对排列组合染色问题的探究

©1994-2010ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.(,,678400),,,,;;,,,,,,,,,,,,,,1,n(11,12),()(1)n=6,?(2)120,n[1]1112:,,,:(1)6,5,5,4,4,1,4,:2008-12-15:(1961),,,092009118No112009vol118©1994-2010ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.=480(2)(1),,:n(n-1)(n-2)(n-3)=120(n2-3n)(n2-3n+2)-120=0,n=52,,,4,[2](13)13::SAB432=24:CDSAB,123;C2,D34;C4,D3,CD3243=723,,6(14)4,,(2003)14:,123,432=24456,123,;4,56,2;4,56,3,123,4565:245=120123(21,22,23):212223123,,[1],23,,,,,,2:(31):19:©1994-2010ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.:43222-4321=72()3:(32):32,4,3,3;2,2,2,2,,,,,4,2,:432222-(43222-4321)=120(),(41)41n(n1),,m(m1),(1)m=2,n1,n(2)m2nT1T2T3T(n-1)Tn,T1m;T1,T2m-1;T1T2,T3m-1;,m(m-1)n-1,Tn-1Tn,Tn-1Tn,Tn-1Tn,n-1,,n,an,n=2,a2=m(m-1)=m2-mn=345,an+an-1=m(m-1)n-1:{an},an+an-1=m(m-1)n-1,anan+an-1=m(m-1)n-1a3=m(m-1)2-a2=m(m-1)2-m(m-1)=m[(m-1)2-(m-1)]a4=m(m-1)3-a3=m(m-1)3-m=m[(m-1)3-(m-1)2-+(m-1)]a5=m(m-1)4-a4=m[(m-1)4-(m-1)3+(m-1)2-(m-1)]29:©1994-2010ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.=m[(m-1)n-1-(m-1)n-2+(m-1)n-3-+(m-1)(-1)n]an=m(m-1)n-11-(-1m-1)n-11-(-1m-1)=(m-1)n[1-(-1m-1)n-1]=(m-1)n-(-1)n-1(m-1)=(m-1)n+(-1)n(m-1)(m2),n(n1),,,,,,,:[1].[M].,2002.[2].[M].,20021(101)106178,87,8116%,5768,01663%,,50%,,,,:[1](http:11nes)1[2].[Z].,2008.11[3].[J].,2007.121[4].(http:11nes).,2006121211[5][N]..200711117139: