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Cyclic Complete Mappings Counting Problems

Cyclic Complete Mappings Counting Problems. Jieh Hsiang, YuhPyng Shieh, and YaoChiang Chen National Taiwan University 2014/8/21. Outline. CM(n),SCM(n),Q(n),TQ(n),TSQ(n) Symmetry operators Three types of counting problems Solutions counting problems Fixpoints counting problems

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Cyclic Complete Mappings Counting Problems

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  1. Cyclic Complete Mappings Counting Problems Jieh Hsiang, YuhPyng Shieh, and YaoChiang Chen National Taiwan University 2014/8/21

  2. Outline • CM(n),SCM(n),Q(n),TQ(n),TSQ(n) • Symmetry operators • Three types of counting problems • Solutions counting problems • Fixpoints counting problems • Equivalence classes counting problems • Demo

  3. Complete Mappings • A complete mapping of a group (G,+) is a permutation f(x) of G with f(0)=0 where 0 is the identity of (G,+) such that f(x)-x is also a permutation. • For example, (1,6,2,3,5,4) is a complete mapping of (Z7,+).

  4. History of complete mappings • Complete mappings was first introduced by Mann (1942) and used to construct orthogonal Latin squares. • Orthogonal mapping was called by Bose, Chakravarti, and Knuth (1960). • Orthomorphism was called by Johnson, Dulmage and Mendelsohn (1961). • Paige (1947) solved the existence problem of complete mappings for finite Abelian groups. • Singer (1960) found R,A,Tc,H, symmetry operators. • Jungnickel (1978) constructed complete mappings by “directly production” method. • D. F. Hsu and A. D. Keedwell (1984) used complete mappings to construct left-neofields.

  5. Cyclic Complete Mappings Counting Problems • Given a cyclic group (Zn,+), how many complete mappings are there?(=CM(n))

  6. CM(n) • The followings are all known nonzero CM(n)’s. • n=9: Johnson, Dulmage, and Mendelsohn(1961) • n=11: Singer (1961) • n=13,15: Hsu (1991) • n=17: Wanless (<2000) • n=19, 21, 23: Hsiang, Hsu, Shieh (2001)

  7. f 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 n-Queen Problem: Q(n) • A solution of n-queen problem is a permutation f(x) of Zn, such that ij, f(i)-i  f(j)-j and ij, f(i)+i  f(j)+j, under thenatural number addition. • Let Q(n) be the number of solutions of the n-queen problem.

  8. f 0 1 2 3 4 5 6 0 1 2 3 4 5 6 Toroidal N-Queen Problem: TQ(n) • A solution of the toroidal n-queen problem is a permutation f(x) of Zn, such that (under the cyclic group (Zn,+)), f(x)-x and f(x)+x are both permutations. • Let TQ(n) be the number of solutions of the toroidal n-queen problem.

  9. f 0 1 2 3 4 5 6 0 1 2 3 4 5 6 Strong complete mappings • A strong complete mapping of Zn is a solution f(x) of toroidal n-queen problem such that f(0)=0. • Let SCM(n) be the number of strong complete mappings of Zn.

  10. Toroidal-Semi N-Queen problem TSQ(n) • A solution of the toroidal-semi n-queen problem is a permutation f(x) of Zn, such that (under the cyclic group (Zn,+)), f(x)-x is also a permutation. • Let TSQ(n) be the number of solutions of the toroidal-semi n-queen problem.

  11. Board view of complete mappings • A complete mapping of a group (G,+) is a permutationf(x) of G with f(0)=0 where 0 is the identity of (G,+) such that f(x)-x is also a permutation. Function Permutation Complete mapping constraints

  12. TSQ(n) = n  CM(n)    Q(n) TQ(n) = n  SCM(n) Comparison of 5 problems

  13. TSQ(n) = n  CM(n)    Q(n) TQ(n)= n  SCM(n) Increasing rates

  14. Outline • CM(n),SCM(n),Q(n),TQ(n),TSQ(n) • Symmetry operators • Three types of counting problems • Solutions counting problems • Fixpoints counting problems • Equivalence classes counting problems • Demo

  15. f 0 1 2 3 4 5 6 7 0 1 2 Reflection 3 4 5 90o 6 7 N-Queen Problem • Reflection: • Rotation 90o • 8 distinct operators totally

  16. Symmetry operators • A symmetry operator  of some problem is a bijection on the solutions. 

  17. Symmetry operators • Symmetry operators are important to define interesting counting problems. • QR90o(n):Rotational n-queen problem. • A group of symmetry operators can define a notation of equivalence classes and reduce the search space of a counting problem. • Without these symmetry operators and more, we don’t think that CM(23), and QR90o(61) can be computed.

