Some new and old results regarding Room squares and related designs Jeff Dinitz University of Vermont Wallapolooza

Wal wrote the book on Room Squares in 1972 (only 4 years after getting his Ph.D) A Beginners Guide to Room Squares

Wal’s first visit to Burlington

First Vermont Summer Workshop, 1987

Lecturing at the Workshop Hiking Mt. Mansfield

Wal and Spyros Magilveras, Prague 1990

Wal’s current interest

Room squares A Room square of side n is an n x n array filled with the symbols from a set S with |S|= n +1 such that the following properties hold: 1) Every cell in the array is either empty or contains an unordered pair of symbols from S. 2) Each column and each row contain every symbol from S exactly once. 3) Every unordered pair of symbols appears exactly once in the array. We see that n is necessarily odd.

Example of a Room Square 0        n = 7, S = {0,1,…6,  }

After much work the following was proven in 1975: Theorem: (Mullin, Wallis) There exists a Room square of side n for when n =1 and for all odd n  7. There does not exist a Room square of side 3 or 5. But this really just opened the door for many questions about generalizations of Room squares and special types of Room squares.

An (n, s)-incomplete Room square (IRS) F is an n × n array for which every cell of F is empty or contains an unordered pair of symbols from an (n + 1 )-set N; there is an empty s × s subarray G of F; every symbol of N \ S occurs once in each row and column of F, where S  N is an (s + 1 )-set; each symbol of S occurs once in each row and each column not meeting G but in no row or column meeting G; the pairs in F are precisely those {x, y}  (N × N)\(S × S). If there exists a Room square of side s, this is just a Room square of side n with a sub Room square of side s.

An (11,3) incomplete Room square. S = {0,1,2,Y}

Necessarily n  3 s + 2 First existence result is due to Wallis in 1974 who showed that for all odd s, there is a constant n s such that a (n,s)-IRS for all odd n  n s. In 1981 Stinson showed that n s  max{s+ 644, 6 s+ 9 }. Subsequent papers by Wallis (Congr. Numer ) Stinson (Lec. Notes Math 1983 ) Dinitz, Stinson, Wallis (Disc. Math 1983 ) ( s = 3,5,7) Dinitz, Stinson (AMS Contemp. Math ) ( n s  3s+240 ) Dinitz, Stinson, Zhu (Electron J. Math #1, 1994 ) The current knowledge is the following:

Theorem (D,S,Z, 1994 ): For every pair of odd positive integers (n,s) with n  3 s + 2, there exists an (n, s) incomplete Room square, except when (n, s) = ( 5, 1 ) and with the possible exception of (n, s) = ( 67, 21 ).

The extremal case of ( 3 s + 2, s)-IRS is interesting. In 1974 Wal conjectured that a ( 23,7 )-IRS does not exist and offered a prize of $10.00 for a proof or disproof. In 1980, I constructed the first example of a ( 3 s + 2, s)-IRS. I made an ( 11, 3 )-IRS (by hill-climbing) on the computer. Using the ( 11, 3 )-IRS, Stinson proved in 1981 that there exists a ( 3 s + 2, s)-IRS for all s  3 (mod 8 ). In 1982, Wallis disproved his own conjecture by constructing a ( 23, 7 )-IRS. Did you pay yourself the $10.00?? Theorem (Wallis 1983): For all odd s  3, there is a ( 3 s + 2, s)-IRS.

Search for the (67,21)-IRS This is the last unsolved case for an (n,s)-IRS There are 68 symbols that need to be placed in the array.

Hill-climbing Using a hill-climbing algorithm, many examples of (n, s) incomplete Room squares were found. For example: (n, 5 )-IRS for all odd n, 19 ≤ n ≤ 61 (n, 7 )-IRS for all odd n, 25 ≤ n ≤ 53 (n, 9 )-IRS for all odd n, 35 ≤ n ≤ 77 Hill-climbing is the standard method now for finding Steiner triple systems (Stinson’s Algorithm). First successful hill-climbing algorithm for finding a combinatorial design was the hill-climb for strong starters -- to make Room cubes ( 1981, Dinitz and Stinson)

The setup for Room squares

0        The rows of a Room square of side n form a one-factorization of K n+1 The columns also form a one-factorization of K n+1 These two one-factorizations are orthogonal. (i.e. any row factor and any column factor have at most one edge in common)

