totally unimodular matrices - Devi Ahilya … unimodular...Basic examples of totally unimodular...
Transcript of totally unimodular matrices - Devi Ahilya … unimodular...Basic examples of totally unimodular...
1
Introduction
This dissertation is a reading of chapters 16 (Introduction to Integer Liner Programming)
and 19 (Totally Unimodular Matrices: fundamental properties and examples) in the book
: Theory of Linear and Integer Programming, Alexander Schrijver, John Wiley & Sons ©
1986.
The chapter one is a collection of basic definitions (polyhedron, polyhedral
cone, polytope etc.) and the statement of the decomposition theorem of polyhedra.
Chapter two is “Introduction to Integer Linear Programming”. A finite set
of vectors a1 ,… ,at is a Hilbert basis if each integral vector b in cone {a1 ,… ,at} is a non-
negative integral combination of a1 ,… ,at. We shall prove Hilbert basis theorem: Each
rational polyhedral cone C is generated by an integral Hilbert basis.
Further, an analogue of Caratheodory’s theorem is proved: If a system
a1x � β1, … , amx � β
m has no integral solution then there are 2
n or less constraints
among above inequalities which already have no integral solution.
Chapter three contains some basic result on totally unimodular matrices.
The main theorem is due to Hoffman and Kruskal: Let A be an integral matrix then A is
totally unimodular if and only if for each integral vector b the polyhedron �x| x � 0;Ax � b is integral.
Next, seven equivalent characterization of total unimodularity are proved.
These characterizations are due to Hoffman & Kruskal, Ghouila-Houri, Camion and
R.E.Gomory.
Basic examples of totally unimodular matrices are incidence matrices of
bipartite graphs & Directed graphs and Network matrices. We prove Konig-Egervary
theorem for bipartite graph.
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Chapter 1
Preliminaries
Definition 1.1: (Polyhedron) A polyhedron P �� is the set of points that satisfies
finite number of linear inequalities i.e., P = {x �� | Ax ≤ b} where (A, b) is an
m � (n + 1) matrix.
Definition 1.2: (Polyhedral Cone) A cone C is polyhedral if C = {x | Ax ≤ 0} for some
matrix A, i.e., C is the intersection of finitely many linear half spaces.
Definition 1.3: (Rational Polyhedral Cone) A cone C is rational polyhedral if
C = {x |Ax ≤ 0} for some rational matrix A.
Definition 1.4: (Characteristic Cone) The Characteristic cone of P, denoted by
Char.cone P, is the polyhedral cone
Char.cone P: = �y | x � y P for all x in P = {y | Ay ≤ 0} (1)
The non-zero vectors in char.cone P are called the infinite directions of P.
Definition 1.5: (Polytope) A bounded polyhedron is called a polytope.
Definition 1.6: (Pointed Cone) The linearity space of P is {y | Ay = 0} which is
Char.cone P � - Char.cone P.
Clearly it is a linear space, as the kernel of A. If the dimension of this space is zero then
P is called pointed.
Definition 1.7: (Characteristic Vector) Let S be a finite set. If T S, the characteristic
vector of T is the {0, 1} - vector in ��, denoted by z�, satisfies z��s! = 1 if s T
z��s! = 0 if s S\T
Theorem 1.8: (Farkas’-Minkowski-Weyl theorem) A convex cone is polyhedral if
and only if it is finitely generated.
Theorem 1.9: (Decomposition theorem for Polyhedra ): P is a polyhedron in �� if and
only if P = Q + C for some polytope Q and some polyhedral cone C.
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Theorem 1.10: (Farkas’ Lemma): Let A be a real m � n matrix and let c be a real
nonzero n- vector. Then either the primal system
Ax � 0 and c#x $ 0
has a solution for x �� or the dual system
A#y = c and y � 0
has a solution for, y �� but never both.
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Chapter 2
Integer Linear Programming
Definition 2.1: (Integer Linear Programming) Given rational matrix A, and rational
vectors b and c, determine,
max {cx | Ax ≤ b; x integral} (1)
Another definition is
Definition 2.2: Given rational matrix A, and rational vectors b and c,
determine,
max {cx | x ≥ 0 Ax = b; x integral} (2)
Remark 2.3: It is easy to see that we can obtain from one formulation to another one.
Note 2.4: We have the duality relation:
max {cx | Ax ≤ b ; x integral} � min {yb | y ≥ 0, yA = c; y integral} (3)
[Since, Ax ≤ b yAx ≤ yb cx ≤ yb]
We can have strict inequality. For example, Take A = �2!, b = �1!, c = �1!. Thus the primal problem is
max {x | 2x ≤ 1; x integral}
Clearly, the maximum is 0. [x {0,-1,-2,-------}]
The dual is
min {y | y ≥ 0, 2y = 1; y integral}
So the problem is infeasible. But the corresponding LP – optimal both are )* . Note 2.5: We may write an analogous statement for Farkas’ lemma.
