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# The inverse football pool problem

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## Abstract

The minimal number of spheres (without "interior") of radius n required to cover the finite set {0,…, q – 1}^n equipped with the Hamming distance is denoted by T(n, q). The only hitherto known values of T(n, q) are T(n, 3) for n = 1,…, 6. These were all given in the 1950s in the Finnish football pool magazine Veikkaaja along with upper and lower bounds on T(n, 3) for n ≥ 7. Recently, Östergård and Riihonen found improved upper bounds on T(n, 3) for n = 9, 10, 11, 13 using tabu search. In the present paper, a new method to determine T(n, q) is presented. This method is used to find the next two values of T(n, 3) as well as six non-trivial values of T(n, q) with q > 3. It is also shown that, modulo equivalence, there is only one minimal covering of {0, 1, 2}^n for each n ≤ 7, thereby proving a conjecture of Östergård and Riihonen. For reasons discussed in the paper, it is proposed to denote the problem of determining the values of T(n, 3) as the inverse football pool problem.
23
11
Article 11.8.8
Journal of Integer Sequences, Vol. 14 (2011),
2
3
6
1
47
The Inverse Football Pool Problem
David Brink
Institut for Matematiske Fag
Københavns Universitet
Universitetsparken 5
DK-2100
Denmark
brink@math.ku.dk
Abstract
The minimal number of spheres (without “interior”) of radius n required to cover
the ﬁnite set {0, . . . , q1}
n
equipped with the Hamming distance is denoted by T (n, q).
The only hitherto known values of T (n, q) are T (n, 3) for n = 1, . . . , 6. These were all
given in the 1950s in the Finnish football po ol magazine Veikkaaja along with upper
and lower bounds for T (n, 3) for n 7. Recently,
¨
Osterg˚ard and Riihonen found
improved upper bounds for T (n, 3) for n = 9, 10, 11, 13 using tabu search. In the
present paper, a new method to determine T (n, q) is presented. This method is used
to ﬁnd the next two values of T (n, 3) as well as six non-trivial values of T (n, q) with
q > 3. It is also shown that, modulo equivalence, there is only one minimal covering of
{0, 1, 2}
n
for each n 7, thereby proving a conjecture of
¨
Osterg˚ard and Riihonen. For
reasons discussed in the paper, it is proposed to denote the problem of determining
the values of T (n, 3) as the inverse football pool problem.
1 Introduction
Consider the ﬁnite set Q = {0, . . . , q 1}
n
equipped with the Hamming distance
d(x, y) = |{i : x
i
6= y
i
}|
for x = (x
1
, . . . , x
n
) and y = (y
1
, . . . , y
n
) in Q. Let
B(x
0
, r) = {x Q : d(x, x
0
) r}, S (x
0
, r) = {x Q : d(x, x
0
) = r}
be the ball and sphere, respectively, with centre x
0
1
1
Note that the word “sphere” is often used in the literature for what is here called a “ball.”
1
These simple concepts give rise to a plethora of combinatorial problems. The most famous
is the football pool problem which asks for the minimal number K
3
(n, 1) of balls of radius 1
required to cover {0, 1, 2}
n
. In the football pools, one bets simultaneously on the outcomes of
a number of football games, typically n = 12 or 13. Each game has th ree possible outcomes:
win, draw or loss. A bet can thus be thought of as a point in {0, 1, 2}
n
, and K
3
(n, 1) is equal
to the minimal number of bets on n games required to guarantee at least one bet with at
most one wrong outcome.
The complexity of such problems increases drastically with n. Indeed, when n is not of the
form (3
m
1)/2 and the existence of a perfect Hamming code shows K
3
(n, 1) = 3
n
/(2n + 1),
then K
3
(n, 1) is only known for n = 2, 3, 5, cf. Sloane’s A004044 [
interesting discussion of this and related problems.
In the present paper, we study coverings of Q with spheres of radius n. S uch spheres are
n-dimensional grid graphs, i.e., they are of the form
Q
n
i=1
A
i
where A
i
{0, . . . , q 1} with
|A
i
| = q 1. The function deﬁned below therefore equals the special case T (n, q, q 1) of
the function T (n, q, p) introduced by
¨
Osterg˚ard and Riihonen [
3].
Deﬁnition 1. Let T (n, q) be the minimal number of spheres of radius n required to cover Q.