The algorithm There are two heuristics – this is the first. Start with a partial one-factorization F: Choose a live point x (x does not occur in some 1 -factor) Choose a partial one-factor f i not containing x Choose any point y such that y and x do not occur together If y does not occur in f i, then add {x,y} to f i else there is a pair {z,y} with z ≠ x in f i replace {z,y} with {x,y} in f i The second heuristic is similar (picks a live factor and two points missing from it).

f 1 {1,2}{3,7}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {1,3} f 4 {4,8}{2,5}deficit = 11 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5} Choose a live point and a partial one-factor not containing it Choose any point not occurring with the first point ok? Yes – add the pair (deficit decreases by one) 2,6,6

f 1 {1,2}{3,7}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {1,3} {2,6} f 4 {4,8}{2,5}deficit = 10 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5} Choose a live point and a partial one-factor not containing it

f 1 {1,2}{3,7}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {1,3} {2,6} 7 f 4 {4,8}{2,5}deficit = 10 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5} Choose a live point and a partial one-factor not containing it

f 1 {1,2}{3,7}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {1,3} {2,6} 7 f 4 {4,8}{2,5}deficit = 10 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5} Choose a live point and a partial one-factor not containing it Choose any point not occurring with the first point

f 1 {1,2}{3,7}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {1,3} {2,6} 7,1 f 4 {4,8}{2,5}deficit = 10 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5} Choose a live point and a partial one-factor not containing it Choose any point not occurring with the first point

f 1 {1,2}{3,7}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {2,6} 7,1 f 4 {4,8}{2,5}deficit = 10 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5} Choose a live point and a partial one-factor not containing it Choose any point not occurring with the first point ok? No – make the switch (deficit remains the same) {1,3}

f 1 {1,2}{3,7}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {2,6}{7,1} f 4 {4,8}{2,5}deficit = 10 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5}

An example of Heuristic 2

f 1 {1,2}{3,7}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {2,6}{7,1} f 4 {4,8}{2,5}deficit = 10 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5} Choose a live partial one-factor and two points not in it

f 1 {1,2}{3,7}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {2,6}{7,1} f 4 {4,8}{2,5}{3,7}deficit = 10 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5} Choose a live partial one-factor and two points not in it

f 1 {1,2}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {2,6}{7,1} f 4 {4,8}{2,5}{3,7}deficit = 10 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5} Choose a live partial one-factor and two points not in it ok? Yes – add the pair (deficit decreases by one) No – make the switch (deficit remains the same) {3,7}

f 1 {1,2}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {2,6}{7,1} f 4 {4,8}{2,5}{3,7} deficit = 10 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5} Choose a live partial one-factor and two points not in it ok? Yes – add the pair (deficit decreases by one) No – make the switch (deficit remains the same)

f 1 {1,2}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {2,6}{7,1} f 4 {4,8}{2,5}{3,7} deficit = 10 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5} Choose a live partial one-factor and two points not in it ok? Yes – add the pair (deficit decreases by one) {2,8}

f 1 {1,2}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {2,6}{7,1} f 4 {4,8}{2,5}{3,7} deficit = 10 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5}{2,8} Choose a live partial one-factor and two points not in it ok? Yes – add the pair (deficit decreases by one)

f 1 {1,2}{4,6} f 2 {2,4} {3,5} {7,8 } {1,6} f 3 {4,5} {2,6}{7,1} f 4 {4,8}{2,5}{3,7} deficit = 9 f 5 {5,8} {1,4} {2,7} f 6 {5,7} {2,3} f 7 {1,5}{2,8} Choose a live partial one-factor and two points not in it ok? Yes – add the pair (deficit decreases by one)

The algorithm is very effective at finding one-factorizations. It can be easily modified to find orthogonal one-factorizations. So it can find Room squares. It can also easily be modified for different underlying graphs other than K n.

My notebook from it describes a successful search for an (61,19)-IRS and some of my unsuccessful attempts at finding a (67,21)-IRS.

