitrary order.

$(\mathrm{Step}_{(14)} {1})$ :

If $|S_1\cup S_2| \ge 5$ we know $\{1, \ldots , 5\} \cap I = \emptyset$. We set ${\bar{n}} = 5$ and can use the same construction as in Fischer and Helmberg (2012) building tours ${\bar{C}}_{dim}^{{\bar{n}},1} = \{ K \in \mathcal{K }_n :\{\langle {\bar{n}}+1,{\bar{n}} +2,{\bar{n}}+3\rangle , \langle {\bar{n}}+2,{\bar{n}}+3,{\bar{n}}+4\rangle , \ldots , \langle n-2,n-1,n\rangle \} \subset K, \{n-1,n\}\in K^{\{2\}}\}$. A largest affinely independent subset of ${\bar{C}}_{dim}^{{\bar{n}},1}$ contains 54 tours that are collect in set $C_{dim}^{{\bar{n}},1}$. In the case $|S_1|=1, |S_2|=3$ it holds $5\in I$ and so setting ${\bar{n}} = 5$ we have to restrict to tours ${\tilde{C}} _{dim, (14)}^{{\bar{n}}, 1} = \{K\in \mathcal{K }_n :\{\langle {\bar{n}} +1, {\bar{n}}+2, {\bar{n}} +3\rangle ,\langle {\bar{n}} +2, {\bar{n}}+3, {\bar{n}} +4\rangle , \ldots , \langle n-2, n-1, n\rangle \} \subset K, \{n-1,n\} \in K^{\{2\}}, (\{5,n\}\in K^{\{2\}} \vee \{5,6\}\in K^{\{2\}} \vee \langle n,1,5\rangle \in K \vee \langle 5,1,6\rangle \in K \vee \langle 2,5,3\rangle \in K \vee \langle 2,5,4\rangle \in K \vee \langle 3,5,4\rangle \in K) \}$. The rank reduces by one to 53 in this case.

$(\mathrm{Step}_{(14)} {2})$ :

The set $C_{dim}^{{\bar{n}},2} = \bigcup _{{\bar{n}} < k< n-1} T_k$ is formed iteratively. For each $k \in \{{\bar{n}}+1, \ldots , n-2\}$ we build a set of tours $T_k$ that contains $n_k$ tours $t_k^{1}, \ldots , t_k^{n_k}$. The tour construction uses five substeps. During each substep the order of the tours is arbitrary. In each substep we append new rows of incidence vectors of tours to a large matrix built by the affinely independent tours. At the end we have to check that the tours indeed fulfill the described matrix structure (a lower triangular matrix with ones on the main diagonal).

Let $k$ be fixed with ${\bar{n}}<k<n-1$. All tours presented next are represented by the order of the nodes, i.e., a tour $t=\{ v_1v_2v_3, v_2v_3v_4, \ldots , v_{n-1}v_nv_1, v_nv_1v_2\}$ is represented by $v_1\,v_2\,v_3\ldots v_{n-1}\, v_n$. Only the relevant parts of the tours are specified. The node sequence $(k+2)\,(k+3)\ldots (n-2)\,(n-1)$ is subsumed and denoted by the symbol $\varpi _k$. If some nodes are not explicitly mentioned and the completion of the tour is arbitrary we denote this by “...”. We underline the decisive 2-edge (the three corresponding nodes) $e_k^i, i=1, \ldots , n_k,$ that is used for forming the triangular structure. It belongs to one of the four types

• (Type-I1) $\langle a,k,b\rangle , a,b\in \{1,\ldots , k-1\}, a<b$,

• (Type-I2) $\langle k,a,k+1\rangle , a\in \{2,\ldots ,k-1\}$,

• (Type-I3) $\langle a,b,k+1\rangle , a,b\in \{1,\ldots , k-1\}, a\ne b$,

• (Type-I4) $\langle n,a,k\rangle , \langle n,k,a\rangle , a\in \{1,\ldots , k-1\}$.

In Fischer and Helmberg (2012) the standard construction for fixed $k$ is

• (I1) $\ldots \underline{a\,k\,1}\,(k+1) \, \varpi _k\, n \ldots$, for $a\in \{2, \ldots , k-1\}$ (the 2-edge $\langle k, 1,k+1\rangle$ is not used as an $e_k^i$),

• (I2) $\ldots 1\,\underline{k\,a\,(k+1)} \, \varpi _k\, n \ldots$, for $a\in \{2, \ldots , k-1\}$,

• (I3) $\ldots \underline{a\,k\,b}\,(k+1) \, \varpi _k\, n\ldots$, for $a,b\in \{2, \ldots , k-1\}, a<b$,

• (I4) $\ldots k\,\underline{a\,b\,(k+1)}\, \varpi _k \, n\ldots$, for $a,b\in \{1, \ldots , k-1\},\, a\ne b$,

• (I5) $\ldots (k+1)\,\varpi _k\, \underline{n\,a\,b} \ldots ,$ for $a,b\in \{1,\ldots , k\}, a\ne b, k\in \{a,b\}$.

These substeps fulfill the desired triangular structure (proof of Claim 1 in the proof of Theorem 2.3 in Fischer and Helmberg (2012)).

As long as $k\in S_1 \cup S_2$ the nodes in $I$ lie next to each other and so the corresponding tours define roots of (14). Adaptations of (I1)–(I5) are needed for the case $k\in I$. We start with a specific ordering for $k=i_1$ for the case $|S_1\cup S_2|\ge 5$.

• $(\mathrm{I}_{(14)}^{i_1}{1})$ $\ldots \underline{a\, i_1\,1}\, i_2 \,\varpi _k \, n\ldots ,$ for $a\in (S_1\cup S_2)\setminus \{1\}$ (the 2-edge $\langle i_1, 1, i_2\rangle$ is not used as an $e_k^{\hat{\imath }}$; the same 2-edge is not used in (I1), too),

• $(\mathrm{I}_{(14)}^{i_1}\mathrm{2a})$ $\ldots 1 \, \underline{i_1 \, a\, i_2}\, \varpi _k\, n\ldots ,$ for $a\in S_1\setminus \{1\}$,

• $(\mathrm{I}_{(14)}^{i_1}\mathrm{3a})$ $\ldots \underline{a\, i_1\, b}\, i_2 \,\varpi _k \, n\ldots ,$ for $a\in (S_1\cup S_2)\setminus \{1\}, b\in S_1\setminus \{1\}, b<a$,

• $(\mathrm{I}_{(14)}^{i_1}\mathrm{4a})$ $\ldots \underline{a\, b\, i_2} \,\varpi _k \, n\, 1 \, i_1\ldots ,$ for $a,b\in ( S_1\cup S_2)\setminus \{1\}$ (the 2-edge $\langle n, 1, i_1\rangle$ is not used as an $e_k^{\hat{\imath }}$; it is the one specific tour that is lost in comparison to the dimension proof in Fischer and Helmberg (2012)),

• $(\mathrm{I}_{(14)}^{i_1}\mathrm{5a})$ $\left\{ \begin{array}{ll} \ldots m\, o\, i_2\, \varpi _k \, \underline{n\, i_1 \, a}\ldots , &{}\text {for }a\in S_1\cup S_2,\\ \ldots m\, o\, i_2\, \varpi _k \, \underline{n\, a \, i_1}\ldots , &{}\text {for }a\in S_1\setminus \{1\}, \end{array}\right.$ with $m,o\in (S_1\cup S_2)\setminus \{1\}, |\{a,m,o\}|=3$,

• $(\mathrm{I}_{(14)}^{i_1}\mathrm{4b})$ $\ldots \underline{a\, b\, i_2}\, \varpi _k \, n\, i_1\ldots$, for $a,b\in S_1\cup S_2, 1\in \{a,b\}, a\ne b$,

• $(\mathrm{I}_{(14)}^{i_1}\mathrm{3b})$ $\ldots i_2\, \varpi _k \, n\, 1\, \underline{a\, i_1\, b}\ldots ,$ for $a,b\in S_2, a<b$,

• $(\mathrm{I}_{(14)}^{i_1}\mathrm{5b})$ $\ldots i_2\, \varpi _k \, \underline{n\, a\, i_1}\, m\ldots ,$ for $a\in S_2$ with $m\in S_2, m\ne a$,

• $(\mathrm{I}_{(14)}^{i_1}\mathrm{2b})$ $\ldots m\, \underline{i_1 \, a\, i_2}\,\varpi _k \, n\ldots$, for $a\in S_2$ with $m\in S_2, m \ne a$.

All tours in $(\mathrm{I}_{(14)}^{i_1}{1})$–$(\mathrm{I}_{(14)}^{i_1}\mathrm{2b})$ define roots of (14) because the nodes $i_2$ to $n$ lie next to each other and for $i_1$ it holds that either $i_1$ lies next to node $n$, or there is exactly one node between $i_1$ and $i_2$ resp. $n$ and this node belongs to $S_1$ or $i_1$ lies between two nodes in $S_2$. Furthermore we have to show that all underlined 2-edges are not used in a tour of a previous substep. It suffices to look only at previous substeps for the same $k=i_1$. Note, all tours contain the edge $\{n-1,n\}$.

• Tours in $(\mathrm{I}_{(14)}^{i_1}\mathrm{2a})$: All tours in $(\mathrm{I}_{(14)}^{i_1}{1})$ contain the 2-edge $\langle i_1, 1,i_2\rangle$.

• Tours in $(\mathrm{I}_{(14)}^{i_1}\mathrm{3a})$: All tours in $(\mathrm{I}_{(14)}^{i_1}{1})$–$(\mathrm{I}_{(14)}^{i_1}\mathrm{2a})$ contain the edge $\{ i_1, 1 \}$.

• Tours in $(\mathrm{I}_{(14)}^{i_1}\mathrm{4a})$: All tours in $(\mathrm{I}_{(14)}^{i_1} \mathrm{1})$–$(\mathrm{I}_{(14)}^{i_1}\mathrm{3a})$ contain a 2-edge $\langle i_1, {\tilde{a}},i_2\rangle \in V^{\langle 3 \rangle }$.

• Tours in $(\mathrm{I}_{(14)}^{i_1}\mathrm{5a})$: All tours in $(\mathrm{I}_{(14)}^{i_1}{1})$–$(\mathrm{I}_{(14)}^{i_1}\mathrm{3a})$ contain a 2-edge $\langle n, {\tilde{a}},\tilde{b}\rangle \in V^{\langle 3 \rangle }$ with ${\tilde{a}}, \tilde{b} \in S_1 \cup S_2$ and the tours in $(\mathrm{I}_{(14)}^{i_1} \mathrm{4a})$ contain the 2-edge $\langle n,1,i_1\rangle$.

• Tours in $(\mathrm{I}_{(14)}^{i_1}\mathrm{4b})$: All tours in $(\mathrm{I}_{(14)}^{i_1}\mathrm{1})$–$(\mathrm{I}_{(14)}^{i_1}\mathrm{3a})$ contain a 2-edge $\langle i_1, \tilde{a},i_2\rangle \in V^{\langle 3 \rangle }$ and in $(\mathrm{I}_{(14)}^{i_1}\mathrm{4a})$ the nodes $a,b$ and in $(\mathrm{I}_{(14)}^{i_1}\mathrm{5a})$ the nodes $m,o$ are not allowed to be 1.

• Tours in $(\mathrm{I}_{(14)}^{i_1}\mathrm{3b})$: All tours $(\mathrm{I}_{(14)}^{i_1}{1})$–$(\mathrm{I}_{(14)}^{i_1}\mathrm{4b})$ contain an edge $\{n,i_1\}$ or an edge $\{i_1,\tilde{a}\}, \tilde{a}\in S_1$.

