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Matroid optimisation problems with nested non-linear monomials in the objective function

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Abstract

Recently, Buchheim and Klein (Discrete Appl Math 177:34–52, 2014) suggested to study polynomial-time solvable optimisation problems with linear objective functions combined with exactly one additional quadratic monomial. They concentrated on special quadratic spanning tree or forest problems. We extend their results to general matroid optimisation problems with a set of nested monomials in the objective function. We study polytopes arising from the standard linearisation of the monomials. Our results provide insight on the polyhedral structure of matroid optimisation problems with arbitrary polynomial objective function, with a focus on separation algorithms and strengthened cutting planes. Extending results by Edmonds (Comb Struct Appl, 69–87, 1970) for the matroid polytope we present a complete description for the linearised polytope. Indeed, apart from the standard linearisation one needs appropriately strengthened rank inequalities satisfying certain non-separability conditions. The separation problem of these extended rank inequalities reduces to a submodular function minimisation problem. In the case of exactly one additional non-linear monomial we completely characterise the facetial structure of the associated polytope.

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References

  1. Berstein, Y., Lee, J., Maruri-Aguilar, H., Onn, S., Riccomagno, E., Weismantel, R., Wynn, H.: Nonlinear matroid optimization and experimental design. SIAM J. Discrete Math. 22(3), 901–919 (2008)

    Article  MathSciNet  MATH  Google Scholar 

  2. Brualdi, R.A.: Comments on bases in dependence structures. Bull. Aust. Math. Soc. 1(2), 161–167 (1969)

    Article  MathSciNet  MATH  Google Scholar 

  3. Buchheim, C., Klein, L.: The spanning tree problem with one quadratic term. In: Cornelissen, K., Hoeksma, R., Hurink, J., Manthey, B. (eds.) CTW, Volume WP 13-01 of CTIT Workshop Proceedings, pp. 31–34 (2013)

  4. Buchheim, C., Klein, L.: Combinatorial optimization with one quadratic term: spanning trees and forests. Discrete Appl. Math. 177, 34–52 (2014)

    Article  MathSciNet  MATH  Google Scholar 

  5. Buchheim, C., Rinaldi, G.: Efficient reduction of polynomial zero-one optimization to the quadratic case. SIAM J. Optim. 18(4), 1398–1413 (2008)

    Article  MathSciNet  MATH  Google Scholar 

  6. Crama, Y., Rodríguez-Heck, E.: A class of valid inequalities for multilinear 0–1 optimization problems. Discrete Optim. (2017). doi:10.1016/j.disopt.2017.02.001

  7. De Loera, J.A., Haws, D.C., Lee, J., O’Hair, A.: Computation in multicriteria matroid optimization. J. Exp. Algorithmics 14, 8:1.8–8:1.33 (2010)

    MATH  Google Scholar 

  8. Del Pia, A., Khajavirad, A.: A polyhedral study of binary polynomial programs. Math. Oper. Res. (2016). doi:10.1287/moor.2016.0804

  9. Edmonds, J.: Optimum branchings. J. Res. Natl. Bur. Stand. 71B, 233–240 (1967)

    Article  MathSciNet  MATH  Google Scholar 

  10. Edmonds, J.: Submodular functions, matroids, and certain polyhedra. In: Guy, R.K. (ed.) Proceedings of Combinatorial Structures and their Applications, pp. 69–87. Gordon and Breach, New York (1970)

  11. Edmonds, J.: Matroids and the greedy algorithm. Math. Progr. 1(1), 127–136 (1971)

    Article  MathSciNet  MATH  Google Scholar 

  12. Fischer, A., Fischer, F.: Complete description for the spanning tree problem with one linearised quadratic term. Oper. Res. Lett. 41(6), 701–705 (2013)

    Article  MathSciNet  MATH  Google Scholar 

  13. Fortet, R.: L’Algèbre de Boole et ses applications en recherche opérationnelle. Cahiers du Centre d’Etudes de Recherche Opérationelle 4, 5–36 (1959)

    MATH  Google Scholar 

  14. Fujishige, S.: Submodular Functions and Optimization. Elsevier, Amsterdam (2005)

    MATH  Google Scholar 

  15. Glover, F., Woolsey, E.: Further reduction of zero-one polynomial programming problems to zero-one linear programming problems. Oper. Res. 21(1), 156–161 (1973)

