Perfect 1-factorizations
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1 Perfect 1-factorizations Alexander Rosa Department of Mathematics and Statistics, McMaster University Rosa (McMaster) Perfect 1-factorizations 1 / 17
2 Smolenice 1963 Rosa (McMaster) Perfect 1-factorizations 2 / 17
3 Problem No.20 Does there exist an integer n > 1 such that the complete 2n-gon is not Hamiltonian? 3 3 Hamiltonian in the sense of Kotzig, see p.63. Translated into contemporary terminology: Does a perfect 1-factorization of the complete graph K 2n for all integers n > 1? exist Rosa (McMaster) Perfect 1-factorizations 3 / 17
4 Problem No.20 Does there exist an integer n > 1 such that the complete 2n-gon is not Hamiltonian? 3 3 Hamiltonian in the sense of Kotzig, see p.63. Translated into contemporary terminology: Does a perfect 1-factorization of the complete graph K 2n for all integers n > 1? exist Rosa (McMaster) Perfect 1-factorizations 3 / 17
5 Two results of Kotzig on perfect 1-factorizations A 1-factorization of K 2n is perfect if the union of any two of its 1-factors is a Hamiltonian cycle. Theorem 1. A perfect 1-factorization of K 2n exists whenever 2n 1 is a prime. Theorem 2. A perfect 1-factorization of K 2n exists whenever n is a prime. Rosa (McMaster) Perfect 1-factorizations 4 / 17
6 Small orders The values of 2n 100 for which neither 2n 1 nor n is a prime, are n = 16, 28, 36, 40, 50, 52, and 56, 64, 66, 70, 76, 78, 88, 92, 96, 100. After considerable effort, both Kotzig and (independently) Bruce Anderson succeeded in finding a perfect 1-factorization of K 16. Bruce Anderson subsequently constructed a perfect 1-factorization of K 28. A discovery of a perfect 1-factorization for each of the next four orders (n = 36, 40, 50, 52) merited a separate paper! And n = 56 is currently the first open order... Rosa (McMaster) Perfect 1-factorizations 5 / 17
7 Small orders The values of 2n 100 for which neither 2n 1 nor n is a prime, are n = 16, 28, 36, 40, 50, 52, and 56, 64, 66, 70, 76, 78, 88, 92, 96, 100. After considerable effort, both Kotzig and (independently) Bruce Anderson succeeded in finding a perfect 1-factorization of K 16. Bruce Anderson subsequently constructed a perfect 1-factorization of K 28. A discovery of a perfect 1-factorization for each of the next four orders (n = 36, 40, 50, 52) merited a separate paper! And n = 56 is currently the first open order... Rosa (McMaster) Perfect 1-factorizations 5 / 17
8 Small orders The values of 2n 100 for which neither 2n 1 nor n is a prime, are n = 16, 28, 36, 40, 50, 52, and 56, 64, 66, 70, 76, 78, 88, 92, 96, 100. After considerable effort, both Kotzig and (independently) Bruce Anderson succeeded in finding a perfect 1-factorization of K 16. Bruce Anderson subsequently constructed a perfect 1-factorization of K 28. A discovery of a perfect 1-factorization for each of the next four orders (n = 36, 40, 50, 52) merited a separate paper! And n = 56 is currently the first open order... Rosa (McMaster) Perfect 1-factorizations 5 / 17
9 Small orders The values of 2n 100 for which neither 2n 1 nor n is a prime, are n = 16, 28, 36, 40, 50, 52, and 56, 64, 66, 70, 76, 78, 88, 92, 96, 100. After considerable effort, both Kotzig and (independently) Bruce Anderson succeeded in finding a perfect 1-factorization of K 16. Bruce Anderson subsequently constructed a perfect 1-factorization of K 28. A discovery of a perfect 1-factorization for each of the next four orders (n = 36, 40, 50, 52) merited a separate paper! And n = 56 is currently the first open order... Rosa (McMaster) Perfect 1-factorizations 5 / 17
10 Small orders The values of 2n 100 for which neither 2n 1 nor n is a prime, are n = 16, 28, 36, 40, 50, 52, and 56, 64, 66, 70, 76, 78, 88, 92, 96, 100. After considerable effort, both Kotzig and (independently) Bruce Anderson succeeded in finding a perfect 1-factorization of K 16. Bruce Anderson subsequently constructed a perfect 1-factorization of K 28. A discovery of a perfect 1-factorization for each of the next four orders (n = 36, 40, 50, 52) merited a separate paper! And n = 56 is currently the first open order... Rosa (McMaster) Perfect 1-factorizations 5 / 17