• Tours in $(\mathrm{I}_{(14)}^{i_1}\mathrm{5b})$: All tours in $(\mathrm{I}_{(14)}^{i_1}{1})$–$(\mathrm{I}_{(14)}^{i_1}\mathrm{3a})$ contain a 2-edge $\langle n, {\tilde{a}},{\tilde{b}}\rangle \in V^{\langle 3 \rangle }$ with ${\tilde{a}}, {\tilde{b}} \in S_1 \cup S_2$ and the tours in $(\mathrm{I}_{(14)}^{i_1}\mathrm{4a})$–$(\mathrm{I}_{(14)}^{i_1}\mathrm{3b})$ contain the edge $\{n,i_1\}$ or an edge $\{n,\tilde{a}\}, {\tilde{a}}\in S_1$.

• Tours in $(\mathrm{I}_{(14)}^{i_1}\mathrm{2b})$: All tours in $(\mathrm{I}_{(14)}^{i_1}{1})$–$(\mathrm{I}_{(14)}^{i_1}\mathrm{3a})$ contain a 2-edge $\langle i_1, {\tilde{a}}, i_2\rangle \in V^{\langle 3 \rangle }$ with ${\tilde{a}}\in S_1$ and there are at least two nodes between $i_1$ and $i_2$ in the tours in $(\mathrm{I}_{(14)}^{i_1}\mathrm{4a})$–$(\mathrm{I}_{(14)}^{i_1}\mathrm{5b})$ because $i_1 \ge 6$ or the tours contain $\langle i_2,i_3,i_1 \rangle$.

All in all, we constructed exactly one tour less than described in (Type-I1)–(Type-I4) for this $k$.

For $k\in I, k \ne i_1, k \le n-2$ the substeps presented next provide tours having the desired root structure.

• $(\mathrm{I}_{(14)}{1})$ $\ldots \underline{a\, k \, 1}\, (k+1)\, \varpi _k \, n\, {\bar{I}}\ldots ,$ for $a\in \{2, \ldots , k-1\}$ (the 2-edge $\langle k,1,k+1\rangle$ is not used as an $e_k^{\hat{\imath }}$, see (I1)),

• $(\mathrm{I}_{(14)}\mathrm{2a})$ $\ldots 1\, \underline{k\, a\, (k+1)}\, \varpi _k \, n\, {\bar{I}}\ldots ,$ for $a\in \{2,\ldots , k-1\}\setminus S_2$,

• $(\mathrm{I}_{(14)}\mathrm{3a})$ $\ldots \underline{a\, k\, b}\, (k+1)\, \varpi _k\, n\, {\bar{I}}\ldots ,\text { for }a\in \{2, \ldots , k-1\}, b\in S_1\setminus \{1\}, b<a,$

• $(\mathrm{I}_{(14)}\mathrm{3b})$ $\ldots \underline{a\, k\, b}\, (k+1)\, \varpi _k\, n\, {\bar{I}}\ldots ,\text { for }a\in \{1, \ldots , k-1\}\setminus S_1, b\in \{1, \ldots , k-1\}\cap I, a<b,$

• $(\mathrm{I}_{(14)}\mathrm{4a})$ $\ldots \underline{a\, b\, (k+1)} \, \varpi _k\, n\, 1\, {\bar{I}} \, k\ldots ,$ for $a,b\in S_2, a\ne b,$

• $(\mathrm{I}_{(14)}\mathrm{5a})$ $\ldots m\, o\, (k+1)\, \varpi _k \, \underline{n\, a\, b}\, {\bar{I}}\ldots ,$ for $a,b\in \{1, \ldots , k\}\cap (I \cup S_1), k\in \{a,b\},$ with $m,o\in S_2, |\{a,b,m,o\}|=4$,

• $(\mathrm{I}_{(14)}\mathrm{4b})$ $\ldots \underline{a\, b\, (k+1)} \, \varpi _k\, n\, k \,{\bar{I}} \ldots ,$ for $a,b\in S_1\cup S_2,\{a,b\}\cap S_1 \ne \emptyset , a\ne b,$

• $(\mathrm{I}_{(14)}\mathrm{4c})$ $\ldots {\bar{I}}\, k\,\underline{a\, b\, (k+1)}\, \varpi _k \, n\ldots , \text { for } a,b\in \{1, \ldots , k-1\}\setminus S_2, \{a,b\}\cap I \ne \emptyset ,a\ne b,$

• $(\mathrm{I}_{(14)}\mathrm{4d})$ $\ldots \underline{a\, b\, (k+1)}\, \varpi _k\, n\, k\, {\bar{I}}\,1\ldots , \text { for } a\in S_2,b\in \{1, \ldots , k-1\}\cap I,$

• $(\mathrm{I}_{(14)}\mathrm{4e})$ $\ldots m\, \underline{a\, b\, (k+1)}\, \varpi _k\, n\, k\, {\bar{I}} \, 1\ldots , \text { for } a\in \{1, \ldots , k-1\}\cap I, b\in S_2 \text {with }m\in S_2, m\ne b,$

• $(\mathrm{I}_{(14)}\mathrm{5b})$ $\ldots {\bar{I}} \, (k+1)\, \varpi _k\, \underline{n\, k\, a}\ldots ,$ for $a\in S_2$,

• $(\mathrm{I}_{(14)}\mathrm{3c})$ $\ldots (k+1)\, \varpi _k \, n\, {\bar{I}} \, \underline{a\, k\, b}\ldots ,$ for $a,b\in S_2, a<b$,

• $(\mathrm{I}_{(14)}\mathrm{5c})$ $\ldots {\bar{I}}\,(k+1)\, \varpi _k \, \underline{n\, a\, k}\, m\ldots ,$ for $a\in S_2$ with $m\in S_2, m\ne a$,

• $(\mathrm{I}_{(14)}\mathrm{2b})$ $\ldots m\, \underline{k\, a\, (k+1)}\, \varpi _k\, n\, {\bar{I}}\ldots ,$ for $a\in S_2$ with $m\in S_2, m\ne a$.