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The rational system Ax = b has a nonnegative integral solution x if and only if yb is a
nonnegative integer whenever yA is a nonnegative integral vector.
But, the statement is not true. For Example, Take A =�2, 3!, b = �1!. The rational system
is
2x1 + 3x2 = 1.
Clearly, yA = �2y, 3y! ≥ 0 implies y ≥ 0. As 2x)+ 3x*= 1 has no non-negative integral
solution, converse is not true.
Note 2.6: We restrict to integer linear programming with rational input data. Otherwise,
there may not be an optimum to the given problem. For Example,
Sup ,ξ � η√2 | ξ � η√2 � )* , ξ, η, integer1 2 )*
But no ξ, η attain the supremum. (Since √2 is irrational)
Definition 2.7: The LP-relaxation of the integer linear programming problem in the
definition 2.1 is the following LP problem:
max {cx | Ax ≤ b} (5)
Clearly, LP-relaxation gives an upper bound for corresponding integer linear
programming.
Definition 2.8: (The Integer Hull of a Polyhedron) For any polyhedron P, the integer
hull PI of P is
PI = the convex hull of the integral vectors in P. (6)
Note 2.9: The ILP problem (1) is equivalent to determine,
max {cx | x PI} for P = {x | Ax ≤ b}
Remark 2.10: For any rational polyhedral cone C,
CI = C (7)
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(as C is generated by rational and hence by integral, vectors.)
Theorem 2.11: [Meyer] For any rational polyhedron P, the set PI is again a polyhedron.
If PI is nonempty then
Char. cone (P) = Char. cone (PI) Proof: Consider a rational polyhedron P with a decomposition
P = Q + C,
Where Q is a Polytope and C is the characteristic cone of P. As C = CI, let C be generated
by the integral vectors y),… , y� and let B be the polytope, defined by,
B = 5∑ 7898:8;) | 0 � µi � 1 for i 2 1,… , s< (8)
It is enough to show that
PI = (Q � B!I+ C
Note that (Q � B!I is a polytope as both Q & B are polytopes.
Observe,
(Q � B!I + C PI + C
= PI + CI ( Remark 2.10)
(P � C!I
= PI
i.e., (Q � B!I + C PI
Now to show reverse inclusion. Take p be any integral vector in P. i.e., p PI. Now,
p = q + c for some q Q and c C. We have,
C = ∑ µiyi ( µi� 0)
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= ∑@µiAy
i + ∑� µ
i - @µ
iA )y
i,
Denote first term by c′ and the second by b (= c –c′). Clearly c′ C � BC and b D.
Hence,
p = q + c′ + b = (q+b) + c′
q + b = p – c′
q + b is an integral vector as p and c′ are integral.
p (Q � B!I + C
i.e., PI = (Q � B!I + C. ∎
Remark 2.12: The above theorem implies that any integer linear programming problem
can be written as max {cx | x Q} for some polyhedron Q which is again a linear
programming problem.
This means we can represent PI by linear inequalities. But generally this is
a difficult task. The theorem (2.11) can be extended to: For each rational matrix A, there
exists a integral matrix M such that for each column vector b there exists a column vector
d such that
�x | Ax � b I = {x | Mx ≤ b} (9)
So the coefficients of the inequalities defining PI can be described by the
coefficients of the inequalities defining P.
Definition 2.13: (Integral Polyhedron) A rational polyhedron with property P = PI is
called an integral polyhedron.
Remark 2.14: It is easy to see that for a rational polyhedron P the following are
equivalent.
(i) P is integral i.e., P = PI i.e., P is the convex hull of the integral vectors in P.
(ii) Each face of P contains integral vectors.
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(iii) Each minimal face of P contains integral vectors.
(iv) max {cx | x F} is attained by an integral vector for each C for which the
maximum is finite.
Definition 2.15: (Hilbert Basis) A finite set of vectors a1 ,… ,at is a Hilbert basis if each
integral vector b in cone 5a1 ,… ,at< is a nonnegative integral combination of a1 ,… ,at.
Note 2.16: When each vector in a Hilbert basis is integral, it is called Integral Hilbert
Basis.
Theorem 2.17: [Gordan] Each rational polyhedral cone C is generated by an integral
Hilbert basis.
