Document Sample

Sudoku, the Golden Ratio, Determinants, Eigenvectors and Matrices In Z Merciadri Luca⋆ Luca.Merciadri@student.ulg.ac.be Abstract. The Sudoku game has gained much interest for a dozen years. It is now put in various mag- azines and mathematical research is more and more interested in it. This document aims at providing some newer pieces of information about the mathematical properties of the Sudoku, but not according to graphs’ theory. We do not speak about the “minimum number of clues,” but about (erroneous or not) Sudokus’ matrix interpretation: general properties of its determinant, in relation to eigenvalues and eigenvectors; transpose; neither symmetric nor antisymmetric character; non-Hermitian character; non-normal character and non-orthogonal character; trace and antitrace. A part is also devoted to the proposal of Prof. Steven Leon, which states that, given a 9×9 Sudoku matrix, the maximum eigenvalue of a given 3 × 3 submatrix extracted from it, equals 22. Last sections give results about Numerical Analysis of Sudoku matrices and the representation of a Sudoku when its matrix is taken modulo 2, and then considered as the adjacency matrix of a graph. Keywords: Sudoku, Determinants, Matrix, Matrices, Eigenvalues, Eigenvectors. REVISION 1: 08/05/2010 1 Introduction The Sudoku game has gained much interest for a dozen years. It is now put in various magazines and mathematical research is more and more interested in it. This document aims at providing some newer pieces of information about the mathematical properties of the Sudoku, but not according to graphs’ theory. We do not speak about the “minimum number of clues,” but about (erroneous or not) Sudokus’ matrix interpretation: general properties of its determinant, in relation to eigenvalues and eigenvectors; transpose; neither symmetric nor antisymmetric character; non-Hermitian character; non-normal character and non- orthogonal character. A part is also devoted to the proposal of Prof. Steven Leon, which states that, given a 9 × 9 Sudoku matrix, the maximum eigenvalue of a given 3 × 3 submatrix extracted from it, equals 22. 2 Conventions Through this paper, we use the following conventions: – S denotes the set of all the (solved) Sudokus. We have |S| ≈ 6.771 · 1021 , according to [3], – Si,j denotes the element at the ith row and the jth column of the general S Sudoku square matrix. We shall use 1 ≤ i ≤ n, 1 ≤ j ≤ n, where n denotes the number of rows (or columns, as it is always a square matrix) of the Sudoku (we always consider n ≥ 4). S has thus the following form: S1,1 S1,2 · · · S1,n S2,1 S2,2 · · · S2,n S= . . .. . , . . . . . .. Sn,1 Sn,2 · · · Sn,n ⋆ The author is thankful to Prof. Steven Leon, University of Massachussets, Dartmouth, for his idea of the maximum eigenvalue in a submatrix of a Sudoku 9 × 9 matrix. n – j=1 Si,j = n(n+1) , 1 ≤ i ≤ n (i being free in this interval) denotes the sum of all the elements on one 2 line i of S. Evidently, n n n(n + 1) n(n + 1) Si,j = = Si,j = , j=1 2 i=1 2 because of the fundamental properties of the Sudoku, – Z∗ denotes the set of the integers without 0; that is, Z \ {0}, – C is the transpose of the matrix C, – C ∗ is the adjoint of the matrix C, i.e. C, n n – Tr(C) = i=1 Ci,i , antiTr(C) = i=1 Ci,n−i+1 , – M i,j is the algebraic minor of the element at the ith row and the jth column of a given matrix, – cofact(Ci,j ) = (−1)i+j · M i,j , – The dimension of a Sudoku matrix is expressed as a product n × n, where n denotes its dimension, and n2 the number of elements it is ﬁlled with, – When taking matrices with a n > 9 (i.e. 16 × 16 or 25 × 25), we assume there exists a suﬃciently complete alphabet A which has n distinct symbols. For example, when speaking about 16 × 16 Sudokus, “A” is used for “10,” “B” is used for “11,” . . . , until “G” is used for “16.” Evidently, we assume the reader is familiar with the notions of Sudoku, Latin Square, and related concepts. 3 Determinants 3.1 General Case As S is a square matrix, one can wonder about the value of det(S), for a given n. Based on many experiments, det S has the following properties: – It is in Z∗ or in Z, – The assertion (det(S) (mod 2) = 0) ∨ (det(S) (mod 2) = 0) (1) is always true, but det S is sometimes even, and sometimes odd. 3.2 Why it Sometimes Does Not Equal Zero If det(S) = 0, S −1 cannot exist, and 0 must be an eigenvalue of the spectrum of S. Thus, proving that, for a given n, (det(S) = 0) is sometimes a true assertion is the same as proving that 0 is sometimes an eigenvalue of the spectrum of S. However, proving that 0 is sometimes an eigenvalue of the spectrum of S would be less straightforward. The rows of the Sudoku are sometimes linearly dependent, and we can only