11 Difficulties Why is it so difficult to find perfect 1-factorizations? Part of the answer: n of (n) A pof (n) A of (n) = number of nonisomorphic 1-factorizations of K 2n, pof (n) = number of nonisomorphic perfect 1-factorizations of K 2n Rosa (McMaster) Perfect 1-factorizations 6 / 17
12 Difficulties Why is it so difficult to find perfect 1-factorizations? Part of the answer: n of (n) A pof (n) A of (n) = number of nonisomorphic 1-factorizations of K 2n, pof (n) = number of nonisomorphic perfect 1-factorizations of K 2n Rosa (McMaster) Perfect 1-factorizations 6 / 17
13 Ihrig s results In a series of papers, Ed Ihrig proves several results on the structure of automorphism groups of perfect 1-factorizations. E.g.: If F is a starter-induced perfect 1-factorization of K 2n then the automorphism group of F is a semidirect product of Z 2n 1 with H where H is a subgroup of the automorphism group of Z 2n 1, H divides n 1 and H is odd. For example, Ihrig, Seah and Stinson find a perfect 1-factorization of K 50 by searching for a starter in Z 49 which is fixed by a multiplicative subgroup {1, 18, 30} (i.e., assuming as an automorphism group the semidirect product of Z 49 with Z 3 ). Rosa (McMaster) Perfect 1-factorizations 7 / 17
14 Ihrig s results In a series of papers, Ed Ihrig proves several results on the structure of automorphism groups of perfect 1-factorizations. E.g.: If F is a starter-induced perfect 1-factorization of K 2n then the automorphism group of F is a semidirect product of Z 2n 1 with H where H is a subgroup of the automorphism group of Z 2n 1, H divides n 1 and H is odd. For example, Ihrig, Seah and Stinson find a perfect 1-factorization of K 50 by searching for a starter in Z 49 which is fixed by a multiplicative subgroup {1, 18, 30} (i.e., assuming as an automorphism group the semidirect product of Z 49 with Z 3 ). Rosa (McMaster) Perfect 1-factorizations 7 / 17
15 Recursive constructions? Another drawback: all known "sporadic" P1Fs were obtained by direct constructions, mostly using properties of finite fields. No recursive methods are known. Attempts to obtain a 2n 4n 2 construction using P1Fs generated by starters in cyclic groups have not been successful so far. Rosa (McMaster) Perfect 1-factorizations 8 / 17
16 Conclusion Kotzig s conjecture A PERFECT 1-FACTORIZATION OF K 2n EXISTS FOR ALL n > 1 is likely to remain open for quite some time... Rosa (McMaster) Perfect 1-factorizations 9 / 17
17 Perfect 1-factorizations of cubic graphs We consider only cubic graphs of Class 1. These fall into three categories. Category N. Category P. Cubic graphs without a P1F. Cubic graphs admitting only P1Fs. Category PN. Cubic graphs admitting a P1F and also a non-p1f. Rosa (McMaster) Perfect 1-factorizations 10 / 17
18 Forbidden subgraphs Some subgraphs cannot occur in cubic graphs admitting P1F: K 4 e, G 1, G 2 G 1 G 2 Rosa (McMaster) Perfect 1-factorizations 11 / 17
19 Category P: Cubic graphs admitting only P1Fs. Rosa (McMaster) Perfect 1-factorizations 12 / 17
20 Category PN. Cubic graphs admitting a P1F and also a non-p1f. We may assume that the number of vertices is at least 8. There are some forbidden subgraphs of cubic graphs admitting a P1F. Rosa (McMaster) Perfect 1-factorizations 13 / 17
21 Kotzig s theorem. A bipartite cubic graph with 2n vertices admits a perfect 1-factorization only if n is odd. For example, the prism GP(2k, 1) (prism with 4k vertices) cannot have a P1F (but obviously has a 1-factorization) Cubic graphs of category P exist for all orders 2n 4. (Examples of cubic graphs with a unique 1-factorization.). Number of vertices Class N P PN Total Rosa (McMaster) Perfect 1-factorizations 14 / 17
