# About the inverse football pool problem for 9 games 1

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1 Seventh International Workshop on Optimal Codes and Related Topics September 6-1, 013, Albena, Bulgaria pp About the inverse football pool problem for 9 games 1 Emil Kolev Tsonka Baicheva Institute of Mathematics and Informatics Bulgarian Academy of Sciences, BULGARIA Dedicated to the memory of Professor Stefan Dodunekov Abstract. Consider the set Q n = {0, 1, } n equipped with the usual Hamming distance. Denote by T (n) the minimal number of spheres of radius n needed to cover Q n. The exact values of T (n) are known for n 8. The first undecided case is n = 9 and it is known that 67 T (9) 68. We settle the case by showing that T (9) = 68. The inequality T (9) = 68 implies T (10) 10, T (11) 153, T (1) 30 and T (13) 345 thus improving the best known lower bounds for 10 n Introduction In football pools one bets over 13 games. For each game he chooses between three possible outcomes win, draw or loss. The goal is to predict correctly as many games as possible. Finding the minimal number of bets in order to guarantee certain number of correctly predicted games is known as the football pool problem. To put this into mathematical terms consider the set Q n = {0, 1, } n with the usual Hamming distance. For x = (x 1, x,..., x n ) and y = (y 1, y,..., y n ) we define the Hamming distance d(x, y) as d(x, y) = {i x i y i }. A ball B(x, r) and a sphere S(x, r) with center x and radius r are defined as the sets B(x, r) = {y d(x, y) r}, S(x, r) = {y d(x, y) = r}. When we consider a bet as a point in Q n the football pool problem is defined as: For R = 1, or 3 find a set A Q n of minimal cardinality such that for every y Q n there exists x A for which d(x, y) R. In other words we cover the space Q n with balls of radius R. For more information on this topic the reader is referred to []. 1 This research is partially supported by the Bulgarian NSF under the contract I01/0003

2 16 OC013 We may consider football pools from the opposite point of view. That is, how to make certain number of bets in order to ensure that there exists a bet for which none of the games will be correctly predicted. This problem was first considered in Finish football pool magazine Veikkaaja. A subset A of Q n is called covering if the spheres of radius n centered at the elements of A cover Q n. In other words A Q n is a covering if for any y Q n there exists x A such that d(x, y) = n. The minimal cardinality of a covering of Q n is denoted by T (n). Finally, a covering A of Q n is called optimal if A = T (n). The problem of finding T (n) and all optimal coverings of Q n is known as the inverse football pool problem (see [1]). The sequence T (n) is part of the on-line encyclopedia of integer sequences [5], htpp://www.research.att.com/ njas/sequences/ number A After having the exact value of T (n) we are interested in finding the number of distinct optimal coverings. Two coverings A and B are equivalent if A is obtained from B by a permutation of the coordinates followed by a permutation of the elements of every coordinate. More precisely we have Definition 1. Two coverings A and B are equivalent if there exists a permutation σ S n and n permutations s 1, s,..., s n of {0, 1, } such that (x 1, x,..., x n ) A (s 1 (x σ(1) ), s (x σ() ),..., s n (x σ(n) ) B. Call a pair (σ, (s 1, s,..., s n )) equivalence transformation. There exist n!6 n equivalence transformations and therefore, in general there exist that many copies of every covering. The full automorphism group of A consists of all equivalence transformations that map A onto itself. The next proposition is straightforward and gives an important recursive bound on T (n). Lemma 1. The following inequality holds T (n) 3 T (n 1). Proof. Consider an optimal covering A of Q n, i.e. A is covering and A = T (n). For each i {0, 1, } denote by A i the set of vectors x = (x 1, x,..., x n 1 ) such that (x 1, x,..., x n 1, i) A. Therefore A = {x0 x A 0 } {x1 x A 1 } {x x A }. For {i, j, k} = {0, 1, } it is clear that A i A j is a covering of Q n 1. Therefore A i + A j T (n 1). The same argument implies that A i + A k T (n 1) and A j + A k T (n 1). Summing up these inequalities gives hence T (n) 3 T (n 1). T (n) = ( A i + A j + A k ) 3T (n 1),

