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On \(\Delta \)-modular integer linear problems in the canonical form and equivalent problems

Abstract

Many papers in the field of integer linear programming (ILP, for short) are devoted to problems of the type \(\max \{c^\top x :A x = b,\, x \in {{\,\mathrm{\mathbb {Z}}\,}}^n_{\ge 0}\}\), where all the entries of Abc are integer, parameterized by the number of rows of A and \(\Vert A\Vert _{\max }\). This class of problems is known under the name of ILP problems in the standard form, adding the word ”bounded” if \(x \le u\), for some integer vector u. Recently, many new sparsity, proximity, and complexity results were obtained for bounded and unbounded ILP problems in the standard form. In this paper, we consider ILP problems in the canonical form

$$\begin{aligned} \max \{c^\top x :b_l \le A x \le b_r,\, x \in {{\,\mathrm{\mathbb {Z}}\,}}^n\}, \end{aligned}$$

where \(b_l\) and \(b_r\) are integer vectors. We assume that the integer matrix A has the rank n, \((n + m)\) rows, n columns, and parameterize the problem by m and \(\Delta (A)\), where \(\Delta (A)\) is the maximum of \(n \times n\) sub-determinants of A, taken in the absolute value. We show that any ILP problem in the standard form can be polynomially reduced to some ILP problem in the canonical form, preserving m and \(\Delta (A)\), but the reverse reduction is not always possible. More precisely, we define the class of generalized ILP problems in the standard form, which includes an additional group constraint, and prove the equivalence to ILP problems in the canonical form. We generalize known sparsity, proximity, and complexity bounds for ILP problems in the canonical form. Additionally, sometimes, we strengthen previously known results for ILP problems in the canonical form, and, sometimes, we give shorter proofs. Finally, we consider the special cases of \(m \in \{0,1\}\). By this way, we give specialised sparsity, proximity, and complexity bounds for the problems on simplices, Knapsack problems and Subset-Sum problems.

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Acknowledgements

Results of the Sects. 3, 4 were prepared within the framework of the Basic Research Program at the National Research University Higher School of Economics (HSE). Results of the Sect. 5 were prepared under financial support of Russian Science Foundation Grant No. 21-11-00194. The authors would like to thank J. Paat for useful discussions during preparation of this article. Additionally, the authors would like to thank anonymous reviewers for their useful comments and remarks.

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Appendix

Appendix

Proof of the Lemma 3

The structure of \(A_{{{\,\mathrm{\mathcal {B}}\,}}}^*\) directly follows from the triangular structure of \(A_{{{\,\mathrm{\mathcal {B}}\,}}}\) and from the definition of \(A_{{{\,\mathrm{\mathcal {B}}\,}}}^*\).

Let the matrix H be obtained from \(A_{{{\,\mathrm{\mathcal {B}}\,}}}\) by deleting any row and any column. The value of \(\vert \det (H) \vert \) corresponds to some element of \(A_{{{\,\mathrm{\mathcal {B}}\,}}}^*\). It is easy to see that H is a lower triangular matrix with at most one additional diagonal. We can expand the determinant of H by the first row, using the Laplace rule. Then, \(\vert \det (H) \vert \le 2^{k-1} d_1 d_2 \dots d_{k}\), where k is the number of diagonal elements in \(A_{{{\,\mathrm{\mathcal {B}}\,}}}\) that are not equal to 1, and \((d_1, d_2, \dots , d_{k})\) is a sequence of diagonal elements. Since \(k \le \log _2(\delta )\), we have \(\vert \det (H) \vert \le \delta ^2/2\).