22 Generalized Petersen graphs GP(n, k) These are cubic graphs with vertex set Z n {0, 1} and edge-set {i 0, (i + 1) 0 }, {i 1, (i + k) 1 }, {i 0, i 1 }, i Z n (n 5, k < n 2 ). Every generalized Petersen graph, except GP(5, 2) (the Petersen graph itself), is Class 1. Bonvicini-Mazzuoccolo: GP(n, 2) has a P1F if and only if n 3 or 4 (mod 6). GP(n, 3) has a P1F if and only if n = 9. GP(n, k) does not admit P1F when n is even and k is odd. For k 4, to determine whether GP(n, k) admits a P1F remains open. Rosa (McMaster) Perfect 1-factorizations 15 / 17
23 GP(n, 4) and GP(n, 5) The graph GP(9, 4) is the well-known Tutte s example of a uniquely 3-edge-colourable cubic graph, and so is category P. When n is odd and 9 n 65, then GP(n, 4) has a P1F, except when n {11, 13, 17, 35}. When n is even and 10 n 64, then GP(n, 4) has a P1F if and only if n {28, 30} or 46 n 64. GP(n, 5) does not have a P1F when n is even. But when n is odd, and 11 n 51 then GP(n, 5) admits a P1F except when n {11, 13, 29, 41}. Moreover, GP(15, 5) admits only perfect 1-factorizations! It is difficult to even formulate an existence conjecture for GP(n, 4) or GP(n, 5). Rosa (McMaster) Perfect 1-factorizations 16 / 17
24 GP(n, 4) and GP(n, 5) The graph GP(9, 4) is the well-known Tutte s example of a uniquely 3-edge-colourable cubic graph, and so is category P. When n is odd and 9 n 65, then GP(n, 4) has a P1F, except when n {11, 13, 17, 35}. When n is even and 10 n 64, then GP(n, 4) has a P1F if and only if n {28, 30} or 46 n 64. GP(n, 5) does not have a P1F when n is even. But when n is odd, and 11 n 51 then GP(n, 5) admits a P1F except when n {11, 13, 29, 41}. Moreover, GP(15, 5) admits only perfect 1-factorizations! It is difficult to even formulate an existence conjecture for GP(n, 4) or GP(n, 5). Rosa (McMaster) Perfect 1-factorizations 16 / 17
25 GP(n, 4) and GP(n, 5) The graph GP(9, 4) is the well-known Tutte s example of a uniquely 3-edge-colourable cubic graph, and so is category P. When n is odd and 9 n 65, then GP(n, 4) has a P1F, except when n {11, 13, 17, 35}. When n is even and 10 n 64, then GP(n, 4) has a P1F if and only if n {28, 30} or 46 n 64. GP(n, 5) does not have a P1F when n is even. But when n is odd, and 11 n 51 then GP(n, 5) admits a P1F except when n {11, 13, 29, 41}. Moreover, GP(15, 5) admits only perfect 1-factorizations! It is difficult to even formulate an existence conjecture for GP(n, 4) or GP(n, 5). Rosa (McMaster) Perfect 1-factorizations 16 / 17
26 GP(n, 4) and GP(n, 5) The graph GP(9, 4) is the well-known Tutte s example of a uniquely 3-edge-colourable cubic graph, and so is category P. When n is odd and 9 n 65, then GP(n, 4) has a P1F, except when n {11, 13, 17, 35}. When n is even and 10 n 64, then GP(n, 4) has a P1F if and only if n {28, 30} or 46 n 64. GP(n, 5) does not have a P1F when n is even. But when n is odd, and 11 n 51 then GP(n, 5) admits a P1F except when n {11, 13, 29, 41}. Moreover, GP(15, 5) admits only perfect 1-factorizations! It is difficult to even formulate an existence conjecture for GP(n, 4) or GP(n, 5). Rosa (McMaster) Perfect 1-factorizations 16 / 17
27 GP(n, 4) and GP(n, 5) The graph GP(9, 4) is the well-known Tutte s example of a uniquely 3-edge-colourable cubic graph, and so is category P. When n is odd and 9 n 65, then GP(n, 4) has a P1F, except when n {11, 13, 17, 35}. When n is even and 10 n 64, then GP(n, 4) has a P1F if and only if n {28, 30} or 46 n 64. GP(n, 5) does not have a P1F when n is even. But when n is odd, and 11 n 51 then GP(n, 5) admits a P1F except when n {11, 13, 29, 41}. Moreover, GP(15, 5) admits only perfect 1-factorizations! It is difficult to even formulate an existence conjecture for GP(n, 4) or GP(n, 5). Rosa (McMaster) Perfect 1-factorizations 16 / 17
28 THANK YOU FOR YOUR ATTENTION! Rosa (McMaster) Perfect 1-factorizations 17 / 17
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