Massachusetts Institute of Technology
Michel X. Goemans
18.433: Combinatorial Optimization
Spring 2015
Solutions to problem set 3
3-8 We know that every extreme point of a polytope is the unique solution for a system
of equations obtained by selecting some of the inequalities defining the polytope and
setting them as equalities. In particular, if x∗ is a vertex of Q, then there exists a
subset I of {1, . . . , m}, where m is the number of rows of A, such that x∗ is the unique
solution of the system
(Ax)i = bi , for all i ∈ I
Cx = d.
But this implies that x∗ is the unique solution for a system of equations obtained by
selecting some of the inequalities that define P and setting them as equalities. Since
x∗ ∈ P , it follows that x∗ is an extreme point of P . Therefore, all the vertices of Q are
vertices of P .
3-9 First consider the situation in which M1 and M2 are such that M1 ∆M2 have more than
one connected component. Consider one of these connected components, say S ⊆ V ,
and partition M1 and M2 into M1 = M1s ∪ M1t and M2 = M2s ∪ M2t where M1s and
M2s correspond to the edges within S. By definition M1s ∪ M2s 6= ∅. Now define two
other matchings by M3 = M1s ∪ M2t and M4 = M2s ∪ M1t . Observe that
χ(M1 ) + χ(M2 ) = χ(M3 ) + χ(M4 )
which implies that any face that contains M1 and M2 will also contain M3 and M4 ,
and thus cannot be an edge.
Conversely, suppose that M1 ∆M2 has only one connected component, and say that
this component has k1 edges from M1 and k2 edges from M2 . We must have that
|k1 − k2 | ≤ 1. Now consider the following cost function:

1
e ∈ M1 ∩ M2



−1 e ∈
/ (M1 ∪ M2 )
ce =
k2 e ∈ M1 \ M2



k1 e ∈ M2 \ M1 .
Notice that c(M1 ) = c(M2 ) = b where b := |M1 ∩ M2 | + 2k1 k2 and for any other
matching M we have that c(M ) < b. Thus the valid inequality cT x ≤ b induces a
face with only the incidence vectors of M1 and M2 has vertices. Thus the line segment
between M1 and M2 defines an edge.
3-11 Consider the family of permutations {σ1 , . . . , σn } given by:


σi (1) = i,
σi : σi (i) = 1,


σi (j) = j, j ∈
/ {1, i}.
Solutions to problem set 3
Spring 2015
2
In other words, σ1 is the identity permutation and σi is the permutation that transpose
1 and i. Let us prove that these n permutations are affinely independent. For that
suppose that λ1 , . . . , λn are real numbers such that:
n
X
n
X
λi σi = 0,
i=1
λi = 0.
i=1
Then for every j 6= 1:
0=
n
X
!
λi σi (j) =
X
i=1
λi · j
+ λj · 1
i6=j
=
n
X
!
λi · j
− λj · j + λj · 1
i=1
=
X
n
λi
· j + λj · (1 − j).
i=1
| {z }
=0
Hence,
for every j 6= 1, λj (1 − j) = 0, which implies that λj = 0. Using that
Pn
i=1 λi = 0 we deduce that λ1 is also 0. Therefore, the family {σ1 , . . . , σn } is a set of
n affinely independent permutations.
P
P
3-13 (a) Fix a ∈ {−1, 0, 1}n satisfying ni=1 ai = 1 and 0 ≤ ji=1 ai ≤ 1 for each j =
1, . . . , n − 1. It is enough to show that
−1 ≤
n
X
ai (ek )i ≤ 1
i=1
for each k = 0, . . . , n − 1 (it is symmetric for −ek ’s).
Note that (ek )i = 1 if i ≤ k and (ek )i = −1 if i > k. We have
n
X
i=1
Since
Pk
i=1
ai (ek )i =
k
X
i=1
ai −
n
X
ai = 2
i=k+1
k
X
ai − 1.
i=1
ai is 0 or 1, it is between −1 and 1.
P
(b) Fix a ∈ {−1, 0, 1}n as in the previous part. Let bj = ji=1 ai for j = 1, . . . , n.
Then, bj ∈ {0, 1} for any j = 1, . . . , n − 1 and bn = 1 by definition of a. On the
other hand, if we are given b ∈ {0, 1}n with bn = 1, we can find the corresponding
a ∈ {−1, 0, 1}n by letting a1 = b1 and ai = bi − bi−1 for i = 2, . . . , n. This is a
bijection between a’s and b’s. Hence, there are 2n−1 such a’s and 2n inequalities.
Solutions to problem set 3
Spring 2015
3
P
(c) First note that aT ek is either −1 or 1, since aT ek = 2 ki=1 ai − 1. Let b as defined
in (b). Then, aT ek = 1 if and only if bk = 1 (we say b0 = 0). Thus,
{x ∈ S | aT x = 1} = {ek | bk = 1} ∪ {−ek | bk = 0}
{x ∈ S | aT x = −1} = {ek | bk = 0} ∪ {−ek | bk = 1}.
So each inequality defines distinct hyperplanes, because they contain different set
of extreme points. Moreover, if we choose exactly one vector from each {ek , −ek },
then they are affinely independent. For, note that it is enough to show that
{e0 , . . . , en−1 } are linearly independent, and they are indeed linearly independent
since { 12 (e1 − e0 ), 12 (e2 − e1 ), . . . , 21 (en−1 − en−2 ), − 21 (en−1 + e0 )} is the standard
basis of Rn .
(d) Note that 0 is in the interior of conv(S). Hence, no facet can contain {ek , −ek }
for any k = 0, . . . , n − 1 (otherwise it will contain 0). Since conv(S) is fulldimensional, any facet should contain at least n extreme points, i.e., it contains
exactly one from each {ek , −ek }. So there are at most 2n facets of conv(S). On
the other hand, we showed in (c) that each of 2n inequalities defines distinct facet.
Hence they define conv(S).