The tours in $(\mathrm{I}_{(14)}{1})$–$(\mathrm{I}_{(14)}\mathrm{2b})$ define roots of (14) because in $(\mathrm{I}_{(14)}\mathrm{1})$–$(\mathrm{I}_{(14)}\mathrm{5b})$ all nodes in $I$ lie next to each other, partially with exactly one node from $S_1$ between them. In $(\mathrm{I}_{(14)}\mathrm{3c})$–$(\mathrm{I}_{(14)}\mathrm{2b})$ the nodes $I\setminus \{k\}$ lie next to each other and node $k$ lies between two nodes of $S_2$. It remains to prove that all underlined 2-edges are not used in a tour of a previous substep. Note, all tours contain the edge $\{n-1,n\}$.

• Tours in $(\mathrm{I}_{(14)}\mathrm{2a})$: All tours in $(\mathrm{I}_{(14)}{1})$ contain the 2-edge ${\langle k,1,k+1\rangle }$.

• Tours in $(\mathrm{I}_{(14)}\mathrm{3a})$, $(\mathrm{I}_{(14)}\mathrm{3b})$: The two substeps use different 2-edges of type (Type-I2). So we can treat them together. All tours in $(\mathrm{I}_{(14)}{1})$–$(\mathrm{I}_{(14)}\mathrm{2a})$ contain the edge $\{k,1\}$.

• Tours in $(\mathrm{I}_{(14)}\mathrm{4a})$: All tours in $(\mathrm{I}_{(14)}{1})$–$(\mathrm{I}_{(14)}\mathrm{3b})$ contain a 2-edge ${\langle k, {\tilde{a}}, k+1\rangle }\in V^{\langle 3\rangle }$.

• Tours in $(\mathrm{I}_{(14)}\mathrm{5a})$: In all tours in $(\mathrm{I}_{(14)}{1})$–$(\mathrm{I}_{(14)}\mathrm{4a})$ there are at least two nodes between node $n$ and node $k$ on both sides. Note, ${\bar{I}}$ represents at least one node in $(\mathrm{I}_{(14)}\mathrm{4a})$.

• Tours in $(\mathrm{I}_{(14)}\mathrm{4b})$–$(\mathrm{I}_{(14)}\mathrm{4e})$: The four substeps use different 2-edges of type (Type-I3). So we can treat them together. All tours in $(\mathrm{I}_{(14)}{1})$–$(\mathrm{I}_{(14)}\mathrm{3b})$ contain a 2-edge ${\langle k, {\tilde{a}}, k+1\rangle }\in V^{\langle 3\rangle }$. The tours in $(\mathrm{I}_{(14)}\mathrm{4a})$–$(\mathrm{I}_{(14)}\mathrm{5a})$ contain a 2-edge ${\langle {\tilde{a}}, {\tilde{b}}, k+1\rangle }\in V^{\langle 3 \rangle }, {\tilde{a}}, {\tilde{b}} \in S_2, {\tilde{a}} \ne {\tilde{b}}$.

• Tours in $(\mathrm{I}_{(14)}\mathrm{5b})$: In all tours in $(\mathrm{I}_{(14)}{1})$–$(\mathrm{I}_{(14)}\mathrm{4a})$, $(\mathrm{I}_{(14)}\mathrm{4c})$ there are at least two nodes between node $n$ and node $k$ on both sides. All tours in $(\mathrm{I}_{(14)}\mathrm{5a})$–$(\mathrm{I}_{(14)}\mathrm{4b})$, $(\mathrm{I}_{(14)}\mathrm{4d})$–$(\mathrm{I}_{(14)}\mathrm{4e})$ contain a 2-edge ${\langle n,{\tilde{a}}, {\tilde{b}}\rangle } \in V^{\langle 3 \rangle }, {\tilde{a}}, {\tilde{b}} \in \{k\}\cup I \cup S_1$.

• Tours in $(\mathrm{I}_{(14)}\mathrm{3c})$: All tours in $(\mathrm{I}_{(14)}{1})$–$(\mathrm{I}_{(14)}\mathrm{3a})$ contain an edge $\{k,{\tilde{a}}\}$, ${\tilde{a}} \in S_1$ and all tours in $(\mathrm{I}_{(14)}\mathrm{3b})$–$(\mathrm{I}_{(14)}\mathrm{5b})$ contain an edge $\{k,{\tilde{b}}\}, \tilde{b} \in I\cup S_1$ (note $n\in I$).

• Tours in $(\mathrm{I}_{(14)}\mathrm{5c})$: In all tours in $(\mathrm{I}_{(14)}{1})$–$(\mathrm{I}_{(14)}\mathrm{4a})$, $(\mathrm{I}_{(14)}\mathrm{4c})$ there are at least two nodes between node $n$ and node $k$ on both sides. All tours in $(\mathrm{I}_{(14)}\mathrm{5a})$–$(\mathrm{I}_{(14)}\mathrm{4b})$, $(\mathrm{I}_{(14)}\mathrm{4d})$–$(\mathrm{I}_{(14)}\mathrm{4e})$ contain a 2-edge ${\langle n,{\tilde{a}}, {\tilde{b}}\rangle }\in V^{\langle 3 \rangle }, {\tilde{a}}, {\tilde{b}} \in \{k\}\cup I \cup S_1$. The tours in $(\mathrm{I}_{(14)}\mathrm{5b})$ contain the edge $\{n,k\}$ and each tour in $(\mathrm{I}_{(14)}\mathrm{3c})$ contains an edge $\{n, {\tilde{a}}\}, \tilde{a} \in I$.

• Tours in $(\mathrm{I}_{(14)}\mathrm{2b})$: All tours in $(\mathrm{I}_{(14)}{1})$–$(\mathrm{I}_{(14)}\mathrm{3b})$ contain a 2-edge ${\langle k, {\tilde{a}}, k+1\rangle }\in V^{\langle 3 \rangle }, {\tilde{a}} \in S_1\cup I,$ and in the tours in $(\mathrm{I}_{(14)}\mathrm{4a})$–$(\mathrm{I}_{(14)}\mathrm{5c})$ there are at least two nodes between nodes $k+1$ and $k$ on both sides or the tours contain the 2-edge ${\langle k+1, k+2, k \rangle }$.