    Article  MathSciNet  MATH  Google Scholar 

  16. Glover, F., Woolsey, E.: Converting the 0–1 polynomial programming problem to a 0–1 linear program. Oper. Res. 22(1), 180–182 (1974)

    Article  MATH  Google Scholar 

  17. Grötschel, M.: Polyedrische Charakterisierungen kombinatorischer Optimierungsprobleme, volume 36 of Mathematical Systems in Economics. Verlag Anton Hain, Meisenheim am Glan (1977)

  18. Grötschel, M., Lovász, L., Schrijver, A.: Geometric Algorithms and Combinatorial Optimization, Volume 2 of Algorithms and Combinatorics, second corrected edition. Springer, Berlin (1993)

  19. Hupp, L., Klein, L., Liers, F.: An exact solution method for quadratic matching: the one-quadratic-term technique and generalisations. Discrete Optim. 18, 193–216 (2015)

    Article  MathSciNet  Google Scholar 

  20. Klein, L.: Combinatorial Optimization with One Quadratic Term. Ph.D. thesis, TU Dortmund (2014)

  21. Lee, J., Leung, J.: On the Boolean quadric forest polytope. INFOR 42(2), 125–141 (2004)

    MathSciNet  Google Scholar 

  22. McCormick, S.T.: Submodular function minimization. In: Aardal, K., Nemhauser, G., Weismantel, R. (eds.) Handbook on Discrete Optimization, pp. 321–391. Elsevier, Amsterdam (2006)

    Google Scholar 

  23. Onn, S.: Convex matroid optimization. SIAM J. Discrete Math. 17(2), 249–253 (2004)

    Article  MathSciNet  MATH  Google Scholar 

  24. Oxley, J.G.: Matroid Theory. Oxford University Press, Oxford (1992)

    MATH  Google Scholar 

  25. Padberg, M.: The Boolean quadric polytope: some characteristics, facets and relatives. Math. Program. 45(1–3), 139–172 (1989)

    Article  MathSciNet  MATH  Google Scholar 

  26. Schrijver, A.: A combinatorial algorithm minimizing submodular functions in strongly polynomial time. J. Comb. Theory Ser. B 80(2), 346–355 (2000)

    Article  MathSciNet  MATH  Google Scholar 

  27. Schrijver, A.: Combinatorial Optimization-Polyhedra and Efficiency. Springer, Berlin (2003)

    MATH  Google Scholar 

  28. Sherali, H.D., Adams, W.P.: A hierarchy of relaxations between the continuous and convex hull representations for zero-one programming problems. SIAM J. Discrete Math. 3(3), 411–430 (1990)

    Article  MathSciNet  MATH  Google Scholar 

  29. Sherali, H.D., Adams, W.P.: Reformulation-Linearization Technique for Solving Discrete and Continuous Nonconvex Problems. Springer, Berlin (1998)

    MATH  Google Scholar 

  30. Walter, M.: Complete Description of Matching Polytopes with One Linearized Quadratic Term for Bipartite Graphs. ArXiv e-prints (2016)

  31. Welsh, D.J.A.: Matroid Theory. Academic Press, London (1976)

    MATH  Google Scholar 

Download references

Acknowledgements

We thank two anonymous referees for their valuable comments and suggestions that helped to improve the paper.

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Correspondence to Anja Fischer.

Appendix

Appendix

In the following we present several detailed, rather technical proofs of the results stated in the main part of this paper.

Proof of Observation 5

Let \(e \in {{\mathrm{cl}}}(T)\), then \(r(T+e) = r(T)\).

$$\begin{aligned} r(T{+}S){+}r(T)&\mathop {\le }\limits ^{(R2)} r(T{+}S{+}e){+} r((T{+}S)\cap (T+e))\mathop {\le }\limits ^{(R3)} r(T+S)+r(T+e) \\&= r(T+S)+r(T). \end{aligned}$$

\(\square \)

Proof of Observation 7

Let \(T\subseteq E\).

(A1) :

Let T be closed. By the definition of \(\alpha _i(\cdot )\) we can assume that \(e_i\notin T\). Then \(\alpha _1(T)=1+r(T)-(r(T)+1)=0\).