assume that (det(S) = 0) is sometimes a true assertion. Theorem 1 (Determinant of S sometimes equals 0). The determinant of the S matrix, det S, formed by the elements of a complete Sudoku, can equal 0. ⊔ ⊓ Proof. If det(S) = 0, the rows of S are linearly dependent. We prove ﬁrst that there exists at least a situation where det(S) = 0, S still being a valid Sudoku matrix. It is shown above for at least two rows (it is the same for columns, just transpose). There are n! possible orders of the elements of a row of S. From these possibilities, – Let’s take 947258136 123467985 658193724 895642371 det(B) = 7 6 4 9 3 1 2 5 8 = 0, 312875469 489726513 236519847 571384692 – It is shown in [2] that abcd abdc cdab cdba = = a4 + b4 + c4 + d4 − 2a2 b2 − 2a2 c2 − 2b2 c2 − 2a2 d2 − 2b2 d2 − 2c2 d2 + 8abcd, dcba bacd badc dcab which is logical, as there is an even number of permutations of rows (and columns) between the ﬁrst and the second matrix. In the standard case, i.e. if (a, b, c, d) are mapped bijectively to (1, 2, 3, 4), we have a4 + b4 + c4 + d4 − 2a2 b2 − 2a2 c2 − 2b2 c2 − 2a2 d2 − 2b2 d2 − 2c2 d2 + 8abcd = 0. Note that 673514982 921368754 584792136 896237415 2 1 5 9 4 6 3 7 8 = 0. 347185269 452871693 169453827 738629541 This result will be used later. ⊔ ⊓ Remark 1 (Possible conﬁgurations). Given a matrix S, [at least] two rows in S will never be formed by the same elements in the same order than another row. If this happens, S is not a valid Sudoku anymore, as it leads to a part of S (the example is for the rows of S, but it is as trivial for the columns of S) like ··· ··· ··· ··· ··· 1 2 3 ··· n 1 2 3 ··· n , ··· ··· ··· ··· ··· where 1, 2, · · · , n can be in a diﬀerent order. This would also be shown by the determinant of a submatrix of k k + 1 k + 2 ··· n , k k + 1 k + 2 ··· n which always equals 0 (thus leading to a rank ρ less than 2), Beginning by the end, i.e. using the following conﬁguration: ··· ··· ··· ··· ··· 1 2 3 ··· n n n − 1 n − 2 ··· 1 ··· ··· ··· ··· ··· can only lead to linearly independent rows. The rank ρ of a matrix A being deﬁned as the biggest dimension of the square submatrices of = 0 det extracted from A, we know that ··· ··· ··· ··· ··· 1 2 3 · · · n ρ n n − 1 n − 2 · · · 1 = 2, ··· ··· ··· ··· ··· under some given conditions which will be determined now. For k ∈ {1, · · · , n − 1}, we have k k+1 = k(n − k) − n(k + 1) = kn − k 2 − kn − n = −k 2 − n; nn−k That is, −k 2 − n ≤ 2 iﬀ −n ≤ 2 + k 2 , leading to n ≥ −2 − k 2 . As −2 − k 2 = −3 iﬀ k = 1 and k=(n−1) −2 − k 2 = −2 − (n − 1)2 = −n2 + 2n + 3, we must ask n ≥ −n2 + 2n + 3 , n ≥ −3 which is equivalent to √ √ n ∈ − ∞, 1−2 5 1+ 5 2 , +∞ , n ≥ −3 thus giving √ √ n ∈ − ∞, 1−2 5 1+ 5 2 , +∞ , n ≥ −3 and √ 1+ 5 n∈ , +∞ , 2 which is always the case, as √ 1+ 5 ≃ 1.6180339887498948482045868343656, 2 and the dimension of a Sudoku is always the square of a number. Thus, the ﬁrst interesting Sudoku’s dimension would be 4 = 22 (thus greater than the golden ratio). ⊔ ⊓ 4 Erroneous Sudokus Theorem 2 (Finding if a Sudoku matrix is erroneous or not). There is no way to be sure about the correctness of a Sudoku matrix by only computing its determinant. ⊔ ⊓ Proof. An erroneous Sudoku can have a determinant of zero, but it is not always the case. Furthermore, having a determinant of zero cannot even be a suﬃcient condition to be an erroneous Sudoku matrix, as there are correct Sudoku matrices which have a determinant of 0. Let’s take two erroneous Sudokus: 1. The ﬁrst will verify, ··· ··· ··· ··· ··· ··· ··· ··· ··· 1 2 3 4 5 6 7 8 9 = 0, 1 2 6 5 4 3 7 9 8 ··· ··· ··· ··· ··· ··· ··· ··· ··· assuming the · · · can be ﬁlled correctly to make no other mistake. At least a given Sudoku can thus have a non-zero determinant, 2. The second is “more” erroneous: ··· ··· ··· ··· ··· ··· ··· ··· ··· 1 2 3 4 5 6 7 8 9 = 0, 1 2 3 4 5 6 7 8 9 ··· ··· ··· ··· ··· ··· ··· ··· ··· which is very clear, as 123456789 ρ = 1. 123456789 We have also seen before that abcd abdc cdab cdba = = a4 + b4 + c4 + d4 − 2a2 b2 − 2a2 c2 − 2b2 c2 − 2a2 d2 − 2b2 d2 − 2c2 d2 + 8abcd, dcba bacd badc dcab which equals 0 iﬀ (a, b, c, d) are mapped bijectively to (1, 2, 3, 4), leading however to a correct Sudoku. ⊔ ⊓ 5 Sudoku Eigenvalues 5.1 General Case Let’s take the eigenvalues of S; that is, let’s compute det(S −λI). Depending on the dimension of S, denoted by n, det(S − λI) will be a polynom of a diﬀerent order: this order will always be of n. That is, n det(S − λI) = αλn + βλn−1 + · · · + ζλ0 = αi λi (2) i=0 for a general S matrix of dimension n. 5.2 Example: Dimension Nine As a 9 × 9 Sudoku is also a magic square, det(S − λI) = 0 for at least a λi = 45 (also found in [3]). The spectrum of S (restricted to a real spectrum, as there is no order in C, if it is composed of some complex values, and this is a situation which is possible and current), composed of the λi , veriﬁes thus max λi = 45. 1 ≤ i ≤ n, λi ∈ R Equation 2 can be rewritten (for a 9 × 9 Sudoku) as n−1 det(S − λI) = C(λ − 45) (λ − λi ), (3) i=1 C being a constant in Z∗ , as we are in a suﬃciently big structure to have n roots, counted with their multiplicities. Why 45? One can notice easily two facts: – 9 × 5 = 45, 9(9+1) 90 – 1 + 2 + 3 + ··· + 9 = 2 = 2 = 45. Thus, the dominant eigenvalue in the spectrum of S, with a size of 9, equals the sum of all the elements of a given line (the ﬁrst fact cannot be extended to other dimensions). It is said in [3] that the determinant of a Sudoku 9 × 9 is always divisible by 45, and it is even the case if at least one eigenvalue is in C. It can be seen very easily, as λ=0 det(S − λI) = det(S). (4) We then have n−1 λ=0 D(45) (λ − λi ) = det(S), (5) i=1 the 45 factor multiplying every other λi , with D = −C, |D| = |C|. 5.3 Return to General Case Theorem 3 (Perron). If all of the entries of a matrix are positive, then the matrix has a dominant eigenvalue that is real and has multiplicity 1. ⊔ ⊓ The proof of this theorem is not given here. Theorem 4 (Max eigenvalue of a square and positive matrix). The dominant eigenvalue of any square and positive matrix where each row and column have the same sum, will equal that sum. ⊔ ⊓ Proof. Easily stated by using Theorem 3. ⊔ ⊓ Corollary 1 (Eigenvalues of S). A Sudoku matrix S with a dimension of n always veriﬁes n n(n + 1) max λi = Si,j = . 1 ≤ i ≤ n, j=1 2 λi ∈ R ⊔ ⊓ Proof. Trivial using Theorem 4. ⊔ ⊓ Corollary 2. A Sudoku square matrix S, of dimension n, and of determinant det(S), always veriﬁes det(S) mod max λi = 0, 1 ≤ i ≤ n, λi ∈ R which is equivalent to ask n(n + 1) det(Si ) mod = 0. 2 ⊔ ⊓ Proof. Trivial, as det(S − λI) = det(S) when λ = 0, and using Corollary 2. ⊔ ⊓ Example 1 (Divisibility of the determinant of S if it represents a 4 × 4 Sudoku). Whatever the 4 × 4 Sudoku, its determinant is always divisible by 4(4+1) = 10. Evidently, it can be divisible by 45, for example, but it 2 might not always be the case. ⊔ ⊓ 5.4 Subsudoku Eigenvalues Prof. Steven Leon from the University of Massachussets Dartmouth said in an e-mail conversation we had, that for each of the 9 submatrices of a (9 × 9) Sudoku matrix, the absolute value of any (real) eigenvalue of the submatrix is less than or equal to 22. This statement is generalized in Theorem 6, but we ﬁrst have to speak about the submatrices which can be extracted from a n × n given Sudoku matrix. They are never Sudoku ones. Theorem 5 (Submatrices of a Sudoku matrix are not Sudoku matrices anymore). The subma- trices of a given Sudoku matrix are never Sudoku matrices. ⊔ ⊓ √ √ Proof. Let’s take a n × n Sudoku matrix, n being the square of an integer. Its submatrices will be n × n. √ √ The number n is √ always in Z. (If it was not, the submatrix would not be a matrix!) If n is not the square √ of an integer, the n × n submatrices cannot be, by deﬁnition, Sudoku ones (an example is the 9 × 9 √ Sudoku case: 3 × 3 submatrices are not Sudoku matrices anymore). If n is the square of an integer (e.g. √ √ 16 × 16 Sudoku matrices, resulting in 4 × 4 submatrices), it won’t prevent the n × n submatrices from being partially ﬁlled with elements in {1, · · · , n}, thus leading to a non-Sudoku matrix, by deﬁnition. ⊔ ⊓ We now give the important theorem about the maximum eigenvalue of a given Sudoku’s submatrix. Theorem 6 (Maximum eigenvalue of a Sudoku submatrix). If S is the matrix of a n × n Sudoku, with n being a square number (i.e. there exists one a ∈ N∗ so that a × a = n), the maximum (bounding) ′ ′ eigenvalue of one of its submatrices Sk , k ∈ {1, · · · , n}, is, if we consider that Sk = S ′ : √ √ n ′ n min max Si,j , max√ Si,j . ′ (6) i∈{1,··· ,√n} j∈{1,··· , n} j=1 i=1 = S ∞ = S 1 This is an upper bound, and there is not necessarily an eigenvalue of these submatrices which equals this value. ⊔ ⊓ ′ Proof. If λ1 , · · · , λp are the eigenvalues of any Sk submatrix taken from S, it is known (this is property of matrix norms) that ′ |λm | ≤ Sk , ′ whatever the Sk (k having the same value in this proof, from the beginning to the end) and the m ∈ ′ ′ ′ ′ {1, · · · , p}, where Sk is the norm of Sk . As, if, to simplify writings, we consider locally Si,j = Ski,j , √ √ n n max ′ Si,j and max√ j∈{1,··· , n} ′ Si,j i∈{1,··· ,√n} j=1 i=1 = S ∞ = S 1 are both norms (the ﬁrst is the maximal sum on a row of S ′ , and the second is the maximal sum on a column of S ′ ), they verify ′ |λm | ≤ Sk . ⊔ ⊓ What it actually means is that we have two maximal1 sums: the one which can be computed on a row of S ′ , and the one which