Thank God for Moore’s Law!! Using the hill-climbing algorithm Greg Warrington and I found a ( 67,21 )-incomplete Room square. He did it overnight !!! more details to come

A ( 67,21 ) - Incomplete Room Square !!! In honor of Wal’s 68 th birthday

Greg generated several such squares and compiled some data. The expected number of restarts before finding a ( 67,21 )- IRS is 480,000. A restart begins at around 500,000 attempted switches. The expected time ( 2009 hours) for a single process to finish on a single 3 GHz cpu is 16 hours. In 1992 hours (probably a Sparc 2 workstation) this would have taken roughly 3200 hours – over 4 months.

Theorem: ( 2009 ) For every pair of odd positive integers (n,s) with n  3 s + 2, there exists an (n, s) incomplete Room square, except when (n, s) = ( 5, 1 ).

Frames A Room frame of type

Closely related to Room squares and incomplete Room squares. A RS(n) is a frame of type 1 n An (n,s)-IRS is a frame of type s 1 1 n-s. They are still two orthogonal one-factorizations, but now the underlying graph is the complete graph missing a spanning set of disjoint complete graphs. First mentioned in Wallis-Street-Wallis (but not formally defined). First defined in paper by Mullin, Schellenberg, Vanstone, and Wallis in 1979 (but only when skew). Defined and studied in paper by Stinson and D in Generalized and extended to almost every kind of design in the past 30 years as they are extremely useful in recursive constructions. Kirkman frames, triplewhist frames, frame SOLS …

A frame is uniform if all the holes have the same size. A Room frame of type 2 5

Again much work was done to find the spectrum of uniform Room frames. In the 1994 paper by Dinitz Stinson and Zhu, all exceptional cases were found except for one. The following theorem gives our current knowledge. Theorem(D,S,Z, 1994 ): Suppose that t and u are positive integers with u  4 and (t,u) ≠ (1,5) or (2,4). There exists a Room frame of type t u if and only if t(u-1) is even, except possibly when t = 14 and u = 4 (i.e. type 14 4 ).

Thank God for Moore’s Law! Greg and I found a Room frame of type 14 4 using hill climbing. It took on average 5,260,000 restarts to find one. 10 times harder than the ( 67,21 )- IRS Couldn’t even get close to finding this in 1992.

Room frame of type 14 4

One more new Room frame The following theorem is also from the 1994 paper with Stinson and Zhu. Theorem (D,S,Z): Suppose that t and u are positive integers. If t  4 there exists a Room frame of type 2 u t 1 if and only if t is even and u  t + 1, except possibly when u = 19 and t = 18 (i.e. type ).

Thank God for Moore’s Law! Greg and I found a Room frame of type using hill climbing. It took on average 550,000 restarts about the same as for the ( 67,21)- IRS

Room frame of type

Howell cubes A Howell design, H(s, 2 n), is equivalent to a pair of orthogonal one-factorizations of an s-regular graph on 2 n vertices. A Howell design H(4,6) joint work with Greg Warrington and Esther Lamken

Theorem: (Stinson 1982, Anderson, Schellenberg, Stinson 1984 ) Let s and n be positive integers, where n is even and s+ 1  2 n  2 s. Then there exists an H(s, 2 n) if and only if (s, 2 n) ≠ ( 2, 4), (3, 4), (5, 6 ), or ( 5, 8 ). Three special classes of Howell designs: An H( 2 n- 1, 2 n) is a Room square of side 2 n- 1 (so the underlying graph is the complete graph K 2n ). An H(n, 2 n) can be constructed by superimposing two orthogonal Latin squares of side n (underlying graph is K n,n ). The underlying graph of an H( 2 n- 2, 2 n) is the cocktail party graph K 2n − f, where f is a 1 -factor.

Extending the definition, a Howell cube H 3 (s, 2 n), is equivalent to three orthogonal one-factorizations of an s-regular graph on 2 n vertices. To get an actual cube, index each side of the cube by one of the one-factorizations and put pair {x,y} in cell (i,j,k) if {x,y} is in the i th factor, the j th factor and the k th factor of the three factorizations, respectively. So an H 3 ( 2 n- 1, 2 n), is a Room cube of side n and is equivalent to three orthogonal one-factorizations of the complete graph K 2n.

Room cube of side 7 or a Howell cube H 3 (7,8)

Theorem (Dinitz, Stinson, 1981 ) There exists a Room cube of side n (a H 3 (n, n+ 1 )) for all odd n  7. Theorem. There exists a H 3 (n, 2 n) if and only if n  4 with the possible exceptions of H 3 ( 6, 12 ) and H 3 ( 10, 20 ) This is a consequence of the existence of three MOLS(n) for all n  4 except 6 and possibly 10. The H 2 ( 6, 12 ) was found by Brickell in The underlying graph is the icosahedron with antipodal points joined.