(A2) :

Let \(j\in \{1, \cdots , k\}\). Then \(\alpha _{1,j}(T) = \sum _{i=1}^j ( |\{e_i\}{\setminus } T| + r(T+\bar{E}_{1,i-1}) - r(T+\bar{E}_{1,i}))= |\bar{E}_{1,j}{\setminus } T| + r(T+\bar{E}_{1,0}) - r(T+\bar{E}_{1,j})=|\bar{E}_{1,j}{\setminus } T|+r(T)-r(T+\bar{E}_{1,j})\).

(A3) :

Let \(i\in \{1, \cdots , k\}\). If \(e\in T\), the statement is clear. So let \(e\in ({{\mathrm{cl}}}(T){\setminus } T){\setminus } \{e_i\}\). Then \(\alpha _i(T)=|\{e_i\}{\setminus } T|+r(T+\bar{E}_{1,i-1})-r(T+\bar{E}_{1,i}) = |\{e_i\}{\setminus } (T+e)| + r(T+e+\bar{E}_{1,i-1}) - r(T+e+\bar{E}_{1,i})=\alpha _i(T+e)\) by Observation 5.

(A4) :

Let \(i\in \{1, \cdots , k\}\) and \(e_i\in {{\mathrm{cl}}}(T){\setminus } T\). Then Observation 5 and the definition of \(\alpha _i(\cdot )\) show \(\alpha _i(T)=|\{e_i\}{\setminus } T|+r(T+\bar{E}_{1,i-1})-r(T+\bar{E}_{1,i})=1+r(T+e_i+\bar{E}_{1,i-1})-r(T+\bar{E}_{1,i})=1=\alpha _i(T+e_i)+1\). \(\square \)

Proof of Lemma 14

Let \(a^T x + \alpha ^T y\le b\) be a facet defining inequality of \(P_{M}^{\bar{E},K}\) , \(K=\{2, \cdots , k\}\), satisfying the requirements. For \(m=2\) the existence of a set S with the desired properties follows directly from the fact that \(a^T x +\alpha ^T y \le b \) is not a positive multiple of (5). For \(m > 2\) there exists a root S of \(a^T x + \alpha ^T y \le b\) with \(e_m \notin S\) and \(\bar{E}_{1,m-1} \nsubseteq S\) because \(a^T x + \alpha ^T y\le b\) is not a positive multiple of (6). So we may assume that

$$\begin{aligned} \mu := |S \cap \bar{E}_{1,m-1}| {\text { is maximum.}} \end{aligned}$$
(21)

If \(|S\cap \bar{E}_{1,m}| = |\bar{E}_{1,m}|-2\), then S satisfies the requirements. Otherwise there must exist two elements \(e_i, e_j \notin S\) with \(i< j < m\) and \(\bar{E}_{1,j-1} {\setminus } \{e_i\} \subseteq S\). Let \(S_{j-1}\) be a root of \(a^T x + \alpha ^T y\le b\) with \(\bar{E}_{1,j-1} \subseteq S_{j-1}\) and \(e_j \notin S_{j-1}\) (exists because \(a^T x + \alpha ^T y\le b\) is not a positive multiple of (4) or of (7) in the case \(j=2\)).

Now assume in addition that S and \(S_{j-1}\) are chosen so that

$$\begin{aligned} |S \cap \bar{E}_{1,m-1}|&=\mu \quad {\text {and}}\quad |S \cap S_{j-1}|\quad {\text { is maximum}}. \end{aligned}$$
(22)
  1. 1.

    If \(r(S) < r(S_{j-1})\), then there exists an \(e \in S_{j-1} {\setminus } S\) so that \(e+S \in {\mathcal {I}}\). By Observation 13 we get \(\tilde{a}(S+e,e) \le 0 \le \tilde{a}(S_{j-1},e)\). If \(e \ne e_i\), then \(e \notin \bar{E}_{1,j}\) and so \(\tilde{a}(S_{j-1},e) = a_e = \tilde{a}(S+e,e)\) proving \(a_e = 0\). Thus \(S+e\) is also a root of \(a^T x + \alpha ^T y \le b\) contradicting (22). Otherwise we have \(\tilde{a}(S_{j-1},e) = a_e + \alpha _i + \cdots + \alpha _{j-1} = \tilde{a}(S+e,e)=0\). As before \(S+e\) is also root of \(a^T x + \alpha ^T y \le b\), a contradiction to (21).

  2. 2.

