第3周起每周一交作业,作业成绩占总成绩的15%; 平时不定期的进行小测验,占总成绩的 1503 期中考试成绩占总成绩的20%;期终考试成绩占总成绩的 50% zhym@fudan.edu.cn 张宓13212010027 fudan. edu,cn BBS id: abchjsabc软件楼103 杨侃10302010007@ fudan,edu,cn 程义婷11302010050 fudan.,edu,cn BBS id: chengyiting 刘雨阳13212010013 fudan. edu,cn,软件楼405 liy(fudan. edu,cn李弋
第3周起每周一交作业，作业成绩占总成绩的15%； 平时不定期的进行小测验，占总成绩的 15%； 期中考试成绩占总成绩的20%；期终考试成绩占总成绩的 50% zhym@fudan.edu.cn 张宓 13212010027@fudan.edu.cn BBS id:abchjsabc 软件楼103 杨侃 10302010007@fudan.edu.cn 程义婷 11302010050@fudan.edu.cn BBS id:chengyiting 刘雨阳 13212010013@fudan.edu.cn，软件楼405 liy@fudan.edu.cn 李弋
2. Composition Definition 2.14: Let r be a relation from a to B, and r be a relation from B to c. The composition of R and r,, we write R,R1, is a relation from a to c. and is defined R,=la, c there exist some beB so that (a,b)∈R1and(b,c)∈R2, where a∈ A and ceo (R, is a relation from A to B, and R, is a relation from b to c (2)commutative law? X R1={(a1,b1),(a2b3),(a1,b2) R2={(b4a1),(b4C1),(b2,a2,(b3C2)
2.Composition Definition 2.14: Let R1 be a relation from A to B, and R2 be a relation from B to C. The composition of R1 and R2 , we write R2 R1 , is a relation from A to C, and is defined R2 R1={(a,c)|there exist some bB so that (a,b)R1 and (b,c)R2 , where aA and cC}. (1)R1 is a relation from A to B, and R2 is a relation from B to C (2)commutative law? R1={(a1 ,b1 ), (a2 ,b3 ), (a1 ,b2 )} R2={(b4 ,a1 ), (b4 ,c1 ), (b2 ,a2 ), (b3 ,c2 )}
Associative law? Forr caB. rcbxc and rccXd R3(R2R1)=(R3R2)R1 subset ofAXd For any(a2d)∈R3(R2R1),(a,d)∈?(R3R2)R1, Similarity,(R3oRyR,ER3(rori Theorem 2.3: Letr be a relation from a to B, R be a relation from b to C, Ra be a relation from c to D. Then R3(R2or=(r3oR)R,(Associative law
Associative law? For R1 A×B, R2B×C, and R3C×D R3 (R2 R1 )=?(R3 R2 )R1 subset of A×D For any (a,d)R3 (R2 R1 ), (a,d)?(R3 R2 )R1 , Similarity, (R3 R2 )R1R3 (R2 R1 ) Theorem 2.3：Let R1 be a relation from A to B, R2 be a relation from B to C, R3 be a relation from C to D. Then R3 (R2 R1 )=(R3 R2 )R1 (Associative law)
Definition 2. 15: Let r be a relation on A, and neN. The relation Rn is defined as follows (1)RO=(a, a)laEA)), we write IA (2R叶+=RRn, Theorem 2, 4 Let r be a relation on A. and m,n∈N.Then (RmoR=Rmtn (2)(Rm=Rmm
Definition 2.15: Let R be a relation on A, and nN. The relation Rn is defined as follows. (1)R0 ={(a,a)|aA}), we write IA . (2)Rn+1=RRn . Theorem 2.4: Let R be a relation on A, and m,nN. Then (1)RmRn=Rm+n (2)(Rm) n=Rmn
A={a1,a2,…,anB,B={b1,b2…,bm I and kg be relations from A to B R RIX i, mR2=(i M RIUR2(X MRnR2=(xr∧y) V01∧01 001000 111101 Example:A={2,3,4},B={1,3,5,7 R1={(2,3),(2,5),2,7),(3,5)、3,7),(45),(4,7) R2={(25)3,3),(4,1)4,7) Inverse relation R- of R: Mp-=M., MpT is the transpose of M R
A={a1 ,a2 ,,an },B={b1 ,b2 ,,bm} R1 and R2 be relations from A to B. MR1=(xij), MR2=(yij) MR1∪R2=(xijyij) MR1∩R2=(xijyij) 0 1 0 1 0 0 1 0 0 0 1 1 1 1 0 1 Example:A={2,3,4},B={1,3,5,7} R1={(2,3),(2,5),(2,7),(3,5),(3,7),(4,5),(4,7)} R2={(2,5),(3,3),(4,1),(4,7)} Inverse relation R-1 of R : MR-1=MR T , MR T is the transpose of MR