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The Problem of the 36 Officers Kalei Titcomb

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1780: No mutual pair of orthogonal Latin squares of order n=4k+2. k=0, : 6x6 case. 812,851,200 possible reduced to 9,408 pairs 1984: short, four page, noncomputer proof.

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Latin Square of order n n x n array of n different symbols Each symbol occurs once in each row and once in each column Example n=2 {1,2}

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Example: n=3 {1,2,3}{a,b,c} cba bac acb cba bac acb Orthogonal

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Orthogonal Latin squares Two Latin squares of order n Every ordered pair of symbols occurs once when superimposed Example: n=3 {1,a} {1,b} {1,c} {2,a} {2,b} {2,c} {3,a} {3,b} {3,c} cba bac acb

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Exampill n=2 Impossible Have the pairs {1,2} and {2,1} twice Don’t have {1,1} and {2,2}

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M for n=3 RCNL r11111 r21111 r31111 c11111 c21111 c a1111 b1111 c1111 1c2b3a 3b1a2c 2a3c1b r1 r2 r3 c1c2c R: {r1,r2,r3}C: {c1,c2,c3} N: {1,2,3}L: {a,b,c}

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M for n=6 RCNL1 … 36 r1 : r6 c1 : c6 1:61:6 a:fa:f 7 ones per row 6 or 4 ones per column

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Lemma I Our 24 x 40 matrix must have dependencies in our rows (r1,…,r6)+(c1,…,c6)=0 (r1,…,r6)+(1,…,6)=0 (r1,…,r6)+(a,…,f)=0 And one more in addition to these! (proof uses properties of the Latin square) Call that set of rows Y.

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Lemma II Must have submatrices Y and Y’ of M Must still have same amount of ones in each row (and if you sum them, each column) By a counting argument, Y must have 8 or 12 rows. So… Y’ has 16 or 12 rows

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Y=8 and Y’=16 By some more counting, we find that - Y has two from each group {r1, r2, c1, c2, 1, 2, a, b} So Y’ has four from each group {r3, r4, r5, r6, c3, c4, c5, c6, 3, 4, 5, 6, c, d, e, f}

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a1 b1 c1 d1 e1 f1 g1 h

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Y=8 and Y’=16 By some more counting, we find that - Y has 28 columns with 2 ones and the rest have no ones. This splits our matrix into 6 parts….

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a1 b1 c1 d1 e1 f1 g1 h

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G and H Y={a,b,c,d,e,f,g,h} Q={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}

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Lemma III Everyone needs to be friends with everyone else in H Everyone is friends only once in H Neighbors in G can’t be a group of friends in H

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WLOG 1 is friends with 5,9,13 {1,5,9,13} are neighbors Blocks {1,6,10,14} {1,7,11,15} {1,8,12,16}

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Lemma IV These graphs show Y cannot be of size 8. Therefore Y must be of size 12 according to Lemma II. We have five cases of how these 12 rows are spread among the 4 groups: {6,6,0,0} {6,4,2,0} {6,2,2,2} {4,4,4,0} {4,4,2,2}

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Lemma V We can’t have Y=12! This is because: {6,6,0,0} is already a dependency {6,4,2,0}, {4,4,4,0}, {6,2,2,2}, {4,4,2,2} Sum of two even subsets (mod 2) is even Namely, of size 4 and 8, which is not possible

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Lemma II through V gives us our non- existence of a pair of 6 x 6 orthogonal Latin squares!

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So beware the puzzle 36 Cube by Thinkfun!

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Curriculum Latin Squares and examples Orthogonal Latin Squares and examples Informed them of the 6 x 6 case Magic Squares Constant Construct magic squares of odd order

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Activity I Latin squares definition and examples 2 x 2 2 cases 3 x 3 12 cases Had them find all 12, or as many as they could Gave an example of how to prove all 12 had been found

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Two of each case

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Activity II Orthogonal Latin Squares definition and examples 2 x 2 exampill Used playing cards 3 x 3 case 4 x 4 case Mentioned the 6 x 6 exampill

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Activity III Magic Squares definition and examples Constant using 1+…+k=k(k+1)/2 got a little lost Gave an example using the formula

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Activity IV How to construct a magic square of odd order method of up one and over one Gave a 3 x 3 example 5 x 5 example in groups got a little lost Ran out of time for construction of even order magic square for multiples of 4.

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Reference Stinson, D.R.. “A Short Proof of the Nonexistence of a Pair of Orthogonal Latin Squares of Order Six”. Journal of Combinatorial Theory, Series A, Volume 36, pg , 1984.

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Thank You John Caughman Joe Ediger Karen Marrongelle PSU Mathematics Department Faculty and Staff Eileen Mitchell-Babbitt

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