Because, in total, the same 2-edges are underlined and used for building the triangular structure we get exactly $\tfrac{3}{2}k^2-\tfrac{3}{2}k-1$ tours for $k\in \{{\bar{n}}+1, \ldots ,n -2\}\setminus \{i_1\}$ ($\tfrac{1}{2}(k-1)(k-2)$ with (Type-I1), $k-2$ with (Type-I2), $(k-1)(k-2)$ with (Type-I3) and $2(k-1)$ with (Type-I4)), see proof of Claim 3 in the proof of Theorem 2.3 in Fischer and Helmberg (2012).

$(\mathrm{Step}_{(14)} {3})$ :

In all tours in $(\mathrm{Step}_{(14)} {1})$ and $(\mathrm{Step}_{(14)} {2})$ the nodes $n-1$ and $n$ are adjacent. Now we construct tours $C_{dim, (14)}^{{\bar{n}}, 3}=\{t_L^{1}, \ldots , t_L^{n_L}\}$ in which $n-1$ and $n$ do not lie next to each other. Each tour will contain a 2-edge $e_L^i, i=1, \ldots , n_L,$ of one of the types

• (Type-L1) ${\langle a,n-1,b\rangle }, a,b\in \{1,\ldots ,n-2\},a<b$,

• (Type-L2) ${\langle a,n,b\rangle }, a,b\in \{1,\ldots , n-2\}, a<b$,

• (Type-L3)${\langle n-1,a,n\rangle }, a\in \{1,\ldots , n-2\}$.

Except for one all of these 2-edges are used as $e_L^i$. During the following substeps we will ensure that each underlined $e_L^{i}$ is not used in a previous substep (the order in each substep is arbitrary) and not in $(\mathrm{Step}_{(14)} {1})$ and $(\mathrm{Step}_{(14)} {2})$. Indeed, the substeps are only slightly modified in comparison to the ones used in the third step in the original dimension proof in Fischer and Helmberg (2012). We specify the position of ${\bar{I}}$ here and split up some of the original substeps in several successive ones in order to simplify the presentation. We set $w_1=1,w_2,w_3\in S_2, w_2\ne w_3$, with $W=\{w_1,w_2,w_3\}$.

• $(\mathrm{L}_{(14)}\mathrm{1a})$ $\ldots \underline{a\, (n-1)\, b}\, {\bar{I}} \, w_1 \, n\, w_2\ldots ,$ for $a\in \{1, \ldots , n-2\}\setminus \{w_1,w_2\}, b\in I\setminus \{n-1,n\}, a<b,$ (the 2-edge ${\langle w_1, n, w_2\rangle }$ is not used as an $e_L^{\hat{\imath }}$),

• $(\mathrm{L}_{(14)}\mathrm{1b})$ $\ldots \underline{a\, (n-1)\, b}\, {\bar{I}} \, w_1 \, n\, w_2\ldots ,$ for $a\in (S_1\cup S_2)\setminus \{w_1,w_2\}, b\in S_1\setminus \{w_1\}$, $a>b,$

• $(\mathrm{L}_{(14)}\mathrm{1c})$ $\ldots \underline{a\, (n-1)\, b}\, {\bar{I}} \, w_1 \, n\, w_2\ldots ,$ for $a,b\in S_2\setminus \{w_2\}, a<b,$

• $(\mathrm{L}_{(14)}{2})$ $\left\{ \begin{array}{ll} \ldots m\, (n-1)\, {\bar{I}} \, \underline{w_1 \, n\, w_3}\ldots ,&{}\text {with } m\in (S_1 \cup S_2)\setminus W, \\ \ldots m\, (n-1)\, {\bar{I}} \, \underline{w_2 \, n\, w_3}\ldots ,&{}\text {with } m\in (S_1 \cup S_2)\setminus W, \end{array} \right.$

• $(\mathrm{L}_{(14)}{3})$ $\ldots \underline{a\, (n-1)\, w_1}\,{\bar{I}}\, w_2 \, n\, w_3\ldots , \text { for }a\in \{1, \ldots , n-2\}\setminus W,$

• $(\mathrm{L}_{(14)}{4})$ $\ldots \underline{w_2\, (n-1)\, a}\,{\bar{I}}\, w_1 \, n\, w_3\ldots , \text { for }a\in \{1, \ldots , n-2\}\setminus W,$

• $(\mathrm{L}_{(14)}\mathrm{5a})$ $\ldots \underline{a\, n\, w_1}\, {\bar{I}}\, (n-1)\, w_2\ldots$, for $a\in (S_1\cup S_2)\setminus W,$

• $(\mathrm{L}_{(14)}\mathrm{5b})$ $\ldots \underline{a\, n\, b}\, {\bar{I}}\, (n-1)\, w_1\ldots$, for $a\in \{w_2,w_3\},b\in (S_1\cup S_2)\setminus W,$

• $(\mathrm{L}_{(14)}\mathrm{5c})$ $\ldots \underline{w_1\, n\, a}\, {\bar{I}} \, m \, (n-1)\, o\ldots ,$ for $a\in I\setminus \{n-1,n\} \text { with }m,o\in \{1, \ldots , n-2\}, |\{a,m,o\}|=3,$ and $((m,o\in S_2, \{m,o\}\not \subset W) \vee (m\in I\cup S_1))$

• $(\mathrm{L}_{(14)}\mathrm{5d})$ $\ldots \underline{a\, n\, b}\, w_1 \, (n-1)\,{\bar{I}}\, m\ldots ,$ for $a\in \{w_2,w_3\}, b\in I\setminus \{n-1,n\} \text { with }m\in \{1, \ldots , n-2\}\setminus W, |\{a,b,m\}|=3,$

• $(\mathrm{L}_{(14)}{6})$ $\left\{ \begin{array}{l} \ldots \underline{w_2 \, (n-1)\, w_1}\, {\bar{I}}\, n\, w_3\ldots , \\ \ldots \underline{w_3 \, (n-1)\, w_1}\, {\bar{I}}\, n\, w_2\ldots , \\ \ldots \underline{w_2 \, (n-1)\, w_3}\, {\bar{I}}\, n\, w_1\ldots , \end{array} \right.$