    If \(r(S) > r(S_{j-1})\), then there exists an \(e \in S {\setminus } S_{j-1}\) so that \(S_{j-1} + e \in {\mathcal {I}}\). Certainly \(e \notin \bar{E}_{1,j}\), so as above it follows \(a_e = 0\). So \(S, (S_{j-1}+e)\) contradict to assumption (22).

  3. 3.

    If \(r(S) = r(S_{j-1})\), then we know by Theorem 1 that there exists an \(e \in S {\setminus } S_{j-1}\) so that \(S-e+e_i\), \(S_{j-1}+e-e_i \in {\mathcal {I}}\). Note, \(e \notin \bar{E}_{1,j}\), which implies \(\tilde{a}(S,e) = a_e = \tilde{a}(S_{j-1}+e-e_i,e)\) as well as \(\tilde{a}(S_{j-1},e_i) = a_{e_i} + \alpha _i + \cdots + \alpha _{j-1} = \tilde{a}(S-e+e_i,e_i)\). Applying Observation 13 twice then shows \(\tilde{a}(S-e+e_i,e_i) \le \tilde{a}(S,e) = \tilde{a}(S_{j-1}+e-e_i,e) \le \tilde{a}(S_{j-1},e_i) = \tilde{a}(S-e+e_i,e_i)\) and so \(S-e+e_i\) is also a root of \(a^T x + \alpha ^T y \le b\), contradicting (21). \(\square \)

Proof of Lemma 15

Let \(a^T x+\alpha ^T y\le b\) be a facet defining inequality of \(P_{M}^{\bar{E},K}\)  with \(K=\{2, \ldots , k\}\) that is not a positive multiple of one of (3)–(8). Assume, for a contradiction, the assertion is false. Then there must exist an \(m \in \{2, \cdots , k\}\) and a minimum \(h \in \{m, \cdots , k\}\) so that \(\alpha _{m} + \cdots + \alpha _h < 0\) and there exists a root S of \(a^T x + \alpha ^T y \le b\) satisfying \(\bar{E}_{1,m} {\setminus } \{e_j,e_{m}\} = S \cap \bar{E}_{1,m}\), \(j < m\), and \(h=k\) or \(e_{h+1} \notin S\). Furthermore, there exists a root \(S_h\) with \(\bar{E}_{1,h} \subseteq S_h\) and \(h=k\) or \(e_{h+1} \notin S\) because \(a^T x+\alpha ^T y\le b\) is not a positive multiple of (4). We may assume that

$$\begin{aligned} |S \cap S_h| {\text { is maximum.}} \end{aligned}$$
(23)
  1. 1.

    If \(r(S) < r(S_h)\), then there exists an \(e \in S_h {\setminus } S\) so that \(S+e \in {\mathcal {I}}\). We consider four cases depending on e and apply Observation 13.

    • If \(e \notin \bar{E}_{1,h}\), then \(\tilde{a}(S+e,e) = a_e\) and \(\tilde{a}(S_h,e) = a_e\). Observation 13 shows that \(a_e = 0\), so \(S+e\) is also a root of \(a^T x+\alpha ^T y\le b\), contradicting (23).

    • If \(e=e_m\), then \(0 \ge \tilde{a}(S+e_m,e_m) = a_{e_m}\) and \(0 \le \tilde{a}(S_h,e_m) = a_{e_m} + \alpha _m + \cdots + \alpha _h\), hence \(\alpha _m + \cdots + \alpha _h \ge 0\), a contradiction.

    • If \(e=e_j\), then we similarly get \(0 \ge \tilde{a}(S+e_j,e_j) = a_{e_j} + \alpha _j + \cdots + \alpha _{m-1}\) and \(0 \le \tilde{a}(S_h,e_j) = a_{e_j} + \alpha _j + \cdots + \alpha _h\), hence \(\alpha _m + \cdots + \alpha _h \ge 0\), a contradiction.

    • If \(e=e_i\), \(m < i \le h\), then \(0 \ge \tilde{a}(S+e_i,e_i) = a_{e_i}\) and \(0 \le \tilde{a}(S_h,e_i) = a_{e_i} + \alpha _i + \cdots + \alpha _h\), hence we derive \(\alpha _i + \cdots + \alpha _h \ge 0\). Because \(e_i \notin S\) we know by minimality of h that \(\alpha _m + \cdots + \alpha _{i-1} \ge 0\). Together we have \(\alpha _m + \cdots + \alpha _h \ge 0\), a contradiction.

