5.2 From Cubes to Equalities
5.2.1 Finding Equalities
The first step in computing an equality basis for a polyhedron Ax≤ b is to detect whether the system contains any equalities. At the beginning of this section, we have already stated a criterion that detects this:
Lemma 5.2.1 (Cube-Equality). Let Ax≤ b be a polyhedron. Then exactly one of the following statements is true:
(1)Ax≤ b implies an equality hTx = g with h6= 0n, or (2)Ax≤ b contains a cube with edge length e > 0.
Proof. This proof is a case distinction over the sign of xe for the following slightly simplified version of the largest cube test (see Chapter 4.3.1 for the original definition):
maximize xe
subject to Ax + a0xe≤ b, where a0i= 12kaik1 . (5.1) If the maximum objective value is positive, Ax≤ b contains a cube with edge length e > 0. Therefore, we have to prove that Ax≤ b contains no equality hTx = g with h 6= 0n, which we will do by contradiction. Assume Ax ≤ b contains an equality hTx = g with h 6= 0n. Then, by transitivity of the subset relation, the polyhedron consisting of the inequalities hTx ≤ g and −hTx ≤ −g must also contain a cube of edge length e. However, applying the transformation from Corollary 4.2.2 to this new polyhedron results in two contradicting inequalities: hTx≤ g −khk
1·2e and−hTx≤ −g −khk1·e2. Thus, (1) and (2) cannot hold at the same time.
If the maximum objective value is zero, then Ax ≤ b is satisfiable but contains no cube with edge length e > 0. Therefore, we have to prove that Ax≤ b contains an equality hTx = g with h6= 0. Consider the dual linear program [129] of (5.1): minimize yTb subject to yTA = (0n)T , yTa0= 1 , where a0 i = 12kaik1 , y ≥ 0 . (5.2) Due to strong duality, the objectives of the dual and primal linear programs are equal [129]. Therefore, there exists a y∈ Qm that has objective yTb = 0 and that satisfies the dual (5.2). Since yTa0 = 1 and a0
i ≥ 0 and yi≥ 0 holds, there exists a k∈ {1, . . . , m} such that yk> 0. By multiplying yTA = (0n)T with an x∈ Qδ(Ax≤ b) and isolating aTkx, we get:
aT kx =− Pm i=1,i6=k yi yka T ix .
Using yi ≥ 0, and our original inequalities aTi x ≤ bi, we get a finite lower bound for aT kx: aT kx =− Pm i=1,i6=k yi yka T i x ≥ −Pm i=1,i6=k yi ykbi . Now, we reformulate yTb = 0 analogously and get: b
k=−Pmi=1,i6=k yi ykbi . Thus, aT
kx = bk is an equality contained in the original inequalities Ax≤ b. If the maximum objective value is negative, Ax≤ b is unsatisfiable and contains no cube with edge length e > 0. Since Qδ(Ax≤ b) is now empty, Ax≤ b contains all equalities.
A cube with positive edge length is enough to prove that there exists no implied equality. The actual edge length e of this cube is not relevant. Therefore, we can assume that the edge length e is arbitrarily small. We can even assume that our edge length is so small that we can ignore the different multipleskaik1 and any infinitesimals introduced by strict inequalities. We just have to turn all of our inequalities into strict inequalities.
Lemma 5.2.2 (Strict-Cube). Let Ax ≤ b be a polyhedron, where ai 6= 0n, bi= (pi, qi), qi ≤ 0, and bδi = (pi,−1) be the strict versions of the bounds bi for alli∈ {1, . . . , m}. Then the following statements are equivalent:
(1)Ax≤ b contains a cube with edge length e > 0, and (2)Ax≤ bδ is satisfiable.
Proof. (1) ⇒ (2): If Ax ≤ b contains a cube of edge length e > 0, then Ax≤ b − a0 is satisfiable, where a0i = e
2kaik1. By Lemma 2.3.1, we know that there exists a δ∈ Q such that Ax ≤ p + qδ − a0. Now, let
δ0 = min{a0
i− qiδ : i = 1, . . . , m} .
Since a0i− qiδ ≥ δ0, it holds that Ax≤ p − δ01m. Since a0i =kaik1 > 0 and qi ≤ 0, it also holds that δ0 > 0. By Lemma 2.3.1, we deduce that Ax < p and, therefore, Ax≤ bδ holds.
(2)⇒ (1): If Ax ≤ bδ is satisfiable, then we know by Lemma 2.3.1 that there must exist a δ > 0 such that Ax≤ p − δ1m holds. Let
amax= max{kaik1 : i = 1, . . . , m} , δ0 = δ
2, and e = amaxδ . Then pi− δ = pi− δ
0− e
2amax ≤ bi−e2kaik1. Thus, Ax≤ b contains a cube with edge length e > 0.
In case Ax ≤ bδ is unsatisfiable, Ax ≤ b contains no cube with positive edge length and, therefore by Lemma 5.2.1, an equality. In case Ax≤ bδ is unsatisfiable, the algorithm returns an explanation, i.e., a minimal set C of unsatisfiable constraints aT
i x≤ bδi from Ax≤ bδ (see also Chapter 2.5.1). If Ax≤ b itself is satisfiable, we can extract equalities from this explanation:
Lemma 5.2.3 (Equality Explanation). Let Ax ≤ b be a satisfiable poly- hedron, where ai 6= 0n, bi = (pi, qi), qi ≤ 0, and bδi = (pi,−1) for all i ∈ {1, . . . , m}. Let Ax ≤ bδ be unsatisfiable. Let C be a minimal set of unsatisfiable constraints aT
ix ≤ bδi from Ax ≤ bδ. Then it holds for every aT
ix≤ bδi ∈ C that aTi x = bi is an equality implied by Ax≤ b.
Proof. Because a set of linear inequalities inherits all implied (in)equalities of a subset of its inequalities, we can assume that Ax ≤ b and Ax ≤ bδ contain only the inequalities associated with the explanation C. Therefore, C = {aT
1x ≤ bδ1, . . . , aTmx ≤ bδm}. By Lemma 2.5.4 and Ax ≤ bδ being unsatisfiable, we know that there exists a y∈ Qm with y≥ 0, yTA = (0n)T, and yTbδ < 0. By Lemma 2.5.4 and Ax ≤ b being satisfiable, we know that yTb≥ 0 is also true. By Lemma 2.5.6, we know that y
k > 0 for every k∈ {1, . . . , m}.
Now, we use yTbδ< 0, yTb≥ 0, and the definitions of < and ≤ for Q δto prove that yTb = 0 and b = p. Since yTbδ < 0, we get that yTp≤ 0. Since yTb≥ 0, we get that yTp≥ 0. If we combine yTp ≤ 0 and yTp≥ 0, we get that yTp = 0. From yTp = 0 and yTb≥ 0, we get yTq ≥ 0. Since y > 0 and qi ≤ 0, we get that yTq = 0 and qi = 0. Since qi = 0, b = p.
Next, we multiply yTA = (0n)T with an x∈ Q
δ(Ax≤ b) to get yTAx = 0. Since yk > 0 for every k ∈ {1, . . . , m}, we can solve yTAx = 0 for every aT kx and get: aT kx =− Pm i=1,i6=k yi yka T i x . Likewise, we solve yTb = 0 for every b
k to get: bk =− Pm i=1,i6=k yi ykbi . Since x∈ Qδ(Ax≤ b) satisfies all aTi x ≤ bi, we can deduce bk as the lower bound of aT kx: aT kx =− Pm i=1,i6=k yi yka T i x ≥ −Pm i=1,i6=k yi ykbi = bk, which proves that Ax≤ b implies aT
kx = bk.
Lemma 5.2.3 justifies simplifications on Ax ≤ bδ. We can eliminate all inequalities in Ax≤ bδ that cannot appear in the explanation of unsatisfia- bility, i.e., all inequalities aT
i x ≤ bδi that cannot form an equality aTi x = bi that is implied by Ax≤ b. For example, if we have an assignment v ∈ Qn δ such that Av≤ b is true, then we can eliminate every inequality aT
i x≤ bδi for which aT
i v = bi is false. According to this argument, we can also eliminate all inequalities aT
i x≤ bδi that were already strict inequalities in Ax≤ b.