• $(\mathrm{L}_{(14)}\mathrm{7a})$ $\ldots \underline{a\, n\, b}\, {\bar{I}} \, w_1 \, (n-1)\ldots ,\text { for } a\in \{1, \ldots , n-2\}\setminus W, b\in I\setminus \{n-1,n\}, a<b,$

• $(\mathrm{L}_{(14)}\mathrm{7b})$ $\ldots \underline{a\, n\, b}\, {\bar{I}} \, (n-1)\ldots ,\text { for } a\in \{1, \ldots , n-2\}\setminus ( I\cup W), b\in S_1\setminus W, a>b,$

• $(\mathrm{L}_{(14)}\mathrm{7c})$ $\ldots \underline{a\, n\, b}\, {\bar{I}} \, (n-1)\ldots ,\text { for } a,b\in S_2\setminus W, a<b,$

• $(\mathrm{L}_{(14)}\mathrm{8a})$ $\ldots \underline{(n-1)\, a\, n}\, {\bar{I}}\ldots ,\text { for }a\in (S_1\cup I)\setminus \{n-1,n\},$

• $(\mathrm{L}_{(14)}\mathrm{8b})$ $\ldots {\bar{I}} \, \underline{(n-1)\, a\, n}\, m \ldots , \text { for }a\in S_2 \text { with }m\in S_2, m\ne a.$

It follows from the proof of Claim 2 in the proof of Theorem 2.3 in Fischer and Helmberg (2012) (and is indeed easy to check) that all underlined 2-edges are not used in a previous substep and that we build exactly $n^2-4n+3$ tours in $(\mathrm{Step}_{(14)} 3)$ ($\tfrac{1}{2}(n-2)(n-3)$ of type (Type-L1), $\tfrac{1}{2}(n-2)(n-3)-1$ of type (Type-L2) and $(n-2)$ of type (Type-L3)).

• All 2-edges underlined in substeps with the same number belong to the same type and are in pairwise conflict. So we subsume all substeps with the same number to one in the following investigations.

• Tours in $(\mathrm{L}_{(14)}\mathrm{2})$: all tours created in $(\mathrm{L}_{(14)}\mathrm{1a})$–$(\mathrm{L}_{(14)}\mathrm{1c})$ contain the 2-edge ${\langle w_1, n, w_2\rangle }$.

• Tours in $(\mathrm{L}_{(14)}{3})$, $(\mathrm{L}_{(14)}{4})$: all tours created in $(\mathrm{L}_{(14)}\mathrm{1a})$–$(\mathrm{L}_{(14)}{2})$ contain a 2-edge ${\langle a, n-1,b\rangle }\in V^{\langle 3 \rangle }, a,b\in \{1,\ldots ,n-2\}\setminus \{w_1,w_2\}$.

• Tours in $(\mathrm{L}_{(14)}\mathrm{5a})$–$(\mathrm{L}_{(14)}\mathrm{5d})$: all tours created in $(\mathrm{L}_{(14)}\mathrm{1a})$–$(\mathrm{L}_{(14)}{4})$ contain a 2-edge $c \in \{{\langle w_1, n, w_2\rangle }, {\langle w_1, n, w_3\rangle }, {\langle w_2, n, w_3\rangle }\}$.

• Tours in $(\mathrm{L}_{(14)}{6})$: all tours created in $(\mathrm{L}_{(14)}\mathrm{1a})$–$(\mathrm{L}_{(14)}\mathrm{5d})$ contain none of the three 2-edges ${\langle w_1,n-1,w_2\rangle },{\langle w_1,n-1,w_3\rangle },{\langle w_2,n-1,w_3\rangle }$.

• Tours in $(\mathrm{L}_{(14)}\mathrm{7a})$–$(\mathrm{L}_{(14)}\mathrm{7c})$: all tours created in $(\mathrm{L}_{(14)}\mathrm{1a})$–$(\mathrm{L}_{(14)}{4})$ contain a 2-edge $c \in \{{\langle w_1, n, w_2\rangle }, {\langle w_1, n, w_3\rangle }, {\langle w_2, n, w_3\rangle }\}$. In $(\mathrm{L}_{(14)}\mathrm{5a})$–$(\mathrm{L}_{(14)}{6})$ node $n$ is adjacent to one of the nodes $w_1,w_2,w_3\in W$.

• Tours in $(\mathrm{L}_{(14)}\mathrm{8a})$, $(\mathrm{L}_{(14)}\mathrm{8b})$: in all tours created in $(\mathrm{L}_{(14)}\mathrm{1a})$–$(\mathrm{L}_{(14)}\mathrm{7c})$ there are at least two nodes between nodes $n-1$ and $n$ on both sides.

It remains to check the root property of all tours constructed. There is one large block of nodes in $I$, partially with one node of $S_1$ between two of these nodes, in all substeps except for $(\mathrm{L}_{(14)}\mathrm{1c})$, one tour in $(\mathrm{L}_{(14)}{2})$, $(\mathrm{L}_{(14)}{3})$, some tours in $(\mathrm{L}_{(14)}\mathrm{5b})$–$(\mathrm{L}_{(14)}\mathrm{5c})$, one tours in $(\mathrm{L}_{(14)}{6})$, $(\mathrm{L}_{(14)}\mathrm{7c})$ and $(\mathrm{L}_{(14)}\mathrm{8b})$. In these node $n-1$ or node $n$ does not belong to that block but lies between two nodes in $S_2$.

All in all we created exactly $f(n)$ tours and so one tour less than in the proof of the dimension in Fischer and Helmberg (2012). If $|S_1\cup S_2|=4$ we get one tour less in $(\mathrm{Step}_{(14)} 1)$ by the special structure of the tours and for $|S_1\cup S_2| \ge 5$ we lost one tour in $(\mathrm{Step}_{(14)} 2)$ for $k=i_1$. Thus, inequalities (14) define facets of $P_{\mathbf{SQTSP }_n}, n\ge 7$. $\square$

Theorem 19

Inequalities (15) define facets of $P_{\mathbf{SQTSP }_n}$ if $I, S_1, S_2 \subset V, V=I \dot{\cup }S_1 \dot{\cup }S_2,$ $I \cap S_1 = \emptyset$, $I\cap S_2 = \emptyset$, $S_1\cap S_2 = \emptyset ,$ $S_1 \ne \emptyset ,$ $|S_2|> |I|\ge 3$.