  2. 2.

    If \(r(S) > r(S_h)\), then there exists an \(e \in S {\setminus } S_h\) so that \(S_h + e \in {\mathcal {I}}\). Note that \(e \notin \bar{E}_{1,h+1}\) (if \(h < k\)), so by Observation 13 \(0 \ge \tilde{a}(S_h+e,e) = a_e \ge 0\). Therefore \(S_h + e\) is also a root of \(a^T x+\alpha ^T y\le b\), contradicting (23).

  3. 3.

    If \(r(S) = r(S_h)\), then by Theorem 1 there exists an \(e \in S {\setminus } S_h\) so that \(S-e+e_m, S_h+e-e_m \in {\mathcal {I}}\). By assumption \(h=k\) or \(e \notin \bar{E}_{1,h+1}\), thus we have \(\tilde{a}(S,e) = a_e = \tilde{a}(S_h+e-e_m,e)\) as well as \(\tilde{a}(S_h,e_m) = a_{e_m} + \alpha _m +\cdots + \alpha _h\) and \(\tilde{a}(S-e+e_m,e_m) = a_{e_m}\). We know by Observation 13 that \(a_{e_m} = \tilde{a}(S-e+e_m,e_m) \le \tilde{a}(S,e) = \tilde{a}(S_h - e_m+e,e) \le \tilde{a}(S_h,e_m) = a_{e_m} + \alpha _m + \cdots + \alpha _h\), so \(\alpha _m + \cdots + \alpha _h \ge 0\), a contradiction. \(\square \)

Proof of Lemma 16

Let \(a^T x+\alpha ^T y\le b\) be a facet defining inequality of \(P_{M}^{\bar{E},K}\) , \(K=\{2, \cdots , k\}\), satisfying the requirements defined above. Fix some \(m \in \{2, \cdots , k\}\) and suppose that the claim is false. Then there exists a maximum \(h \in \{m, \cdots , k\}\) so that \(\alpha _m + \cdots + \alpha _h < 0\). Choose a root S with \(\bar{E}_{1,m} {\setminus } \{e_j, e_m\} = S \cap \bar{E}_{1,m}\) (exists by Lemma 14) and a root \(S_h\) with \(\bar{E}_{1,h} \subseteq S_h\), \(h=k\) or \(e_{h+1} \notin S_h\) [exists because \(a^T x+\alpha ^T y\le b\) is not a positive multiple of (3) and (4)] so that \(|S \cap S_h|\) is maximum. If \(e_{h+1} \notin S\) or \(h=k\), then the claim follows from Lemma 15, so we may assume \(e_{h+1} \in S\). We use similar arguments as in the previous proof.

  1. 1.

    If \(r(S) < r(S_h)\), then there is an \(e \in S_h {\setminus } S\) so that \(S+e \in {\mathcal {I}}\). If additionally \(e \ne e_i\), \(m < i \le h\), then all arguments of the previous proof apply (in particular, we do not require the assumption of h being minimal). We only have to consider the case \(e = e_i\), \(m < i \le h\). Then we get \(\tilde{a}(S+e_i,e_i) = a_{e_i}\) and \(\tilde{a}(S_h,e_i) = a_{e_i} + \alpha _i + \cdots + \alpha _h\). So by Observation 13 we have \(\alpha _i+ \cdots +\alpha _h\ge 0\). Note that \(e_i \notin S\) by construction, so we may apply Lemma 15 with this S, j and m and \(h = i-1\) to derive \(\alpha _m + \cdots + \alpha _{i-1} \ge 0\). Hence, we conclude that \(\alpha _m + \cdots + \alpha _h \ge 0\), a contradiction.

  2. 2.

    If \(r(S) > r(S_h)\), then there is an \(e \in S {\setminus } S_h\) so that \(S_h + e \in {\mathcal {I}}\). The case \(e \ne e_{h+1}\) works as in the previous proof, so it remains \(e=e_{h+1}\). Then \(\tilde{a}(S_h+e_{h+1},e_{h+1}) = a_{e_{h+1}} + \alpha _{h+1} + \cdots + \alpha _{o}\) for some \(o \in \{h+1, \cdots , k\}\) and \(\tilde{a}(S,e_{h+1}) = a_{e_{h+1}}\). Therefore, applying Observation 13 yields \(\alpha _{h+1} + \cdots + \alpha _{o} \le 0\). By maximality of h we know \(\alpha _m + \cdots + \alpha _{o} \ge 0\) and can conclude \(\alpha _m + \cdots + \alpha _h \ge 0\), a contradiction.

