Chapter 6. Integrality of Polyhedra
Background
 Def:
Polyhedron 𝑃 = {𝑥 ∈ 𝑅𝑛: 𝐴𝑥 ≤ 𝑏}, .i.e. a set which can be described as the
solution set of finite number of linear inequalities. Note that there can be
many descriptions of the same polyhedron.
Polytope is a bounded polyhedron.
An inequality 𝑤𝑇𝑥 ≤ 𝑡 is valid for a polyhedron 𝑃 if 𝑃 ⊆ {𝑥: 𝑤𝑇𝑥 ≤ 𝑡}
Hyperplane: {𝑥𝑅𝑛 ∶ 𝑤𝑇𝑥 = 𝑡}
Supporting hyperplane: 𝑤𝑇𝑥 ≤ 𝑡 is valid for 𝑃 and 𝑃 ∩ {𝑥: 𝑤𝑇𝑥 = 𝑡} ≠ ∅
Face of 𝑃: 𝑃 ∩ {𝑥: 𝑤𝑇𝑥 = 𝑡} for a valid inequality 𝑤𝑇𝑥 ≤ 𝑡
Proper face: face except ∅ and 𝑃 itself
 Prop 6.6: A nonempty set 𝐹 ⊆ 𝑃 = {𝑥: 𝐴𝑥 ≤ 𝑏} is a face of 𝑃 if and only if
for some subsystem 𝐴0𝑥 ≤ 𝑏0 of 𝐴𝑥 ≤ 𝑏 we have 𝐹 = {𝑥 ∈ 𝑃: 𝐴0𝑥 = 𝑏0}
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 Prop 6.7: Let 𝐹 be a minimal (by inclusion) nonempty face of 𝑃 = {𝑥: 𝐴𝑥 ≤
𝑏}. Then 𝐹 = {𝑥 ∶ 𝐴0𝑥 = 𝑏0} for some subsystem 𝐴0𝑥 ≤ 𝑏0 of 𝐴𝑥 ≤ 𝑏.
Moreover, the rank of the matrix 𝐴0 is equal to the rank of 𝐴.
 Def:
A vector 𝑣 ∈ 𝑃 is called a vertex if {𝑣} is a face of 𝑃.
Polyhedron 𝑃 is pointed if it has at least one vertex.
 Prop 6.7 implies that a vertex 𝑣 is the unique solution for some 𝐴0𝑥 = 𝑏0.
Furthermore, since the rank of 𝐴0 is equal to the rank of 𝐴 for any minimal
face, 𝑃 has a vertex if and only if 𝐴 has full column rank.
 Prop 6.8: If a polyhedron 𝑃 is pointed then every minimal nonempty face of
𝑃 is a vertex.
 A maximal proper face of a polyhedron is called a facet.
 Prop 6.16: Let 𝐹 be a nonempty proper face of a polyhedron 𝑃. Then 𝐹 is a
facet if and only if dim(𝐹) = dim(𝑃) - 1.
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6.4 Integral Polytopes
 Assume rational polyhedron.
A rational polyhedron is called integral if every nonempty face contains an
integral vector.
A pointed rational polyhedron is integral if and only if all its vertices are
integral.
 Thm 6.22:
Suppose 𝑃 is nonempty and pointed. Consider max {𝑤𝑇𝑥: 𝑥 ∈ 𝑃}. Then the
following statements are equivalent.
1. 𝑃 is integral
2. LP over 𝑃 has an integral optimal solution for all 𝑤 ∈ 𝑅𝑛 for which it has
an optimal solution.
3. LP over 𝑃 has an integral optimal solution for all 𝑤 ∈ 𝑍𝑛 for which it has
an optimal solution.
4. Optimal value of LP over 𝑃 is integral for all 𝑤 ∈ 𝑍𝑛 for which it has an
optimal solution.
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(Pf) 1 → 2 If LP has an optimal solution, it has an optimal solution at a
vertex of P.
2 → 3, 3 → 4 are obvious.
