d) There is a Web page that includes links to both Web page A and Web page B.

Transcription

1 P Determine whether the relation R on the set of all eb pages is reflexive( 自反 ), symmetric( 对 称 ), antisymmetric( 反对称 ), and/or transitive( 传递 ), where (a, b) R if and only if a) Everyone who has visited eb page A has also visited eb page B. Reflexive( 自反 ), transitive( 传递 ) b) There are no common links found on both eb page A and eb page B. Symmetric( 对称 ) c) There is at least one common link on eb page A and eb page B. Symmetric( 对称 ) d) There is a eb page that includes links to both eb page A and eb page B. Symmetric( 对称 ) 47. Find the error in the proof of the following theorem. Theorem : Let R be a relation on a set A that is symmetric and transitive. Then R is reflexive. Proof : Let a A. Take an element b A such that (a, b) R. Because R is symmetric, we also have (b, a) R. Now using the transitive property, we can conclude that (a, a) R because (a, b) R and (b, a) R. There may be no such b(b 不存在 ) 55. Let R be a reflexive relation on a set A. Show that R n is reflexive for all positive integers n. Use mathematical induction. The result is trivial for n = 1. Assume R n is reflexive. Then (a, a) R n for all a A and (a, a) R. Thus (a, a) R n R = R n+1 for all a A. 56. Let R be a symmetric relation. Show that R n is symmetric for all positive integers n. 使用数学归纳法 : R 对称, 所以对于 (a, c) R, (c, a) R.

2 假设 R n 对称, 则 (b, d) R n, (b, d) R n. 对于 (x, y) R n+1 = R n R, c, (x, c) R, (c, y) R n. 因为 R 和 R n 对称, 所以 (y, c) R n, (c, x) R. 因为 (y, x) R R n = R n+1 P Let R1 and R2 be relations on a set A represented by the matrices M R1 = and M R2 = Find the matrices that represent a) R 1 R 2 = b) R 1 R 2 = c) R 2 R 1 = d) R 1 R 1 = e) R 2 R 1 = Determine whether the relations represented by the directed graphs shown in Exercises are reflexive, irreflexive, symmetric, antisymmetric, asymmetric, and/or transitive. 自反的

3 对称的, 非传递的 [ 反例 :(c, d), (b, a), (a, c). 没有 (c, c)] 自反, 对称, 传递 34. Let R be a relation on a set A. Explain how to use the directed graph representing R to obtain the directed graph representing the complementary relation R. (x, y) R, 去掉 (x, y) (x, y) R, 添加 (x, y) ( 假设 x y) (x, x) R, 去掉 (x, x) (x, x) R, 添加 (x, x) P Suppose that the relation R is symmetric. Show that R is symmetric. RR nn 是对称的 The result follows from (R ) 1 = ( n=1 RR nn ) 1 = n=1 (RR nn ) 1 = n=1 RR nn = R. 28. Use arshall s algorithm to find the transitive closures of the relations in Exercise 26. a) {(a, c), (b, d), (c, a), (d, b), (e, d)} aa 0 0 aa = bb 0 0 cc = bb 1 0 cc 1 0 dd 0 0 dd 1 0 ee 0 0 ee 1 0 b) {(b, c), (b, e), (c, e), (d, a), (e, b), (e, c)}

4 aa aa = bb 0 1 cc = bb cc dd dd ee ee c) {(a, b), (a, c), (a, e), (b, a), (b, c), (c, a), (c, b), (d, a), (e, d)} aa aa = bb 1 0 cc = bb cc dd dd ee 0 0 ee d) {(a, e), (b, a), (b, d), (c,d), (d,a), (d, c), (e,a), (e,b), (e, c), (e, e)} aa aa = bb 1 0 cc = bb cc dd 1 0 dd ee ee P Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) R if and only if a + d = b + c. Show that R is an equivalence relation. For reflexivity, ((a, b), (a, b)) R because a +b = b + a. For symmetry, if ((a, b), (c, d)) R, then a + d = b + c, so c + b = d + a, so ((c, d), (a, b)) R. For transitivity, if ((a, b), (c, d)) R and ((c, d), (e, f )) R, then a + d = b + c and c + e = d + f, so a + d + c + e = b + c + d + f, so a + e = b + f, so ((a, b), (e, f )) R. 20. Let R be the relation on the set of all people who have visited a particular eb page such that x Ry if and only if person x and person y have followed the same set of links starting at this eb page (going from eb page to eb page until they stop using the eb). Show that R is an equivalence relation. 对于自反, 用户自己的浏览轨迹相同 对于对称,x 与 y 的浏览轨迹相同, 那么 y 与 x 的浏览轨迹也相同 对于传递,x 与 y 的浏览轨迹相同,y 与则 z 的浏览轨迹相同, 那么 x 与 z 的浏览轨迹也相同 35. hat is the congruence class [n] 5 (that is, the equivalence class of n with respect to congruence modulo 5) when n is a) 2? [2] 5 = {i i 2 (mod 5)} = {..., 8, 3, 2, 7, 12,...}