So what about H 3 ( 2 n- 2, 2 n) ? Remember the underlying graph is the cocktail party graph on 2 n points. H 3 ( 2, 4 ), H 3 ( 4, 6 ), and H 3 ( 6, 8 ) do not exist. H 3 ( 2 n, 2 n + 2 ) exist for all n  3 (Lamken and Vanstone, ‘84) so there exist H 3 ( 8, 10 ) and H 3 ( 16, 18 ). H 3 ( 10, 12 ) and H 3 ( 12, 14 ) and H 3 ( 14, 16 ) exist. we very recently constructed them using hill-climbing Thank God for Moore’s Law!

( 1, 2) ( 3, 4) ( 5, 6) ( 7, 8) ( 9,10) (11,12) ( 1, 3) ( 2, 4) ( 5, 7) ( 6, 8) ( 9,11) (10,12) ( 1, 4) ( 2, 3) ( 5, 9) ( 6,10) ( 7,11) ( 8,12) ( 1, 5) ( 2, 6) ( 3, 7) ( 4, 8) ( 9,12) (10,11) ( 1, 6) ( 2, 5) ( 3, 9) ( 4,10) ( 7,12) ( 8,11) ( 1, 7) ( 2, 8) ( 3,11) ( 4,12) ( 5,10) ( 6, 9) ( 1, 8) ( 2, 9) ( 3, 5) ( 4,11) ( 6,12) ( 7,10) ( 1, 9) ( 2, 7) ( 3,12) ( 4, 5) ( 6,11) ( 8,10) ( 1,10) ( 2,12) ( 3, 6) ( 4, 7) ( 5,11) ( 8, 9) ( 1,11) ( 2,10) ( 3, 8) ( 4, 6) ( 5,12) ( 7, 9) ( 1, 2) ( 3,11) ( 4, 7) ( 5, 9) ( 6,12) ( 8,10) ( 1, 3) ( 2, 6) ( 4, 5) ( 7,10) ( 8, 9) (11,12) ( 1, 4) ( 2, 8) ( 3, 5) ( 6,11) ( 7,12) ( 9,10) ( 1, 5) ( 2,12) ( 3, 8) ( 4,10) ( 6, 9) ( 7,11) ( 1, 6) ( 2, 3) ( 4,11) ( 5,10) ( 7, 8) ( 9,12) ( 1, 7) ( 2, 9) ( 3, 4) ( 5,12) ( 6, 8) (10,11) ( 1, 8) ( 2, 4) ( 3,12) ( 5,11) ( 6,10) ( 7, 9) ( 1, 9) ( 2,10) ( 3, 6) ( 4,12) ( 5, 7) ( 8,11) ( 1,10) ( 2, 5) ( 3, 7) ( 4, 6) ( 8,12) ( 9,11) ( 1,11) ( 2, 7) ( 3, 9) ( 4, 8) ( 5, 6) (10,12) ( 1, 2) ( 3, 5) ( 4, 6) ( 7,11) ( 8, 9) (10,12) ( 1, 3) ( 2,12) ( 4, 8) ( 5,10) ( 6,11) ( 7, 9) ( 1, 4) ( 2, 5) ( 3,12) ( 6, 9) ( 7, 8) (10,11) ( 1, 5) ( 2,10) ( 3, 9) ( 4, 7) ( 6, 8) (11,12) ( 1, 6) ( 2, 9) ( 3, 7) ( 4,12) ( 5,11) ( 8,10) ( 1, 7) ( 2, 3) ( 4, 5) ( 6,12) ( 8,11) ( 9,10) ( 1, 8) ( 2, 7) ( 3, 6) ( 4,10) ( 5,12) ( 9,11) ( 1, 9) ( 2, 4) ( 3,11) ( 5, 6) ( 7,10) ( 8,12) ( 1,10) ( 2, 6) ( 3, 8) ( 4,11) ( 5, 9) ( 7,12) ( 1,11) ( 2, 8) ( 3, 4) ( 5, 7) ( 6,10) ( 9,12) Three orthogonal one-factorizations of the cocktail party graph on 12 vertices F1F1 F3F3 F2F2 An H 3 (10,12)

A recursive construction for Howell cubes Fundamental Frame construction and filling in the holes. Ingredients: A pairwise balanced design PBD(v,{ 5,9}). These exists for all v  1 mod 4 except for v = 13, 17, 29, 33 and possibly 113. ( See the Handbook ) 3 -dimensional Room frames of types 2 5 and 2 9. Found in 1981 by Dinitz and Stinson H 3 ( 8, 10 ) and H 3 ( 16, 18 ). Found in 1984 by Lamken and Vanstone.