We shall be particularly concerned with the problem of determining the values of T (n, 3),
Sloane’s A086676. We propose to call this the inverse football pool problem since it means
asking for the minimal number of bets needed to guarantee that at least one is totally wrong.
The inverse football pool problem was ﬁrst considered in the Finnish football pool mag-
azine Veikkaaja in the 1950s.
2
The principal goal then was to determine T(12, 3), a value
which remains unknown today. The only hitherto known values of T (n, q) are T(1, 3) = 2,
T (2, 3) = 3, T(3, 3) = 5, T (4, 3) = 8, T (5, 3) = 12, and T (6, 3) = 18. The history of these
results is discussed below.
The results in Veikkaaja on the inverse football pool problem seem to have been unknown
to the mathematical community until they were discovered by
¨
Osterg˚ard and Riihonen in
2002 and presented in [
3] and [4]. All references to Veikkaaja here come from these two
papers.
2 General bounds
First we state some easy or well-known bounds. If a sphere covering of {0, . . . , q 1}
n
is
given, a covering of {0, . . . , q 1}
n1
is obtained by discarding the last coordinate as well
as all spheres whose centres have any given value in that coordinate, cf. Veikkaaja no. 23,
1956, and [
3, Theorem 3.3]. Hence
T (n, q)
q
q 1
· T (n 1, q). (1)
2
As was the ordinary football pool problem in the previous decade. Indeed, Veikkaaja holds a place of
the ﬁrst rank in the history of combinatorics. It was here that Juhani Virtakallio gave the ternary Golay
code (a perfect covering of {0, 1, 2}
11
with 729 balls of radius 2) in 1947, two years before its rediscovery by
Marcel Golay.
2
In particular, it follows inductively from T (1, q) = 2 that T (n, q) n + 1. If n < q, the
spheres with centres (0, . . . , 0), . . . , (n, . . . , n) cover Q. Thus
T (n, q) = n + 1 for n < q. (2)
Taking direct product gives
T (n + n
, q) T (n, q) · T (n
, q). (3)
Finally, an exceedingly elegant and ingenious construction for q = 3, described in Veikkaaja
no. 15, 1960, and [
3, Corollary 3.5], gives the bound
T (n + n
2, 3)
1
3
· T (n, 3) · T (n
, 3) (4)
(and sometimes a little more, cf. [
3, Theorem 3.4]) which is considerably better than (3).
Deﬁnition 2. (a) We represent a covering of {0, . . . , q 1}
n
with m spheres S (x
1
, n), . . . ,
S (x
m
, n) by the n × m covering matrix A whose columns are the centres x
1
, . . . , x
m
.
(b) Two covering matrices of the same size (and the sphere coverings they represent) are
called equivalent if one can be transformed into the other by permuting rows and columns
and by renaming the symbols in any given row.
3
(c) We call a covering matrix canonical if it is minimal among all equivalent matrices
with resp ect to the lexicographic ordering (reading ﬁrst row 1 from left to right, then row 2
from left to right, etc.).
The deﬁnition of equivalence is standard. If n < q, for example, the matrix consisting of
n identical rows (0, 1, . . . , n) is the unique canonical n×(n+1) covering matrix, cf. (
2). Note
that there is a 1–1 correspondence between equivalence classes of coverings and canonical
covering matrices. The following lemma is a generalization of (
1).
Lemma 3. Suppose A is an n × m covering matrix. Let there be given b
i
{0, . . . , q 1}
for all i in some proper subset I {1, . . . , n}. Then the number of columns (a
1
, . . . , a
n
)
t
of
A with a
i
6= b
i
for all i I is at least T (n |I|, q).
Proof. For each i I, delete all columns (a
1
, . . . , a
n
)
t
of A having a
i
= b
i
. Then delete row i
for each i I. The resulting matrix A
is a covering matrix for {0, . . . , q 1}
n−|I|
and hence
has at least T (n |I|, q) columns.
3 Main results
Theorem 4. We have the following two values of T (n, 3), i.e., the minimal number of
spheres of radius n needed to cover {0, 1, 2}
n
: T (7, 3) = 29 and T (8, 3) = 44.
3
By “renaming the symbols,” we mean that given any row a = (a
1
, . . . , a
m
) and given any permutation
σ of 0, . . . , q 1, we may replace a by (σ(a
1
), . . . , σ(a
m
)).
