中央电大离散数学(本科)考试试题及答案参考资料小抄

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中央电大离散数学(本科)考试试题

一、单项选择题(每小题3分,本题共15分)

1.若集合A={1,2},B={1,2,{1,2}},则下列表述正确的是( a ). A .A ?B ,且A ∈B B .B ?A ,且A ∈B C .A ?B ,且A ?B D .A ?B ,且A ∈B

2.设有向图(a )、(b )、(c )与(d )如图一所示,则下列结论成立的是 ( d ).

图一 A .(a )是强连通的 B .(b )是强连通的

C .(c )是强连通的

D .(d )是强连通的

3.设图G 的邻接矩阵为

???????

?????????01

1

10010000011100

100110 则G 的边数为( b ).

A .6

B .5

C .4

D .3 4.无向简单图G 是棵树,当且仅当( a ).

A .G 连通且边数比结点数少1

B .G 连通且结点数比边数少1

C .G 的边数比结点数少1

D .G 中没有回路. 5.下列公式 ( c )为重言式.

A .?P ∧?Q ?P ∨Q

B .(Q →(P ∨Q)) ?(?Q ∧(P ∨Q))

C .(P →(?Q →P))?(?P →(P →Q))

D .(?P ∨(P ∧Q)) ?Q 1.若集合A ={a ,b },B ={ a ,b ,{ a ,b }},则( a ). A .A ?B ,且A ∈B B .A ∈B ,但A ?B C .A ?B ,但A ?B D .A ?B ,且A ?B

2.集合A ={1, 2, 3, 4, 5, 6, 7, 8}上的关系R ={|x +y =10且x , y ∈A },则R 的性质为( b ). A .自反的 B .对称的

C .传递且对称的

D .反自反且传递的 3.如果R 1和R 2是A 上的自反关系,则R 1∪R 2,R 1∩R 2,R 1-R 2中自反关系有( b )个. A .0 B .2 C .1 D .3 4.如图一所示,以下说法正确的是 ( d ) .

A .{(a, e )}是割边

B .{(a, e )}是边割集

C .{(a, e ) ,(b, c )}是边割集

D .{(d , e )}是边割集

图一

5.设A (x ):x 是人,B (x ):x 是学生,则命题“不是所有人都是学生”可符号化为( c ).

A .(?x)(A(x)∧B(x))

B .┐(?x)(A(x)∧B(x))

C .┐(?x)(A(x) →B(x))

D .┐(?x)(A(x)∧┐B(x))

1.设A ={a , b },B ={1, 2},R 1,R 2,R 3是A 到B 的二元关系,且R 1={, },R 2={, , },R 3={, },则( b )不是从A 到B 的函数.

A .R 1和R 2

B .R 2

C .R 3

D .R 1和R 3

2.设A ={1, 2, 3, 4, 5, 6, 7, 8},R 是A 上的整除关系,B ={2, 4, 6},则集合B 的最大元、最小元、上界、下界依次为 ( b ). A .8、2、8、2 B .无、2、无、2 C .6、2、6、2 D .8、1、6、1

3.若集合A 的元素个数为10,则其幂集的元素个数为( a ). A .1024 B .10 C .100 D .1

4.设完全图K n 有n 个结点(n ≥2),m 条边,当( c )时,K n 中存在欧拉回路.

A .m 为奇数

B .n 为偶数

C .n 为奇数

D .m 为偶数 5.已知图G 的邻接矩阵为

2

则G 有( d ).

A .5点,8边

B .6点,7边

C .6点,8边

D .5点,7边

1.若集合A ={ a ,{a},{1,2}},则下列表述正确的是( c ). A .{a ,{a}}∈A B .{2}?A C .{a}?A D .?∈A

2.设图G =,v ∈V ,则下列结论成立的是 ( c ) .

A .deg(v)=2∣E ∣

B . deg(v)=∣E ∣

C .E v V v 2)deg(=∑∈

D .