Proof

We set, w. l. o. g., $I=\{i_1=n-|I|+1, \ldots , i_{|I|}=n\}$, $\bar{\imath }=n-1$, $S_1=\{1, \ldots , |S_1|\}$. Again we use the proof-framework of Theorem 2.3 in Fischer and Helmberg (2012), similar to the proof of Theorem 18, with its notation and explain the differences only. Additionally, we denote by ${\bar{I}}$ all nodes of $I$ and by ${\bar{S}}_1$ all nodes of $S_1$ that are not explicitly mentioned, in arbitrary order.

$(\mathrm{Step}_{(15)}{1})$ :

By $|S_1| \ge 1$ and $|S_2| \ge 4$ we know $\{1, \ldots , 5\} \cap I = \emptyset$. So setting ${\bar{n}} = 5$ we can use the same construction as in $(\mathrm{Step}_{(14)} {1})$ taking a largest affinely independent subset $C_{dim}^{{\bar{n}}, 1}$ of set ${\bar{C}}_{dim}^{{\bar{n}}, 1}$ containing 54 tours.

$(\mathrm{Step}_{(15)}{2})$ :

As long as $k\in S_1 \cup S_2$ the nodes of $I$ lie next to each other and so the corresponding tours define roots of (14). Adaptations are needed for the case $k\in I$. We start with a specific ordering for $k=i_1$. Here we can use $(\mathrm{I}_{(14)}^{i_1}{1})$– $(\mathrm{I}_{(14)}^{i_1}\mathrm{2b})$ because all corresponding tours define also roots of (15) by the same arguments as in the proof of Theorem 18. Similarly for $n-2 \ge k > i_1$, constructing substeps $(\mathrm{I}_{(14)}{1})$–$(\mathrm{I}_{(14)}{2b})$ in the proof of Theorem 18 provide roots of (15) and can be applied here.

$(\mathrm{Step}_{(15)}{3})$ :

Some adaptations of the construction in step three are needed specifying the position of ${\bar{I}}$ and splitting up some of the substeps in several successive ones. We set $W=\{w_1,w_2,w_3\}\subset S_2, |W|=3$.

• $(\mathrm{L}_{(15)}\mathrm{1a})$ $\ldots \underline{a\, (n-1)\, b}\, {\bar{I}} \, w_1 \, n\, w_2\ldots ,$ for $a\in \{1, \ldots , n-2\}\setminus \{w_1,w_2\}, b\in I\setminus \{n-1,n\}, a<b,$ (the 2-edge ${\langle w_1, n, w_2\rangle }$ is not used as an $e_L^{\hat{\imath }}$),

• $(\mathrm{L}_{(15)}\mathrm{1b})$ $\ldots \underline{a\, (n-1)\, b}\, {\bar{I}} \, w_1 \, n\, w_2\ldots ,$ for $a\in (S_1\cup S_2)\setminus \{w_1,w_2\}, b\in S_1$, $a>b,$

• $(\mathrm{L}_{(15)}\mathrm{1c})$ $\ldots \underline{a\, (n-1)\, b}\, \eta _{{\bar{I}}, {\bar{S}}_2} \, w_1 \, n\, w_2 \, {\bar{S}}_1\ldots ,$ for $a,b\in S_2\setminus \{w_1,w_2\}, a<b,$ with $\eta _{{\bar{I}},{\bar{S}}_2}$ denoting a path of all nodes of ${\bar{I}} = I \setminus \{n-1,n\}$ and of ${\bar{S}}_2= S_2\setminus \{a,b,w_1,w_2\}$. This path starts with a node $v\in {\bar{I}}$, then an alternating sequence of the remaining nodes of ${\bar{S}}_2$ and of ${\bar{I}} \setminus \{v\},$ (so that each node $w\in I$ lies between two nodes of $S_2$) and depending on the size of $S_2$ in comparison to $I$ a block of nodes in $S_2$.

• $(\mathrm{L}_{(15)}{2})$ $\left\{ \begin{array}{l}\ldots 1\, (n-1)\, {\bar{I}} \, \underline{w_1 \, n\, w_3}\ldots ,\\ \ldots 1\, (n-1)\, {\bar{I}} \, \underline{w_2 \, n\, w_3}\ldots , \end{array} \right.$

• $(\mathrm{L}_{(15)}\mathrm{3a})$ $\ldots \underline{w_1\, (n-1)\, a}\,{\bar{I}}\, w_2 \, n\, w_3\ldots , \text { for }a\in \{1, \ldots , n-2\}\cap (I\cup S_1),$

• $(\mathrm{L}_{(15)}\mathrm{3b})$ $\ldots \underline{w_1\, (n-1)\, a}\,\eta _{ {\bar{I}}, {\bar{S}}_2}\, w_2 \, n\, w_3\, {\bar{S}}_1\ldots , \text { for }a\in \{1, \ldots , n-2\}\cap (S_2 \setminus W)$ with $\eta _{{\bar{I}}, {\bar{S}}_2}$ as above with ${\bar{I}} = I\setminus \{n-1,n\}, {\bar{S}}_2 = S_2\setminus (\{a\}\cup W)$,

• $(\mathrm{L}_{(15)}\mathrm{4a})$ $\ldots \underline{w_2\, (n-1)\, a}\,{\bar{I}}\, w_1 \, n\, w_3\ldots , \text { for }a\in \{1, \ldots , n-2\}\cap (I\cup S_1),$