  18. Symmetry operators

  19. f -3 -2 -1 0 1 2 3 -3 f -3 -2 -1 0 1 2 3 f -3 -2 -1 0 1 2 3 f -3 -2 -1 0 1 2 3 -2 -3 -3 -3 -1 -2 45o -2 -2 0 -1 45o -1 -1 1 0 0 0 2 1 1 1 3 2 2 2 3 3 3 Toroidal N-Queen Problem Reflection: Rotation 45o Toroidal shifting: Homology: 8nn(n)distinct operators totally. TS(2,3) R45o H2

  20. f -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 Strong Complete Mapping Problem Reflection: Rotation 45o Translation: Homology: 8n(n)distinct operators totally. T1

  21. Toroidal-semi N-Queen Problem Reflection: Operator A Toroidal Shifting: Homology: 6nn(n)distinct operators totally.

  22. Complete Mapping Problem Reflection: Operator A Translation: Homology: 6n(n)distinct operators totally.

  23. Outline • CM(n),SCM(n),Q(n),TQ(n),TSQ(n) • Symmetry operators • Three types of counting problems • Solutions counting problems • Fixpoints counting problems • Equivalence classes counting problems • Demo

  24. Three types of counting problems • Let P be a problem in {Q,TQ,TSQ,CM,SCM}. • Let  be a symmetry operator of P. • For example, =R90o. • Let  be a symmetry operator group spanned by some symmetry operators. • For example, =span{R,R90o} • P(n): the number of solutions. • Q(23),CM(23),SCM(29), and etc. • P(n): the number of fixpoints under . • P(n): the number of equivalence classes.

  25. R: R90o: QR90o(5)=2 QR180o(5)=2 Qspan{R,R90o}=2

  26. Results and challenges (type 1) • Q(n),TQ(n),TSQ(n) : Best nonzero known results • Q(23)= 24,233,937,684,440 [1](A000170) • TQ(29)= 605,917,055,356 [1](A007705) • SCM(29)= 20,893,691,564 [1](A071607) • TSQ(23)= 452,794,797,220,965 [1](A006717) • CM(23)= 19,686,730,313,955 [1](A003111) • Q(24) can be done using our method in 1266 days using one P4 1.8G machine [1] http://www.research.att.com/~njas/sequences/

  27. Results and challenges (type 2) • QR90o(n), QR180o(n) • QR90o(61) =291,826,098,503,680 [1](A033148) • QR180o(32)= 181,254,386,312 [1](A032522) • SCMR45o(n), SCMR90o(n), SCMR180o(n) • Note: R45o is a newly known operator • CMR(n)), CMR180o(n), CMA(n), TSQA(n). • CMR(29)= 1,317,606,101 [1](A006204) • CMR180o(37)= 131,777,883,431,119 [1](A071706) • CMA(49)= 430,415,593,603,072 [1](A071608)

  28. Results and challenges (type 3) • Q{R,R90o}(n), TQ{R,R90,TS(c,d)}(n). • Q{R,R90o}(23)=3029242658210 [1](A002562) • TQ{R,R90,TS(c,d)}(29)=90120677 [1](A054500) • TQ{R,R45}(n) • SCM{R,R45,Tc,H}(n) • CM{R,A,Tc,H}(n)

  29. Our results We develop symmetry cutting strategies and partition strategies to solve these problems.

  30. Our results (cont.)

  31. Q(24) project

  32. CM(25) project

  33. Demo • Counting program. • Translation program: • We translate the proposed problems into propositional logic. • http://turing.csie.ntu.edu.tw/~arping/cm

  34. The EndThank you very much.

  35. R: R90o: f 0 1 2 3 4 f 0 1 2 3 4 0 ♛ 0 ♛ 1 ♛ 1 ♛ QR90o(5)=2 QR180o(5)=2 Qspan{R,R90o}=2 2 ♛ 2 ♛ 3 ♛ 3 ♛ 4 ♛ 4 ♛ f 0 1 2 3 4 f 0 1 2 3 4 f 0 1 2 3 4 f 0 1 2 3 4 0 ♛ 0 ♛ 0 ♛ 0 ♛ 1 ♛ 1 ♛ 1 ♛ 1 ♛ 2 ♛ 2 ♛ 2 ♛ 2 ♛ 3 ♛ 3 ♛ 3 ♛ 3 ♛ 4 ♛ 4 ♛ 4 ♛ 4 ♛ f 0 1 2 3 4 f 0 1 2 3 4 f 0 1 2 3 4 f 0 1 2 3 4 0 ♛ 0 ♛ 0 ♛ 0 ♛ 1 ♛ 1 ♛ 1 ♛ 1 ♛ 2 ♛ 2 ♛ 2 ♛ 2 ♛ 3 ♛ 3 ♛ 3 ♛ 3 ♛ 4 ♛ 4 ♛ 4 ♛ 4 ♛

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