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One way to show that checking the feasibility of a linear system of inequalities is as hard as linear programming is via the reduction given by the ellipsoid method. An even easier way is to guess the optimal solution and introduce it as a constraint via binary search.

Both of these reductions are polynomial, but not strongly polynomial (i.e they depend on the number of bits in the coefficients of the inequalities).

Is there a strongly polynomial reduction from LP optimization to LP feasibility ?

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    $\begingroup$ actually no. It is as you say. I realize that LP optimization solves LP feasibility. I'm asking for the opposite reduction. $\endgroup$ – Suresh Venkat Jan 24 '14 at 23:44
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    $\begingroup$ Well, the output for optimization can have as many bits as "the number of bits in the coefficients", while feasibility is yes/no. Thus, if by reduction you mean something "black-box"-ey then the answer must be negative. $\endgroup$ – Noam Jan 25 '14 at 6:53
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    $\begingroup$ But, if the feasibility check does not only give a yes/no answer as discussed by Noam above, but rather in the case of feasibility provides a feasible solution, then the answer is yes, by LP duality. $\endgroup$ – Kristoffer Arnsfelt Hansen Jan 25 '14 at 8:22
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    $\begingroup$ @SureshVenkat: Suppose the primal is a maximization program in variables $x$, with the dual then being a minimization program in variables $y$. Then form the system of inequalities in variables $x,y$, taking the constraints from both the primal and the dual, together with an inequality stating that the value of the primal solution is at least the value of the dual solution. The cases of the LP being infeasible and unbounded can also be dealt with. $\endgroup$ – Kristoffer Arnsfelt Hansen Jan 25 '14 at 18:17
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    $\begingroup$ What about polytopes/polyhedra defined by implicit constraints? $\endgroup$ – Chandra Chekuri Jan 26 '14 at 2:23
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The answer is yes, and in fact one can even reduce to the decision problem of linear inequalities feasibility!

We are as input given a LP instance P: $\max c^Tx\ \text{s.t.}\ Ax \leq b\ ;\ x\geq 0$.

We furthermore have access to an oracle that given a system of inequalities $S=\{Bz \leq d\}$ returns yes/no, whether the system is feasible.

The reduction now proceeds as follows:

  1. Test if $S_1=\{Ax \leq b\ ;\ x \geq 0\}$ is feasible. If not, we can report that P is INFEASIBLE.
  2. Form the dual program D: $\min b^Ty\ \text{s.t.}\ A^Ty \geq c\ ;\ y \geq 0$.
  3. Test if $S_2=\{Ax \leq b\ ;\ x\geq 0\ ;\ A^Ty\geq c\ ;\ y\geq0 \ ;\ b^Ty \leq c^Tx\}$ is feasible. If not, we can report that P is UNBOUNDED.
  4. Iterate over the inequalities of $S_1$ and try to add them one-by-one as equalities (i.e. add the reverse inequality) to the system $S_2$. If the system remains feasible we keep the constraint in $S_2$, and otherwise remove it again. Let $S_3$ be the system of constraints (linear equalities) that gets added in this way. The system $S_3$ will now completely determine an optimal basic solution to P.
  5. Using Gaussian Elimination on the system $S_3$ compute an optimal solution $x$ to P.
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  • $\begingroup$ The steps 4 and 5 are not needed. If $S_2$ is feasible, then we have obtained the optimal solution to $P$. $\endgroup$ – hengxin Feb 19 '17 at 7:39
  • $\begingroup$ @hengxin. It write in the first line of my answer that the answer is yes even when considering reducing to the decision problem. Below I obviously make that assumption, and hence steps 4 and 5 are required. $\endgroup$ – Kristoffer Arnsfelt Hansen Feb 20 '17 at 8:36

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