电大离散数学综合复习资料（小抄参考） .doc

专业好文档离散数学综合复习资料一、判断题1. （ ）命题联结词Ø，Ù，Ú是最小联结词组。2. （ ）（PÙQ）ÙØP为矛盾式。3. （ ）（ØPÚQ）Ù（Q®R）®（P®R）为重言式。4. （ ）A、B、C是任意命题公式，如果AÚCÛBÚC，一定有AÛB。5. （ ）若集合A上的二元关系R是对称的，RC一定是对称的。6. （ ）R是A上的二元关系，R是自反的，当且仅当r(R)=R。7. （ ）集合A上的等价关系确定了A的一个划分。8. （ ）有理数集是可数的。9. （ ）若函数f，g为入射则其复合函数也为入射。10. （ ）R是集合A上的关系，R有传递性的充要条件是RoRÍR。11. （ ）设<A，*>是一个代数系统，且集合A中元素的个数大于1。如果该代数系统中存在幺元e和零元q，则e¹q。12. （ ）交换群必是循环群。13. （ ）一个群可以有多个等幂元。14. （ ）模格一定是分配格。15. （ ）每个有向图中，结点入度数总和等于结点出度总和。16. （ ）图G的邻接矩阵A，Al中的i行j列表示结点vi到vj长度为l路的数目。17. （ ）任何图中必有偶数个度数为奇数的结点。18. （ ）有向图中，它的每一个结点位于且只位于一个单侧分图中。19. （ ）任意平面图最多是四色的。20. （ ）不存在既有欧拉回路又有汉密尔顿回路的图。二、填空题1 设P：“天下雨”，Q：“他骑自行车上班”，R：“他乘公共汽车上班”。则命题“除非下雨，否则他就骑自行车上班”可符号化为 。“他或者骑自行车，或者乘公共汽车上班”可符号化为 2 设N(x)：x是自然数；J(x)：x是奇数；Q(x)：x是偶数，用谓词公式符号化命题“任何自然数不是偶数就是奇数”。3 设P(x)：x是运动员，Q(x)：x是教练。则命题“不是所有运动员都是教练”可符号化为。4 设D=a,b；P(a,a)=P(b,b)=T；P(a,b)=P(b,a)=F。则公式("x)($y)(P(x,y)®P(y,x)的真值是。5 集合A=Æ,Æ的幂集P(A)为6 集合A=1,2，B=a,b,c,d，C=c,d,e，则A´(B-C)为7 试用空集Æ构成集合A（A¹Æ）= 和B= ，使得AÎB且AÍB都成立。并且A´B=。8 设A=1,2,3，R=<1,2>,<2,1>,<1,3>,<1,1>，传递闭包t(R)为 。9 设A=1,2,3，B=x,y，f：A®B，则不同的函数个数为 个。10 Q为有理数集，Q上定义运算*为a*b=a+b-ab，则<Q,*>的幺元为 。11 代数系统<Sk,+>，其中Sk=x|xÎZÙx>=K，+为普通加法，则<Sk,+>是一个半群的必要条件是 。12 设G为v个结点e条边的连通平面图，则面r等于 。13 一棵树有n2个结点度数为2，n3个结点度数为3，nk个结点度数为k，则度数为1的结点的个数为 。14 设T为根树，若每个结点的出度都小于等于m，则T称为 树，若除 外，每个结点的出度都等于m，则T称为完全m叉树。15 设<A，£>是偏序集，如果A中任意两个元素都有 和 ，则称<A，£>为格。三、解答题1. 将公式(P®Q) Ù (Q®R)®(P®R)化成与之等价且仅含Ø、Ú、Ù的公式。2. 将下列命题符号化：（1）他虽聪明但不用功。（2）除非你努力否则你将失败。（3）我们不能既划船又跑步（4）仅当你走我才留下。3. 用谓词表达式符号化下列命题：（1）所有老的国家选手都是运动员。（2）某些教练是年老的，但是健壮的。（3）任何自然数不是偶数就是奇数。（4）不是所有运动员都是教练。4. 求命题公式Ø(P®Q)的主合取范式。5. 求命题公式PÙ(P®Q)的主析取范式。6. 设集合A1, 2, 3，A上的关系R<1, 1>,<1, 2>,<2, 2>,<3, 2>,<3, 3>, （1）画出R的关系图；（2）写出R的关系矩阵；（2）问R具有关系的哪几种性质(自反、反自反、对称、反对称、传递)。7. 构造一非空偏序集，它存在一子集有上界，但没有最小上界。它还有一子集，存在最大下界但没有最小元。8. 以下哪些是函数？哪些是入射？哪些是满射？对任意一个双射，写出它们的逆函数。a) f: Z®N, f(x)=x2+1b) f: N®Q, f(x) = 1/xc) f: 1,2,3®a,b,c, f=<1,b>,<2,c>,<3,a>d) f: N®N, f(x)=2xe) f: R´R®R´R, f(x,y)=<y+1,x+1>9. 设S=1,2,3,4,6,12，D为S上的整除关系，（1）试写出该关系并画出哈斯图；（2）设子集B=2,3,6，试求B的最大元、最小元、极大元和极小元；（3）试求B的上界、上确界、下界和下确界。10. 设集合A有m个元素，B有n个元素，则A到B的关系有多少个？A到B的函数有多少个？11. 判定下列代数系统是否为群，请说明原因。（1）<R,+>，其中R为实数集，+为普通加法；（2）<I,´>，其中I为整数集，´为普通乘法 12. 设群<G,*>的运算表如下：*eabeeabaabebbea试写出<G,*>的所有子群，及其相应的左陪集。13. 设G=<V,E>，V=V1,V2,V3,V4的邻接矩阵：0 1 0 11 0 1 1 1 1 0 0 1 0 0 0 A(G)=（1）试画出该图。（2）V2的入度d-(V2)和出度d+(V2)是多少？（3）从V2到V4长度为2的路有几条？v1v3v2v5v414. 试求下面有向图的强分图、单侧分图和弱分图15. （1）画一个有欧拉回路和一条汉密尔顿回路的图。（2）画一个有欧拉回路，但没有汉密尔顿回路的图。（3）画一个没有欧拉回路，但有汉密尔顿回路的图。V1V2V3V4V54325112216. 下图给出的赋权图表示五个城市及对应两个城镇间公路的长度。是给出一个最优的设计方案使各城市间有公路连通。17. 