5 b) 3? [3] 5 = {i i 3 (mod 5)} = {..., 7, 2, 3, 8, 13,...} c) 6? [6] 5 = {i i 6 (mod 5)} = {..., 9, 4, 1, 6, 11,...} d) 3? [ 3] 5 = {i i 3 (mod 5)} = {..., 8, 3, 2, 7, 12,...} 41. hich of these collections of subsets are partitions of {1, 2, 3, 4, 5, 6}? a) {1, 2}, {2, 3, 4}, {4, 5, 6} No b) {1}, {2, 3, 6}, {4}, {5} Yes c) {2, 4, 6}, {1, 3, 5} Yes d) {1, 4, 5}, {2, 6} No P Show that the reflexive closure of the symmetric closure of a relation is the same as the symmetric closure of its reflexive closure. 1 (R 1 R) = 1 R 1 R = 1 R 1 R = ( R) 1 ( R) P hat is the covering relation of the partial ordering {(A, B) A B} on the power set of S, where S = {a, b, c}? (, {a}), (, {b}), (, {c}), ({a}, {a, b}), ({a}, {a, c}), ({b}, {a, b}), ({b}, {b, c}), ({c}, {a, c}), ({c}, {b, c}), ({a, b}, {a, b, c}), ({a, c}, {a, b, c})({b, c}, {a, b, c}) 31. Show that a finite poset can be reconstructed from its covering relation. [Hint: Show that the poset is the reflexive transitive closure of its covering relation.] 设 (S, ) 是一个有穷偏序集 我们的目标是验证这个有穷偏序集合时它的覆盖关系的传递闭包 那么我们需要证明两件事,1 对于(S, ) 中的 (a, b) 它的覆盖关系的自反传递闭包 ;2 对于覆盖关系的传递闭包中的 (a, b) (S, ) 首先我们证明 1, (a, b) (S, ), 则 a = b, 则 (a, a) 它的覆盖关系自反传递闭包 ; 或 a b 且! z a z b, 则 (a, a) 它的覆盖关系 ; 或 a 1, a 2,... a n 使得 a a 1 a 2 a n b 且 a a 1 它的覆盖关系 则 (a, b) 它的覆盖关系自反传递闭包 对于 (S, ) 中的 (a, b) 它的覆盖关系的自反传递闭包 现在证明 2, (a, b) 覆盖关系的自反传递闭包, 则 a=b, (a, a) (S, ); 或 a b, (a, b) (S, ); 或者 a a 1 a 2 a n b, 因为 是传递的, 所以 (a, b) (S, ) 43. Determine whether the posets with these Hasse diagrams are lattices.

6 a)yes b)no c)yes 45. Show that every nonempty finite subset of a lattice has a least upper bound and a greatest lower bound. Use mathematical induction. 基本情况 :{x} 显然成立,lub({x}) = x, glb({x}) = x. 递归步骤 : 设对于一个格中的 k 个元素的集合, 存在最小上界和最大下界 现在证明 k+1 个元素的集合成立, 假设 S = S {x}, S 中含有 k 个元素 因为 k 个元素的集合成立,lub(S ) = a, glb(s ) = b. 因为 a, x 均为格中元素, 则存在 m = lub(a, x), 则 a m, 则 w S,w m 则 m 为 S 的上界 现在证明 m 是 S 的最小上界, 假设存在一个元素 n,n m, 且 n 是 S 的最小上界 因为 a 为 S 的最小上界, 那么 a n,x n,m n; 所以 m = n. 所以 m 是最小上界 同理 t = glb(b, x) 为最大下界 综上可得一个格中的每个有限非空子集有最小上界和最大下界 51. Show that every finite lattice has a least element and a greatest element. By Exercise 45 there is a least upper bound and a greatest lower bound for the entire finite lattice. By definition these elements are the greatest and least elements, respectively. 60. Show that a finite nonempty poset has a maximal element. a is maximal in the poset (S, ) if there is no b S such that a b. 反证, 假设没有最大, 那么 b, a 比它大, 所以集合必然无穷 但是集合有穷, 所以矛盾