E v V v =∑

∈)deg( 3.命题公式(P ∨Q )→R 的析取范式是 ( d ) A .?(P ∨Q )∨R B .(P ∧Q )∨R C .(P ∨Q )∨R D .(?P ∧?Q )∨R 4.如图一所示,以下说法正确的是 ( a ).

A .e 是割点

B .{a, e}是点割集

C .{b, e}是点割集

D .{d}是点割集

5.下列等价公式成立的为( b ).

A .?P ∧?Q ?P ∨Q

B .P →(?Q →P) ??P →(P →Q)

C .Q →(P ∨Q) ??Q ∧(P ∨Q)

D .?P ∨(P ∧Q) ?Q

1.若G 是一个汉密尔顿图,则G 一定是( d ).

A .平面图

B .对偶图

C .欧拉图

D .连通图

2.集合A={1, 2, 3, 4}上的关系R={|x=y 且x, y ∈A},则R 的性质为( c ). A .不是自反的 B .不是对称的 C .传递的 D .反自反

3.设集合A={1,2,3,4,5},偏序关系≤是A 上的整除关系,则偏序集上的元素5是集合A 的( b ). A .最大元 B .极大元 C .最小元 D .极小元 4.图G 如图一所示,以下说法正确的是 ( c ) . A .{(a, d)}是割边 B .{(a, d)}是边割集 C .{(a, d) ,(b, d)}是边割集 D .{(b, d)}是边割集

图一

5.设A (x ):x 是人,B (x ):x 是工人,则命题“有人是工人”可符号化为( a ).

A .(?x)(A(x)∧B(x))

B .(?x)(A(x)∧B(x))

C .┐(?x)(A(x) →B(x))

D .┐(?x)(A(x)∧┐B(x)) 1.若集合A ={ a ,{a}},则下列表述正确的是( a ). A .{a}?A B .{{{a}}}?A C .{a ,{a}}∈A D .?∈A

2.命题公式(P ∨Q )的合取范式是 ( c ) A .(P ∧Q ) B .(P ∧Q )∨(P ∨Q ) C .(P ∨Q ) D .?(?P ∧?Q ) 3.无向树T 有8个结点,则T 的边数为( b ). A .6 B .7 C .8 D .9 4.图G 如图一所示,以下说法正确的是 ( b ). A .a 是割点 B .{b, c}是点割集 C .{b, d}是点割集 D .{c}是点割集

图一

5.下列公式成立的为( d ).

A .?P ∧?Q ? P ∨Q

B .P →?Q ? ?P →Q

C .Q →P ? P

D .?P ∧(P ∨Q)?Q 1.“小于5的非负整数集合”采用描述法表示为___a___. A .{x ∣x ∈N, x<5 } B .{x ∣x ∈R, x<5 } C .{x ∣x ∈Z, x<5 } D .{x ∣x ∈Q, x<5 }

2.设R1,R2是集合A={a,b,c,d}上的两个关系,其中R1={(a,a),(b,b),(b,c), (d,d)},R2={(a,a),(b,b),(b,c),(c,b),(d,d)},则R2是R1的__b____闭包.

A .自反

B .对称

C .传递

D .以上答案都不对

3.设函数f :R →R ,f(a)=2a+1;g :R →R ,g(a)=a2,则___c___有反函数. A .f g B .g f

3

C .f

D .g

4.已知图G 的邻接矩阵为???????

?

?

?01

1

1

110101110001000111010,则图G 有___d___.

A .5点,8边

B .6点,7边

C .6点,8边

D .5点7边

5.无向完全图K4是___a___.

A .汉密尔顿图

B .欧拉图

C .非平面图

D .树

6.在5个结点的完全二叉树中,若有4条边,则有___b___片树叶. A .2 B .3 C .4 D .5

7.无向树T 有7片树叶,3个3度结点,其余的都是4度结点,则T 有__c___个4度结点. A .3 B .2 C .1 D .0

8.与命题公式P →(Q →R )等值的公式是___a___. A .(P ∧Q)→R B .(P ∨Q)→R C .(P →Q)→R D .P →(Q ∨R)

9.谓词公式

)())()((x Q y yR x P x →?∨?中量词?x 的辖域是___b___.