• $(\mathrm{L}_{(15)}\mathrm{4b})$ $\ldots \underline{w_2\, (n-1)\, a}\,\eta _{ {\bar{I}}, {\bar{S}}_2}\, w_1 \, n\, w_3\, {\bar{S}}_1\ldots , \text { for }a\in \{1, \ldots , n-2\}\cap (S_2 \setminus W)$ with $\eta _{{\bar{I}}, {\bar{S}}_2}$ as above with ${\bar{I}} = I\setminus \{n-1,n\}, {\bar{S}}_2 = S_2\setminus (\{a\}\cup W),$

• $(\mathrm{L}_{(15)}\mathrm{5a})$ $\ldots \underline{a\, n\, b}\, {\bar{I}}\,1\, (n-1)\ldots$, for $a\in W, b\in I\setminus \{n-1,n\},$

• $(\mathrm{L}_{(15)}\mathrm{5b})$ $\ldots \underline{a\, n\, b}\, {\bar{I}}\, (n-1)\ldots$, for $a\in W, b\in S_1,$

• $(\mathrm{L}_{(15)}\mathrm{5c})$ $\ldots \underline{a\, n\, b}\, {\bar{I}} \, 1 \, (n-1)\ldots ,$ for $a\in W, b\in S_2\setminus W$

• $(\mathrm{L}_{(15)}{6})$ $\left\{ \begin{array}{l} \ldots \underline{w_1 \, (n-1)\, w_2}\, \eta _{ {\bar{I}}, {\bar{S}}_2}\, n\, w_3\, {\bar{S}}_1\ldots , \\ \ldots \underline{w_1 \, (n-1)\, w_3}\, \eta _{ {\bar{I}}, {\bar{S}}_2}\, n\, w_2, {\bar{S}}_1\ldots , \\ \ldots \underline{w_2 \, (n-1)\, w_3}\, \eta _{ {\bar{I}}, {\bar{S}}_2}\, n\, w_1, {\bar{S}}_1\ldots , \end{array} \right.$ with $\eta _{{\bar{I}}, {\bar{S}}_2}$ as above with ${\bar{I}} = I\setminus \{n-1,n\}, {\bar{S}}_2 = S_2\setminus W$,

• $(\mathrm{L}_{(15)}\mathrm{7a})$ $\ldots \underline{a\, n\, b}\, {\bar{I}} \, 1 \, (n-1)\ldots ,\text { for } a\in \{1, \ldots , n-2\}\setminus (S_1\cup W), b\in I\setminus \{n-1,n\}, a<b,$

• $(\mathrm{L}_{(15)}\mathrm{7b})$ $\ldots \underline{a\, n\, b}\, {\bar{I}} \, (n-1)\ldots ,\text { for } a\in \{1, \ldots , n-2\}\setminus (I\cup W), b\in S_1, a>b,$

• $(\mathrm{L}_{(15)}\mathrm{7c})$ $\ldots \underline{a\, n\, b}\, {\bar{I}} \, (n-1)\ldots ,\text { for } a,b\in S_2\setminus W, a<b,$

• $(\mathrm{L}_{(15)}\mathrm{8a})$ $\ldots {\bar{S}}_1\,{\bar{I}}\, \underline{(n-1)\, a\, n}\ldots ,\text { for }a\in (S_1\cup I)\setminus \{n-1,n\},$

• $(\mathrm{L}_{(15)}\mathrm{8b})$ $\ldots {\bar{I}} \, \underline{(n-1)\, a\, n}\, m \ldots , \text { for }a\in S_2 \text { with }m\in S_2, m\ne a,$

• $(\mathrm{L}_{(15)}\mathrm{7d})$ $\ldots \underline{a\, n\, b}\, {\bar{I}} \,(n-1)\ldots ,\text { for } a\in \{1, \ldots , n-2\}\cap S_1, b\in I\setminus \{n-1,n\}$.

It follows from the proof of Claim 2 in the proof of Theorem 2.3 in Fischer and Helmberg (2012) and from the proof of Theorem 18 (or is easy to check) in combination with the fact that no 2-edge ${\langle {\tilde{a}},n,{\tilde{b}}\rangle }, {\tilde{a}}\in S_1, {\tilde{b}} \in I,$ is contained in the tours in $(\mathrm{L}_{(15)}\mathrm{8a})$–$(\mathrm{L}_{(15)}\mathrm{8b})$ that all underlined 2-edges are not used in a previous substep and that we build exactly $n^2-4n+3$ tours in $(\mathrm{Step}_{(15)}{3})$. It remains to check the root property of all tours constructed.

• Tours in $(\mathrm{L}_{(15)}\mathrm{1a})$, $(\mathrm{L}_{(15)}\mathrm{1b})$, $(\mathrm{L}_{(15)}\mathrm{2})$, $(\mathrm{L}_{(15)}\mathrm{3a})$, $(\mathrm{L}_{(15)}\mathrm{4a})$, $(\mathrm{L}_{(15)}\mathrm{5c})$, $(\mathrm{L}_{(15)}\mathrm{7c})$, $(\mathrm{L}_{(15)}\mathrm{8b})$: The tours contain one large block of nodes of $I\setminus \{n\}$, partially with one node of $S_1$ between two of these nodes, and $n$ lies between two nodes that belong to $S_2$.

• Tours in $(\mathrm{L}_{(15)}\mathrm{1c})$, $(\mathrm{L}_{(15)}\mathrm{3b})$, $(\mathrm{L}_{(15)}\mathrm{4b})$, $(\mathrm{L}_{(15)}{6})$: Each node of $I$ lies between two nodes that belong to $S_2$.

• Tours in $(\mathrm{L}_{(15)}\mathrm{5a})$, $(\mathrm{L}_{(15)}\mathrm{5b})$, $(\mathrm{L}_{(15)}\mathrm{7a})$, $(\mathrm{L}_{(15)}\mathrm{7b})$, $(\mathrm{L}_{(15)}\mathrm{7d})$,$(\mathrm{L}_{(15)}\mathrm{8a})$: The tours contain one large block of nodes of $I$, partially with one node of $S_1$ between two of these nodes.

All in all we created the same number of tour as in the proof of Theorem 18, more precisely $f(n)$ tours. So inequalities (15) define facets of $P_{\mathbf{SQTSP }_n}, n\ge 8$. $\square$