离散数学期末考试试题(配答案)

由天下分享时间：2024/5/19 4:09:03 加入收藏我要投稿点赞

relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \o\p, Then idge (3)-review of solid n-line solution. Category fraction multiplication word problem score Division applications engineering problem problem XV, a subject: review of the measurof cuboiement of the amount of capacity, measurement and units of measurement of common units of measurement and their significance in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowlecontent category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features ds and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1, size 2, table ...和广东技术师范学院

模拟试题

科目：离散数学

考试形式：闭卷考试时间: 120 分钟系别、班级：姓名：学号：

一．填空题（每小题2分，共10分）

1. 谓词公式?xP(x)??xQ(x)的前束范式是__ ?x?y?P(x)∨Q(y) __________。 2. 设全集E??1,2,3,4,5?,A??1,2,3?,B??2,5?,则A∩B =__{2}__，A?_{4,5}____，

A?B?__ {1,3,4,5} _____ 3. 设A??a,b,c?,B??a,b?，则?(A)??(B)?__ {{c},{a,c},{b,c},{a,b,c}} __________，

?(B)??(A)?_____Φ_______。

4. 在代数系统（N，+）中，其单位元是0，仅有 _1___ 有逆元。 5．如果连通平面图G有n个顶点，e条边，则G有___e+2-n____个面。二．选择题（每小题2分，共10分）

1. 与命题公式P?(Q?R)等价的公式是（）

（A）(P?Q)?R （B）(P?Q)?R （C）P?(Q?R) （D）P?(Q?R) 2. 设集合A??a,b,c?,A上的二元关系R???a,a?,?b,b??不具备关系( )性质（A） (A)传递性 (B)反对称性 (C)对称性 (D)自反性 3. 在图G??V,E?中,结点总度数与边数的关系是( ) (A)deg(vi)?2E (B) deg(vi)?E(C)

?deg(v)?2E(D) ?deg(v)?E

iiv?Vv?V4. 设D是有n个结点的有向完全图,则图D的边数为( ) (A)n(n?1) (B)n(n?1) (C)n(n?1)/2 (D)n(n?1)/2 5. 无向图G是欧拉图,当且仅当( )

(A) G的所有结点的度数都是偶数 (B)G的所有结点的度数都是奇数

1. 求命题公式p?q?r的主合取范式与主析取范式。（6分）

解：主合取方式：p∧q∨r?(p∨q∨r)∧(p∨?q∨r)∧(?p∨q∨r)= ∏0.2.4

主析取范式：p∧q∨r?(p∧q∧r) ∨(p∧q∧?r) ∨(?p∧q∧r) ∨(?p∧?q∧r) ∨(p∧?q∧r)= ∑1.3.5.6.7

?1??12. 设集合A??a,b,c,d?上的二元关系R的关系矩阵为MR??0??0?r(R),s(R),t(R)的关系矩阵，并画出R，r(R),s(R),t(R)的关系图。（10分）

000001000??1?,求?0?1??

3 无向图G有12条边，G中有6个3度结点，其余结点的度数均小于3，问G中至少有多少个结点？（10分）

relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \2) find associate \o\p, Then idge (3)-review of solid n-line solution. Category fraction multiplication word problem score Division applications engineering problem problem XV, a subject: review of the measurof cuboiement of the amount of capacity, measurement and units of measurement of common units of measurement and their significance in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowlecontent category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features ds and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1, size 2, table ...和解：∵G（V,E），| E |=V，d（Vi）<3, 设至少有x个节点，由握手定理得： 2×12=∑d（Vi）<6×3+(x-6)×3 2<(x-6) =＞ x>8 故G中至少有9个节点。

4 求下面两个图的最小生成树。（12分）

5. 试判断(z,?)是否为格？说明理由。（5分）

解：（Z,≤）是格，理由如下：

对于任意a∈Z，a≤a成立，满足自反性；

对于任意a∈Z，b∈Z，若a≤b且b≤a，则a=b，满足反对称性；对于任意a，b，c∈Z，若a≤b，b≤c，则a≤c，满足传递性；

而对于任意a，b∈Z，a≤b，b为最小上界，a为最大下界，故（Z，≤）是格。

（注：什么是格？

第 3 页共 5 页

）