[Vander Corput] If C is pointed there is a unique minimal Hilbert basis generating C,
(minimal relative to taking subset).
Proof: Let C be a rational polyhedral cone generated by say b1,… , bk [Theorem 1.8].
Without loss of generality, we can assume that b1,… , bk are integral vectors. Let a1,… , at
be all integral vectors in the polytope
{λ1b1 � … � λkbk | 0 � λ8 � 1, i = 1,…, k} (10)
We claim that a1,… , at form an integral Hilbert basis. Inparticular those b1,… , bk among
a1,… , at shall generate C. Let b be any integral point in C. We have
b = ∑ µiy
iki;1 , µ
i� 0 (11)
Then
b = @µ1Ab1 + + @µ
kAbk+ GHµ
1I @µ
1AJb1 � K � Hµ
kI @µ
kAJbkL (12)
b - @µ1Ab1 + + @µ
kAbk = GHµ
1I @µ
1AJb1 � K � Hµ
kI @µ
kAJbkL (13)
The left hand side vector, as it is integral, occurs among, a1,… , at. Observe that the R.H.S.
of (13) clearly belongs to (11) because 0 � µ8 I @µMA $ 1.
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Since b1,… , bk occur among a1,… , at, it follows that b is a nonnegative integral
combination of a1,… , at. So a1,… , at form a Hilbert basis.
Now, assume that the cone C is pointed. Define
H = {a C | a N 0, a integral, a is not the sum of two other integral
vectors in C}. (14)
It is clear that any integral Hilbert basis generating C must contain H. So H is finite as H
is contained in (10).
We claim that H itself a Hilbert basis generating C. Let b be a vector such that
bx 0 if x C – {0} (b exists as C is pointed). Suppose not every integral vector in C is
a nonnegative integral combination of vectors in H. Let c be such a vector, with bc as
small as possible this exists as, c must be in the set (10). Then c is not in H. Hence
c = c1 � c2
for certain nonzero integral vectors c1 and c2 in C. Then
0 $ Oc1 $ OP and 0 $ Oc2 $ OP
Hence both c1 and c2 are nonnegative integral combinations of vectors in H and therefore
c is also. Therefore H is a Hilbert basis. As H is contained in any Hilbert basis of C, it is
minimal. ∎
Remarks 2.18:
(i) Combining the methods of theorem 2.11 and 2.17 for any rational polyhedron
P there exist integral vectors x1, … , xt, y1, … , y
s such that
�x| x P, x integral =5λ1x1 � K� λtxt � µ
1y
1� K�
µsy
s | λ1, … , λt,µ1, … ,µ
s nonnegative integers with ∑ λi 2 1< (15)
(ii) If the cone is not pointed there is no unique minimal integral Hilbert basis
generating the cone.
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(iii) If a vector c belongs to a minimal integral Hilbert basis generating a pointed
cone then the components of c are relatively prime integers.
Theorem 2.19: [A Theorem of Doignon] Let a system
a1x � β1, … , amx � β
m (16)
of linear inequalities in n variables be given. If (16) has no integral solution, then there
are 2n or less constraints among (16) which already have no integral solution.
Proof: Suppose (16) has no integral solution. We may assume that if we delete one of
the constrains in (16), the remaining system has an integral solution. This means there
exists integral vectors x1, … , xm so that, for j = 1, … , m,
ajxj Q β
j
aixj � β
i i j
We must show m 2n. So assume m > 2
n. Let
Z = Bn � Conv.hull {x1, … , xm} (17)
Choose γ1, … , γ
m so that:
(i) γj� min ,ajz Rz Z; ajz Q β
j1
(ii) the system a1x � γ1, … , amx � γ
m has no solution in Z.
(iii) γ1� K � γ
m is as large as possible. (18)
We claim that such γ1, … , γ
m exists. This is proved by showing the set of
Hγ1, … , γ
m J satisfying (18) is nonempty, bounded and closed. Note that
xj ,z Rz Z; ajz Q βj1
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If we take
γj2 min ,ajz Rz Z; ajz Q β
j1
such γ1, … , γ
m exist. Note also that γ
jQ β
j. As ajx
j Q βj, the system in (ii) has no
solution in Z. This shows that (18) is nonempty. Next, if
γj� ajx
j
Then as aixj $ β
i i j and β
i� γ
i we get that system in (ii) has a solution. Therefore
γj� ajx
j j
i.e., the set of Hγ1, … , γ
mJ, satisfying (18), is bounded.
Now the complement of the set of Hγ1, … , γ
m J satisfied (18) is
,z Rajz $ γj, S j1
which is a finite intersection of open half spaces ajz $ γj and hence it is an open set.