  3. 3.

    If \(r(S) = r(S_h)\), then by Theorem 1 there exists an \(e \in S {\setminus } S_h\) so that \(S-e+e_m, S_h+e-e_m \in {\mathcal {I}}\). Note that \(e \notin \bar{E}_{1,h}\). We have \(\tilde{a}(S,e) = a_e = \tilde{a}(S_h - e_m+e,e)\), \(\tilde{a}(S_h,e_m) = a_{e_m} + \alpha _m +\cdots + \alpha _h\) and \(\tilde{a}(S-e+e_m,e_m) = a_{e_m}\). Observation 13 shows \(a_{e_m} = \tilde{a}(S-e+e_m,e_m) \le \tilde{a}(S,e) = \tilde{a}(S_h+e-e_m,e) \le \tilde{a}(S_h, e_m) = a_{e_m} + \alpha _m + \cdots + \alpha _h\). This implies \(\alpha _m + \cdots + \alpha _h \ge 0\), a contradiction. \(\square \)

Proof of Observation 21

The polytope \(P_{M}^{\bar{E},K}\)  is full-dimensional with dimension \(|E|+l\). We prove this by explicitly constructing \(|E|+l+1\) independent sets whose incidence vectors are affinely independent. Indeed, the sets \(\emptyset \in {\mathcal {I}}, \{e\}\in {\mathcal {I}}\) for all \(e\in E\) and \(\{e_1, \cdots , e_{k_j}\}\in {\mathcal {I}}\) for all \(j=\{1, \cdots , l\}\) have affinely independent incidence vectors.

Similarly for the facet defining inequalities, we explicitly present the \(|E|+l\) respective sets.

  • \(y_{k_l}\ge 0\) (3): We can use the independent sets of the dimension proof except for \(\{e_1, \cdots , e_{k_l}\}\).

  • \(y_{k_j}-y_{k_{j-1}}\le 0, j=2, \cdots , l\) (4): We can use the independent sets of the dimension proof except for \(\{e_1, \cdots , e_{k_{j-1}}\}\).

  • \( x_{\bar{E}_{1,k_1}} -y_{k_1}\le k_1-1 \) (5): We use \(\bar{E}_{1,k_j}\) for all \(j=1, \cdots , l\), and \(\bar{E}_{1,k_1}{\setminus } \{e\}\) for all \(e\in \bar{E}_{1,k_1}\). By assumption (1) we know that for each \(e\in E{\setminus } \bar{E}_{1,k_1}\) there exists an \(f\in \bar{E}_{1,k_1}\) such that \((\bar{E}_{1,k_1}+e)-f\in {\mathcal {I}}\). So we use one such set for each \(e\in \bar{E}{\setminus } \bar{E}_{1,k_1}\).

  • \( x_{\bar{E}_{k_{j-1}+1,k_{j}}} + y_{k_{j-1}}-y_{k_j}\le k_j-k_{j-1}, j=2, \cdots , l\), (6): We use the independent sets \(\bar{E}_{k_{j-1}+1,k_j}\), \(\bar{E}_{k_{j-1}+1,k_j}+e\) for all \(e\in \bar{E}_{1,k_{j-1}}\), and \(\bar{E}_{k_{j-1}+1,k_j}+\bar{E}_{1,k_m}\) for all \(m\in \{1, \cdots , j-2\}\), as well as \(\bar{E}_{1,k_j}-e\) for all \(e\in \bar{E}_{k_{j-1}+1,k_j}\). For each \(e\in E{\setminus } \bar{E}_{1,k_j}\) there exists an \(f\in \bar{E}_{1,k_j}\) such that \(\bar{E}_{1,k_j}+e-f\in {\mathcal {I}}\). We use one such set for each \(e\in E{\setminus } \bar{E}_{1,k_j}\). Furthermore we take all the sets \(\bar{E}_{1,k_m}\) with \(m\in \{j, \cdots , l\}\).