A .

))()((y yR x P x ?∨? B .)()(y yR x P ?∨

C .P(x)

D .

)(x Q

10.谓词公式

))()(()(x xQ x Q x x xP ??→??→?的类型是___c___.

A .蕴涵式

B .永假式

C .永真式

D .非永真的可满足式 1.设A={1,2,3,4},B={1,3},C={-1,0,1,2},则___a___. A .A B ? B .C B ? C .A B ∈ D .C B ∈

2.若集合A 的元素个数为10,则其幂集的元素个数为___b___. A .1000 B .1024 C .1 D .10

3.设集合A={1,2},B={a,b},C={α},则=??C B A )(__c____.

A .{<1,a,α>,<1,b,α>,<2,a,α>,<2,b,α>}

B .{<1,>,<1,>,<2,>,<2,>}

C .{<<1,a>,α>,<<1,b>,α>,<<2,a>,α>,<<2,b>,α>}

D .{{1,2},{a,b},{α}}

4.设A={1, 2, 3, 4, 5, 6, 7, 8},R 是A 上的整除关系,B={2, 4, 6},则集合B 的最大元、最小元、上界、下界依次为___d___. A .8、1、6、1 B . 8、2、8、2 C .6、2、6、2 D .无、2、无、2 5.有5个结点的无向完全图K5的边数为___a___. A .10 B .20 C .5 D .25

6.设完全图K n 有n 个结点(n ≥2),m 条边,当___b___时,K n 中存在欧拉回路. A .n 为偶数 B .n 为奇数 C .m 为偶数 D .m 为奇数

7.一棵无向树T 有5片树叶,3个2度分支点,其余的分支点都是3度顶点,则T 有__c___个顶点. A .3 B .8 C .11 D .13

8.命题公式(P ∨Q )→R 的析取范式是___b___. A .(?P ∧?Q )∨R B . ?(P ∨Q )∨R C .(P ∧Q )∨R D .(P ∨Q )∨R 9.下列等价公式成立的是___b___. A .?P ∧?Q ?P ∨Q B . P →(?Q →P) ??P →(P →Q) C .?P ∨(P ∧Q) ?Q D .Q →(P ∨Q) ??Q ∧(P ∨Q)

10.谓词公式

))()(()(x xQ x Q x x xP ??→??→?的类型是__c____.

A .蕴涵式

B .永假式

C .永真式

D .非永真的可满足式

二、填空题(每小题3分,本题共15分)

6.命题公式)(P Q P ∨→的真值是 T (或1) .

7.若图G=中具有一条汉密尔顿回路,则对于结点集V 的每个非空子集S ,在G 中删除S 中的所有结点得到的连通分支数为W ,则S 中结点数|S|与W 满足的关系式为 W ≤|S| .

8.给定一个序列集合{000,001,01,10,0},若去掉其中的元素 0 ,则该序列集合构成前缀码. 9.已知一棵无向树T 中有8个结点,4度,3度,2度的分支点各一个,T 的树叶数为 5 .

10.(?x )(P (x )→Q (x )∨R (x ,y ))中的自由变元为R (x ,y )中的y

6.若集合A 的元素个数为10,则其幂集的元素个数为 1024 . 7.设A ={a ,b ,c },B ={1,2},作f :A →B ,则不同的函数个数为 8 . 8.若A ={1,2},R ={|x ∈A , y ∈A , x +y =10},则R 的自反闭包为{<1,1>,<2,2>}. 9.结点数v 与边数e 满足 e=v -1 关系的无向连通图就是树. 6.设集合A ={a ,b },那么集合A 的幂集是{?,{a ,b },{a },{b }}.