Since γ1� K � γ
m is as large as possible, for each j =1, … , m, there exists yj Z. So that
ajyj = γ
j and aiy
j $ γi �i N j! (19)
As m Q 2n, there exists k, l �k N l!so that
yk = yl mod�2! i.e. , either both are even or both are odd. [n = 1 m > 2, n = 2 m > 4]. Thus,
1
2Hyk � ylJ belongs to Z and in view of (19), satisfies the system in (ii). This contradicts
(ii). Therefore m 2n. ∎
Corollary 2.20: [Scarf] Let Ax b, be a system of linear inequalities in n variables, and
let c �n. If max {cx | Ax ≤ b ; x integral} is finite then
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max {cx | Ax ≤ b; x integral} = max {cx | A′x ≤ b′; x integral} (20)
for some subsystem A′x ≤ b′ of Ax
b with at most 2
n - 1 inequalities.
Proof: Let
µ = max {cx | Ax ≤ b ; x integral}
Hence for each t �n., the system
Ax ≤ b, cx � µ� 1U (21)
has no integral solution. Therefore, by theorem (2.19) for each t N there is a subsystem
of (21) of at most 2n constraints having no integral solution.
Since Ax � b does have an integral solution (as µ
is finite), each such
subsystem must contain the constraint cx � µ� 1U. Hence there is a subsystem A′x ≤ b
′ of atmost 2
n - 1 constraints so that the
system (21) has no integral solution for infinitely many values of t V. Therefore, A′x
≤ b′, cx > µ has no integral solution. This gives (20). ∎
Note 2.21: The bound 2n in theorem (2.19) is best possible. This is shown by the system
∑ xii I - ∑ xiiWI � |I| - 1 (I {1, …, n}) (22)
of 2n constraints in the n variables x1, … , xn. Observe that for n= 1, the above system is
x1 � 1 �I 2 X! x1 � 0 �I 2 �1!
which is clearly infeasible.
Next, for n = 2, the system is
x1 � x2 � 1 �I 2 X!
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x) I x* � 0 �I 2 �1!
x2 I x1 � 0 �I 2 �2!
x1 � x2 � 1 �I 2 �1,2!
In particular, we have
x1 � x2 2 1
x1 I x2 2 0
Clearly, the above system has no integral solution (x1 = 1
2 , x2 2 1
2 ).
Now, take n + 1 variables. Put IY = {1, 2, … ,n, n + 1} = {1, …, n} Z {n + 1}.
Observe that 2�[) 2 2� � 2�. Hence we can arrange 2�[) inequalities in the system (22)
as
∑ xii I - ∑ xiiWI - x�[) � |I| - 1 $ |I| ( I {1, …, n})
∑ xii I + x�[) - ∑ xiiWI � |I| � 1 - 1 = |I| If this system has a solution then adding we get a solution of the system
∑ xii I - ∑ xiiWI $ |I| This means
∑ xii I - ∑ xiiWI � |I| – 1
which has no solution by induction hypothesis.
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Chapter 3
Totally Unimodular Matrices
Definition 3.1: (Totally Unimodular Matrix) A matrix A is totally unimodular if each
sub determinant of A is 0, +1 or -1.
Note3.2: In particular each entry in a totally unimodular matrix is 0, +1 or -1.
Remark 3.3: It is easy to see that if A is totally unimodular then all following matrices
are totally unimodular.
AT, -A, HA
IJ, H A\A
J, H A
ATJ , ] III AIA
^
Further if A is a square totally unimodular matrix then A\1
is also totally unimodular.
A relation between totally unimodularity and integer linear programming is
given by following result.
Theorem 3.4: Let A be totally unimodular matrix and let b be an integral vector. Then
the polyhedron P = �x| Ax � b is integral.
Proof: Consider a minimal face F of P
F = 5x_ A′x 2 b′<
where A′x � b′
is a subsystem of Ax � b with A′ having full row rank. Then we may
permute the columns of A in such a way that,
A′ 2 `U Va where U is a nonsingular matrix and let det U = 1. A basic feasible solution of A
′x 2
b′ is
= GUb1b
0L (1)
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Since U is totally unimodular U\1 is also. Hence each entry in U\1 is 0, 1. Thus x is
integral. Thus every minimal face is an integral vector. Hence P is an
integer polyhedron. ∎ Note 3.5: Following corollary makes clear that each linear program with integer data and
totally unimodular constraints matrix has an integral optimum solution.
Corollary 3.6: Let A be totally unimodular matrix and let b and c be integral vectors.
Then both problems in the LP-duality equation
max {cx| Ax � b = min {yb | y ≥ 0, yA = c} (2)
have integral optimum solution.