3 Kolev, Baicheva 17 Remark 1. Suppose we know the exact value of T (n) for particular n and let T (n) be even. Moreover assume we know all optimal coverings of Q n. It follows from Lemma 1 that the least feasible value of T (n + 1) is T (n + 1) = 3 T (n). The proof of Lemma 1 implies that for any i, j {0, 1, }, i j the set A i A j is a covering of Q n and A 0 = A 1 = A = 1 T (n). Therefore A 0 is contained in the intersection of two coverings of Q n (these two coverings are A 0 A 1 and A 0 A ). This observation prompts the following approach. Since A 0 A 1 is a covering of Q n we may consider A = A 0 A 1 as one of the known coverings of Q n. Go through all copies of all known coverings of Q n and find a copy (denote this copy by B) that intersects A in at least 1 T (n) elements. If A B > A 0 = 1 T (n) then we have A 1 A 1 and therefore A 1 A < T (n), a contradiction. Thus, we have A B = A 0 = 1 T (n), A 1 = A\A 0 and A = B\A 0. What remains to be checked is whether A 1 A is a covering of Q n. If this is the case we have a covering of Q n+1 with cardinality 3 T (n). If one of the described steps (finding B and checking whether A 1 A is a covering of Q n ) fails then T (n + 1) > 3 T (n). Known results It is straightforward to show that T (1) = and using the inequality from Lemma 1 we obtain T () 3, T (3) 5, T (4) 8, T (5) 1, T (6) 18. Using the observations from Lemma 1 and Remark 1 it is not difficult to show that T (6) = 18 by finding a covering of Q 6 with 18 elements. This covering is given in the following table: Table 1: Optimal covering of Q 6.

4 18 OC013 Note that all optimal coverings for n = 1,, 3, 4, 5 are contained as substructures of the above covering. For example, to find a covering of Q 5 first choose any coordinate t and any two i, j {0, 1, }, i j. Second, take all elements from the covering of Q 6 having i or j in coordinate t and then delete this coordinate. As a result we obtain optimal covering of Q 5. Therefore we have first six exact values T (1) =, T () = 3, T (3) = 5, T (4) = 8, T (5) = 1, T (6) = 18. All these results and the bounds T (7) 9 and T (8) 44 are due to the Finish football pool magazine Veikkaaja. It is shown in [1] that for any n = 1,, 3, 4, 5, 6 there exists unique optimal covering of Q n. Observe that for all n 6 we have T (n) = 3 The first value of n for which T (n) 3 T (n 1) T (n 1). is n = 7. It has been shown in [1] by computer search that T (7) = 9 while the bound from Lemma 1 implies T (7) 7. An optimal covering of Q 7 is given in the following table. It is shown in [1] that this covering is unique Table : The unique optimal covering of Q Lemma 1 implies T (8) 44 and since a covering of Q 8 with 44 elements exists we conclude that T (8) = 44. The known results concerning T (n) are summarized in Table 3 and are taken from [1]. n T(n) Table 3. Results on T (n).

5 Kolev, Baicheva 19 The upper bounds for n = 9 and n = 10 were found in [4] using so-called tabu search. It is shown in [3] that up to equivalence there exist two optimal coverings of Q 8. They are given in Table 4 and Table Table 4: Optimal covering A 1 of Q Table 5. Optimal covering A of Q 8.

6 130 OC013 We continue by examining some properties of the two optimal coverings of Q 8. For obtaining the main result in this paper it is important to know the pairs distance distribution for both coverings. Those are given in the following table. t pairs from A 1 at distance t pairs from A at distance t Note that in both coverings the distance between any two elements is at least 4. The full automorphism group of A 1 has order 384 and the full automorphism group of A has order 4. 3 Main results The main result of this paper is given in the following Theorem. Theorem 1. It is true that T (9) 68. Proof. Suppose there exists a covering A of Q 9 with cardinality 67. For any t, 1 t 9 and any i {0, 1, } denote by A t i the set of elements of A having i in coordinate number t without this coordinate. Let also a t i = At i. It follows from T (8) = 44 that for any coordinate t we have {a t 0, a t 1, a t } = {1, 3, 3} or {,, 3}. In both cases there exists a special element i {0, 1, } such that a t j + at k = 44 for {i, j, k} = {0, 1, }. Note that in the case {1, 3, 3} there exist two special elements. Without loss of generality assume that a a9 = 44. Hence, the set A9 1 A9 is equivalent to A 1 or A. The following Lemma provides an important property of the elements of A 9 1 and A 9. Lemma. Let A be a covering of Q 9 and u = (u 1,..., u 8 ) and v = (v 1,..., v 8 ) be two elements from A 9 1 A9 such that d(u, v) = 4. If there exists t, 1 t 8, such that u t v t and {0, 1, } \ {u t, v t } is a special element for coordinate t, then u and v are not simultaneous elements of A 9 i for i = 1,. Proof. Let u and v be vectors satisfying the given properties. Denote the extensions of u and v in A without coordinate t by u and v. Since {0, 1, } \