设有一组权3、4、13、5、6、12，（1）求相应的最优树（要求构造的过程中，每个分支点的左儿子的权小于右儿子的权）。（2）设上述权值分别对应英文字母b、d、e、g、o、y，试根据求得的最优树构造前缀码，并对二进制序列0100110110010001011译码。四、证明题1. A ®(BÙC),(E®ØF)®ØC,B®(AÙØS)ÞB®E2. 试证明命题公式为永真式。3. 试证明：(PÚQ) (P®R) (Q®S) ÞSÚR4. 用推理规则证明：("x)(P(x)®Q(x) Þ($x) P(x)®($y)(P(y)ÙQ(y)5. 对所有集合A、B和C，有(AÇB)ÈC=AÇ(BÈC)，当且仅当CÍA。6. 若R和S是集合A上的等价关系，试证明RÇS也是A上的等价关系。7. 证明集合0,1和(0,1)是等势的。8. 设f: X->Y和g: Y->Z是函数，使得g°f是一个满射，且g是一个入射。证明f是满射。9. 设<G1,*>,<G2,°>是两个群，在G1´G2上定义运算为：<a1,b1><a2,b2>=<a1*a2,b1°b2>，证明< G1´G2,>是一个群。10. f是群<G,°>到群<G,*>的同态映射，e是G中的幺元则，f的同态核K=x|xÎG且f(x)=e构成的代数系统<K,°>是<G,°>的子群。11. 证明在格中，若a£b£c，则（1）aÚb=bÙc（2）(aÙb)Ú(bÙc)=b=(aÚb)Ù(aÚc)12. 若有n个人，每个人恰有三个朋友，证明n必为偶数。13. 证明当且仅当G的一条边e不包含在G的回路中时，e才是G的割边。14. 画出K3,3图，并证明其不是欧拉图，也不是平面图。15. 设G为连通图，证明当且仅当边e是G的割边时，e才在G的每颗生成树中。16. 设T是非平凡的无向树，T中度数最大的结点有2个，它们的度数为k（k>=2），证明：T中至少有2k-2片树叶。17. 设G=<V,E>有11个结点，m条边，证明G或者其补图G是非平面图。部分参考答案一、判断题1. （错误）2. （正确）3. （正确）4. （错误）5. （正确）6. （正确）7. （正确）8. （正确）9. （正确）10. （正确）11. （正确）12. （错误）13. （错误）14. （错误）15. （正确）16. （正确）17. （正确）18. （正确）19. （正确）20. （错误）"If we don't do that it will go on and go on. We have to stop it; we need the courage to do it."His comments came hours after Fifa vice-president Jeffrey Webb - also in London for the FA's celebrations - said he wanted to meet Ivory Coast international Toure to discuss his complaint.CSKA general director Roman Babaev says the matter has been "exaggerated" by the Ivorian and the British media.Blatter, 77, said: "It has been decided by the Fifa congress that it is a nonsense for racism to be dealt with with fines. You can always find money from somebody to pay them."It is a nonsense to have matches played without spectators because it is against the spirit of football and against the visiting team. It is all nonsense."We can do something better to fight racism and discrimination."This is one of the villains we have today in our game. But it is only with harsh sanctions that racism and discrimination can be washed out of football."The (lack of) air up there Watch mCayman Islands-based Webb, the head of Fifa's anti-racism taskforce, is in London for the Football Association's 150th anniversary celebrations and will attend City's Premier League match at Chelsea on Sunday."I am going to be at the match tomorrow and I have asked to meet Yaya Toure," he told BBC Sport."For me it's about how he felt and I would like to speak to him first to find out what his experience was."Uefa has opened disciplinary proceedings against CSKA for the "racist behaviour of their