find a column c in A having maximum weight delete all rows of A that contain a “1” in column c delete column c from A
output: Returns a submatrix of A with no all-zero row
At each state, a new column is added to the submatrix that maximizes the number of
“new” rows that are yet uncovered. When all rows are covered, the algorithm stops. It seems quite interesting that we can use the Stein-Lov´asz theorem to derive bounds for some combinatorial array [8].
3.2 Extension of The Stein-Lov´asz Theorem
The Stein-Lov´asz theorem can be further extended from rows of the resulting submatrix with weight at least 1 to the case of rows of the resulting submatrix with weight at least z ≥ 1. The bound can be further improved when A is a matrix with constant row weight and column weight as well, i.e., in the language of hypergraphs, uniform and regular.
Theorem 3.2.1. Let A be a (0,1) matrix of order N × M , and let v, a, z be positive integers.
Assume that each row contains at least v ones, and each column at most a ones. Then there exists a submatrix C of order N × K with
K < v
v − (z − 1) z(M
v )(1 + ln a), such that each row of C has weight at least z.
More specifically, if the matrix is v-uniform and a-regular, the upper bound can then be reduced to
K < z(M
v )(1 + ln a).
The strategy for the proof of Theorem 3.2.1 is as follows:
1. Use the Stein-Lov´asz theorem to obtain a submatrix C1 with each row has weight at least 1.
2. Choose some columns in the matrix A\C1 to combine with the submatrix C1 to form a submatrix C2 with each row has weight at least 2.
3. Choose some columns in the matrix A\C2 to combine with the submatrix C2 to form a submatrix C3 with each row has weight at least 3.
4. Step by step, and finally we obtain the desired submatrix C = Cz with each row has weight at least z.
Note that this upper bound makes sense only if v
v − (z − 1) z(M
v )(1 + ln a) < M, i.e.,
z < v + 1 2 + ln a
in general, or if
z(M
v )(1 + ln a) < M,
i.e.,
z < v 1 + ln a
for the case of uniform and regular.
Proof. A constructive approach for producing C is presented. Let A1 = A. By the Stein-Lov´asz theorem, there exists an N × M1 submatrix C1(= B10 = B1) of A1 with M1 <
M
v (1 + ln a) such that each row of C1 has weight at least 1.
The algorithm used in the proof of the Stein-Lov´asz theorem shows that some rows of C1
have weight exactly 1. Let R1 be the set of indices of those rows and |R1| = r1. Let A2 be the submatrix of order r1× (M − M1) obtained from A1 by deleting the submatrix C1 and the i-th row, i /∈ R1 as well. Then each row of A2 contains at least v − 1 ones, and each column at most a ones. Again, by the Stein-Lov´asz theorem, there exists an r1× M2 submatrix B20 with M2 < M −Mv−11(1 + ln a) such that each row of B20 has weight at least 1. Let B2 be the matrix of order N × M2 obtained from B20 by adding the i-th row, i /∈ R1. Let C2 be the matrix of order N × (M1+ M2) obtained by the union of B1 and B2. Then C2 is a submatrix of A with each row weight at least 2.
Similarly, there exist some rows of C2 that have weight exactly 2. Let R2 be the set of indices of those rows and |R2| = r2. Continue in this way, we have:
For 2 ≤ i ≤ z,
1. Ai is a matrix of order ri−1× (M −
i−1
X
j=1
Mj), and each row contains at least v − (i − 1) ones, and each column at most a ones.
2. Bi0 is an ri−1× Mi submatrix of Ai with Mi <
M −
i−1
X
j=1
Mj
v − (i − 1) (1 + ln a), and each row has
weight at least 1.
For 1 ≤ i ≤ z,
3. Bi is a matrix of order N × Mi.
4. Ci is an N ×
i
X
j=1
Mj submatrix of A, and each row has weight at least i.
Hence, C = Cz is the submatrix required, that is,
K =
z
X
j=1
Mj = M1+ M2+ · · · + Mz
< M
v (1 + ln a) +M − M1
v − 1 (1 + ln a) + · · · + M −
z−1
X
j=1
Mj
v − (z − 1) (1 + ln a)
< M
v (1 + ln a) + M
v − 1(1 + ln a) + · · · + M
v − (z − 1)(1 + ln a)
= M (1 + ln a)(1 v + 1
v − 1+ · · · + 1 v − (z − 1))
< M (1 + ln a)( 1
v − (z − 1) + 1
v − (z − 1) + · · · + 1 v − (z − 1))
= z
v − (z − 1) M (1 + ln a)
= v
v − (z − 1) z(M
v )(1 + ln a), thus gives the result.