Proof: By above theorem the polyhedron Ax � b is integral and hence
max {cx| Ax � b} is integral. Further as A is unimodular,
c II AT IAT d (3)
is also unimodular which is a constraints matrix for the minimization problem. We again
use above theorem to conclude that min {yb | y ≥ 0, yA = c} is integral. ∎
Remark 3.7: Hoffman & Kruskal theorem characterizes totally unimodularity which is
similar to above characterization.
Definition 3.8: Let A be any m n matrix of row full rank. A is called unimodular if A
is integral, and each basis of A has determinant e 1.
Proposition 3.9: The matrix A Zm�n is totally unimodular if and only if ` I A a is
unimodular.
Proof: [f] If a basis of ` I A a contains columns from A then its determinant is e 1.
Otherwise given basis can be rearranged (if necessary) in following form
gI B1
0 B2h (4)
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Note that det B2 N 0, as columns form basis. But A is totally unimodular, so det B2 = e 1
and hence determinant of the basis is e 1.
`ia Let ` I A a be unimodular. Consider a submatrix B of A. If rk�B!= m then by
unimodularity of ` I A a, det B = e 1.
Suppose rk�B! = k m. Now we can complete the basis using columns of
and taking columns in A corresponding to columns in B. Further, we can rearrange
these columns to have form
B1 2 jI Bk0 B
l
Now, by unimodularity of ` I A a, det B1 = e1 = det B. ∎
Theorem 3.10: Let A be an integral matrix of full row rank then the polyhedron �x| x � 0;Ax 2 b is integral for each integral vector b, if and only if A is unimodular.
Proof: `ia Let A be m � n matrix. First suppose that A is unimodular. Let b be an
integral vector, and let x be a vertex of the polyhedron �x| x � 0;Ax 2 b. Then there are
n linearly independent constraints satisfied by x with equality. Therefore the columns of
A corresponging to the nonzero components of x are linearly independent.
We can extend these columns to a basis B of A. Then x restricted to the
coordinates corresponding to B is equal B\1b, which is integral as det B = e 1. Since
outside B, x is zero, it follows that x is integral.
[f] Suppose that �x| x � 0;Ax 2 b is integral for each integral vector b.
Let B be a basis of A. To prove that B is unimodular it suffices to show
that B\1t is integral for each integral vector t. Then there exist an integral vector y such
that
z = y + B\1t � 0
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Then b = Bz is integral. Now, extend z by adding zero components. Let this vector be z′. Then
Az′ = Bz = b
and z′ is a vertex of the polyhedron �x| x � 0;Ax 2 b. (As it is in the polyhedron, and
satisfies n linearly independent constraints with equality.) Therefore, z′ is integral, so z
and z – y = B\1t is integral. ∎
Corollary 3.11: (Hoffman and Kruskal’s
theorem) Let A be an integral matrix. Then A
is totally unimodular if and only if for each integral vector b the polyhedron �x | x � 0;Ax � b is integral.
Proof: Note that, for any integral vector b, the vertices of the polyhedron �x | x � 0;Ax � b are integral if and only if the vertices of the polyhedron �z |z � 0; `I Aaz 2 b are integral. (Transform Ax � b into Ax � y 2 b and put
(x,y) = z). By the proposition (3.9) A is totally unimodular if and only if ` I A a is
unimodular. Hence, the theorem proves the corollary. ∎
Remark 3.12: An integral matrix A is totally unimodular if only if for all integral vectors
a,b,c,d the vertices of the polytope �x|c � x � d; a � Ax � b are integral.
Observe that the constraints can be written as x � d, Ix � c, Ax � b, I Ax � Ia. Hence
the corresponding matrix has the form
] IIIAIA
^ (5)
Note 3.13: It is clear from the Hoffman and Kruskal’s theorem that an integral matrix A
is totally unimodular if and only if one of the following polyhedron has all vertices
integral, for each integral vector b and for some integral vector c.
�x| x � c;Ax � b
�x| x � c;Ax � b
18
�x| x � c;Ax � b �x| x � c;Ax � b Corollary 3.14: An integral matrix A is totally unimodular if and only if for all integral
vectors b and c both sides of the linear programming duality equation
max{cx | x � 0,Ax � b} = min {yb | y ≥ 0, yA � c } (7)
are achieved by integral vectors x and y (if they are finite).
Proof: Clear from above corollary and noting that A is totally unimodular if and only if
AT is totally unimodular. ∎
Theorem 3.15: Let A be a matrix with entries 0, +1 or -1. Then the following are
equivalent:
(i) A is totally unimodular, i.e. each square submatrix of A has determinant 0, +1,
or -1.