3
Proof. The lower bounds T (7, 3) 27 and T (8, 3) 41 follow from (1) and (2). The upper
bounds T (7, 3) 29 and T (8, 3) 44 are known, cf. [
3] and the discussion below. After
having shown T (7, 3) = 29, it will follow that T (8, 3) = 44 by (
1). The rest of the proof
consists of three parts. F irst we describe in general our method for ﬁnding canonical covering
matrices. Then we illustrate this method by proving in detail the bound T (3, 7) > 27. Finally
we discuss how the method can be used to prove T (3, 7) > 28.
Suppose A is a canonical n × m covering matrix. Then, by Deﬁnition 2 and Lemma 3,
the following two conditions hold for each i = 1, . . . , n:
(I) The i × m submatrix A
i
consisting of the ﬁrst i rows of A is canonical.
(II) Given b
1
, . . . , b
i
{0, . . . , q 1}, there are at least T (n i, q) columns (a
1
, . . . , a
i
)
t
of
A
i
satisfying a
j
6= b
j
for all j = 1, . . . , i.
In order to ﬁnd all canonical n × m matrices, we proceed as follows: First ﬁnd the set R
1
of
all rows r
1
such that the 1×m matrix A
1
consisting only of r
1
satisﬁes (I) and (II). Then, for
each r
1
R
1
, ﬁnd the set R
2
(r
1
) of all rows r
2
such that the 2 × m matrix A
2
consisting of r
1
and r
2
satisﬁes (I) and (II). Next, for each r
1
R
1
and r
2
R
2
(r
1
), ﬁnd the set R
3
(r
1
, r
2
),
and so forth. Every time a non-empty set R
n
(r
1
, . . . , r
n1
) is obtained in this way, each row
r
n
in that set gives rise to a canonical n × m covering matrix, namely the matrix consisting
of r
1
, . . . , r
n
.
To illustrate the method just described, we prove in detail the bound T (7, 3) > 27.
Suppose indirectly that a covering of {0, 1, 2}
7
is given as a canonical 7 × 27 covering matrix
A. First we ﬁnd R
1
. By condition (I), every r
1
R
1
is of the form
r
1
= (
l
0
times
z
}| {
0, . . . , 0,
l
1
times
z }| {
1, . . . , 1,
l
2
times
z }| {
2, . . . , 2)
with
l
0
+ l
1
+ l
2
= 27. (5)
By specifying b
1
{0, 1, 2} and using T (6, 3) = 18, condition (II) gives the following three
linear inequalities:
l
1
+ l
2
18, l
0
+ l
2
18, l
0
+ l
1
18.
(6)
It follows immediate from (
5) and (6) that l
0
= l
1
= l
2
= 9 and hence
r
1
= (0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2).
Then we ﬁnd R
2
(r
1
). By condition (I), every r
2
R
2
(r
1
) is of the form
r
2
= (
l
0
times
z
}| {
0, . . . , 0,
l
1
times
z
}| {
1, . . . , 1,
l
2
times
z
}| {
2, . . . , 2,
l
0
times
z
}| {
0, . . . , 0,
l
1
times
z
}| {
1, . . . , 1,
l
2
times
z
}| {
2, . . . , 2,
l
′′
0
times
z
}| {
0, . . . , 0,
l
′′
1
times
z
}| {
1, . . . , 1,
l
′′
2
times
z
}| {
2, . . . , 2)
with
l
0
+ l
1
+ l
2
= l
0
+ l
1
+ l
2
= l
′′
0
+ l
′′
1
+ l
′′
2
= 9. (7)
4
By specifying b
1
, b
2
{0, 1, 2} and using T (5, 3) = 12, condition (II) now gives the following
nine linear inequalities:
l
1
+ l
2
+ l
′′
1
+ l
′′
2
12, l
0
+ l
2
+ l
′′
0
+ l
′′
2
12, l
0
+ l
1
+ l
′′
0
+ l
′′
1
12,
l
1
+ l
2
+ l
′′
1
+ l
′′
2
12, l
0
+ l
2
+ l
′′
0
+ l
′′
2
12, l
0
+ l
1
+ l
′′
0
+ l
′′
1
12,
l
1
+ l
2
+ l
1
+ l
2
12, l
0
+ l
2
+ l
0
+ l
2
12, l
0
+ l
1
+ l
0
+ l
1
12.
(8)
It is not diﬃcult to show that (
7) and (8) imply l
0
= l
1
= l
2
= l
0
= l
1
= l
2
= l
′′
0
= l
′′
1
= l
′′
2
= 3
and consequently
r
2
= (0, 0, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 1, 1, 1, 2, 2, 2).