  • \(-x_{e_i}+y_{k_j}\le 0,j=1, \cdots , l, i=k_{j-1}+1, \cdots , k_j\), (7): We can use the independent sets of the dimension proof except for \(\{e_i\}\).

  • \(x_e\ge 0, e\in E{\setminus } \bar{E}\) (8): We can use the independent sets of the dimension proof except for \(\{e\}\).

Further note, constraints \(-x_e\le 0, e\in \bar{E}\), are implied by (3), (4) and (7). \(\square \)

Proof of Observation 26

  1. 1.

    Let \(T\subseteq \bar{E}\). Then T is \((\bar{E},\{k\})\)-separable with \(T_1=T{\setminus } \{e\}\ne \emptyset \) and \(T_2=\{e\}\) for each \(e \in T\).

  2. 2.

    Let \(T\subseteq E, \bar{E}\cap T\ne \emptyset \) and \(\alpha _{1,k}(T{\setminus } \bar{E})=0\). If \(T \subseteq \bar{E}\), we are in case 1. Otherwise \(\alpha _{1,k}(T {\setminus } \bar{E}) = 0\) and (A2) imply \(r(T {\setminus } \bar{E}) + |\bar{E}| = r(T + \bar{E})\), hence \(r(T {\setminus } \bar{E}) + |T \cap \bar{E}| = r(T)\) (by \(r(T)\le r(T{\setminus } \bar{E})+r(T\cap \bar{E})=r(T + \bar{E})+r(T\cap \bar{E})-|\bar{E}|\le r(T)\)). Furthermore, this implies \(\alpha _{1,k}(T)=0\) as well. Consequently T is \((\bar{E},\{k\})\)-separable with \(T_1 = T {\setminus } \bar{E}\ne \emptyset \) and \(T_2 = T \cap \bar{E}\).\(\square \)

Proof of Observation 28

We will prove this result by showing that if \((\mathbf{P1}^k)\) and \((\mathbf{P2}^k)\) are not satisfied, we can easily derive the inequality as a combination of other inequalities. So let \(T \subseteq E\) be a closed and \((\bar{E},\{k\})\)-non-separable set that satisfies neither \((\mathbf{P1}^k)\) nor \((\mathbf{P2}^k)\). Then either \(|\bar{E}|=2\) and \(T=\{e\}\) for some \(e \in \bar{E}\) or \(\bar{E}\subseteq T\) and \(\alpha _{1,k}(T {\setminus } \bar{E}) \le 1\).

In the first case, because T is closed and \((\bar{E},\{k\})\)-non-separable, the rank inequality associated with T can be derived by adding the two constraints \(x_{\bar{E}}-y\le 1\) and \(-x_{\bar{e}}+y\le 0\) for \(\{\bar{e}\} = \bar{E}{\setminus } \{e\}\). In the second case, we can assume \(\alpha _{1,k}(T {\setminus } \bar{E}) =1\) by Observation 26. Then we can derive (14) for T by adding (11) and (14) for \(T':=T {\setminus } \bar{E}\). \(\square \)

Proof of Observation 29

Let \(T\subseteq E\) be a closed set with \(\bar{E}\subseteq T\) and \(\alpha _{1,k}(T{\setminus } \bar{E}) = r(T{\setminus } \bar{E})+|\bar{E}|-r(T)\ge 2\) (by (A2)). Then we take a basis B of \(T{\setminus } \bar{E}\) and extend B to a basis of T by adding elements of \(\bar{E}\). By the assumption on \(\alpha _{1,k}(T{\setminus } \bar{E})\) at most \(r(T)-r(T{\setminus } \bar{E})\le |\bar{E}|-2\) elements are added to B. \(\square \)

Proof of Claim 1 in the Proof of Lemma 30

We want to prove that inequalities \(x_T+\alpha _{1,k}(T)y\le r(T)\) for a closed and \((\bar{E},\{k\})\)-non-separable set \(T\subseteq E\) satisfying \((\mathbf{P1}^k)\) or \((\mathbf{P2}^k)\) are not implied by (12)–(13). For this, we determine in each of the cases an independent set \(J\in {\mathcal {I}}\) such that J is a root of (14), but J is not a root of the considered other constraint out of (11)–(13).

  • \(x_{\bar{E}}-y \le |\bar{E}|-1\) (11): We consider three cases.

    • \(|T\cap \bar{E}|<|\bar{E}|-1\): Let B be a basis of T. Then we set \(J:=B\).