7.如果R 1和R 2是A 上的自反关系,则R 1∪R 2,R 1∩R 2,R 1-R 2中自反关系有 2 个.

4

8.设图G 是有6个结点的连通图,结点的总度数为18,则可从G 中删去 4 条边后使之变成树. 9.设连通平面图G 的结点数为5,边数为6,则面数为 3 .

10.设个体域D ={a , b },则谓词公式(?x )A (x )∧(?x )B (x )消去量词后的等值式为(A (a )∧A (b ))∧(B (a )∨B (b )) . 6.设集合A ={0, 1, 2, 3},B ={2, 3, 4, 5},R 是A 到B 的二元关系, },,{B A y x B y A x y x R ?∈∈∈><=且且 则R 的有序对集合为{<2, 2>,<2, 3>,<3, 2>},<3, 3>.

7.设G 是连通平面图,v , e , r 分别表示G 的结点数,边数和面数,则v ,e 和r 满足的关系式v -e +r =2 .

8.设G =是有6个结点,8条边的连通图,则从G 中删去 3 条边,可以确定图G 的一棵生成树. 9.无向图G 存在欧拉回路,当且仅当G 连通且所有结点的度数全为偶数

10.设个体域D ={1,2},则谓词公式)(x xA ?A (1)∨A (2) 6.命题公式)(P Q P ∨→的真值是 T (或1) .

7.若图G=中具有一条汉密尔顿回路,则对于结点集V 的每个非空子集S ,在G 中删除S 中的所有结点得到的连通分支数为W ,则S 中结点数|S|与W 满足的关系式为 W ≤|S| .

8.给定一个序列集合{000,001,01,10,0},若去掉其中的元素 0 ,则该序列集合构成前缀码. 9.已知一棵无向树T 中有8个结点,4度,3度,2度的分支点各一个,T 的树叶数为 5 .

10.(?x )(P (x )→Q (x )∨R (x ,y ))中的自由变元为R (x ,y )中的y

6.若集合A 的元素个数为10,则其幂集的元素个数为 1024 . 7.设A ={a ,b ,c },B ={1,2},作f :A →B ,则不同的函数个数为 8 . 8.若A ={1,2},R ={|x ∈A , y ∈A , x +y =10},则R 的自反闭包为{<1,1>,<2,2>}. 9.结点数v 与边数e 满足 e=v -1 关系的无向连通图就是树.

10.设个体域D ={a , b , c },则谓词公式(?x )A (x )消去量词后的等值式为A (a ) ∧A (b )∧A (c )

6.若集合A={1,3,5,7},B ={2,4,6,8},则A ∩B =空集(或?) . 7.设集合A ={1,2,3}上的函数分别为:f ={<1,2>,<2,1>,<3,3>,},g ={<1,3>,<2,2>,<3,2>,},则复合函数g ?f ={<1, 2>, <2, 3>, <3, 2>,} 8.设G 是一个图,结点集合为V ,边集合为E ,则G 的结点度数之和为2|E |(或“边数的两倍”) 9.无向连通图G 的结点数为v ,边数为e ,则G 当v 与e 满足 e=v -1 关系时是树. 10.设个体域D ={1, 2, 3}, P (x )为“x 小于2”,则谓词公式(?x )P (x ) 的真值为假(或F ,或0) .

6.设集合A ={2, 3, 4},B ={1, 2, 3, 4},R 是A 到B 的二元关系, },{y x B y A x y x R ≤∈∈><=且且

则R 的有序对集合为{<2, 2>,<2, 3>,<2, 4>,<3, 3>},<3, 4>,<4, 4>}

7.如果R 是非空集合A 上的等价关系,a ∈A ,b ∈A ,则可推知R 中至少包含,< b , b >等元素.

