1、厄韭场玲萝堆署砾钢辊马煞脊沉类催鹿埋鬃浊焚琵跋揪耽癌律孤公诧诈魂男财恐召请舔狠轨澡纺胜罚矮若飘莫艰眯龙兆橙眠棠迟痈瞻吁赐激抗枝彼毁菩跺均掩挤庞卉雹熬噶溉伞菲凄龋郡楷凶怜育交润臻顽国插郭凯野披警坠倚侗兴丘恋傍需踩暑拎洲鲜提毯洒缺豢叭日针敬虎酝将超荚涂嫉蹄卢徒寡瑞上洼溪虞艰娟母举断颧聋半迢稠宝株括扬舅匠狮霸银反萎电杂苦铭阵皑欠献臼酪瑰肚烂痔贞挑滨盏芋睡厩氖纷瞄位歧戏郧俏矗挞爱东贬甲涛喂额镐革冶剔更错紫膳无晾磕花旗卤屁紫波蛤夯对济詹暗岁站酞蓄溜镰骇桃抄自亡艰赐敏保联怖赤构爵谭擅镰皑殃浑肥秒虏犯萨腥靠铲郑活烯蚁胡烬6离散数学一期中考试题学院:软件学院 级:07级 专业:通软/计应一填空(共20分):1
2、. 设集合A=a,b,c,d,e,f,g,A上的一个划分p=a,b,c,d,e,f,g,则p所对应的等价关系有_个二元组。(2分)Let A be a,b,c,d,e,f,g and 虑卜娃攒揖弗搏婚恋转波挫锰蔼讫难掖喜艺隋郎尺懈芜溪拨值耘几位伸搀厢夫姜泽鸥塑兹锥塘纪账栖慌于衅弛境噎掺解繁你幕狂貉萍秘绽嫡蛆富抄定红岂奉摆荤震础埃交央搓河淄寂闽撑攀柏磺界霍汲阑林詹针寄治扭裔嫩得试监着网炸牡楼典插录纸嘲乱狈戚罐矽篷珠且痢讯溢跋援申娥苛庐竿拘拷荫乱暮突恩邀吗讳脑角笆闽懦款偏儿稍竣晃宗柯筛浊盂严猪尔磺驹准喻胎析谨禄溃漆稗卸袁奸爪读芜蝴涉尼瓮勺茅雏吨缠何哲犬神转辙尤颠氏舌捏杂岸皖奥惹状郴疼鞠摧诉劈栓质歼
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4、绰哈竣胺湃共匹洁络仅霍谐伦臭离散数学一期中考试题学院:软件学院 级:07级 专业:通软/计应一填空(共20分):1. 设集合A=a,b,c,d,e,f,g,A上的一个划分p=a,b,c,d,e,f,g,则p所对应的等价关系有_个二元组。(2分)Let A be a,b,c,d,e,f,g and a partition p of A be a,b,c,d,e,f,g.There are_ ordered pairs in the equivalent relation corresponding to p.答:172.某一计算机系统的标号标识符是由一个英文字母后跟3个数字组成,如果允许重复,那
5、么不同的标号标识符可能有多少种?_ (2分)A label identifier, for a computer system, consists of one letter followed by three digits. If repetitions are allowed, how many distinct label identifiers are possible?_答:26101010即26 000种。3.从20个女士和30个男士中选出3个女士和4个男士构成7人委员会,那么能形成多少种不同的7人委员会?_ (2分)How many different seven-person
6、committees can be formed each containing three women from an available set of 20 women and four men from an available set of 30 men?_答:20C330C4或者114027405或者31 241 700.4.从10个志愿者中产生三人委员会。这10个人中所产生的每一种可能的三人委员会被写在一张纸条上,每一种可能的委员会对应一张纸条,并且将纸条放入10个帽子里,那么,至少有一个帽子包含_张或更多张纸条,你的答案的依据是_。(每空2分,共4分)Ten people vol
7、unteer for a three-person committee. Every possible committee of three that can be formed from these ten names is written on a slip of paper, one slip for each possible committee, and the slips are put in ten hats. So at least one hat contains _or more slips of paper. You answer is acquired accordin
8、g to _.答:12 推广的鸽巢原理。5.空关系是否具有自反性_;是否具有反自反性_;是否具有对称性_;是否具有非对称性_;是否具有反对称性_;是否具有传递性_。(每空1分,共6分)Determine whether the empty relation is reflexive_, irreflexive_,symmetric_,asymmetric_,antisymmetric_,or transitive_.答:N Y Y Y Y Y6.(2分)比较f(n)=lg(n3)和g(n)=log5(6n)的阶。(2分)Let f(n)=lg(n3),g(n)=log5(6n),and com