can be computed on a column of S ′ . As these two maximums are norms, we have to take the lowest value between them, to ﬁnd the maximal bound for the eigenvalue of S ′ . √ √ Remark 2. If S ′ is n × n (n being a square number), we do not always have √ √ √ √ n−1 √ m= n n (n−i) = (n)+(n−1)+· · · +(n−( n−1)) = max ′ Si,j or max√ j∈{1,··· , n} ′ Si,j , i∈{1,··· ,√n} i=0 j=1 i=1 = S ∞ = S 1 √ (7) as there are valid Sudoku where at least one of the submatrices of size n where √ (n) + (n − 1) + · · · + (n − ( n − 1)) equals neither S ′ 1 nor S ′ ∞. ⊔ ⊓ Corollary 3 (Most general eigenvalue of a Sudoku submatrix). If S is the matrix of a n×n Sudoku, with n being a square number (i.e. there exists one a ∈ N∗ so that a × a = n), the most general maximum ′ ′ (bounding) eigenvalue of one of its submatrices Sk , k ∈ {1, · · · , n}, veriﬁes, if we consider that Sk = S ′ : √ √ √ n−1 n ′ n ′ (n − i) ≥ min max √ Si,j , max√ Si,j , (8) i=0 i∈{1,··· , n} j=1 j∈{1,··· , n} i=1 = S ∞ = S 1 and √ n−1 (n − i) (9) i=0 is thus the most general maximum eigenvalue of a given Sudoku submatrix. ⊔ ⊓ Proof. Trivial. (Just take a counterexample.) ⊔ ⊓ Example 2 (9 × 9 Sudoku matrices). If we take a 9 × 9 Sudoku matrix, having 3 × 3 submatrices (which are not Sudoku), the maximal sum one can encounter on a column or a row of a submatrix is √ n + (n − 1) + n − ( n − 1) = 24. =9 √ =8 =9−( 9−1)=9−(2)=7 ′ Consequently, the maximal eigenvalue one can ﬁnd in a Sk , k ∈ {1, · · · , 9} is 24. ⊔ ⊓ 1 For a given submatrix. Example 3 (16 × 16 Sudoku matrices). Let’s take a 16 × 16 Sudoku matrix. Its submatrices are not Sudoku matrices anymore, and are composed by elements in {1, · · · , 16} ⊂ Z. The maximal sum we can have, when summing elements from one submatrix of a 16 × 16 Sudoku is 16 + 15 + 14 + 13 = 58. √ four elements, as 16 = 4 ⊔ It is the maximum real eigenvalue which can be found with the submatrices of a 16 × 16 Sudoku matrix. ⊓ Example 4 (25 × 25 Sudoku matrices). Let’s take a 25 × 25 Sudoku matrix. Its submatrices will never be Sudoku matrices. They are composed by elements in {1, · · · , 25} ⊂ Z. The maximal sum we can have, when summing elements from one submatrix of a 25 × 25 Sudoku is 25 + 24 + 23 + 22 + 21 = 115. √ ﬁve elements, as 25 = 5 The maximum eigenvalue of a 5 × 5 (non-Sudoku) submatrix extracted from a 25 × 25 Sudoku matrix equals thus 115. ⊔ ⊓ 6 Sudoku Eigenvectors n(n+1) We can compute the corresponding eigenvector for the eigenvalue 2 , as computing n(n + 1) S− I x = 0n 2 is the same as computing S1,1 − n(n+1) 2 S1,2 ··· S1,n S2,1 S2,2 − n(n+1) 2 ··· S2,n . . . xn = 0n , . . .. n(n+1) . . . .− 2 . n(n+1) Sn,1 Sn,2 ··· Sn,n − 2 thus making computing S1,1 − n(n+1) 2 S1,2 ··· S1,n n(n+1) a S2,1 S2,2 − 2 ··· S2,n . . . b = 0n . (10) . . .. n(n+1) . . . . .− 2 . . n(n+1) . Sn,1 Sn,2 ··· Sn,n − 2 7 Other Facts About S 7.1 Transpose General Facts One could ask if S1,1 S1,2 · · · S1,n S1,1 S2,1 · · · Sn,1 S2,1 S2,2 · · · S2,n S1,2 S2,2 · · · Sn,2 S= . . .. . = . . .. . .. . . . . . . . . . . .. Sn,1 Sn,2 · · · Sn,n S1,n S2,n · · · Sn,n is still a valid Sudoku matrix (i.e. it fulﬁlls the rules of Sudoku). An example would be given by 629385147 658421397 5 1 3 7 6 4 8 2 9 2 1 4 9 7 3 8 6 5 8 4 7 1 2 9 6 5 3 9 3 7 6 5 8 1 4 2 4 9 6 2 3 1 7 8 5 3 7 1 2 9 6 4 5 8 B = 2 7 5 9 4 8 3 6 1 = 8 6 2 3 4 5 9 7 1 1 3 8 6 5 7 4 9 2 5 4 9 1 8 7 2 3 6 3 8 1 4 9 2 5 7 6 1 8 6 7 3 4 5 2 9 9 6 4 5 7 3 2 1 8 4 2 5 8 6 9 7 1 3 752816934 793512684 It is noticeable that B is still a valid Sudoku matrix, but B = B. Theorem 7 (Transpose of a Sudoku matrix). The transpose of a Sudoku matrix is still a correct, but diﬀerent, Sudoku matrix (of the same dimension). ⊔ ⊓ Proof. According to the rules of Sudoku, at least one number cannot be repeated at least two times on a 1 1 row, a column, or in a n 2 × n 2 square, where n is the dimension of the Sudoku. The matrix which is the result of the transposition is still a valid Sudoku matrix, as 1. Bi,j = Bi,j if i = j (i.e. (i, j) is a diagonal couple), 2. If we let jold := j, j := i and i := jold , everywhere in B, we obtain B. By reversing rows and columns, rules of Sudoku are still respected, as the rules of Sudoku are applicable on both, 3. For subsquares, rules are still respected, because of the last point. We now have to prove that it is always diﬀerent. We have to prove that there is no Sudoku matrix in S which veriﬁes Si,j = Si,j with 1 ≤ i ≤ n, and 1 ≤ j ≤ n. Let’s reason by contradiction. All the S matrices verify Si,j = Si,j for i = j. Let 1 ≤ i ≤ n, 1 ≤ j ≤ n. Is the same equality possible with these conventions? We shall now try to construct such a matrix. Let’s reason ﬁrst with a 4 × 4 matrix. We deﬁne B1,1 B1,2 B1,3 B1,4 C1,1 C1,2 C1,3 C1,4 B B