More specifically, for the case of uniform and regular, using similar argument as above with a minor modification. First we note that N v = M a by counting the weight of A in two ways. For 2 ≤ i ≤ z , Ai is a matrix of order ri−1× (M −
i−1
X
j=1
Mj), and each row contains exactly v − (i − 1) ones, and each column at most a ones. Moreover, a lower bound for
i−1
X
j=1
Mj is derived by counting the weight of the submatrix Ci−1in two ways; each row of Ci−1
contains at least i − 1 ones, and each column exactly a ones, thus N (i − 1) ≤ (
i−1
X
j=1
Mj)a, and
hence M
v (i − 1) ≤
i−1
X
j=1
Mj for 2 ≤ i ≤ z. Furthermore,
K =
z
X
j=1
Mj = M1+ M2+ · · · + Mz
< M
v (1 + ln a) +M − M1
v − 1 (1 + ln a) + · · · + M −
z−1
X
j=1
Mj
v − (z − 1) (1 + ln a)
≤ M
v (1 + ln a) +M − Mv
v − 1 (1 + ln a) + · · · +
M − (z − 1) · M v
v − (z − 1) (1 + ln a)
= M
v (1 + ln a) +M
v (1 + ln a) + · · · + M
v (1 + ln a)
= z(M
v )(1 + ln a), thus gives the result.
Remark 3.2.1. Since this upper bound makes sense only if
z < v + 1 2 + ln a in general,
z < v + 1
2 + ln a < v + 1
2 < v + 2 2 = v
2 + 1.
Then
z − 1 < v 2
and thus
v
v − (z − 1) = 1 + z − 1
v − (z − 1) < 1 + 1 = 2.
Hence
K < v
v − (z − 1)zM
v (1 + ln a) < 2zM
v (1 + ln a)
for general case.
Similarly, Theorem 3.2.1 can be restated in the language of hypergraphs in the following corollary. Recall that a subset T ⊆ X such that |TT E| ≥ z for any hyperedge E is called a z-cover of the hypergarph H, and the minimum size of a z-cover of the hypergraph H is denoted by τz(H).
Corollary 3.2.1. For a hypergraph H = (X, Γ) and a positive integer z ≥ 2,
τz(H) < 2z|X|
minE∈Γ|E|(1 + ln 4), where 4 = maxx∈X|{E : E ∈ Γ with x ∈ E}|.
More specifically, for the case of uniform and regular, we have the following corollary.
Corollary 3.2.2. Let H = (X, Γ) be a v-uniform and a-regular hypergraph with vertex set X and edge set Γ, then
τz(H) < z|X|
v (1 + ln a).
We conjecture that τz(H) ≤ zτ1(H) holds for hypergraphs which are uniform and regular.
However, it is not true in general as shown in the following example. For the hypergraph H with X = {1, 2, 3, ..., 8} and Γ = {{1, 2, 3}, {4, 5, 6}, {1, 7, 8}}. It is easy to see that {1, 4}
is a 1-cover with minimum size, hence τ1(H) = 2. Similarly, {1, 2, 4, 5, 7} is a 2-cover with minimum size, hence τ2(H) = 5. This shows that τ2(H) = 5 > 2 · 2 = 2τ1(H).
Chapter 4
Some Applications of the extended Stein-Lov´ asz Theorem
The Stein-Lov´asz theorem was first used in dealing with the upper bounds for the sizes in the model of (k, m, n)-selecters [2]. Inspired by this work, it was also used in dealing with the upper bounds for the sizes of (d, r; z]-disjunct matrices [5]. In Section 4.1 and Section 4.2, the extended Stein-Lov´asz theorem will be used in dealing the upper bounds for the sizes of several disjunct matrices (Theorem 4.1.1∼ 4.1.8) and selectors (Theorem 4.2.1 ∼ 4.2.4), respectively. Those upper bounds for the sizes of uniform splitting systems, uniform separating systems, covering designs and lotto designs are given in Section 4.3 (Theorem 4.3.1 ∼ 4.3.8) respectively.
4.1 Bounds for several disjunct matrices
Note that upper bounds for the sizes of d-disjunct matrices, (d, r]-disjunct matrices, (d, r)-disjunct matrices and (d, s out of r]-r)-disjunct matrices were given in [13, 14] by the Lov´asz Local Lemma.
Recall that t(n, d) is the minimum size over all d-disjunct matrices with n columns.
Theorem 4.1.1. For any positive integers n and d, if k = d + 1 ≤ n, then
t(n, d) < k(k
d)d{1 + k[1 + ln(n
k + 1)]}.
Proof. For 1 ≤ w ≤ n − d, let A be the binary matrix of order [n
, and each column of A has weight
n − w d
w 1
. By the Stein-Lov´asz theorem, there exists a submatrix M of A of order [n
Note that the equality is obtained by counting the weight of A in two ways. It is straight-forward to show that the columns of M form a d-disjunct matrix of order t × n. We then have
k is an integer. We have
n0
by Lemma 2.4.6 (taking s = r = 1). Therefore,
Recall that t(n, d; z] is the minimum size over all (d; z]-disjunct matrices with n columns.
Theorem 4.1.2. For any positive integers n, d and z, if k = d + 1 ≤ n, then
, and each column of A has weight
n − w d
w 1
. By the extended Stein-Lov´asz theorem, there exists a submatrix M of A of order [n
Note that the equality is obtained by counting the weight of A in two ways. It is straightfor-ward to show that the columns of M form a (d; z]-disjunct matrix of order t × n. We then
have
k is an integer. We have
n0
Recall that t(n, d, r] is the minimum size over all (d, r]-disjunct matrices with n columns.