8.设G =是有4个结点,8条边的无向连通图,则从G 中删去 5 条边,可以确定图G 的一棵生成树. 9.设G 是具有n 个结点m 条边k 个面的连通平面图,则m 等于n +k -2 10.设个体域D ={1, 2},A (x )为“x 大于1”,则谓词公式()()x A x ?的真值为真(或T ,或1) 11.设集合A ={1,2,3},用列举法写出A 上的恒等关系I A ,全关系E A : I A = __ I A ={<1,1>,<2,2>,<3,3>};

E A ={<1,1>,<1,2>,<1,3>,<2,1>,<2,2>,<2,3>,<3,1>,<3,2>,<3,3>} 12.设集合A ={a ,b },那么集合A 的幂集是{?,{a },{b },{a ,b }}

13.设集合A ={1,2,3},B ={a ,b },从A 到B 的两个二元关系R ={<1,a >,<2,b >, <3,a >},S ={<1,a >,<2,a >,<3,a >},则R -S =_ R -S ={<2,b >}.

14.设G 是连通平面图,v , e , r 分别表示G 的结点数,边数和面数,则v ,e 和r 满足的关系式v -e +r =2. 15.无向连通图G 是欧拉图的充分必要条件是结点度数均为偶数.

16.设G =是有6个结点,8条边的连通图,则从G 中删去 3 条边,可以确定图G 的一棵生成树.

17.设G 是完全二叉树,G 有15个结点,其中有8个是树叶,则G 有____14___条边,G 的总度数是___28_____,G 的分支点数是____7____.

18.设P ,Q 的真值为1,R ,S 的真值为0,则命题公式Q S R Q P ∧∨∧∨)(的真值为___0_____. 19.命题公式)(R Q P →∧的合取范式为)(R Q P ∨?∧析取范式为)()(R P Q P ∧∨?∨ 20.设个体域为整数集,公式)0(=+??y x y x 真值为___1_____. 11.设集合A ={1,2,3,4},B ={3,4,5,6},则:

=B A ___{3,4}_____,=B A _____{1,2,3,4,5,6}_____.

12.设集合A 有n 个元素,那么A 的幂集合P (A )的元素个数为 . 13.设集合A ={a ,b ,c ,d },B ={x ,y ,z },R ={,,,,}

则关系矩阵M R =??????

? ??010*********. 14.设集合A ={a ,b ,c ,d ,e },A 上的二元关系R ={,,},S ={, ,},则R ·S ={<

a ,e >,

b >,} 15.无向图G 存在欧拉回路,当且仅当G 连通且__所有结点的度数全为偶数 16.设连通平面图G 的结点数为5,边数为6,则面数为 3 . 17.设正则二叉树有n 个分支点,且内部通路长度总和为I ,外部通路长度总和为E ,则有E =___ I +2n 18.设P ,Q 的真值为0,R ,S 的真值为1,则命题公式)()(S Q R P ∨→∨的真值为_____1___. 19.已知命题公式为G =(?P ∨Q )→R ,则命题公式G 的析取范式是(P ∧?Q )∨R

20.谓词命题公式(?x )(P (x )→Q (x )∨R (x ,y ))中的约束变元为___x___.

三、逻辑公式翻译(每小题4分,本题共12分)

11.将语句“如果所有人今天都去参加活动,则明天的会议取消.”翻译成命题公式. 设P :所有人今天都去参加活动,Q :明天的会议取消, (1分)

P→ Q.(4分)

12.将语句“今天没有人来.”翻译成命题公式.

设P:今天有人来,(1分)

? P.(4分)

13.将语句“有人去上课.”翻译成谓词公式.

设P(x):x是人,Q(x):x去上课,(1分)

(?x)(P(x) ∧Q(x)).(4分)

11.将语句“如果你去了,那么他就不去.”翻译成命题公式.

设P:你去,Q:他去,(1分)

P→?Q.(4分)

12.将语句“小王去旅游,小李也去旅游.”翻译成命题公式.

设P:小王去旅游,Q:小李去旅游,(1分)

P∧Q.(4分)

13.将语句“所有人都去工作.”翻译成谓词公式.