Start with the pairwise balanced design PBD(v,{ 5,9}). Delete one point to construct a group divisible design GDD with blocks of size { 5, 9} and groups of size { 4, 8}. Give each point weight 2 and use the Fundamental Frame construction filling in the blocks of size 5 and 9 with the frames of types 2 5 and 2 9, respectively. Then fill in the holes with the H 3 ( 8, 10 ) and H 3 ( 16, 18 ).

The result is that there exists a Howell cube H 3 (8n, 8 n+ 2 ) for all n except possibly for n = 3, 4, 7, 8 and 28. We’ll get them all pretty soon. (at least the H 3 ( 2 n, 2 n+ 2 ))

Part 3 – Some beautiful music

Tom Johnson An American composer living in Paris

In her 2008 M.S. thesis Susan Janiszewski studied Tom Johnson’s problem for both Room squares of side 7 and Room squares of side 11.

Patterns A pattern for an n x n Room square is the n x n incidence matrix of the filled cells in the Room square. The zero-pattern is the incidence matrix corresponding to the empty cells in the Room square.

Balanced Incomplete Block Designs (BIBDs) A design is a pair (X,A) where X is a set of points and A is a collection of non-empty subsets of X called blocks. A BIBD is characterized by the parameters v, k, and λ v is the number of points k is the number of points in each block Every distinct pair of points is contained in exactly λ blocks. The incidence matrix for a given BIBD is a v x b matrix with the rows indexed by the points of the design and the columns are indexed by the blocks. In a given row, ones are placed in the columns corresponding to the blocks that contain the given point corresponding to that row and zeros are placed in the remaining cells.

Large Sets of Designs A large set of (v,k,λ) designs is a partition of all the subsets of size k chosen from v points into (v,k,λ)- designs.

Room Squares of Side 7 There are 6 nonisomorphic one-factorizations of K 8. There are 6 inequivalent Room squares of side 7. (transposes are allowed) Wallis, (Gelling too)

Room Squares of Side 7

Theorem: There exists a 7  7 Room square with pattern P  P is the incidence matrix of a ( 7,4,2) -BIBD. Proof: → Need to check that given P, it has column sum of 4, row sum of 4, and inner product of two rows as 2. Inspection of all possible Room squares gives this. ← Note that a ( 7,4,2 )-BIBD is the complement of the ( 7,3,1 )- BIBD, the Fano plane. There is only one Fano plane, up to isomorphism. We can find a permutation σ that takes the ( 7,4,2 )-BIBD that is given to the ( 7,4,2 )-BIBD that corresponds to R 1,1.

The previous theorem gives us that every zero-pattern of a Room square of side seven is the incidence matrix of a ( 7,3,1 )-BIBD – a Fano plane. A solution to Tom Johnson's initial problem would be equivalent to the existence of a large set of ( 7,3,1 )- BIBDs. It was proved in 1850 by Cayley that such a set DOES NOT EXIST! The largest set of Fano planes that contain disjoint blocks is 2. Lets briefly look at how many distinct blocks can be contained in a set of Fano planes and how many Fano planes are required to achieve all 35 distinct blocks.

Classes of Fano Planes Two planes are in the same class if they share exactly one block together. (This can be shown to be well-defined) There are 30 distinct Fano planes and they can be divided into two distinct, well-defined classes. We will refer to the two classes of Fano planes Class A and Class B. Each class contains 15 of the 30 total planes. Two planes in different classes are either disjoint or intersect in exactly three blocks. Each plane is disjoint from exactly eight planes in the other class and has intersection of 3 with seven planes in the other class.