B B C C C C B := 2,1 2,2 2,3 2,4 B3,1 B3,2 B3,3 B3,4 and B := C := 2,1 2,2 2,3 2,4 . C3,1 C3,2 C3,3 C3,4 B4,1 B4,2 B4,3 B4,4 C4,1 C4,2 C4,3 C4,4 It is clear that Bi,j = Bi,j for all 1 ≤ i ≤ n and 1 ≤ j ≤ n iﬀ Bi,j = Ci,j for all i and j. The C matrix has thus to be exactly B. It is the case iﬀ Bi,j = Bj,i , with 1 ≤ i ≤ n and 1 ≤ j ≤ n, thus iﬀ B is symmetric. Let’s try to construct a symmetric B. Such a B would be symmetric: B1,1 B2,1 B3,1 B4,1 B2,1 B2,2 B3,2 B4,2 B3,1 B3,2 B3,3 B4,3 , B4,1 B4,2 B4,3 B4,4 1 1 but it does not respect the rules of Sudoku, as there is at least one subsquare of dimension 4 2 × 4 2 ≡ 2 × 2 where two elements are the same (here, there is only one: B4,3 ). Here is the generalization of this fact. If we take B1,1 B1,2 · · · B1,n B2,1 B2,2 · · · B2,n B= . . .. . , . . . . . .. Bn,1 Bn,2 · · · Bn,n B should verify B1,1 B2,1 · · · Bn,1 B2,1 B2,2 · · · Bn,2 B= . . .. . .. . . . .. Bn,1 Bn,2 · · · Bn,n to be symmetric. It is symmetric, but does not respect the rules of Sudoku anmyore, as, at the place of the . . and the · · · , zooming at the end would lead to the submatrix . .. . ··· ··· Sn,2 . .. . . . . . . . . . . . .. . ··· . . Sn,n−1 Sn,2 · · · Sn,n−1 Sn,n ⊔ ⊓ Determinant Theorem 8 (Determinant of the transpose of a square matrix). For every square matrix C, det(C) = det(C). ⊔ ⊓ This theorem is not proved here, even if it is evident. Corollary 4 (Determinant of the transpose of a Sudoku matrix). If S is the matrix of a given Sudoku, det(S) = det(S). ⊔ ⊓ Corollary 5. If S is the matrix of a given Sudoku, S its transpose, and n their dimension (which is the same), n(n + 1) n(n + 1) det(S) mod = det(S) mod = 0. 2 2 ⊔ ⊓ Proof. Trivial, as, on one hand, according to Theorem 7, S and S – have the same dimension (n), n n(n+1) – are both Sudokus, verifying i=1 Si,j = 2 ; n(n+1) on the other hand, Theorem 4 gives us the maximum eigenvalue of S, which is 2 . Finally, we know that λ=0 det(S) = det(S − λI). ⊔ ⊓ Trace Theorem 9 (Trace of a transpose). Whatever the matrix E, Tr(E) = Tr(E). ⊔ ⊓ This theorem is not proved here, even if it is evident. Corollary 6 (Trace of the transpose of a Sudoku matrix). If S is the matrix of a given Sudoku, Tr(S) = Tr(S). ⊔ ⊓ 7.2 Not Hermitian A matrix C is Hermitian if and only if C = C ∗. As, for any Sudoku matrix S, S = S, as · is the complex conjugate of ·, and the Si,j , 1 ≤ i ≤ n, 1 ≤ j ≤ n, asking if any Sudoku matrix is Hermitian or not is equivalent to ask if, for S, S = S. However, this is not true, as we have S = S, thanks to Theorem 7. Corollary 7 (Every Sudoku matrix is not Hermitian). There is no Hermitian Sudoku matrix. ⊔ ⊓ Proof. Direct using Theorem 7, as Si,j = Si,j for 1 ≤ i ≤ n, 1 ≤ j ≤ n, as Si,j ∈ R. ⊔ ⊓ ⊔ ⊓ Corollary 8 (Every Sudoku matrix is not antisymmetric). S is not antisymmetric, for clear reasons: Si,j = −Sj,i . ⊔ ⊓ Proof. The Si,j are in Z∗ . If we take −Si,j for 1 ≤ i ≤ n and 1 ≤ j ≤ n, all the Si,j will lie in Z∗ , or no − Sudoku matrix can have negative elements. ⊔ ⊓ 7.3 Not Normal A matrix C is normal if and only if CC ∗ = C ∗ C. It is equivalent to ask, for now clear reasons, C C = CC, if C is a Sudoku matrix. Theorem 10 (Every Sudoku matrix is not normal). There is no normal Sudoku matrix. ⊔ ⊓ Proof. Let’s try to build such a matrix. If we have S1,1 S1,2 · · · S1,n S1,1 S2,1 · · · Sn,1 S2,1 S2,2 · · · S2,n S1,2 S2,2 · · · Sn,2 S= . . .. . and S ′ := S = . . .. . , . . . . . . . .. . . . . . Sn,1 Sn,2 · · · Sn,n S1,n S2,n · · · Sn,n we have S S = SS iﬀ S1,1 S1,2 · · · S1,n S1,1 S2,1 · · · Sn,1 S1,1 S2,1 · · · Sn,1 S1,1 S1,2 · · · S1,n S2,1 S2,2 · · · S2,n S1,2 S2,2 · · · Sn,2 S1,2 S2,2 · · · Sn,2 S2,1 S2,2 · · · S2,n S= . . .. . . . .. . = . . .. . . . .. . , .. . . . .. . . . . . .. . . . . . .. . . . . . .. Sn,1 Sn,2 · · · Sn,n S1,n S2,n · · · Sn,n S1,n S2,n · · · Sn,n Sn,1 Sn,2 · · · Sn,n thus asking an equality between 0 2 2 2 S1,1 + S1,2 + · · · + S1,n S1,1 S2,1 + S1,2 S2,2 + · · · + S1,n S2,n · · · S1,1 Sn,1 + S1,2 Sn,2 + · · · + S1,n Sn,n 1 B BS 2 2 2 C B 2,1 S1,1 + S2,2 S1,2 + · · · + S2,n S1,n S2,1 + S2,2 + · · · + S2,n · · · S2,1 Sn,1 + S2,2 Sn,2 + · · · + S2,n Sn,n C C B ··· ··· ··· ··· C @ A Sn,1 S1,1 + Sn,2 S1,2 + · · · + Sn,n S1,n Sn,1 S2,1 + Sn,2 S2,2 + · · · + Sn,n S2,n ··· 2 2 2 Sn,1 + Sn,2 + · · · + Sn,n and 0 2 2 2 1 S1,1 + S1,2 + · · · + Sn,1 S1,1 S1,2 + S2,1 S2,2 + · · · + Sn,1 Sn,2 · · · S1,1 S1,n + S2,1 S2,n + · · · + Sn,1 Sn,n 2 S1,2 2 + S2,2 + · · · + Sn,2 2 B C B 1,2 S1,1 + S2,2 S2,1 + · · · + Sn,2 Sn,1 BS · · · S1,2 S1,n + S2,2 S2,n + · · · + Sn,2 Sn,n C C, B ··· ··· ··· ··· C @ A S1,n S1,1 + S2,n S2,1 + · · · + Sn,n Sn,1 S1,n S1,2 + S2,n S2,2 + · · · + Sn,n Sn,2 ··· 2 Sn,1 + S2,n 2 + · · · + S2 n,n leading to such a system: 2 2 S1,n = Sn,1 S 1,2 = S2,1 Sn,1 = S1,n . Sn,2 = S2,n . . . It is clear that S has to be symmetric to respect these equalities. However, it was proven at Theorem 7 that S cannot be symmetric. ⊔ ⊓ 7.4 Not Orthogonal A matrix C is orthogonal if and only if CC = In . Theorem 11 (Non-orthogonality of S if det(S) = 0). If det(S) = 0, S cannot be orthogonal. ⊔ ⊓ Proof. We have det(S · S) = det(S) det(S), which equals 0 if det(S) = 0, and det(In ) = 1. ⊔ ⊓ If det(S) = 0, S has to coincide with S −1 to make S orthogonal. Using Theorem 7, S = S. But one might ask the question “Can S equal S −1 ?” We know that S = S. Thus, if even S = S −1 , S −1 is diﬀerent from S. Theorem 12 (Non-orthogonality of S if det(S) = 0). If det(S) = 0, S cannot be orthogonal. ⊔ ⊓ Proof. The only way to make S orthogonal is to have S = S −1 . Or this equality is impossible: as S −1 = cofact(S) · (det(A))−1 , it would be equivalent to ask S = cofact(S) · (det(A))−1 , which is never true for a S Sudoku matrix. ⊔ ⊓ 8 Order Theorem 13 (Every invertible matrix has a ﬁnite order). Every invertible matrix has a ﬁnite order. ⊔ ⊓ This theorem is proved in many books, and its proof does not enter in the scope of this paper. Corollary 9. If det(S) = 0, S has a ﬁnite order. ⊔ ⊓ Example 5. The determinant 1234 3412 det(B) = 4321 2143 equalling 0, there is no smallest k > 0 such that B k = In . ⊔ ⊓ However, there is no special remark to do about this and Sudokus: there seems to be no link between the order of a Sudoku matrix S and other concepts related to Sudoku’s matrices. 9 Trace Let S be a (square) Sudoku matrix, and Si,i its diagonal elements (1 ≤ i ≤ n). Theorem 14 (Upper bound on the trace of a Sudoku matrix). The value of Tr(S) veriﬁes n Tr(S) = Si,i i=1 √ √n n−1 ≤ (n − i) . i=1 ⊔ ⊓ Example 6 (9 × 9 Sudoku). For a 9 × 9 S Sudoku matrix, we have =2 3 √ 9−1 9 3 Tr(S) = Si,i ≤ (9 − i) = ((9)(9 − 1)(9 − 2)) = 128024064. i=1 i=0 ⊔ ⊓ Remark 3 (Upper bound on the trace of a Sudoku matrix seen as a permutation). If Ak denotes the number n of combinations of k elements among n, the order of the elements being taken into account, we have √ √n √ n−1 n n! (n − i) = √ . i=1 (n − n)! It can be easily understood in another way than using pure Mathematics: we need to know how many √ diﬀerent conﬁgurations exist to combine n − n elements among n, the order being taken into account. This is achieved using n! √ (n − n)! √ but there are n conﬁgurations which can lead to diﬀerent results, at least if one of these conﬁgurations is diﬀerent from the others. ⊔ ⊓ Example 7 (Upper bound on the trace of a Sudoku matrix seen as a permutation for a 9 × 9 Sudoku matrix). For a 9 × 9 Sudoku matrix, the upper bound on its trace is 3 3 3 9! 9! 9 · 8 · 7 · 6! √ = = = 5043 = 128024064. (9 − 9)! 6! 6! ⊔ ⊓ Corollary 10 (Trace of a Sudoku matrix is similarity-invariant). The trace of a S Sudoku matrix is similarity-invariant, i.e. Tr(S −1 AS) = Tr(A). ⊔ ⊓ Corollary 11 (Trace of a Sudoku matrix equals the sum of the eigenvalues). We have √ √n n n−1 Tr(S) = λi ≤ (n − i) . i=1 i=1 ⊔ ⊓ Corollary 12. We have Tr(S ∗ S) ≥ 0. ⊔ ⊓ Corollary 13. Let k P (λ) = (x − λi )di i=1 be the characteristic polynomial of S. Then, √ √n n n−1 Tr(S) = di λ i ≤ (n − i) . i=1 i=1 ⊔ ⊓ Corollary 14 (Linear map). The trace being a linear map, if S and S ′ are Sudoku matrices, they verify Tr(S + S ′ ) = Tr(S) + Tr(S ′ ) Tr(λS) = λ Tr(S), ∀ λ ∈ K. Corollary 15. Let the biggest eigenvalue of a S Sudoku matrix be the jth eigenvalue, with j = n (that is, the eigenvalues are classed with a ≤ order). Then, n−1 n(n + 1) Tr(S) = di λ i + . i=1 2 ⊔ ⊓ Proof. By deﬁnition, n Tr(S) = di λ i . i=1 As the biggest eigenvalue of S is n(n + 1) λn = , 2 whose dn = 1 (both concepts according to Perron’s theorem (3) and to Theorem 4), n−1 n(n + 1) Tr(S) = di λ i + . i=1 2 ⊔ ⊓ 9.1 Antitrace Deﬁnition 1 (Antitrace of a square matrix). The antitrace of a square matrix C is n antiTr(C) = Ci,n−i+1 . i=1 ⊔ ⊓ Theorem 15. Let S be a Sudoku matrix which the expression √ √n n−1 Tr(S) = (n − i) i=1 holds with. Then, S has its antitrace verifying √ √n n−1 antiTr(S) < (n − i) i=1 and reciprocally for Tr(s) iﬀ √ √n n−1 antiTr(S) = (n − i) . i=1 ⊔ ⊓ Proof. Let’s try to build such a matrix. If Tr(S) is maximal, S, of the form S1,1 S1,2 · · · S1,n S2,1 S2,2 · · · S2,n S= . . .. . , . . . . . . . Sn,1 Sn,2 · · · Sn,n √ √ needs to have the n elements of the set {n, n − 1, · · · , n − n + 1} in each of its submatrices’ diagonals. For example, a submatrix extracted from S must have the following form: Sp,q Sp,q+1 · · · Sp,q+√n Sp+1,q Sp+1,q+1 · · · Sp+1,q+√n . . . . .. . . . . . . . √ √ Sp+ n,q Sp+ n,q+1 · · · Sp+ n,q+ n √ √ √ If Tr(S) is maximal, the Sp+i,q+i , i ∈ {0, 1, · · · , n} must be the biggest numbers in {1, 