fans" during City's 2-1 win.Michel Platini, president of European football's governing body, has also ordered an immediate investigation into the referee's actions.CSKA said they were "surprised and disappointed" by Toure's complaint. In a statement the Russian side added: "We found no racist insults from fans of CSKA."Baumgartner the disappointing news: Mission aborted.The supersonic descent could happen as early as Sunda.The weather plays an important role in this mission. Starting at the ground, conditions have to be very calm - winds less than 2 mph, with no precipitation or humidity and limited cloud cover. The balloon, with capsule attached, will move through the lower level of the atmosphere (the troposphere) where our day-to-day weather lives. It will climb higher than the tip of Mount Everest (5.5 miles/8.85 kilometers), drifting even higher than the cruising altitude of commercial airliners (5.6 miles/9.17 kilometers) and into the stratosphere. As he crosses the boundary layer (called the tropopause),e can expect a lot of turbulence.The balloon will slowly drift to the edge of space at 120,000 feet ( Then, I would assume, he will slowly step out onto something resembling an Olympic diving platform.Below, the Earth becomes the concrete bottom of a swimming pool that he wants to land on, but not too hard. Still, he'll be traveling fast, so despite the distance, it will not be like diving into the deep end of a pool. It will be like he is diving into the shallow end.Skydiver preps for the big jumpWhen he jumps, he is expected to reach the speed of sound - 690 mph (1,110 kph) - in less than 40 seconds. Like hitting the top of the water, he will begin to slow as he approaches the more dense air closer to Earth. But this will not be enough to stop him completely.If he goes too fast or spins out of control, he has a stabilization parachute that can be deployed to slow him down. His team hopes it's not needed. Instead, he plans to deploy his 270-square-foot (25-square-meter) main chute at an altitude of around 5,000 feet (1,524 meters).In order to deploy this chute successfully, he will have to slow to 172 mph (277 kph). He will have a reserve parachute that will open automatically if he loses consciousness at mach speeds.Even if everything goes as planned, it won't. Baumgartner still will free fall at a speed that would cause you and me to pass out, and no parachute is guaranteed to work higher than 25,000 feet (7,620 meters).cause there21.6