Some Lemmas Lemma 1 : There are exactly three planes in each class that contain a given block {a,b,c}. Lemma 2 : If three planes from one class intersect in the same block, there does not exist a plane from the other class that is disjoint from all three. Lemma 3 : Given any pair of planes from the same class, there are at least four planes that are disjoint from both of them.

Sets of 2 Fano Planes with the Maximum Number of Distinct Blocks A set of two Fano Planes contains at most 14 distinct blocks. This set must contain one plane from one class and one plane from the other. There are 120 different pairs of Fano planes that contain 14 distinct blocks, all of which are isomorphic to each other. The smallest of these sets lexicographically: A 1 ={{1,2,3},{1,4,5},{1,6,7},{2,4,6},{2,5,7},{3,4,7},{3,5,6}} B 1 ={{1,2,4},{1,3,6},{1,5,7},{2,3,7},{2,5,6},{3,4,5},{4,6,7}}

Sets of 3 Fano Planes with the Maximum Number of Distinct Blocks A set of three Fano Planes contains at most 20 distinct blocks. This set must contain two planes from one class and one plane from the other. There are 840 different sets of three Fano planes that contain 20 distinct blocks, all of which are isomorphic to each other. The smallest of these sets lexicographically: A 1 ={{1,2,3},{1,4,5},{1,6,7},{2,4,6},{2,5,7},{3,4,7},{3,5,6}} A 2 ={{1,2,3},{1,4,6},{1,5,7},{2,4,7},{2,5,6},{3,4,5},{3,6,7}} B 1 ={{1,2,4},{1,3,7},{1,5,6},{2,3,5},{2,6,7},{3,4,6},{4,5,7}}

Sets of 4 Fano Planes with the Maximum Number of Distinct Blocks A set of four Fano Planes contains at most 26 distinct blocks. This set must contain two planes from each class There are 630 different sets of four Fano planes that contain 26 distinct blocks, all of which are isomorphic to each other. The smallest of these sets lexicographically: A 1 ={{1,2,3},{1,4,5},{1,6,7},{2,4,6},{2,5,7},{3,4,7},{3,5,6}} A 2 ={{1,2,3},{1,4,6},{1,5,7},{2,4,7},{2,5,6},{3,4,5},{3,6,7}} B 1 ={{1,2,4},{1,3,7},{1,5,6},{2,3,5},{2,6,7},{3,4,6},{4,5,7}} B 2 ={{1,2,5},{1,3,6},{1,4,7},{2,3,4},{2,6,7},{3,5,7},{4,5,6}}

Sets of 5 Fano Planes with the Maximum Number of Distinct Blocks A set of five Fano Planes contains at most 31 distinct blocks. This set must contain three planes from one class and two planes from the other class. There are 840 different sets of five Fano planes that contain 31 distinct blocks, all of which are isomorphic to each other. The smallest of these sets lexicographically: A 1 ={{1,2,3},{1,4,5},{1,6,7},{2,4,6},{2,5,7},{3,4,7},{3,5,6}} A 2 ={{1,2,3},{1,4,6},{1,5,7},{2,4,7},{2,5,6},{3,4,5},{3,6,7}} A 3 ={{1,2,4},{1,3,5},{1,6,7},{2,3,7},{2,5,6},{3,4,6},{4,6,7}} B 1 ={{1,2,5},{1,3,6},{1,4,7},{2,3,4},{2,6,7},{3,5,7},{4,5,6}} B 2 ={{1,2,7},{1,3,4},{1,5,6},{2,3,6},{2,4,5},{3,5,7},{4,6,7}}

Sets of 6 Fano Planes with the Maximum Number of Distinct Blocks A set of six Fano Planes can contain all 35 distinct blocks. This set must contain four planes from one class and two planes from the other class. There are 210 different sets of six Fano planes that contain all 35 possible blocks, all of which are isomorphic to each other. The smallest of these sets lexicographically: A 1 ={{1,2,3},{1,4,5},{1,6,7},{2,4,6},{2,5,7},{3,4,7},{3,5,6}} A 2 ={{1,2,3},{1,4,6},{1,5,7},{2,4,7},{2,5,6},{3,4,5},{3,6,7}} A 3 ={{1,2,4},{1,3,5},{1,6,7},{2,3,7},{2,5,6},{3,4,6},{4,6,7}} A 4 ={{1,2,6},{1,3,7},{1,4,5},{2,3,5},{2,4,7},{3,4,6},{5,6,7}} B 1 ={{1,2,5},{1,3,6},{1,4,7},{2,3,4},{2,6,7},{3,5,7},{4,5,6}} B 2 ={{1,2,7},{1,3,4},{1,5,6},{2,3,6},{2,4,5},{3,5,7},{4,6,7}}