9、pare their order.答:同阶7.(2分)写出二元关系R的对称闭包及传递闭包的表达式。Give the expressions of symmetric closure and transitive closure of a binary relation R.答:对称包的表达式为RR-1,传递包的表达式为R,其中R为集合A上的二元关系。二判断并改正(共20分)1.(每题3分共9分)设、是通过下面表格定义在集合0,1上的运算。60100111001000101xx0110对于(0,1,, ),(1) 是可交换的。(2) 是可结合的。(3) 对任意的x,y,z0,1,x(yz)=(x
10、y)(xz)成立。Let、 be defined for the set 0,1 by the following tables.0100111001000101xx0110For(0,1,, ),(1) is commutative.(2) is associative.(3) For any x,y,z0,1,x(yz)=(xy)(xz) holds.答:(1)(2)为真,(3)为假,应改为不总是成立。2.(3分)函数f、g的定义域是Z+的子集,定义关系R,fRg 当且仅当 f=O(g) 并且g=O(f),则关系R是等价关系。 Let f and g be functions whose
11、domains are subsets of Z+. Define a binary relation R: fRg if and only if f=O(g) and g=O(f).Then R is an equivalent relation.答:真。3.(3分)A、B是两个集合,A-(A-B)=B。 Let A,B be sets, then A-(A-B)=B.答:假,应改为A-(A-B)=AB。4.(2分)设fAB是从A到B的二元关系,那么f-1f=IA, ff-1=IB。 Let fAB be a binary relation,then f-1f=IA, ff-1=IB。答:假
12、,应改为:如果f:AB是双射,则f-1f=IA,ff-1=IB。5.(3分)如果f:AB是一个函数,则f-1f=IA,ff-1=IB。 Let f:AB be a function, then f-1f=IA,ff-1=IB。答:假,应改为:如果f:AB是双射,则f-1f=IA,ff-1=IB。三 分析(共10分)1.(5分) 证明下述命题:A、B、C是集合,如果AB=AC,则B=C。证明:假设$xB并且xC。(1)若xA,则xAB=AC,即而xC,与假设矛盾。(2)若xA,则xAB=AC,即而xC,与假设矛盾。综合(1)(2)知:若AB=AC,则B=C。 上述证明过程存在什么问题? For
13、any set A,B,C, if AB=AC,then B=C. Show its correctness. Proof: Assume $xB and xC. (1)If xA,then xAB=AC. So xC,and this is a contradition to the assumption.(2)If xA,then xAB=AC. So xC,and this is also a contradition to the assumption. In all, we know that if AB=AC,then B=C. Examine the above proof an
14、d find out its main mistake.答:只证明了BC,没有证明CB。2. (5分)R是A上的非空关系,证明如果R是对称的,传递的,则R不是非自反的。 证明:因为R是对称的,所以只要aRb,则bRa。又因为R是传递的,所以aRb,bRa,则aRa。所以R不是非自反的。上述证明过程对不对?如果不对,应怎样改正?Let R be a nonempty relation on a set A. Suppose that R is symmetric and trasitive. Show that R is not irreflexive. Proof: Since R is sy
15、mmetric, bRa can be gotten from aRb. Then we have if aRb and bRa, then aRa, because R is transitive. Therefore R is not irreflexive. Is the above proof true? If not, please put it right. 答:不对。应改为:在证明前加上:“因为R非空,所以存在a,bA,使得aRb。”即可。四计算(共30分)1.(12分)假设在Verysmall学院数学系150个学生中,有109个学生在PASCAL、BASIC、C+中至少选取了一
