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Predecessor versions of this paper have been around for more than 15 years. I remember that there were counter-examples to the first versions, then first revisions, counter-examples to the first revisions, second revisions, new counter-examples, further revisions, further counter-examples, and so on.
It would be much better, if the authors were able to ...
11
This problem is called bin covering. It's NP-hard and hard to approximate to better than a factor of two by an easy reduction from subset sum but has an asymptotic approximation scheme (i.e. one that fills $(1-\epsilon)\mathrm{OPT}-O(1)$ bins). See e.g. "Better approximation algorithms for bin covering", Csirik, Johnson, and Kenyon, SODA 2001, or &...
11
OptP-complete. Krentel showed that MAX-SAT, finding the lexicographically maximum satisfying assignment, is OptP-complete and the reduction above reduces Max-SAT to ILP. ILP sits in OptP pretty much by definition.
Note that you need n calls to an NP-oracle to solve ILP via binary search, O(log n) isn't sufficient.
There really isn't much of a connection ...
9
Here are a few examples:
(Most Common) Cutting Stock Problem - determine patterns for which to cut boards in order to meet demand.
(Kinda same problem) - Knapsack with general integer variables - used for column generation in the cutting stock problem
Problems with modular constraints: $x \equiv 3 \pmod 5$ can be written as $x + 5y = 3$, $y \in \mathbb{Z}$....
8
If you look at the original paper by Lenstra, Lenstra, and Lovasz, you will see the following applications:
factoring univariate polynomials over the rationals (the motivation for developing LLL basis reduction)
efficient version of Dirichlet's classical diophantine approximation theorem: for rationals $a_1, \ldots, a_n$ and $\epsilon$, find in polynomial ...
7
An instance of CNF-SAT with $k$ variables can easily be written as a 0/1 integer linear program over the same variable set, since a clause such as $x_1 \vee x_3 \vee \neg x_4 \vee \neg x_6$ naturally corresponds to a constraint $x_1 + x_3 + (1-x_4) + (1-x_6) \geq 1$, when all variables are forced to take values $0$ and $1$.
Hence if integer programming in $...
7
NOTE: My original reduction didn't work. Fixed now.
Can't subset-sum be reduced to this problem fairly easily? Suppose all the profits $v_i$ are the same, say $v_i = 1$, and all the profits $p_i$ are proportional to the weights, so
$p_i = \beta w_i$ with $\beta < 1$.
Now, if there's a set of weights $S$ such that $\sum_{i\in S} w_i = W$, you can get a ...
6
I think $S_n$ can be written in terms of inequalities in the obvious way. Let
$$
Q_n = \{(x, y): x = \sum_{i = 0}^{n-1}{2^i y_i}, \forall i: 0 \leq y_i \leq 1\}.
$$
I claim that $Q_n = S_n$. First, obviously all $(x, y) \in S_n$ are also in $Q_n$, so $S_n \subseteq Q_n$. Second, fix a point $(x^*, y^*) \in Q_n$. Consider the probability distribution over $\{...
6
You can find an answer in the following paper:
Joachim von zur Gathen, Malte Sieveking. A bound on solutions of integer linear equalities and inequalities.
Proc. AMS 72(1) (1978)
(pdf)
A simple corollary of their result follows.
Let's say we have two integer systems, $A x = b$ and $C x \ge d$,
where you want $x$ to be in $\mathbb Z^n$.
Consider ...
6
Here's a proof (sketch) that doesn't explicitly use duality. More precisely, it replaces duality by a seemingly weaker (and hopefully easily believable) geometric fact, in Step 3 below.
EDIT: But, per the comment, the proof applies only to polytopes, not (unbounded) polyhedra!
Let $P\subset \mathbb R^n$ be any polyhedron polytope such that, for all $c\in \...
5
The original algorithm of Lenstra (from 1983) has not been implemented AFAIK. Certainly, no open-source code is known to be available.
Lovasz and Scarf proposed (in 1992) a generalized basis reduction algo that also solves IP in fixed dimensions, but avoids the ellipsoidal approximations required by Lenstra's algorithm. An implementation of this algo was ...
5
You're confusing decision problems (in the classical sense) with parameterized decision problems. Classical decision problems are subsets of $\Sigma^*$, whereas parameterized decision problems are often considered as subsets of $\Sigma^* \times \mathbb{N}$ (see [1, Chapter 2]). Classical complexity classes (such as NC, P and NP) are sets of classical ...
5
If I understood it well, (1) is also NP-complete, a possible reduction is from SUBSET SUM:
Given a set of $m$ positive integers $A = \{a_1, ..., a_m\}$, and a positive integer $B$, is there a subset of $A' \subseteq A$ a such that $\sum_{a_i \in A'} a_i$
You simply pick $n = (m+1)$ and build your set in this way:
add $ a_1, a_2, , ..., a_m$
add $m$ zeros
...
5
It's NP-hard. Given an integer programming problem $P$, add an irrelevant variable $z$ with no constraints; call the resulting problem $P'$. Now if $P$ has no solutions, then $P'$ has no solutions; if $P$ has a solution, then $P'$ has $> 1$ solutions. Consequently distinguishing between $\le 1$ solution vs $> 1 $ solution is at least as hard as ...