    • \(|T\cap \bar{E}|=|\bar{E}|-1\): Observation 26 and \((\mathbf{P1}^k)\) imply \(T \nsubseteq \bar{E}\) and \(|T| \ge 2\). Choose \(f \in T \cap \bar{E}\), then \(r(T - f) = r(T)\) because otherwise T would be \((\bar{E},\{k\})\)-separable with \(T_1 = \{f\}\) and \(T_2 = T - f\). Let B be a basis of \(T - f\), then B is also a basis of T with \(|B \cap \bar{E}| \le |\bar{E}| - 2\). So we set \(J:=B\).

    • \(\bar{E}\subseteq T\): In this case, by \((\mathbf{P2}^k)\) and Observation 29, there exists a basis B of T such that \(|B\cap \bar{E}| \le |\bar{E}|-2\). Then we set \(J:=B\).

  • \(y -x_e\le 0, e\in \bar{E}\) (12): We consider four cases:

    • \(e\in T, \bar{E}\nsubseteq T\): Let B be a basis of T with \(e\in T\). Then we have that \(y^B=0, \chi ^B_e=1\) and \(\sum _{e\in T}\chi ^B_e+0=r(T)\). So we use \(J:=B\).

    • \(e\in T, \bar{E}\subseteq T\): In this case we know by \((\mathbf{P2}^k)\) and Observation 29 that there exists a basis B of T with \(|B\cap \bar{E}|\le |\bar{E}|-2\). If \(e\in B\), we set \(J:=B\), otherwise there exists an \(f\in B\) such that \(B+e-f\in {\mathcal {I}}\) is a basis of T and so we use \(J:=B+e-f\) (note \(y^{B+e-f}=0\)).

    • \(e\notin T, \bar{E}\nsubseteq T+e\): Let B be a basis of T. Then \(B+e=:J\in {\mathcal {I}}\) because T is closed.

    • \(e\notin T, \bar{E}\subseteq T+e\): Observation 26 and \((\mathbf{P1}^k)\) imply \(T \nsubseteq \bar{E}\), in particular \(|T| \ge 2\). Choose \(f \in T \cap \bar{E}\), then \(r(T - f) = r(T)\) because otherwise T would be \((\bar{E},\{k\})\)-separable with \(T_1 = \{f\}\) and \(T_2 = T - f\). Let B be a basis of \(T - f\), then B is also a basis of T, and because T is closed we can use \(J := B + e\), which is a basis of \(T + e\) with \(\bar{E}\nsubseteq J\) by the choice of f.

  • \(-x_e\le 0, e\in E{\setminus } \bar{E}\) (13): First, we consider the case \(e\notin T\). Let B be a basis of T. By T closed we know \(B+e\in {\mathcal {I}}\) and set \(J:=B+e\). Second, if \(e\in T\), there exists a basis B of T with \(e\in B\) and we can use \(J:=B\).

  • \(-y \le 0\) (13): We consider two cases. If \(\bar{E}\subseteq T\), there exists a basis B of T with \(\bar{E}\subseteq B\) (note \(\bar{E}\in {\mathcal {I}}\)) and so we use \(J:=B\). If, otherwise, \(\bar{E}\nsubseteq T\), let B be a basis of \(T\cup \bar{E}\) with \(\bar{E}\subset B\). Then by \(\sum _{e\in T}\chi ^B_e +|\bar{E}{\setminus } T| = |B\cap T|+|\bar{E}{\setminus } T|=|((B{\setminus } \bar{E})\cap T) \cup \bar{E}|= r(B)=r(T\cup \bar{E})\) we get \(\sum _{e\in T}\chi ^B_e+r(T) +|\bar{E}{\setminus } T|-r(T\cup \bar{E}) = r(T)\). So B is a root of \(x_T+\alpha _{1,k}(T) y \le r(T)\) and we can set \(J:=B\).

So \(x_T+\alpha _{1,k}(T)y\le r(T)\) is not a positive multiple of one of (11)–(13). This proves Claim 1 in Lemma 30. \(\square \)

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Fischer, A., Fischer, F. & McCormick, S.T. Matroid optimisation problems with nested non-linear monomials in the objective function. Math. Program. 169, 417–446 (2018). https://doi.org/10.1007/s10107-017-1140-9

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