Similarly, in order to determine the elements r
3
R
3
(r
1
, r
2
), we get from (I), (II), and
T (4, 3) = 8 a system of 27 linear inequalities in 27 unknowns which together imply
r
3
= (0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2).
The determination of R
4
(r
1
, r
2
, r
3
) is somewhat more complicated. It is possible by hand to
proceed as above and show that this set consists only of
r
4
= (0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1, 2, 0).
It is then relatively easy to show that R
5
(r
1
, r
2
, r
3
, r
4
) is empty and that in consequence
no 7 × 27 covering matrix of {0, 1, 2}
7
exists. However, it is also possible at this point to
make a short-cut. The determination of r
1
, r
2
, and r
3
shows that every possible column
vector (b
1
, b
2
, b
3
)
t
with b
i
{0, 1, 2} occurs exactly λ = 1 times in the 3 × 27 matrix A
3
. By
symmetry, the same holds for any 3 × 27 submatrix of A formed by t = 3 arbitrary rows.
This means that the 7 × 27 matrix A is an orthogonal array with q = 3 levels, index λ = 1,
and strength t = 3. But by a result of Bush [
1], a such matrix can have at most t + 1 = 4
rows, thus disproving the existence of A.
The method used to prove T (7, 3) > 28 is exactly the same, but now the row sets R
i
for
the hypothetical 7 × 28 covering matrix become too large to do the demonstration by hand.
For example, R
1
consists of the two rows
r
1
= (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2),
r
1
= (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2),
and R
2
(r
1
) consists of
r
2
= (0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 0, 0, 1, 1, 2, 2, 2, 2),
r
2
= (0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 0, 0, 1, 1, 1, 2, 2, 2),
while R
2
(r
1
) consists of
r
2
= (0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 0, 0, 0, 1, 1, 2, 2, 2, 2),
r
2
= (0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 1, 1, 1, 2, 2, 2).
Instead, the method can be implemented on a computer to show that no such matrix exists.
This computation took 28 seconds on the author’s laptop.
5
n T (n, 3)
1 2
2
3
3
5
4
8
5
12
6
18
7
29
8
44
9
66 68
10
99 104
11
149 172
12
224 264
13
336 408
Table 1: Old and new results on T (n, 3).
Table 1 collects old and new results on T (n, 3). All values for n 6 as well as the bounds
T (7, 3) 29 and T (8, 3) 44 go back to Veikkaaja.
4
All lower bounds for n 9 come
from T(7, 3) = 29 combined with (
1). Note that T (7, 3) = 29 is the ﬁrst counter-example
to the natural conjecture T (n, 3) =
3
2
· T (n 1, 3). The upper bounds for n = 9, 10 were
found in [
3] using so-called t abu search. The upper bounds for n = 11, 12, 13 follow from the
upper bounds for smaller n using (
4). A covering proving T (12, 3) 264 appears explicitly
in Veikkaaja no. 52, 1960. As related in [
4], a writer in fact claims in Veikkaaja no. 15, 1960,
to b e in possession of a covering proving T (12, 3) 242, but states—not unlike Pierre de
Fermat three centuries earlier—that it is to o long to be printed in the magazine!
The case n = 7 of the following theorem was conjectured in [
3].
Theorem 5. Modulo equivalence, there is only one minimal cov ering of {0, 1, 2}
n
with
spheres of radius n for each n = 1, . . . , 7.
Proof. The statement can b e proved by hand for n 6. We exemplify this by sketching the
case n = 6. By the same arguments as in the proof of Theorem
4, the ﬁrst three rows of a
canonical 6 × 18 covering matrix A must be
r
1
= (0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2),
r
2
= (0, 0, 1, 1, 2, 2, 0, 0, 1, 1, 2, 2, 0, 0, 1, 1, 2, 2),
r
3
= (0, 1, 0, 2, 1, 2, 0, 2, 1, 2, 0, 1, 1, 2, 0, 1, 0, 2).
Now there are three candidates for row 4:
r
4
= (0, 1, 1, 2, 2, 0, 2, 1, 2, 0, 1, 0, 0, 2, 0, 1, 2, 1),
r
4
= (0, 1, 2, 0, 2, 1, 2, 0, 1, 2, 1, 0, 2, 1, 1, 0, 0, 2),
r
′′
4
= (0, 1, 2, 1, 0, 2, 1, 2, 2, 0, 0, 1, 2, 0, 1, 0, 2, 1).
