Some new and old results regarding Room squares and related designs

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

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

3. Wal’s first visit to Burlington -- 1982

4. First Vermont Summer Workshop, 1987

5. Hiking Mt. Mansfield Lecturing at the Workshop

6. Wal and Spyros Magilveras, Prague 1990

7. Wal’s current interest

8. Room squares • A Room square of side nis an n x n array filled with the symbols from a set S with |S|= n +1such 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.

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

10. 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.

11. 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.

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

13. Necessarily n 3s+ 2 First existence result is due to Wallis in 1974 who showed that for all odd s, there is a constant ns such that a(n,s)-IRS for all odd n  ns. In 1981 Stinson showed that ns  max{s+644, 6s+9}. Subsequent papers by Wallis (Congr. Numer. 1982) Stinson (Lec. Notes Math1983) Dinitz, Stinson, Wallis (Disc. Math1983) (s = 3,5,7) Dinitz, Stinson (AMS Contemp. Math.1990) (ns  3s+240) Dinitz, Stinson, Zhu (Electron J. Math#1, 1994) The current knowledge is the following:

14. Theorem (D,S,Z,1994): For every pair of odd positive integers (n,s) with n  3s + 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).

15. The extremal case of (3s + 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 (3s + 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 (3s + 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 (3s + 2, s)-IRS.

16. 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.

17. 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)

18. The setup for Room squares

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

20. 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 finot containing x Choose any point y such that y and x do not occur together Ify does not occur in fi , then add {x,y} to fi else there is a pair {z,y} with z ≠ x in fi replace {z,y} with {x,y} in fi The second heuristic is similar (picks a live factor and two points missing from it).

21. f1 {1,2} {3,7} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {1,3} f4 {4,8} {2,5} deficit = 11 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {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

22. f1 {1,2} {3,7} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {1,3} {2,6} f4 {4,8} {2,5} deficit = 10 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {1,5} Choose a live point and a partial one-factor not containing it

23. f1 {1,2} {3,7} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {1,3} {2,6} 7 f4 {4,8} {2,5} deficit = 10 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {1,5} Choose a live point and a partial one-factor not containing it

24. f1 {1,2} {3,7} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {1,3} {2,6} 7 f4 {4,8} {2,5} deficit = 10 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {1,5} Choose a live point and a partial one-factor not containing it Choose any point not occurring with the first point

25. f1 {1,2} {3,7} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {1,3} {2,6} 7,1 f4 {4,8} {2,5} deficit = 10 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {1,5} Choose a live point and a partial one-factor not containing it Choose any point not occurring with the first point

26. f1 {1,2} {3,7} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {2,6} 7,1 f4 {4,8} {2,5} deficit = 10 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {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}

27. f1 {1,2} {3,7} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {2,6} {7,1} f4 {4,8} {2,5} deficit = 10 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {1,5}

28. An example of Heuristic 2

29. f1 {1,2} {3,7} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {2,6} {7,1} f4 {4,8} {2,5} deficit = 10 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {1,5} Choose a live partial one-factor and two points not in it

30. f1 {1,2} {3,7} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {2,6} {7,1} f4 {4,8} {2,5} {3,7} deficit = 10 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {1,5} Choose a live partial one-factor and two points not in it

31. f1 {1,2} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {2,6} {7,1} f4 {4,8} {2,5} {3,7}deficit = 10 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {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}

32. f1 {1,2} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {2,6} {7,1} f4 {4,8} {2,5} {3,7} deficit = 10 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {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)

33. f1 {1,2} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {2,6} {7,1} f4 {4,8} {2,5} {3,7} deficit = 10 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {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}

34. f1 {1,2} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {2,6} {7,1} f4 {4,8} {2,5} {3,7} deficit = 10 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {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)

35. f1 {1,2} {4,6} f2 {2,4} {3,5} {7,8 } {1,6} f3 {4,5} {2,6} {7,1} f4 {4,8} {2,5} {3,7} deficit = 9 f5 {5,8} {1,4} {2,7} f6 {5,7} {2,3} f7 {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)

36. 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 Kn.

37. My notebook from 1992. it describes a successful search for an (61,19)-IRS and some of my unsuccessful attempts at finding a (67,21)-IRS.

38. 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

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

41. 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 Sparc2 workstation)  this would have taken roughly 3200 hours – over 4 months.

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

43. Frames A Room frame of type11 34

44. Closely related to Room squares and incomplete Room squares. A RS(n) is a frame of type 1n An (n,s)-IRS is a frame of type s11n-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 1980. 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 …

45. A frame is uniform if all the holes have the same size. A Room frame of type 25

46. 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 tu if and only if t(u-1) is even, except possibly when t = 14 and u = 4 (i.e. type 144).

47. Thank God for Moore’s Law! Greg and I found a Room frame of type 144using 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.

48. Room frame of type 144

49. 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 2u t1 if and only if t is even and u  t + 1 , except possibly when u = 19 and t = 18 (i.e. type 219 181 ).

50. Thank God for Moore’s Law! Greg and I found a Room frame of type 219 181using hill climbing. It took on average 550,000 restarts about the same as for the (67,21)-IRS