Colorings of k-balanced matrices and integer decomposition property of related polyhedra
Article: Giacomo Zambelli
Presentation: Grigoruta Alexandru
\begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 \\
1 & 0 & 0 & 1
\end{bmatrix}
Balanced 0-1 matrix
0-1 matrix
There is no squared submatrix of odd order, which has all rows sum and all columns sum equals to 2
\begin{bmatrix}
1 & -1 & 1 & -1 \\
1 & -1 & 0 & 0 \\
1 & 0 & -1 & 0 \\
-1 & 0 & 0 & 1
\end{bmatrix}
K-Balanced 0-±1 matrix
0-±1 matrix
There is no squared submatrix B, with at most 2k nonzero entries, such that each row and each column has an even number of nonzero entries and the sum of all entries in B is congruent to 2 modulo 4
Sum is 0; 0 ≢ 2 mod 4
Integer decomposition property of polyhedra
\forall h \in \mathbb{Z}{+}
\forall y \in hP := \{ hx | x \in P\}
\}
y= x{^1}+ ... + x{^h}
\exists h \ integral \ vectors
x{^1}, ... , x{^h} \in P \ so \ that
Theorem 1
A = m x n k-balanced 0-1 matrix
a{^i} = rows, \ i \in [m]
= partitions of [m]
S{_1}, S{_2}
has the integer decomposition property
for \forall b \in \mathbb{Z}{^m} so \ that \ \textbf{0} \leq b \leq \textbf{k}
where 0 is a vector with every element equal to 0, and k is a vector with every element equal to k
Totally unimodular matrices
A matrix is totally unimodular if each of its square nonsingular submatrices has a determinant of (+/-) 1
Theorem 2
A = m x n 0-(+/-)1 matrix
A is totally unimodular if and only if:
B = squared submatrix of A
There is no B with an even number of nonzero entries in each row and each column, such that the sum of all entries in B equals 2 modulo 4
Almost totally unimodular matrices
A matrix is almost totally unimodular if A is not totally unimodular and each of its square nonsingular submatrices is totally unimodular
Based on this statement and Theorem 2, a 0-(+/-)1 matrix is k-balanced if and only if it has no
almost totally unimodular submatrix with at most 2k
nonzero entries in each row
For any m × n 0,±1 matrix A, we denote by n(A) the vector with m components whose ith component is the number of −1’s in the ith row of A, and let p(A) = n(−A).
Theorem 3
Notation:
A = m x n k-balanced 0-1 matrix
a{^i} = rows, \ i \in [m]
b \in \mathbb{Z}{^m} \ so \ that \ -n(A) \leq b \leq \textbf{k} - n(A)
= partitions of [m]
S{_1}, S{_2}
Then the polytope
\begin{cases}
x \in \mathbb{R}{^n}: a{^i}x \leq b{_i}, i \in S{_i} \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a{^i}x \geq b{_i}, i \in S{_2} \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \leq \textbf{1}
\end{cases}
is integral
k-equitable colorings
A = m x n k-balanced 0-1 matrix
a{^i} = rows, \ i \in [m]
\alpha{_i}:=min \{k, \lfloor (p{_i}(A) + n{_i}(A)) / 2 \rfloor \}
We say that A has a k-equitable bicoloring if its columns can be partitioned into blue columns and red columns so that the matrix , obtained from A by multiplying its blue columns by −1, has at least positive entries and at least negative entries in row i, for every i ∈ [m].
\bar A
\alpha{_i}
\alpha{_i}
Theorem 4
Let A be a 0,±1 matrix. A is
k-balanced if and only if every submatrix of A has a
k-equitable bicoloring
λ-coloring
A \in \mathbb{R}{^{m \times n}}
\lambda \geq 2
\}
λ-coloring = partition of columns from A into λ sets
I{_1},...,I{_\lambda}
k-equitable λ-coloring
A \in \mathbb{R}{^{m \times n}} \ 0-\pm 1 \ matrix
k,\lambda \in \mathbb{Z}{_+}
k-equitable λ-coloring of A = λ-coloring
I{_1},...,I{_\lambda}
of A
so that
(I{_j},I{_h})
= k-equitable bicoloring of
A{_{I{_j}I{_h}}}
A{_{I{_j}I{_h}}} = I{_j} \cup I{_h} \ , \forall \ 1 \leq j < h < \lambda
Theorem 5
An m × n 0-1 matrix A is k-balanced if
and only if every submatrix of A has a k-equitable -coloring for every integer 2
\lambda
\lambda \geq
Theorem 5 Algorithm
\mu{_{ij}} = \begin{cases}
max(n{_{ij}} - {\lceil \frac{p{_i}(A)}{\lambda}\rceil}, {\lfloor \frac{p{_i}(A)}{\lambda}\rfloor} - n{_{ij}}), \ for \ i \in S{_1} \\
max(0, k-n{_{ij}}), \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for \ i \in S{_2}
\end{cases}
Suppose \ \mu{_{st}} > 0\ for \ some \ s \in [m], \ t \in [\lambda].