5
(1) As finding a second satisfying assignment to a 3SAT formula is still $\mathsf{FNP}$-complete (indeed, it is $\mathsf{ASP}$-complete, see Theorem 3.5 of [1]), and we can encode 3SAT as an integer program by a parsimonious reduction, finding a second integer point in an integer program is also $\mathsf{NP}$-hard.
(2) Barvinok [2] showed that in fixed ...
4
Kolloiopoulos and Young give an $O(\log m)$ approximation for general covering integer programs. See the paper below.
http://www.sciencedirect.com/science/article/pii/S0022000005000656
4
It seems that we can reduce Subset Sum to your problem (2). Hence, your problem (2) is NP-complete.
Consider the following formulation of Subset Sum.
Instance: A multi-set consisting of $n$ integers.
Question: Does there exist a subset of size at least $1$ that sums to zero?
Now, we reduce Subset Sum to your problem (2). Let a multi-set $X$ consisting ...
4
There's a straightforward way to construct a function $f_z:\{0,1\}^n \to \mathbb{R}$ that is zero at only a single point $z=(z_1,\dots,z_n)$ and strictly positive everywhere else: namely,
$$f_z(x_1,\dots,x_n) = (x_1-z_1)^2 + (x_2-z_2)^2 + \dots + (x_n-z_n)^2.$$
Based on this, we can easily construct a function $g : \{0,1\}^n \to \mathbb{R}$ that is zero at ...
4
I believe that this problem is NP-hard, here is a sketch proof (don't hesitate to ask for more details if needed).
The idea is based on a reduction from the Not-all-equal 3-SAT. For $\varphi$ a 3-SAT formula with $m$ clauses, we will build $G(\varphi)$ a graph that has minimum objective value of $2\times m$ iif $\varphi$ has a solution. Before diving into ...
4
We answer OP's last question: can an approximate solution to IQP be obtained by randomized rounding?
We show that the natural randomized-rounding scheme gives a 2-approximation, and a $(1+1/\overline d)$-approximation in graphs with average degree $\overline d$. (Note that any connected graph has average degree at least $2-2/|V|$.)
For the record, here's the ...
3
EDIT (Jan 2019): Lemma 2 as currently stated below is wrong. (Indeed, given any instance, adding a single edge with a single type of very large cost will not change the instance but will yield $N(I)=1$, in which case the lemma claims that the algorithm will give an optimal solution. The proof is wrong because the function $\Phi$ defined there is not ...
3
Via Rubin:
There is a sequence of polytopes $\{ P_k\}_k$ each defined by two variables and one constraint (plus two nonnegativity constraints) such that the integer hull of $P_k$ has $k+3$ vertices and $k+3$ facets.
$P_k=\{ (x,y)\in \mathbb{R}^2_+ ~|~f_{2k} x + f_{2k+1}y \le f^2_{2k+1}-1\}$
Here, $f_j$ is the $j$-th Fibonacci number.
This shows that the ...
3
Definition: Given an undirected graph $G$ and an edge orientation $\vec{G}$, an unstable path is a directed path that goes from a node $s$ to a node $t$, such that the out-degree of node $s$ is strictly larger than the out-degree of node $t$ plus one. We observe that flipping the orientation of all edges along that path decreases the cost.
Unstable path ...
3
This problem can be solved with dynamic programming in pseudo-polynomial time (proof below). Therefore, it is not possible to show that this problem is strongly NP-hard (unless P=NP).
First, let's restate the problem:
Given: values $N$ and $T$ and positive integer intervals $R_1$, $R_2$, $\ldots$, and $R_n$
Output: the largest possible value of $\sum_{i=1}^...
2
An integer program with only equalities can be solved by linear program.
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The complexity of Lenstra's algorithm for mixed-integer programming in his paper runs as $2^{O(n^3)}*poly(d, \phi)$ where there are $n$ integer variables, $d$ continuos variables, and $\phi$ is the binary encoding size of the problem. This is not stated very explicitly in his paper, but the main thing to notice is that the complexity is primarily guided ...
2
Have a look at the paper "Using a Mixed Integer Quadratic Programming Solver for the Unconstrained Quadratic 0-1 Problem" by Alain Billionnet and Sourour Elloumi. They mention there that the binary quadratic problem
$$\min \left\{ x^T Q x + c^Tx : x \in \{0,1\}^n \right\},$$
where $Q$ is a symmetric matrix with diagonal entries all zero, is equivalent to ...
2
It looks to me like this is a special case of minimum cost flow; introduce one vertex per row and one per column, with an edge for each entry whose cost is the negative of the value of that entry and whose capacity is 1. Then add an edge of capacity $b$ from the source to each row, with cost 0, and similarly for the columns, and solve the resulting minimum ...
2
Base on Komus's constraint, we add another constraint which ensures a Steiner Tree on $G^{'}=(V, E^{'})$, where $E'=\{(i,j): i,j \in V\}$:
$$\sum_{e \in cut(U,V)}x_e \ge 1, \forall u,v \in T, \forall \text{ }u-v\text{ }cut \text{ }(U,V)$$where $cut(U,V)$ denotes the cut set of $(U,V)$.
Together with Komus's constraint, our model is obtained as follows:
$...
2
Given an integral LP, you can use a LP algorithm to compute in polynomial time
the optimal value of the LP,
a vertex of the feasible region (polytope) of the LP where the optimal value is attained.
Thus, if you know that the polytope is integral (i.e., all the vertices are integral), then you can indeed compute in polynomial time an integral maximizer. In ...
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