(ii) [Hoffman & Kruskal] For each integral vector b the polyhedron �x| x � 0,Ax � b has only integral vertices.
(iii) [Hoffman & Kruskal] For all integral vectors a,b,c,d the polyhedron �x| c � x � d, a � Ax � b has only integral vertices.
(iv) [Ghouila-Houri] Each collection of columns of A can be split into two parts
so that the sum of the columns in one part minus the sum of the columns in
the other part is a vector with entries only 0, +1, or -1.
(v) [Camion] Each nonsingular submatrix of A has a row with an odd number of
nonzero components.
(vi) [Camion] The sum of the entries in any square submatrix with even row and
column sums is divisible by four.
(vii) [R.E.Gomory] No square submatrix of A has determinant +2 or -2.
Proof: We shall prove equivalence in following way.
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�i! m �ii! m (iii) f �iv! f (v) f (vii)
o
�vi! f (vii) m (i) The equivalence of (i), (ii) and (iii) is the Hoffman and Kruskal’s theorem (corollary
3.11, remark 3.12 and note 3.13)
�iii! f �iv! [(i) f (iv)]
Let A be totally unimodular, and choose a collection of columns of A. Consider the
polyhedron
P = ,xR 0 � x � d, p1
2Adq � Ax � r1
2Ads1 (8)
where d is the characteristic vector of the collection of chosen column and t u and v w denote component-wise lower and upper integral parts of vectors. Since is nonempty,
as 1
2d P, P has atleast one vertex say x, which is by �iii! a �0,1 vector. Then
y = d – 2
has components only 0,+1, or -1 and
y d �mod2! [ Observe that,
y = 0 d = 0 x = 0
y = 1 d = 1 x = 0
y = -1 d =1 x = 1 ]
Hence Ay has components only +1, -1 or 0. So y yields a partition of the columns as
required.
�iv! f �v!
20
Let B be a square submatrix such that each row has even number of nonzero entries.
Thus, its row sum is even. By �iv! there exists a ��1, I1 vector x such that Bx is a �0, �1, I1 vector. Since B has even row sums we know that Bx = 0. Since xN 0, B is
singular. [The nullity is strictly less than the order of the matrix.]
�iv! f �vi! Let B be a square submatrix of A with each row sum and each column sum even. By �iv! the column of B can be split into two classes B1 and B2 so that the sum of the
columns in B1 is the same as the vector as the sum of the columns in B2 as each row sum
of B is even.
Let σ� Bi! denote the sum of the entries in Bi . Then
σ� B)! = σ� B2! and σ� Bi! is even as each column sum of B is even. Hence Then σ� B)! + σ� B2! is
divisible by four.
�i! f �vii! Obvious
�vii! f �i! Suppose no square sub matrix of A has determinant 2. To show that each
square sub matrix of A has determinant 0, 1, it is sufficient to show that each square
�0, e 1 matrix B with |det B| Q 2 has a square submatrix with determinant 2. Let
order of B be n. Consider the matrix
C = ` B I a Let C
′ arise from C by adding or subtracting rows to or from other rows, and by
multiplication of columns by -1, such that
(i) C′ is a �0, e1 matrix.
(ii) C′ contains among its columns the n unit basis column vectors and
(iii) C′ contains among its first n columns as many unit basis column vector as
possible.
21
Let k be the number of unit basis vectors in the first n column of C′. We may suppose
without loss of generality that
C′ 2 cIx K 0y DY y0 K I�\x
d (10)
for a certain square matrix B′ of order n where Ik and In\k denote the identity matrices of
order k and n-k. Since the first n column of C and hence also of C′ form a matrix with
determinant not equal to 1 (z |det B| Q 2 ), we have k n.
So there is a 1, without loss of generality, occurring in some position �i, j! of
C′ with k + 1 i, j n. By our assumption �iii! we cannot transfer column j to a unit
vector by elementary row operations without violating condition �i!. Hence there is a
pair i′,j′ such that the 2 2 sub matrix with row indices i and i′ and column indices j and
j′ has the form
j 1 I1
1 I1 l or j 1 I1I1 I1
l Now the sub matrix of C
′ formed by the columns j, j′ and the unit column vectors, except
the ith
and i′ th
unit vectors, has determinant 2. So also the corresponding columns of C
form a matrix with determinant 2. This implies that B has a submatrix with determinant
2.
�v! f �vii! Suppose �v! is true for A. i.e., each nonsingular submatrix of A has a row with an odd
number of nonzero components. If �vii! is not true then det A = e 2 and each square
proper submatrix of A has determinant 0 or e 1 [since �i! m �vii!]. Now,
Since, det A 0�mod 2! the columns of A are linearly dependent over B2. As �i! & �vii!
are equivalent, det A = 0 over implies det A 0�mod 2! i.e., for each proper submatrix
of A, linear dependent over coincides with linear dependence over B2.
22
As det A = e 2, the columns of A are linearly independent over . But det A
0�mod 2!. This means columns of A are linearly dependent over B2. Thus sum of all
columns of A is a vector having even components only. Clearly this contradicts �v!. �vi! f �vii! Similarly, as above, we can show that the sum of the rows of A has even components
only. Let B arise from A by deleting the first row of A. Note that we have proved �i! m �ii! m �iii! f �iv! f �v! f �vii! m �i!. Hence �vii! f �iv!. Therefore, there
exists a ��1, I1 vector x such that Bx is a �0, �1, I1 vector. But the matrix B has even
row sums (as A has above). This gives Bx = 0. So
Ax =
{|||} α 0
0′′
0 ~���� (9)
for some integer α. Now
where A′ arises from A by deleting the first column of A. Observe that determinant of the
second matrix on L.H.S. is 1. �x is a ��1, I1 vector!. Hence the determinant of the
matrix on R.H.S. must be e α [Infact, minor of α cannot be zero as, now, determinant of
L.H.S. is 2 ].
Further, as x is ��1, I1 vector, 1 – x has even components only. 1TA is the
sum of the rows which has also even components. Therefore
1TA�1 I x! = 0 �mod 4!.
Next, from (9),
1TAx = α = 2�mod 4! (z |�| = 2)
1TA1 = 2�mod 4!
which contradicts �vi!. ∎
23
Theorem 3.16: [Baum and Trotter] An integral matrix A is totally unimodular if and
only if for all integral vectors b and y, and for each natural number k 1, with y
0, Ay
kb, there are integral vectors x1, … , xk in �x � 0;Ax � b such that y = x1 � … + xk.
Proof: `ia To show that A is totally unimodular, it is enough to show that for each
integral vector b, all vertices of the polyhedron �x| x � 0,Ax � b are integral (above
theorem 3.15(ii)). Suppose x0 is a non integral vertex. Let k be the l.c.m.of the
denominators occurring in x0. Then y = kx0 satisfies y 0, Ay kb. Therefore y = x1 �
… + xk. for certain integral vectors x1, … , xk in P. Hence
x0 = x1[ … [ xk
k
is a convex combination of integrals vectors in P. Contradicting the fact that x0 is a non-
integral vertex of P.
[ ] Let A be totally unimodular. Choose integral vectors b and y and a natural number
k 1, such that y 0, Ay kb.We show that there are integral vectors x1, … , xk in
�x| x � 0,Ax � b with y = x1 � … + xk, by induction on k. the case k = 1 is trivial �y 2 x1!. We know that the polyhedron
�x | 0 � x � y; Ay I kb � b � Ax � b (12)
is nonempty, as k\1
y is in it.
Since A is totally unimodular (12) has an integral vertex [theorem 3.15], call it
xk. Let
y′= y - xk.
Then y′ is integral, and y′ � 0 �0 � xk � y! and
Ay′ = A�y I xk! = Ay - Axk
Ay – �Ay I kb � b!
= �k I 1!b
24
Hence by induction,
y′ = x1 � … + xk\1
for integral vectors x1,… , xk\1 in �x � 0;Ax � b. So, y = x1 � … + xk is a
decomposition as required. ∎
Theorem 3.17: Let A be a matrix of full row rank. Then the following are equivalent.
(i) for each basis B of A, the matrix B\1A is integral.
(ii) for each basis B of A the matrix B\1A is totally unimodular.
(iii) there exists a basis B of A for which B\1A is totally unimodular.
Proof: We can arrange columns of A = `B Ra where B is a non singular matrix. Then
B\1A = `I Ca, C = B\1R. Now observe that �i!, �ii! & �iii! are invariant under
premultiplying A, by a non singular matrix. Hence we can assume that A = `I Ca for
some matrix C. Now, by proposition 3.9 each of (i), (ii) and (iii) is equivalent to each basis of
`I Ca being unimodular. ∎
Theorem 3.18: (Chandrasekaran) A matrix A is totally unimodular if and only if for
each nonsingular submatrix B of A and for each nonzero {0, 1} vector y, the g.c.d. of
the entries in yB is 1.