But also rows 5 and 6 must be chosen among r
4
, r
4
, and r
′′
4
. One can check directly that the
only way to get a covering is to take one of each. Since the rows of a canonical matrix must
come in increasing order, rows 4–6 of A must be r
4
, r
4
, and r
′′
4
in that order.
The case n = 7 is too complicated to settle by hand, but an exhaustive computer search
following the same principles as described in the proof of Theorem 4 ﬁnds only one canonical
matrix (namely the one given at the end of this paper).
4
It remains somewhat of a mystery, though, how the coverings proving these results were found.
6
We do not know if the covering of {0, 1, 2}
8
with 44 spheres is unique.
Next we consider values of T (n, q) with q > 3 (the case q = 2 being trivial).
Theorem 6. We have the following values of T (n, q), i.e., the minimal number of spheres
of radius n needed to cover {0, . . . , q 1}
n
:
n
T (n, 4) T (n, 5) T (n, 6) T (n, 7)
1 2 2 2 2
2
3 3 3 3
3
4 4 4 4
4
7 5 5 5
5
10 8 6 6
6
14 16 11 10 7
7
19 28 14 20 12 18 11
The values typed in boldface are the non-trivial ones, i.e., the ones not following from equation
(2).
Proof. The theorem is shown by a computer search following the same principles as in the
proof of Theorem
4. All results were found within 24 hours of computer time, some of them
within minutes or seconds. A few non-trivial partial results can be proved by hand. For
example, one can see T (4, 4) > 6 by showing that rows 1–3 of a canonical 4 × 6 covering
matrix must necessarily be (0, 0, 1, 1, 2, 3), (0, 1, 0, 1, 2, 3), (0, 1, 1 ,0, 2, 3), but that
this leaves no possibilities for row 4. All lower and upper bounds in the undecided cases are
straightforward consequences of (
1) and (3).
It appears that minimal coverings are more abundant when q is composite.
4 Appendix: Lexicographically minimal covering ma-
trices
Finally, we present for each non-trivial value of T (n, q) given in Theorem
6 as well as for
q = 3 and n = 3, . . . , 7 the unique lexicographically minimal n × T (n, q) covering matrix.
T (3, 3) = 5: T (4, 3) = 8: T (5, 3) = 12:
(0, 0, 1, 1, 2) (0, 0, 0, 1, 1, 1, 2, 2) (0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2)
(0, 1, 0, 1, 2) (0, 0, 1, 0, 1, 1, 2, 2) (0, 0, 1, 2, 0, 1, 1, 2, 0, 1, 2, 2)
(0, 1, 1, 0, 2) (0, 1, 2, 2, 0, 1, 0, 1) (0, 1, 0, 2, 1, 0, 1, 2, 2, 2, 0, 1)
(0, 1, 2, 2, 0, 1, 1, 0) (0, 1, 2, 1, 2, 1, 0, 2, 0, 2, 1, 0)
(0, 1, 2, 2, 2, 1, 0, 1, 2, 0, 0, 1)
7
T (6, 3) = 18:
(0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2)
(0, 0, 1, 1, 2, 2, 0, 0, 1, 1, 2, 2, 0, 0, 1, 1, 2, 2)
(0, 1, 0, 2, 1, 2, 0, 2, 1, 2, 0, 1, 1, 2, 0, 1, 0, 2)
(0, 1, 1, 2, 2, 0, 2, 1, 2, 0, 1, 0, 0, 2, 0, 1, 2, 1)
(0, 1, 2, 0, 2, 1, 2, 0, 1, 2, 1, 0, 2, 1, 1, 0, 0, 2)
(0, 1, 2, 1, 0, 2, 1, 2, 2, 0, 0, 1, 2, 0, 1, 0, 2, 1)
T (7, 3) = 29:
(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2)
(0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 0, 0, 1, 1, 2, 2, 2, 2)