There \ exists \ t' \in [\lambda] \ with:
If \ s \in S{_1} \ and \ n{_{st}} < {\lfloor \frac{p{_s}(A)}{\lambda}\rfloor}, \ then \ n{_{st'}} > {\lfloor \frac{p{_s}(A)}{\lambda}\rfloor}
If \ s \in S{_1} \ and \ n{_{st}} > {\lceil \frac{p{_s}(A)}{\lambda}\rceil}, \ then \ n{_{st'}} < {\lceil \frac{p{_s}(A)}{\lambda}\rceil}
If \ s \in S{_2} , \ then \ n{_{st'}} > k
Notations needed for the algorithm:
Theorem 5 Algorithm
Start \ from \ an \ arbitrary \ partition \ I{_1}, I{_2}, ..., I{_\lambda}
Compute \ the \ coresponding \ \mu
If \ \mu = 0, \ stop
Else \ find \ a \ new \ partition \ with \ smaller \ value \ of \ \mu
Choose \ an \ appropriate \ t\ and \ t' \in [\lambda]
Compute \ a \ k-equitable \ bicoloring \ of \ A{_{I{_t}I{_t'}}}
Computing a k-equitable bicoloring can be done in strongly polynomial time
\mu \leq nm
\}
Computing a k-equitable bicoloring can be done in strongly polynomial time
Theorem 1
A = m x n k-balanced 0-1 matrix
a{^i} = rows, \ i \in [m]
= partitions of [m]
S{_1}, S{_2}
has the integer decomposition property
for \forall b \in \mathbb{Z}{^m} so \ that \ \textbf{0} \leq b \leq \textbf{k}
where 0 is a vector with every element equal to 0, and k is a vector with every element equal to k
Proof of Theorem 1
h \in \mathbb{Z}{_+}
y \in hP := \{ hx | x \in P\}
y= x{^1}+ ... + x{^h}
If \ h \geq 2, \ let \bar A = A \cup j{^{th}} column \ of \ A \cup...(y{_j} \ times) ... \cup j{^{th}} column \ of \ A
if \ y{_j} = 0, \ remove \ j{^{th}} \ column
\forall j \in [n], \ let \ C{_j}= set \ of \ indices\ of\ columns\ of \ \bar A \ that \ are
copy \ of \ column \ j \ of \ A
if \ y{_j} = 0, \ let \ C{_j} = \emptyset; \ Therefore \ |C{_j}| = y{_j}
Proof of Theorem 1
let \ \bar a{^i}, \ i\in [m], \ the \ rows\ of \ \bar A
\bar A \ is \ still \ k-balanced, \ so \ by\ Theorem \ 5\ there\ exists \ a \ k-equitable
h-coloring \ I{_1}, ..., I{_h} \ of \ \bar A
Let \ z{^1}, ..., z{^h} \ be \ the \ characteristic\ vectors\ of\ I{_1},...,I{_h}
Let \ x{^1}, ..., x{^h} \ be \ the \ vectors\ in \ \mathbb{R}{^n} \ := \ x{_{j}^{s}} = {\sum_{t \in C{_j}}z{_{t}^{s}}} \ for \ s \in [h], \ j \in [n]
y = x{^1} + ... \ + x{^h}\ ;\ We \ need \ to \ show\ that \ x{^s} \in P \ , \ \forall s \in[h]
Observe \ p(\bar a{^i}) = a{^i}y, \forall i \in [m]. \ Thus:
if \ i \in S{_1}, \ then \ a{^i}x{^s} = \bar a{^i}z{^s} \leq \lceil p(\bar a{^i})/h \rceil = \lceil (a{^i}y)/h \rceil \leq \lceil (hb{_i})/h \rceil = b{_i}
if \ i \in S{_2}, \ then \ a{^i}x{^s} = \bar a{^i}z{^s} \geq min(\lfloor p(\bar a{^i})/h \rfloor , k) \geq b{_i}, \ since \ k \geq b{_i} \ and
\lfloor p(\bar a{^i})/h \rfloor = \lfloor (a{^i}y)/h \rfloor \geq b{_i}
Theorem 1 Algorithm Complexity
Note that, since a k-equitable -coloring can be
found in polynomial time, the proof of Theorem 1
gives a polynomial time algorithm
\lambda
Input:
h \in \mathbb{Z}{_+}
y \in hP := \{ hx | x \in P\}
Output:
y= x{^1}+ ... + x{^h}
x{^1}, ... , x{^h} \in P \ so \ that