Proof: [ ] Let B be a non-singular submatrix of a totally unimodular matrix A. By
above theorem 3.15 (i), B\1 is integral. Let k be the gcd of the components of yB. As
entries of B are 0, e1 and y is a {0, e 1} vector k\1
yB is integral. Thus
k\1
y = k\1
yBB\1
is integral. As y is a {0, e1}-vector, k = 1.
25
`ia Let B be a nonsingular submatrix of A. Then the components of 1B have g.c.d. 1.
Then one of the columns of B must have an odd number of nonzero entries. Thus the
condition (v) is satisfied in the theorem 3.15. ∎
Remark 3.19: [Tamir] The proof shows that A is totally unimodular if and only if for
each non-singular sub matrix B of A the g.c.d. of the entries in 1B is 1.
Remark 3.20: A nonsingular matrix B is unimodular if and only if g.c.d.(yB) = g.c.d.(y)
for each vector y.
Remark 3.21: With Hoffman and Kruskal’s theorem the total unimodularity of the
incidence matrix of a bipartite graph implies several theorems, like Konig’s theorem for
matching and coverings in bipartite graphs and the Birkoff-von Neumann theorem on
doubly stochastic matrices.
The Basic examples
Example 3.22: (Bipartite Graphs) Let G = (V, E) be an undirected graph. Let M be V
E incidence matrix of G (i.e., M is the {0,1}-matrix with rows and columns indexed by
the vertices and edges of G, respectively, where Mv,e = 1 if and only if v e). It is easy
to see that G is bipartite if and only if the rows of M can be split into two columns so that
each column contains a 1 in each of these classes. By theorem 3.15, Ghouila-Houri’s
characterization (iv),
M is totally unimodular if and only if G is bipartite. (13)
Let M be the V E incidence matrix of the bipartite graph G = (V, E) Then by (13) and
corollary 3.14. we have
max �y1 |y � 0, yM � 1, y integral = min �1x |x � 0;Mx � 1; x integral This is equivalent to Konig’s covering theorem: the maximum cardinality of a co-clique
in a bipartite graph is equal to the minimum number of edges needed to cover all vertices.
(we assume that the graph has no isolated vertex). Similarly,
max �1x|x � 0;Mx � 1; x integral = min �y1|y � 0 yM � 1, y integral
26
This is equivalent to the Konig-Egervary theorem: the maximum cardinality of a
matching in a bipartite graph is equal to the minimum cardinality of a set of vertices
intersecting each edge.
More generally,for each w: E � B[
max �wx |x � 0 Mx � 1, x integral = min �1y |y � 0; yM � 0; 9 integral If we consider w as a profit function, above equation gives a min-max formula for the
optimal assignments problem.
Further, again by the theorem 3.15 (ii), it follows that the polytopes �x|x � 0,Mx � 1 and �x|x � 0,Mx 2 1 are integral. Therefore, a function x: E � �[
is a convex combination of incidence vectors of (perfect) matching in G if and only if
∑ x�e!v e � 1 � ∑ x�e!v e 2 1 !
for each vertex v.
Inparticular, if G is the complete bipartite graph Kn,n. Then the last-result is
equivalent to the theorem of Birkoff and Von Neumann: A doubly stochastic matrix is a
convex combination of permutation matrices. ∎
Example 3.23: (Directed Graphs) Let D = �V,A! be a directed graph, and let M be the
V � A incidence matrix of D defined by
Mv,a = +1 if a enters v
= -1 if a leaves v
= 0 otherwise
Then M is totally unimodular by following theorem.
Theorem 3.24: [Poincare] A {0, 1} matrix with in each column exactly one +1 and
exactly one -1 is totally unimodular.
27
Proof: [Veblen & Franklin] Proof uses induction on size t of a submatrix N of M. The
case t = 1 is trivial. Let t Q 1. There are three cases.
Case (i): If N has a column with only zeros. Then, clearly det N = 0.
Case (ii): Now, suppose a column with exactly one nonzero entry. Then we can write N,
if necessary after permuting rows and columns,
N = ge1 bt
0 N′h
For some matrix b and matrix N′. Then by induction hypothsis.
det N′ {0, 1}
and hence det N = �e1! det N′ {0, 1}.
Case (iii): As a last case, suppose each column of N contains exactly two nonzero
entries. Then each column of N contains one +1 and one -1, while all other entries are
zero. So the rows of N add up to the zero vector. This means rows are linearly
dependent. Therefore det N = 0. ∎
Note 3.25: The theorem also follows from Ghouila-Houri’s characterization (iv) in
theorem 3.15.
28
REFERENCE
1. Theory of Linear and Integer Programing, Alexander Schrijver, John Wiley &
Sons © 1986.