(0, 0, 1, 1, 0, 0, 1, 1, 2, 2, 2, 0, 0, 1, 1, 0, 0, 1, 1, 2, 2, 2, 2, 2, 2, 0, 0, 1, 1)
(0, 1, 0, 2, 0, 2, 1, 2, 0, 1, 2, 1, 2, 0, 2, 0, 2, 0, 1, 0, 2, 0, 1, 1, 2, 0, 1, 1, 2)
(0, 1, 1, 2, 2, 1, 2, 0, 1, 0, 2, 2, 0, 2, 1, 1, 2, 0, 1, 2, 1, 1, 0, 2, 0, 0, 1, 0, 2)
(0, 1, 2, 0, 2, 0, 1, 2, 0, 1, 2, 1, 2, 2, 0, 2, 0, 0, 1, 0, 2, 1, 0, 2, 1, 1, 0, 2, 1)
(0, 1, 2, 1, 1, 2, 2, 0, 1, 0, 2, 2, 0, 1, 2, 2, 1, 0, 1, 2, 1, 0, 1, 0, 2, 1, 0, 2, 0)
T (4, 4) = 7: T (5, 4) = 10: T (5, 5) = 8:
(0, 0, 0, 1, 1, 1, 2) (0, 0, 0, 1, 1, 1, 2, 2, 2, 3) (0, 0, 0, 1, 1, 1, 2, 3)
(0, 1, 2, 0, 1, 2, 3) (0, 0, 0, 1, 1, 1, 2, 2, 2, 3) (0, 1, 2, 0, 1, 2, 3, 4)
(0, 1, 2, 0, 1, 2, 3) (0, 1, 2, 0, 1, 2, 0, 1, 2, 3) (0, 1, 2, 0, 1, 2, 3, 4)
(0, 1, 2, 1, 2, 0, 3) (0, 1, 2, 0, 1, 2, 0, 1, 2, 3) (0, 1, 2, 1, 2, 0, 3, 4)
(0, 1, 2, 1, 2, 0, 2, 0, 1, 3) (0, 1, 2, 1, 2, 0, 3, 4)
T (6, 5) = 11: T (6, 6) = 10:
(0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4) (0, 0, 0, 0, 1, 1, 1, 1, 2, 3)
(0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4) (0, 1, 2, 3, 0, 1, 2, 3, 4, 5)
(0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 4) (0, 1, 2, 3, 0, 1, 2, 3, 4, 5)
(0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 4) (0, 1, 2, 3, 0, 1, 2, 3, 4, 5)
(0, 1, 2, 1, 2, 0, 2, 0, 1, 3, 4) (0, 1, 2, 3, 1, 0, 3, 2, 4, 5)
(0, 1, 2, 1, 2, 0, 2, 0, 1, 3, 4) (0, 1, 2, 3, 2, 3, 0, 1, 4, 5)
T (7, 7) = 11:
(0, 0, 0, 0, 1, 1, 1, 1, 2, 3, 4)
(0, 1, 2, 3, 0, 1, 2, 3, 4, 5, 6)
(0, 1, 2, 3, 0, 1, 2, 3, 4, 5, 6)
(0, 1, 2, 3, 0, 1, 2, 3, 4, 5, 6)
(0, 1, 2, 3, 1, 2, 3, 0, 4, 5, 6)
(0, 1, 2, 3, 1, 2, 3, 0, 4, 5, 6)
(0, 1, 2, 3, 1, 2, 3, 0, 4, 5, 6)
References
[1] K. A. Bush, Orthogonal arrays of index unity, Ann. Math. Statistics 23 (1952), 426–434.
8
[2] H. am¨al¨ainen, I. Honkala, S. Litsyn, and P. R. J.
¨
Osterg˚ard, Football pools—a game
for mathematicians, Amer. Math. Monthly 102 (1995), 579–588.
[3] P. R. J.
¨
Osterg˚ard and T. Riihonen, A covering problem for tori, Ann. Comb. 7 (2003),
357–363.
[4] T. Riihonen, How to gamble 0 correct in football pools, Helsinki University of Technol-
ogy, 2002. Available at
http://users.tkk.fi/priihone/tuotokset.html.
[5] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences. Published electroni-
cally at
http://oeis.org/.
2010 Mathematics Subject Classiﬁcation: Primary 05B40; Secondary 11H31, 52C17.
Keywords: Minimal covering codes, football pool problem.
Concerned with sequences A004044 and A086676.
Received January 14 2011; revised versions received February 19 2011; June 28 2011; Septem-
ber 5 2011. Published in Journal of Integer Sequences, October 16 2011.
9
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How to gamble 0 correct in football pools
• T Riihonen
T. Riihonen, How to gamble 0 correct in football pools, Helsinki University of Technology, 2002. Available at http://users.tkk.fi/priihone/tuotokset.html.