0
$\begingroup$

I'm studying computer programming. Our lecturer in one of the courses gave us the following assignment and told us to learn the material needed to solve it. I'm not a mathematician and would like to minimize effort. How can I solve this problem? Or, rather, what are the steps?

Given a nonnegative $c$, a vector of $1$'s $b$, I need to minimize $c^Tx$ under the following constraints:

  1. $Ax\geq b$
  2. $x_j\in \left\{0,1 \right\}$

Here $A$ is a matrix of $0$s and $1$s such that each row has at most $k$ sequences of $1$s, with $k$ given in advance.

Find a $k$ approximate.

  • 5
    I'm going to minimize my effort and suggest you get a textbook. – 2017-02-16
  • 1
    There is not single procedure. There exist different methods for solving a linear programming problem, and the selection and customization of a method for a particular problem is rather an art than a science. Furthermore, your restriction $x \in \{0, 1\}$ makes the problem a 0-1 Linear Program (a special case of Integer programming), and the parameter $k$ does not appear anywhere in your problem statement. This problem is in the field of Combinatorial Optimization, which is a difficult area of research in and of itself. See the book by Papadimitriou and Steiglitz for known methods. – 2017-02-16
  • 0
    As a broad sketch, one puts a linear program in standard form (if not already so) by certain steps, creating a *tableau* similar to how a linear system of equations is represented by an augmented matrix. Then further steps are made (similar to elementary row operations) that drive the tableau to a state where the solution can be read off by inspection (much as the reduced row echelon form of an augmented matrix tells us the solutions of the original linear system). – 2017-02-16
  • 1
    @avs there might be a polynomial approximation scheme for this particular case. – 2017-02-16
  • 1
    Is there something omitted in the last sentence? "Find a $k$ approximate" doesn't sound to me like an actionable request. – 2017-02-16
  • 0
    @Vincent, I'd be interested to learn more about that. – 2017-02-16
  • 0
    @avs if I understand the question right, this is not general 0-1 programming. If $k=1$ then $A $ verifies the consecutive ones property and is totally unimodular – 2017-02-16
  • 0
    @Vincent, I think the question still contains a typo: "at most $k$ sequences" should be "at most $k$ instances". If so, you are right about this being a special case. – 2017-02-16
  • 0
    @avs sequences looks alright to me. – 2017-02-16

1 Answers 1

0

First, calculate a solution $x$ of the relaxed problem ($x_i \in [0,1]$) of cost $c^*$ thanks to the simplex method. This solution has a cost smaller than the cost of the $\{0,1\}$ optimal solution.

Initiate $x'=0$

Then, for each row, consider the sequence of ones $S_j$ such that $\sum_{i \in S} x_i $ is maximal. Its value is at least $\frac {1}{k}$ (according to the pigeon hole principle).

If $\exists i \in S, x_i=1$, set $x'_i$ to $1$. Otherwise, round up the $x'_i $ with the smallest cost function to $1$.

By construction, you have $cx'\leq k c^*$

You can also think about a more elaborate version of this. For instance, after each time you round up an $x_i$, you can calculate a new relaxed solution with this particular $x_i $ fixed to 1.

Following comment : since $\{x \in \{0,1\}^n|Ax\leq b \} \subset \{x \in [0,1]^n|Ax\leq b \}$, we have

$\min \{cx| x \in \{0,1\}^n,Ax\leq b \} \geq \min \{cx | x \in [0,1]^n,Ax\leq b \}$

  • 0
    Thank you very very much! I'm having trouble following your solution and would very much appreciate additional details. I will try to fill in the gaps in the upcoming few hours, but if you have time later - much appreciated. – 2017-02-16
  • 0
    You minimise something on a larger set. This is why the solution of the relaxed problem is always lower than the solution of the integer problem. – 2017-02-17
  • 0
    Its look fantastic,but yet i think i miss something. Just for better understanding cx' is the approximation returned value. But why can we assume that OPT≤cx' ? while OPT is the cost of the orig problem(with the binary constraints ). we started with smaller cost function(c*) by the solution of the relaxed problem so its seems crucial to show that OPT≤cx' . I understood that by construction, i have cx′≤kc*≤kOPT. Vincent thanks :) its really helpful ! – 2017-02-17