4 → 1:
Let 𝑣 = (𝑣1, … , 𝑣𝑛)𝑇 be a vertex of 𝑃 and let 𝑤 ∈ 𝑍𝑛 be such that 𝑣 is the
unique optimal solution to max {𝑤𝑇𝑥: 𝑥 ∈ 𝑃}. (Existence of such 𝑤 can be
shown by taking linear combination of constraints which are active at 𝑣)
We may assume that 𝑤𝑇𝑣 > 𝑤𝑇𝑢 + 𝑢1 − 𝑣1 for all vertices 𝑢 of 𝑃 other than
𝑣. (If necessary, multiply 𝑤 by a large positive number.)
Let 𝑤 = (𝑤1 + 1, 𝑤2, … , 𝑤𝑛)𝑇. Then 𝑣 is still an optimal solution to max
{𝑤 𝑇 𝑥: 𝑥 ∈ 𝑃}. So 𝑤 𝑇 𝑣 = 𝑤𝑇𝑣 + 𝑣1. By assumption, 𝑤𝑇𝑣 and 𝑤 𝑇 𝑣 are
intergers. Thus 𝑣1 is integer. Repeat this for each component of 𝑣. 
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6.6 Total Dual Integrality
 Consider strong duality equation for primal-dual LP
max {𝑤𝑇𝑥 ∶ 𝐴𝑥 ≤ 𝑏} = min {𝑦𝑇𝑏 ∶ 𝑦𝑇𝐴 = 𝑤𝑇, 𝑦 ≥ 0}
A rational linear system 𝐴𝑥 ≤ 𝑏 is called totally dual integral if the
minimum of dual LP can be achieved by an integral vector 𝑦 for each
integral 𝑤 for which the optima exist.
 Thm 6.29: Let 𝐴𝑥 ≤ 𝑏 be a totally dual integral system such that 𝑃 =
{𝑥: 𝐴𝑥 ≤ 𝑏} is a rational polytope and 𝑏 is integral. Then 𝑃 is an integral
polytope.
(Pf) Since 𝑏 is integral, the optimal dual value is integral. The strong
duality equation implies that max {𝑤𝑇𝑥: 𝐴𝑥 ≤ 𝑏} is integer for all integral
vectors 𝑤. By Theorem 6.22, 𝑃 is integral.

Note that any rational system 𝐴𝑥 ≤ 𝑏 can be made totally dual integral by
changing it to (1/𝑡)𝐴𝑥 ≤ (1/𝑡)𝑏 for a positive integer 𝑡. But we need
integral 𝐴 as well as integral 𝑏.
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 Thm 6.30: Let 𝑃 be a rational polyhedron. Then there exists a totally dual
integral system 𝐴𝑥 ≤ 𝑏, with 𝐴 integral, such that 𝑃 = {𝑥: 𝐴𝑥 ≤ 𝑏}.
Furthermore, if 𝑃 is an integral polyhedron, then 𝑏 can be chosen to be
integral.
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6.8 Separation and Optimization
Separation Problem: Given a bounded rational polyhedron 𝑃 ⊆ 𝑅𝑛 and a
rational vector 𝑣 ∈ 𝑅𝑛, either conclude that 𝑣 belongs to 𝑃 or, if not, find a
rational vector 𝑤 ∈ 𝑅𝑛 such that 𝑤𝑇𝑥 < 𝑤𝑇𝑣 for all 𝑥 ∈ 𝑃.
Optimization Problem: Given a bounded rational polyhedron 𝑃 ⊆ 𝑅𝑛 and
a rational (objective) vector 𝑤 ∈ 𝑅𝑛, either find 𝑥 ∗ ∈ 𝑃 that maximizes
𝑤𝑇𝑥 over all 𝑥 ∈ 𝑃, or conclude that 𝑃 is empty.
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 Thm 6.36: (Grötschel, Lovász, and Schrijver [1988])
For any proper class of polyhedra, the optimization problem is polynomially
solvable if and only if the separation problem is polynomially solvable.
(Pf) By ellipsoid method for LP and lots of technical elaboration.
 The theorem implies that the separation problem for the blossom inequalities
can be solved in polynomial time since the optimization problem over the
matching polyhedra can be solved in polynomial time. (Padberg and Rao
[1982] found such algorithm)
Conversely, the existence of the polynomial time separation algorithm
implies that the matching optimization problem can be solved in polynomial
time.
The theorem can be used to prove the existence of polynomial time
algorithm if separation can be solved in polynomial time (e.g. submodular
function minimization problem)
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