Theorem 4.1.3. For any positive integers n, d and r, if k = d + r ≤ n, then
V = {v | v ∈ {0, 1}n, wt(v) = w} respectively. The entry of A at the row indexed by the pair (D, R) and the column indexed by the vector v ∈ V is 1 if the entries of v over D are all zero and the entries of v over R are all one; and 0 otherwise.
Observe that each row of A has weight n − (d + r) w − r
, and each column of A has weight
n − w d
w r
. By the Stein-Lov´asz theorem, there exists a submatrix M of A of order [n
Note that the equality is obtained by counting the weight of A in two ways. It is straight-forward to show that the columns of M form a (d, r]-disjunct matrix of order t × n. We then have
k is an integer. We have
n0
by Lemma 2.4.6 (taking s = r). Therefore,
Recall that t(n, d, r; z] is the minimum size over all (d, r; z]-disjunct matrices with n columns. and the entries of v over R are all one; and 0 otherwise.
Observe that each row of A has weight n − (d + r) w − r
, and each column of A has weight
n − w d
w r
. By the extended Stein-Lov´asz theorem, there exists a submatrix M of A of order [n
Note that the equality is obtained by counting the weight of A in two ways. It is straight-forward to show that the columns of M form a (d, r; z]-disjunct matrix of order t × n. We
then have
k is an integer. We have
n0
by Lemma 2.4.6 (taking s = r). Therefore,
t(n, d, r; z] ≤ t(n0, d, r; z] <
Recall that t(n, d, r) is the minimum size over all (d, r)-disjunct matrices with n columns.
Theorem 4.1.5. For any positive integers n, d and r, if k = d + r ≤ n, then
V = {v | v ∈ {0, 1}n, wt(v) = w} respectively. The entry of A at the row indexed by the pair (D, R) and the column indexed by the vector v ∈ V is 1 if the entries of v over D are all zero and at least one entry of v over R is one; and 0 otherwise.
Observe that each row of A has weight
min(r,w)
. By the Stein-Lov´asz theorem, there exists a submatrix M of A of order [n
Note that the equality is obtained by counting the weight of A in two ways. It is straightfor-ward to show that the columns of M form a (d, r)-disjunct matrix of order t × n. We then have
k is an integer.
We have
by Lemma 2.4.6 (taking s = 1). Therefore,
t(n, d, r) ≤ t(n0, d, r) <
Recall that t(n, d, r; z) is the minimum size over all (d, r; z)-disjunct matrices with n columns. and at least one entry of v over R is one; and 0 otherwise.
Observe that each row of A has weight
, and each column of
A has weight n − w
. By the extended Stein-Lov´asz theorem,
there exists a submatrix M of A of order [n d
Note that the equality is obtained by counting the weight of A in two ways. It is straight-forward to show that the columns of M form a (d, r; z)-disjunct matrix of order t × n. We then have
k is an integer.
We have
by Lemma 2.4.5 (taking s = 1), and
by Lemma 2.4.6 (taking s = 1). Therefore,
t(n, d, r; z) ≤ t(n0, d, r; z)
Recall that t(n, d, r, s] is the minimum size over all (d, s out of r]-disjunct matrices with n columns. and at least s entries of v over R are one; and 0 otherwise.
Observe that each row of A has weight
min(r,w)
has weight n − w
. By the Stein-Lov´asz theorem, there exists a
submatrix M of A of order [n
Note that the equality is obtained by counting the weight of A in two ways. It is straight-forward to show that the columns of M form a (d, s out of r]-disjunct matrix of order t × n.
We then have
t(n, d, r, s] <
k is an integer.
We have
by Lemma 2.4.5, and
ln(n0− w
by Lemma 2.4.6. Therefore,
Recall that t(n, d, r, s; z] is the minimum size over all (d, s out of r; z]-disjunct matrices with n columns.
Theorem 4.1.8. For any positive integers n, d, r, s and z, with 1 ≤ s ≤ r, if k = d + r ≤ n, and at least s entries of v over R are one; and 0 otherwise.
Observe that each row of A has weight
min(r,w)
, and each column of
A has weight n − w
. By the extended Stein-Lov´asz theorem,
there exists a submatrix M of A of order [n d
n − d r
] × t with each row weight at least z,
where
Note that the equality is obtained by counting the weight of A in two ways. It is straightfor-ward to show that the columns of M form a (d, s out of r; z]-disjunct matrix of order t × n.
We then have
t(n, d, r, s; z] <
k is an integer.
We have
by Lemma 2.4.5, and
ln(n0− w
by Lemma 2.4.6. Therefore,
t(n, d, r, s; z] ≤ t(n0, d, r, s; z]
<
zn0 d
n0− d r
n0− w d
n0− w − d r − s
w s
{1 + ln[
n0− w d
n0− w − d r − s
w s
]}
< z(ks)s(k−sk )k−s
r s
{1 + k[1 + ln(n
k + 1)] + lnk − s d
}
as required.