东北大学离散数学试卷及答案(2)

离散数学试卷（二）第1页共9页一、填空20%（每小题2分）1、P：你努力，Q：你失败。“除非你努力，否则你将失败”的翻译为；“虽然你努力了，但还是失败了”的翻译为。2、论域D={1，2}，指定谓词PP(1,1)P(1,2)P(2,1)P(2,2)TTFF则公式),(xyyPx真值为。2、设S={a1，a2，…，a8}，Bi是S的子集，则由B31所表达的子集是。3、设A={2，3，4，5，6}上的二元关系}|,{是质数xyxyxR，则R=（列举法）。R的关系矩阵MR=。5、设A={1，2，3}，则A上既不是对称的又不是反对称的关系R=；A上既是对称的又是反对称的关系R=。6、设代数系统A，*，其中A={a，b，c},则幺元是；是否有幂等性；是否有对称性。7、4阶群必是群或群。8、下面偏序格是分配格的是。*abcabcabcbbcccb离散数学试卷(二)第2页共9页9、n个结点的无向完全图Kn的边数为，欧拉图的充要条件是。10、公式RQPQPP)(())((的根树表示为。二、选择20%（每小题2分）1、在下述公式中是重言式为（）A．)()(QPQP；B．))()(()(PQQPQP；C．QQP)(；D．)(QPP。2、命题公式)()(PQQP中极小项的个数为（），成真赋值的个数为（）。A．0；B．1；C．2；D．3。3、设}}2,1{},1{,{S，则S2有（）个元素。A．3；B．6；C．7；D．8。4、设}3,2,1{S，定义SS上的等价关系},,,,|,,,{cbdaSSdcSSbadcbaR则由R产生的SS上一个划分共有（）个分块。A．4；B．5；C．6；D．9。5、设}3,2,1{S，S上关系R的关系图为离散数学试卷(二)第3页共9页则R具有（）性质。A．自反性、对称性、传递性；B．反自反性、反对称性；C．反自反性、反对称性、传递性；D．自反性。6、设,为普通加法和乘法，则（）,,S是域。A．},,3|{QbabaxxSB．},,2|{ZbanxxSC．},12|{ZnnxxSD．}0|{xZxxS=N。7、下面偏序集（）能构成格。8、在如下的有向图中，从V1到V4长度为3的道路有（）条。A．1；B．2；C．3；D．4。9、在如下各图中（）欧拉图。10、设R是实数集合，“”为普通乘法，则代数系统R，×是（）。离散数学试卷(二)第4页共9页A．群；B．独异点；C．半群。三、证明46%1、设R是A上一个二元关系，)},,,(),(|,{RbcRcaAcAbabaS且有对于某一个试证明若R是A上一个等价关系，则S也是A上的一个等价关系。（9分）2、用逻辑推理证明：所有的舞蹈者都很有风度，王华是个学生且是个舞蹈者。因此有些学生很有风度。（11分）3、若BAf:是从A到B的函数，定义一个函数ABg2:对任意Bb有)})(()(|{)(bxfAxxbg，证明：若f是A到B的满射，则g是从B到A2的单射。（10分）4、若无向图G中只有两个奇数度结点，则这两个结点一定连通。（8分）5、设G是具有n个结点的无向简单图，其边数2)2)(1(21nnm，则G是Hamilton图（8分）四、计算14%1、设Z6,+6是一个群，这里+6是模6加法，Z6={[0]，[1]，[2]，[3]，[4]，[5]}，试求出Z6,+6的所有子群及其相应左陪集。（7分）2、权数1，4，9，16，25，36，49，64，81，100构造一棵最优二叉树。（7分）离散数学试卷(二)第5页共9页一、填空20%（每小题2分）1、QP；QP2、T3、},,,,{876540001111131aaaaaBB4、R={2,2,2,3,2,4,2,5,2,6,3,2,3,3,3,4,3,5,3,6,4,5,4,6,5,2,5,3,5,4,5,5,5,6}；00000111111100011111111115、R={1,2,1,3,2,1}；R={1,1,2,2,3,3}6、a；否；有7、Klein四元群；循环群8、B9、)1(21nn；图中无奇度结点且连通10、二、选择20%（每小题2分）题目12345678910答案B、DD；DDBDABBBB、C三、证明46%1、（9分）（1）S自反的Aa，由R自反，),(),(RaaRaa，Saa,（2）S对称的传递对称定义RSabRRbcRcaSRbcRcaSbaAba,),(),(),(),(,,（3）S传递的离散数学试卷(二)第6页共9页定义传递SScaRRcbRbaRceRebRbdRdaScbSbaAcba,),(),(),(),(),(),(,,,,由（1）、（2）、（3）得；S是等价关系。2、11分证明：设P(x)：x是个舞蹈者；Q(x)：x很有风度；S(x)：x是个学生；a：王华上述句子符号化为：前提：))()((xQxPx、)()(aPaS结论：))()((xQxSx……3分①)()(aPaSP②))()((xQxPxP③)()(aQaPUS②④)(aPT①I⑤).(aQT③④I⑥)(aST①I⑦)()(aQaST⑤⑥I⑧)()((xQxSxEG⑦……11分３、１0分证明：)(,,2121bbBbbAaaf21,满射21212211,),()(,)(,)(aafafafbafbaf是函数由于且使)()()(),()(),()})(()(|{)()},)(()(|{)(21122122112211bgbgbgabgabgabgabxfAxxbgbxfAxxbg但又为单射任意性知由gbb,,21。4、8分证明：设G中两奇数度结点分别为u和v，若u，v不连通，则G至少有两个连通分支G1、G2，使得u和v分别属于G1和G2，于是G1和G2中各含有1个奇数度结点，这与图论基本定理矛盾，因而u，v一定连通。5、8分证明：证G中任何两结点之和不小于n。离散数学试卷(二)第7页共9页反证法：若存在两结点u，v不相邻且1)()(nvdud,令},{1vuV，则G-V1是具有n-2个结点的简单图，它的边数)1(2)2)(1(21'nnnm,可得1)3)(2(21'nnm，这与G1=G-V1为n-2个结点为简单图的题设矛盾，因而G中任何两个相邻的结点度数和不少于n。所以G为Hamilton图.四、计算14%1、7分解：子群有{[0]},+6；{[0],[3]},+6；{[0],[2],[4]},+6；{Z6},+6{[0]}的左陪集：{[0]}，{[1]}；{[2]}，{[3]}；{[4]}，{[5]}{[0]，[3]}的左陪集：{[0]，[3]}；{[1]，[4]}；{[2]，[5]}{[0]，[2]，[4]}的左陪集：{[0]，[2]，[4]}；{[1]，[3]，[5]}Z6的左陪集：Z6。2、7分IndustrialDesignerDearMr.WangThankyouforgivingmeachancetoapplyforthisjob.Iwritethise-mailtoyouforintroducingmyself,soyoucanknowmewell.ThenyouwillfindIwouldbethebestonforthisjob.Youpositionrequirestopuniversity,BachelororaboveinIndustrialdesignorequivalentfield.IfeelthatIamcompetenttomeettherequ--------------------------.......................................................................................................................................................................................-.++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++离散数学试卷(二)第8页共9页+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++-------------------++