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Proof that Valid Cuts Do Not Remove Any Integer Feasible Solutions

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The Formal Theorem

Let P P be a polyhedron defined by P={x∈Rn∣Ax≤b} P = \{ x \in \mathbb{R}^n \mid Ax \le b \} , where A∈Rm×n A \in \mathbb{R}^{m \times n} and b∈Rm b \in \mathbb{R}^m . Let PIP=conv(P∩Zn) P_{IP} = \text{conv}(P \cap \mathbb{Z}^n) be the convex hull of the integer points in P P . Consider a valid cutting plane πx≤π0 \pi x \le \pi_0 such that πx≤π0 \pi x \le \pi_0 is a valid inequality for PIP P_{IP} . Let P′ P' be the polyhedron obtained by adding this inequality to the definition of P P , i.e., P′={x∈Rn∣Ax≤b,πx≤π0} P' = \{ x \in \mathbb{R}^n \mid Ax \le b, \pi x \le \pi_0 \} . Then, PIP′=conv(P′∩Zn) P'_{IP} = \text{conv}(P' \cap \mathbb{Z}^n) satisfies PIP′=PIP P'_{IP} = P_{IP} . Specifically, if x∗∈P∩Zn x^* \in P \cap \mathbb{Z}^n , then x∗∈P′ x^* \in P' .

Analytical Intuition.

Imagine you're searching for a needle (an integer feasible solution) within a vast haystack (the original feasible region of a linear program). This haystack is defined by a set of fences (constraints). A 'valid cut' is like adding a new fence that doesn't exclude any of your precious needles, but it might narrow down the haystack for the non-integer points. Think of it as drawing a line on a map that goes through all the towns (integer points) you are interested in. This line doesn't cut off any towns, but it might cut off areas between towns that you don't care about anyway. The key is that this new fence is carefully chosen *after* you've identified all the possible town locations. Therefore, any town you were looking for before is still within the fences, just in a smaller, more focused region.
CAUTION

Institutional Warning.

It's easy to conflate adding a cut to the LP relaxation with its effect on the integer feasible set. The proof hinges on the fact that cuts are derived from properties of the *integer* feasible set, not the continuous relaxation.

Academic Inquiries.

01

What is a 'valid cut' in the context of Integer Programming?

A valid cut is an inequality πx≤π0 \pi x \le \pi_0 that is satisfied by all integer feasible solutions of an Integer Program, but may not be satisfied by some non-integer points. It is essentially a facet-defining inequality for the convex hull of the integer feasible solutions.

02

Why is it important that valid cuts do not remove any integer feasible solutions?

This property is fundamental to cutting-plane algorithms. These algorithms iteratively add valid cuts to the LP relaxation of the Integer Program. If a cut removed an integer feasible solution, the algorithm would never find the optimal integer solution.

03

Does this mean adding a valid cut never changes the optimal integer solution?

Not exactly. Adding a valid cut does not remove *any* integer feasible solutions from the feasible set. However, it can eliminate *non-integer* feasible solutions, which might tighten the LP relaxation and lead to finding the optimal integer solution more quickly or at a better objective value in the relaxed problem.

04

What is PIP P_{IP} and PIP′ P'_{IP} ?

PIP=conv(P∩Zn) P_{IP} = \text{conv}(P \cap \mathbb{Z}^n) is the convex hull of the integer points within the original polyhedron P P . PIP′=conv(P′∩Zn) P'_{IP} = \text{conv}(P' \cap \mathbb{Z}^n) is similarly defined for the new polyhedron P′ P' which includes the added cut. The theorem states that PIP′=PIP P'_{IP} = P_{IP} , meaning the convex hull of integer points remains the same after adding a valid cut.

Standardized References.

  • Definitive Institutional SourceConforti, Cornuéjols, and Zambelli, Integer Programming

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Institutional Citation

Reference this proof in your academic research or publications.

NICEFA Visual Mathematics. (2026). Proof that Valid Cuts Do Not Remove Any Integer Feasible Solutions: Visual Proof & Intuition. Retrieved from https://nicefa.org/library/linear-and-integer-programming/proof-that-valid-cuts-do-not-remove-any-integer-feasible-solutions

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