2, · · · , n}, that is, √ n, n − 1, · · · , n − n + 1. Consequently, there cannot be any other such numbers in this submatrix, as it would result in an erroneous Sudoku, at least one of its submatrices having at least twice the same element. The only way to reach √ √n n−1 (n − i) i=1 is thus reserved to exactly one diagonal or no diagonal: the principal one (resulting in a maximal trace), the other one (resulting in a maximal antitrace) or no one. ⊔ ⊓ Example 8 (9 × 9 Sudoku matrix). A 9 × 9 Sudoku matrix has a maximal trace iﬀ all its diagonal elements are in {9, 8, 7}, each one being written only once in each of the diagonal submatrices. For example, S1,1 S1,2 S1,3 S1,4 S1,5 S1,6 S1,7 S1,8 9 S2,1 S2,2 S2,3 S2,4 S2,5 S2,6 S2,7 8 S2,9 S3,1 S3,2 S3,3 S3,4 S3,5 S3,6 7 S3,8 S3,9 S4,1 S4,2 S4,3 S4,4 S4,5 9 S4,7 S4,8 S4,9 S5,1 S5,2 S5,3 S5,4 8 S5,6 S5,7 S5,8 S5,9 S6,1 S6,2 S6,3 7 S6,5 S6,6 S6,7 S6,8 S6,9 S7,1 S7,2 9 S7,4 S7,5 S7,6 S7,7 S7,8 S7,9 S8,1 8 S8,3 S8,4 S8,5 S8,6 S8,7 S8,8 S8,9 7 S9,2 S9,3 S9,4 S9,5 S9,6 S9,7 S9,8 S9,9 is a possible conﬁguration, and S1,1 S1,2 S1,3 S1,4 S1,5 S1,6 S1,7 S1,8 8 S2,1 S2,2 S2,3 S2,4 S2,5 S2,6 S2,7 7 S2,9 S3,1 S3,2 S3,3 S3,4 S3,5 S3,6 9 S3,8 S3,9 S4,1 S4,2 S4,3 S4,4 S4,5 9 S4,7 S4,8 S4,9 S5,1 S5,2 S5,3 S5,4 8 S5,6 S5,7 S5,8 S5,9 S6,1 S6,2 S6,3 7 S6,5 S6,6 S6,7 S6,8 S6,9 S7,1 S7,2 8 S7,4 S7,5 S7,6 S7,7 S7,8 S7,9 S8,1 9 S8,3 S8,4 S8,5 S8,6 S8,7 S8,8 S8,9 7 S9,2 S9,3 S9,4 S9,5 S9,6 S9,7 S9,8 S9,9 is also a good one. They both lead to a maximal antitrace. However, the trace of S cannot equal the maximal value 128024064 as, for example, 9 S1,2 S1,3 S1,4 S1,5 S1,6 S1,7 S1,8 8 S2,1 8 S2,3 S2,4 S2,5 S2,6 S2,7 7 S2,9 S3,1 S3,2 7 S3,4 S3,5 S3,6 9 S3,8 S3,9 S4,1 S4,2 S4,3 9, 7 S4,5 9 S4,7 S4,8 S4,9 S5,1 S5,2 S5,3 S5,4 8 S5,6 S5,7 S5,8 S5,9 S6,1 S6,2 S6,3 7 S6,5 7, 9 S6,7 S6,8 S6,9 S7,1 S7,2 8 S7,4 S7,5 S7,6 9 S7,8 S7,9 S8,1 9 S8,3 S8,4 S8,5 S8,6 S8,7 8 S8,9 7 S9,2 S9,3 S9,4 S9,5 S9,6 S9,7 S9,8 7 is not a valid Sudoku matrix anymore. ⊔ ⊓ 10 Condition Number Theorem 16 (No well-conditioning for S can arise). The system Sx = b can be else than well- conditioned (assuming b = 0n ). ⊔ ⊓ Proof. A deﬁnition of the condition number of S is κ(S) = S −1 ∞ S ∞ √ √ n n = max √ A−1 ij max √ |Aij | . i∈{1,··· , n} i∈{1,··· , n} j=1 j=1 1. If they are not equal, let’s call g the greatest, and s the smallest. We then have κ(S) = g · s . √ n−1 ≤ (n − i) i=0 As g > s by hypothesis, κ(S) > 1 can be true. As log(1) = 0, and that 1 > 0, a system of linear equations where the coeﬃcient matrix is S will never be well-conditioned. 2. If they are equal, that is, if √ √ n n p := max√ A−1 = ij max√ |Aij | , i∈{1,··· , n} i∈{1,··· , n} j=1 j=1 √ √ n n min max A−1 , max |Aij | = p, i∈{1,··· ,√n} ij √ i∈{1,··· , n} j=1 j=1 we learn that 0 < κ(S) = p2 < +∞. That is, the system is never well-conditioned too (except iﬀ p ∈ [0, 1[), but such a case does not arise often. 3. If S is not invertible, i.e. iﬀ det S = 0, S −1 is inﬁnite. Consequently, S −1 ∞ = ∞, and the system is thus ill-conditioned. ⊔ ⊓ Example 9 (Extreme example). Consider the matrix 673514982 9 2 1 3 6 8 7 5 4 5 8 4 7 9 2 1 3 6 8 9 6 2 3 7 4 1 5 P := 2 1 5 9 4 6 3 7 8 . 3 4 7 1 8 5 2 6 9 4 5 2 8 7 1 6 9 3 1 6 9 4 5 3 8 2 7 738629541 It veriﬁes κ(P )∞ ≈ (2.052475887800532) · 1017 ≫ det(P ) = 0. As κ(P ) ≫, trying to solve 673514982 α κ 9 2 1 3 6 8 7 5 4 β λ 5 8 4 7 9 2 1 3 6 γ µ 8 9 6 2 3 7 4 1 5 δ ν 2 1 5 9 4 6 3 7 8 ǫ = ξ 3 4 7 1 8 5 2 6 9 ζ χ 4 5 2 8 7 1 6 9 3 η ρ 1 6 9 4 5 3 8 2 7 θ σ 738629541 ι τ is really error-prone. Let’s take κ=1 λ=2 µ=3 ν=4 ξ=5 χ=6 ρ=7 σ=8 τ = 9. Using numerical analysis, we can now solve 673514982 α 1 9 2 1 3 6 8 7 5 4 β 2 5 8 4 7 9 2 1 3 6 γ 3 8 9 6 2 3 7 4 1 5 δ 4 2 1 5 9 4 6 3 7 8 ǫ = 5 . 3 4 7 1 8 5 2 6 9 ζ 6 4 5 2 8 7 1 6 9 3 η 7 1 6 9 4 5 3 8 2 7 θ 8 738629541 ι 9 The result is roughly −6.259323629462618 3.426654756202998 −2.859392164516785 −0.213786330057330 15 10 · 2.235038905144813 . 6.094125101691045 0.146978101914415 0.070452313314347 −2.640747054230881 Let’s now compute A := A + (0.1)9 and b := b + (0.1)9 . We now want to solve 6.1 7.1 3.1 5.1 1.1 4.1 9.1 8.1 2.1 α 1.1 9.1 2.1 1.1 3.1 6.1 8.1 7.1 5.1 4.1 β 2.1 5.1 8.1 4.1 7.1 9.1 2.1 1.1 3.1 6.1 γ 3.1 8.1 9.1 6.1 2.1 3.1 7.1 4.1 1.1 5.1 δ 4.1 2.1 1.1 5.1 9.1 4.1 6.1 3.1 7.1 8.1 ǫ = 5.1 . 3.1 4.1 7.1 1.1 8.1 5.1 2.1 6.1 9.1 ζ 6.1 4.1 5.1 2.1 8.1 7.1 1.1 6.1 9.1 3.1 η 7.1 1.1 6.1 9.1 4.1 5.1 3.1 8.1 2.1 7.1 θ 8.1 7.1 3.1 8.1 6.1 2.1 9.1 5.1 4.1 1.1 ι 9.1 The result is now −1.201790136856824 0.657917713190976 −0.549003295587223 −0.041046975371007 1017 · 0.429127469787804 . 