7 Kolev, Baicheva 131 {u t, v t } is special element for coordinate t it follows that u and v are elements of one of the two optimal coverings A 1 or A of Q 8. If u, v A t i for i = 1 or then d(u, v) = 3 (since we delete coordinate t where u t v t and add coordinate 9 where u and v have one and the same element). This is a contradiction with d(x, y) 4 for any x, y A i for i = 1 or. Definition. A vector (u 1,..., u 8 ) is called characteristic vector for the covering A of Q 9 if for any t, 1 t 8 the element u t is a special element for coordinate t. Recall that A 9 1 A9 is equivalent to A 1 or A. Suppose that for given characteristic vector there exist vectors u, v, w from A 1 for i = 1 or, any two of which satisfy the conditions of Lemma. Since at least two of them are elements of A 9 i for i = 1 or we have a contradiction to Lemma. Therefore this characteristic vector has to be rejected. For example, consider elements 1,3,4 from covering A 1 of Table 4. Any two of these elements satisfy the conditions of Lemma for all characteristic vectors of the form (u 1, u, u 3, u 4,, u 6, u 7, u 8 ) where u 1 = 0 or u = 1 or u 3 = 0 or u 4 = 1. For all possible 3 8 = 6561 characteristic vectors we try to find three vectors u, v, w from A i for i = 1 or, any two of which satisfy the condition of Lemma. For the first covering for any characteristic vector such a triple always exists. Hence all characteristic vectors are rejected and therefore A 9 1 A9 is not equivalent to A 1. For the second covering A we obtain only 4 characteristic vectors for which such a triple does not exist: (000100); (000010); (10100); (10010). Since the permutation (4)(56) is an automorphism of A we may consider only first and third vectors. Thus, without loss of generality we assume that A 9 1 A9 = A and there are two possible characteristic vectors (000100) and (10100). For a particular characteristic vector the next step is to extend each vector with 1 or, i.e. we have to split the elements of A into A 9 1 and A9. Again we make use of Lemma. Let the extension of the first vector be 1, i.e. x = ( ) is in A 9 1. Consider a vector y A, for which d(x, y) = 4. If there exists a coordinate for which the two entries of x and y in this coordinate and the corresponding special element are pairwise distinct (equivalently they form the set {0, 1, }) then it follows from Lemma that y is extended by. This procedure is applicable to any vector that has already been extended. Eventually we extend all vectors. For each of the two characteristic vectors

8 13 OC013 we obtain only two possible extensions. They are given in the next table (i-th element of the given vector is the extension of the i-th element of A ). Characteristic vector extensions of the elements Up to now we know all elements of A 9 1 and A9. Thus, it remains to find the elements of A 9 0, i.e. the elements with last coordinate 0. Note that A9 0 = 3, so we need 3 vectors. Let xi for i = 1 or and y0 be two elements of the covering A. If there exists a coordinate for which the two entries of xi and y0 and the corresponding special element are pairwise distinct then d(x, y) 3. Therefore in order to find all possible elements of A 9 0 we have to find all vectors y of length 8 with the described property. Direct verification shows that there exist 178 such vectors. Denote this set by B. Furthermore, for any two vectors x and y from A 9 0 we have that if there exists a coordinate for which both x and y differ from the corresponding special element then d(x, y) 4. Therefore we have to find a subset A 9 0 of B with cardinality 3 any two elements of which satisfy the above property. In addition, for any coordinate t, 1 t 9 of the set A = A 9 0 A9 1 A9 we must have {a t 0, at 1, at } = {3, 3, 1} or {3,, }. Computer search finds no such set. Therefore T (9) > 67 and since there exists a covering of Q 9 with 68 elements we conclude that T (9) = 68. In order to verify the computer search results, all the computations have been carried out independently by different programs written on Pascal and C++ developed by the authors. The time needed to perform the steps of the computation ranges from a few minutes to a few hours for the last step. The exact value T (9) = 68 and Lemma 1 imply that References T (10) 10, T (11) 153, T (1) 30, T (13) 345. [1] D. Brink, The inverse Football pool problem, Journal of Integer Sequences Vol. 14, 011, 1 9, article [] H. Hämäläinen, I. Honkala, S. Litsyn, P. R. J. Ostergard, Football pools - a game for mathematicians, Amer. Math. Monthly 10, 1995, [3] E. Kolev, How to have a wrong bet in football pools, accepted for publication in CR Acad. Bulg. Sci.

9 Kolev, Baicheva 133 [4] P. R. J. Östergärd, T. Riihonen, A covering problem for tori, Ann. Comb. 7, (003), [5] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, htpp://www.research.att.com/ njas/sequences/.

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