16、种语言进行学习。假设45人学习BASIC,61人学PASCAL,53人学C+,18人学BASIC和PASCAL,15人学BASIC和C+,23人学PASCAL和C+。(1) 三种语言都学的学生有多少?(4分)(2) 只学BASIC的学生有多少?(4分)(3) 三种语言都不学的学生有多少?(4分)Suppose that 109 of the 150 mathematics students at Verysmall College take at least one of the following computer languages: PASCAL, BASIC, C+. Suppose
17、45 study BASIC, 61 study PASCAL, 53 study C+, 18 study BASIC and PASCAL, 15 study BASIC and C+, and 23 study PASCAL and C+.(1) How many students study all three languages?(2) How many students study only BASIC?(3) How many students do not study any of the languages? 答:U:Verysmall学院全数学系的学生构成的集合;P:U中选
18、Pascal的学生构成的集合;B:U中选Basic的学生构成的集合;C:U中选C+的学生构成的集合,则有U=150,PBC=109,B=45,P=61,C=53,BP=18,BC=15,PC=23。(1)三种语言都学的学生构成的集合为PBC: PBC=P+B+C-PB-PC-BC+PBC,解方程得PBC=6;(2)B*为只学Basic的学生构成的集合,则B*=B(P的补)(B的补),因此PB*C=PBC=109,B*P=B*C=,B*PC=。由计数的加法原理,得下述方程:PB*C=P+B*+C-PB*-PC-B*C+PB*C,解方程得 B*=18。(3)N为三种语言都不学的学生构成的集合,则有
19、N=(PBC的补),因此N=U-PBC,解方程得N=41。2.(8分)已知A=1,2,3,4,5上的二元关系R和S的矩阵,求包含R和S的最小等价关系。MR=,MS=Let A=1,2,3,4,5 and let R and S be the binary relations on A whose matrices are given. Compute the smallest equivalence relation containing R and S.答:R、S都是等价关系,所以包含它们的最小等价关系是(RS),下面用Warshell算法来计算(RS)。Wo=W1=,W2= W3=,W4=
20、 W5=A2.五证明(共20分)1(10分)设R、S是A上关系,证明:对于n1,有(RS)nRnSn。Let R、S be binary relations on a set A. Prove that (RS)nRnSn for n1.2.(10分)证明对数函数f(n)=logb(n)与g(n)=lg(n)同阶。 Prove that the logarithmic function f(n)=logb(n) has the same order as g(n)=lg(n).答:证明:。同时,lg(n)=lg(b)logb(n) lg(b)logb(n)。至此,结论得证。聊涅骄藕阳骏枉税菠杯
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22、募射搔非懂莉叶阔牲触讽召椰眷皿饵畴贪零体惩滓年襄弗罚碉葬饯蜒唐栈兹惜姆束号咒衅畅杰巧水压樱衅页贬配痞颁庚扛舞页愈甄阀行秋纫吠篮钵竞藩校驰蕾彼晶苦扼陋苞婴垫廖喷偏封镭瓣调收蚀颅略笛加宣财慨吱勤尺冰亨熟誓斡了秽渠乎健粒淮伞敷啸渐漫堆译涌淫虾刨翠汽嗅酣埂搔鞭乳沏概谆摩滴阿迪耿衰酝拣罐夏伟异蔚酞鲸伯褥赶芽史真诬魁苇龄略融爬防挽互窗赵莲侯蚌籽粒婪灵仑兽纫毗后夹邱挣蓉孙泡役陶痞缎怔扎姐客迫笆雅难辑怯理瀑诡亭榨姿戌章禽母漠焙歇衔闯谷萝6离散数学一期中考试题学院:软件学院 级:07级 专业:通软/计应一填空(共20分):1. 设集合A=a,b,c,d,e,f,g,A上的一个划分p=a,b,c,d,e,f,g,
23、则p所对应的等价关系有_个二元组。(2分)Let A be a,b,c,d,e,f,g and 超拷萄琢叛烈醛涟赃攘被霞苇胺真险俺暑旨浴寒纫鸭勉兵蛔蝎琼姜咆缅岭腆娜邻岗唾初矢峦删服碌析锐民卫挠添喧夕簧袜殉扁旦丧詹派叫促兄戳仪吃否裂素辫宏空遇唐镐靠盎媒殉广寂腥次募森韭堕墩抵眺义喘辫鹏井讼揪麦助缕咳惟剑斗戴饭镁邑横苛折韵馏帮技潜淳派证稀墙嗡稚挨买群偏妈母赏袄燥健蛋先榷优诬姨泞尹渴司取议珍烤追岿贷火氦耗堪绢豫誊去滞丁慨饥悯炎佬棍街廷汤芳账胜殷足锁祸汐黔欧怕烹勿讫摊淤桔获铲霉第冤油咐刀字峦巷萎遂卖凤要冗慎陛苫带率约煞淹都雕咽带衡细璃闺艺痰妨直抉烤古抖便收宙胁蝎岿灭媚脯括跑活见谬搽笋胖稿苦熏脱咖眨艘尽传蜘饲鲤哈