1.170072019524681 0.028219795567568 0.013526844156355 −0.507023434412329 ⊔ ⊓ 11 Graph Modulo 2 Let’s say that Si,j := Si,j (mod 2), i.e. we now consider S modulo 2, for 1 ≤ i ≤ n, 1 ≤ j ≤ n. We now consider S as the adjacency matrix of a graph G. Theorem 17 (Possible diagonal of zeros). There are correct Sudoku matrices which, taken modulo 2, have all their diagonal elements being equal to 0. ⊔ ⊓ Proof. On one hand, consider the case of n = 4. It is easy to construct a Sudoku using digits 1, 2, 3 and 4 such that, modulo 2, all the diagonal elements equal 0: ab e f c d g h i j m n , k l o p where q means that q is even, when q means that q is odd. This matrix has, modulo 2, all its diagonal elements which equal 0. On the other hand, take 593714286 6 8 2 3 9 5 4 7 1 1 7 4 6 2 8 9 5 3 3 6 9 2 5 7 1 4 8 P := 8 1 7 4 6 9 3 2 5 . 4 2 5 8 3 1 7 6 9 9 4 1 5 8 2 6 3 7 2 3 8 9 7 6 5 1 4 756143892 If we compute Pi,j (mod 2), 1 ≤ i ≤ 9, 1 ≤ j ≤ 9, we have 111110000 0 0 0 1 1 1 0 1 1 1 1 0 0 0 0 1 1 1 1 0 1 0 1 1 1 0 0 0 1 1 0 0 1 1 0 1 . 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 1 1 0 1 1 0 110101010 ⊔ ⊓ Theorem 18 (Possible symmetricity). There are correct Sudoku matrices which, taken modulo 2, are symmetric. Sudoku matrices, taken modulo 2, of an even dimension, can be so, but Sudoku matrices, taken modulo 2, of an odd dimension, are never symmetric. ⊔ ⊓ Proof. On one hand, let’s take an example ﬁrst. Consider the case of n = 4. It is easy to construct a Sudoku using digits 1, 2, 3 and 4 such that, modulo 2, the derived adjacency matrix is symmetric: take for example ab e f c d g h i j m n , k l o p where q means that q is even, when q means that q is odd. This matrix veriﬁes c (mod 2) =b (mod 2) i (mod 2) =e (mod 2) k (mod 2) =f (mod 2) (11) j (mod 2) =g (mod 2) l (mod 2) =h (mod 2) o (mod 2) =n (mod 2) as it veriﬁes (c − b) (mod 2) =0 (i − e) (mod 2) =0 (k − f ) (mod 2) =0 (j − g) (mod 2) =0 (l − h) (mod 2) =0 (o − n) (mod 2) =0 as the diﬀerence of two numbers is even (and thus equals 0 modulo 2) iﬀ these two numbers are either even, or odd. (This matrix was found under the hypothesis that o and n were either even, or odd, which is true, as o = n if the matrix is symmetric.) This can be generalized to any even dimension, and such a construction will always be possible because Sudoku matrices, taken modulo 2, of an even dimension, have an even number of even numbers and an even number of odd numbers, whatever their submatrix. On the other hand, every Sudoku matrix taken modulo 2 of an odd dimension cannot be symmetric, √ as every n submatrix will have an odd number of odd numbers, and an even number of even numbers. Consequently, there is a number, say x, which is a Si,j (mod 2), which make a diﬀerence of 0 modulo 2, i.e. if x is the Sµ,θ and y the Sθ,µ , (x − y) (mod 2) = 0, as x and y cannot be simultaneously odd, and, consequently, x (mod 2) = y (mod 2). An example is given by 593714286 6 8 2 3 9 5 4 7 1 1 7 4 6 2 8 9 5 3 3 6 9 2 5 7 1 4 8 P = 8 1 7 4 6 9 3 2 5 . 4 2 5 8 3 1 7 6 9 9 4 1 5 8 2 6 3 7 2 3 8 9 7 6 5 1 4 756143892 If we compute Pi,j (mod 2), 1 ≤ i ≤ 9, 1 ≤ j ≤ 9, we have 111110000 0 0 0 1 1 1 0 1 1 1 1 0 0 0 0 1 1 1 1 0 1 0 1 1 1 0 0 0 1 1 0 0 1 1 0 1 , 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 1 1 0 1 1 0 110101010 and it is not symmetric. ⊔ ⊓ Corollary 16 (Graph of a Sudoku matrix can be simple or not). A graph which is generated using the adjacency matrix modulo 2 of a Sudoku can be simple, or not. ⊔ ⊓ Proof. 1. It can be not simple: It can have at least one loop, as its diagonal can be ﬁlled with at least one “1;” 2. It can be simple: It can have no multiple links and no loop, as its diagonal can be ﬁlled with zeros only. ⊔ ⊓ Corollary 17 (Graph of a Sudoku matrix modulo 2 of an even dimension is not oriented). The graph of a Sudoku matrix of an even dimension is never oriented. ⊔ ⊓ Corollary 18 (Graph of a Sudoku matrix modulo 2 of an odd dimension is always oriented). The graph of a Sudoku matrix of an odd dimension is always oriented. ⊔ ⊓ References 1. Davis, Tom, The Mathematics of Sudoku, (2008). http://www.geometer.org/mathcircles/sudoku.pdf. 2. Frank, Richard, Mathematics in Sudoku, (2005). http://people.brandeis.edu/~kleinboc/47a/sudoku.pdf. 3. LeBoeuf, Robert, Properties of Sudoku and Sudoku Matrices, (2009). http://compmath.files.wordpress. com/2009/02/rlfreport.pdf. 4. , Sudoku Matrices, (2009). http://compmath.files.wordpress.com/2009/02/rlposter.pdf.

DOCUMENT INFO

Shared By:

Categories:

Tags:
how to, math problem, william elliot, maa session, rob johnson, mariano suárez, sat 2, sat 7, meeting room, robert israel, h. ray, pi mu epsilon, august 1, youngstown state university, union college

Stats:

views: | 108 |

posted: | 6/10/2010 |

language: | English |

pages: | 24 |

OTHER DOCS BY iaj67571

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.