Let us prove the second part of this Lemma. Columns of the matrix \(A_{{{\,\mathrm{\mathcal {B}}\,}}}^{-1}\) have at most \(\log _2(\delta ) + 1\) non-zero components. More precisely, the last \(\lfloor \log _2(\delta ) \rfloor \) columns of \(A_{{{\,\mathrm{\mathcal {B}}\,}}}^{-1}\) have at most \(\lfloor \log _2 \delta \rfloor \) non-zero components that are concentrated in the last \(\lfloor \log _2(\delta ) \rfloor \) coordinates. Consequently, \(\Vert A_{{{\,\mathrm{\mathcal {B}}\,}}}^{-1} y\Vert _0 \le \alpha + \log _2(\delta )\). The second inequality trivially follows from the first part of this lemma. \(\square \)

Proof of the Lemma 6

Let \(y = \frac{2 n }{m}\), then

$$\begin{aligned} y \le 2 + \log _2(y) + \log _2 \frac{e}{2} + \frac{2 \log _2(\Delta )}{m}. \end{aligned}$$

Let \(d = 1 + \log _2 e + \frac{2 \log _2(\Delta )}{m}\), then

$$\begin{aligned} y \le \log _2(y) + d. \end{aligned}$$

Let h be defined by the equality

$$\begin{aligned} y = d + \log _2(d) + h, \end{aligned}$$

then

$$\begin{aligned} d + \log _2(d) + h \le \log _2(d + \log _2(d) + h) + d, \end{aligned}$$

and

$$\begin{aligned} h \le \log _2\frac{d + \log _2(d) + h}{d} = \log _2 \left( 1 + \frac{\log _2(d) + h}{ d} \right) . \end{aligned}$$

The function in the r.h.s. of the previous inequality attains its maximum in the point \(d = \frac{e}{2^h}\), hence

$$\begin{aligned} h \le \log _2 \left( 1 + \frac{2^h \cdot \log _2 e}{e} \right) . \end{aligned}$$

Next,

$$\begin{aligned} 2^h \le 1 + \frac{2^h \cdot \log _2 e}{e}\quad \text {and}\quad 2^h \le \frac{e}{e-\log _2e}, \end{aligned}$$

hence

$$\begin{aligned} h \le {\bar{c}} := \log _2 \frac{e}{e - \log _2 e}. \end{aligned}$$

Returning to y, we have \(y \le d + \log _2(d) + {\bar{c}}\) and

$$\begin{aligned} \frac{2 n}{m} \le 1 + \log _2 e + \frac{2 \log _2(\Delta )}{m} + \log _2\left( 1 + \log _2 e + \frac{2 \log _2(\Delta )}{m} \right) + {\bar{c}}. \end{aligned}$$

Finally,

$$\begin{aligned} n \le \frac{{\bar{c}}+1 + \log _2(2e)}{2} \cdot m + \log _2(\Delta ) + \frac{m}{2} \cdot \log _2 \left( \frac{1}{2} + \frac{1}{2} \log _2 e + \frac{\log _2(\Delta )}{m} \right) . \end{aligned}$$

\(\square \)

Proof of the Lemma 7

Let \(g = \frac{\log _2(\Delta )}{m}\). We are going to use the inequality

$$\begin{aligned} f(x + h) \le f(x) + f^\prime (x) \cdot h, \end{aligned}$$

which is true for concave functions, such as \(\log _2(x)\). Taking \(h = g - k\) and \(x = c_2 + k\), we have

$$\begin{aligned} \log _2(c_2 + g) \le \log _2(c_2 + k) + \frac{1}{(c_2 + k)\cdot \ln 2} \cdot (g - k). \end{aligned}$$

We substitute the last inequality to the inequality in the Lemma’s definition:

$$\begin{aligned} n \le \left( c_1 + \log _2\sqrt{c_2 + k} - \frac{k}{(c_2 + k)\cdot \ln 4} \right) \cdot m + \frac{1}{(c_2 + k)\cdot \ln 4} \cdot \log _2(\Delta ). \end{aligned}$$

Finally, we note that

$$\begin{aligned} \frac{k}{(c_2 + k)\cdot \ln 4} \ge \frac{1}{\ln 4}. \end{aligned}$$

\(\square \)

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Gribanov, D., Shumilov, I., Malyshev, D. et al. On \(\Delta \)-modular integer linear problems in the canonical form and equivalent problems. J Glob Optim (2022). https://doi.org/10.1007/s10898-022-01165-9

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Keywords

  • Integer linear programming
  • Knapsack problem
  • Subset-sum problem
  • Group minimization problem
  • Sparsity & proximity bounds
  • Empty simplex