Classes of Fano Planes Two planes are in the same class if they share exactly one block together. (This can be shown to be well-defined) There are 30 Fano planes that can be divided into two distinct, well-defined classes. We will refer to the two classes of Fano planes Class A and Class B. Each class contains 15 of the 30 total planes. Two planes in different classes are either disjoint or intersect in exactly three blocks. Each plane is disjoint from exactly eight planes in the other class and has intersection of 3 with seven planes in the other class. The seven planes that have intersection of 3 with a plane in the other class can be identified with the points in a Fano plane in a natural way.

Creating a Fano Plane from 7 Fano Planes Example: A 1 ={{a,b,c}, {a,d,e},{a,f,g},{b,d,f},{b,e,g},{c,d,g},{c,e,f}} B 1 ={{a,b,c},{a,d,e},{a,f,g},{b,d,g},{b,e,f},{c,e,g},{c,d,f}} B 2 ={{a,b,c}, {a,d,f},{a,e,g},{b,d,e},{b,f,g},{c,d,g},{c,e,f}} B 3 ={{ a,b,c},{a,d,g},{a,e,f},{b,d,f},{b,e,g},{c,d,e},{c,f,g}} B 4 ={{a,b,d},{a,c,e},{a,f,g},{b,c,f},{b,e,g},{c,d,g},{d,e,f}} B 5 ={{a,b,e},{a,c,d},{a,f,g},{b,c,g},{b,d,f},{c,e,f},{d,e,g}} B 6 ={{a,b,f},{a,c,g},{a,d,e},{b,c,d},{b,e,g},{c,e,f},{d,f,g}} B 7 ={{a,b,g},{a,c,f},{a,d,e},{b,c,e},{b,d,f},{c,d,g},{e,f,g}} D 1 ={B 1,B 2,B 3 }D 2 ={B 1,B 6,B 7 } D 3 ={B 1,B 4,B 5 }D 4 ={B 3,B 5,B 7 } D 5 ={B 3,B 4,B 6 }D 6 ={B 2,B 4,B 7 } D 7 ={B 2,B 5,B 6 }

Now – what about size 11 ??

Hadamard Room Squares A Hadamard design is a symmetric ( 4 m -1, 2 m -1, m -1 )- BIBD. We define a Hadamard Room Square to be a Room square whose zero-pattern is the incidence matrix of a Hadamard design. Every Room square of side 7 is a Hadamard Room square (zero-pattern is a ( 7,3,1 ) design)

Hadamard Room Square of Side 11 The zero-pattern is an (11,5,2) Hadamard design

Large Set of (11,5,2)-BIBDs A large set will contain 42 designs. Kramer, Magliveras, and Stinson (1991) created a large set by first obtaining six starter designs by having permutations A through F act on a single base design. Base design={{ 1,2,3,7,10},{1,2,6,9,11},{1,3,4,5,9},{1,4,6,7,8}, {1,5,8,10,11}, {2,3,4,8,11},{2,4,5,6,10},{2,5,7,8,9},{3,5,6,7,11}, {3,6,8,9,10}, {4,7,9,10,11}} A. ( ) B. ( ) (2 6) (3 9) C. (1) (2) (3) (4) (5) ( ) (9) D. ( ) (5 7) (9) (10) (11) E. (1) (2) (3) (4) (5) ( ) (11) F. (1 3 11) ( ) (4) (6 8) The permutation σ = (1)(2)(3)(4)( ) is then used to obtain seven different (11,5,2)- BIBDs from each of the six starter designs.

Solution to Tom Johnson’s Problem for Room Squares of Side 11. Begin with the Hadamard Room square of side 11. This Room square can be transformed into a Room square whose incidence matrix is the base design through a set of row and column permutations. We can then use the permutations that created the large set of ( 11,5,2 )-BIBDs to create a large set of Room squares of side 11.

Vermont Rhythms by Tom Johnson performed by Klang

Thanks for attending!