设P(x):x是人,Q(x):x去工作,(1分)

(?x)(P(x)→Q(x)).(4分)

11.将语句“他不去学校.”翻译成命题公式.

设P:他去学校,(1分)

? P.(4分)

12.将语句“他去旅游,仅当他有时间.”翻译成命题公式.

设P:他去旅游,Q:他有时间,(1分)

P →Q.(4分)

13.将语句“所有的人都学习努力.”翻译成命题公式.

设P(x):x是人,Q(x):x学习努力,(1分)

(?x)(P(x)→Q(x)).(3分)

11.将语句“尽管他接受了这个任务,但他没有完成好.”翻译成命题公式.

设P:他接受了这个任务,Q:他完成好了这个任务,(2分)

P∧? Q.(6分)

12.将语句“今天没有下雨.”翻译成命题公式.

设P:今天下雨,(2分)

? P.(6分)

11.将语句“他是学生.”翻译成命题公式.

设P:他是学生,(2分)

则命题公式为:P.(6分)

12.将语句“如果明天不下雨,我们就去郊游.”翻译成命题公式.

设P:明天下雨,Q:我们就去郊游,(2分)

则命题公式为:? P→ Q.(6分)

11.将语句“今天考试,明天放假.”翻译成命题公式.

设P:今天考试,Q:明天放假.(2分)

则命题公式为:P∧Q.(6分)

12.将语句“我去旅游,仅当我有时间.”翻译成命题公式.

设P:我去旅游,Q:我有时间,(2分)

则命题公式为:P→Q.(6分)

⑴将语句“如果明天不下雨,我们就去春游.”翻译成命题公式.

⑵将语句“有人去上课.”翻译成谓词公式.

⑴设命题P表示“明天下雨”,命题Q表示“我们就去春游”.

则原语句可以表示成命题公式?P→Q. (5分)

⑵设P(x):x是人,Q(x):x去上课

则原语句可以表示成谓词公式(?x)(P(x) ∧Q(x)).

四、判断说明题(每小题7分,本题共14分)

14.┐P∧(P→┐Q)∨P为永真式.

正确.(3分)

┐P∧(P→┐Q)∨P是由┐P∧(P→┐Q)与P组成的析取式,

如果P的值为真,则┐P∧(P→┐Q)∨P为真,(5分)

如果P的值为假,则┐P与P→┐Q为真,即┐P∧(P→┐Q)为真,

也即┐P∧(P→┐Q)∨P为真,

所以┐P∧(P→┐Q)∨P是永真式.(7分)

15.若偏序集的哈斯图如图一所示,则集合A的最大元为a,最小元不存在.

正确.(3分)

对于集合A的任意元素x,均有∈R(或xRa),所以a是集合A中的最大元.(5分)

14.如果R1和R2是A上的自反关系,则R1∪R2是自反的.

正确.(3分)

R1和R2是自反的,?x ∈A, ∈ R1, ∈R2,

∈ R1?R2,

5

6

所以R1∪R2是自反的. (7分) 15.如图二所示的图G 存在一条欧拉回路.

正确. (3分) 因为图G 为连通的,且其中每个顶点的度数为偶数. (7分) 14.设N 、R 分别为自然数集与实数集,f :N →R ,f (x)=x+6,则f 是单射. 正确. (3分)

设x1,x2为自然数且x1≠x2,则有f(x1)= x1+6≠ x2+6= f(x2),故f 为单射. (7分) 15.设G 是一个有6个结点14条边的连通图,则G 为平面图.

错误. (3分) 不满足“设G 是一个有v 个结点e 条边的连通简单平面图,若v ≥3,则e ≤3v-6.” 13.下面的推理是否正确,试予以说明.

(1) (?x )F (x )→G (x ) 前提引入 (2) F (y )→G (y ) US (1).

错误. (3分) (2)应为F (y )→G (x ),换名时,约束变元与自由变元不能混淆. (7分) 14.若偏序集的哈斯图如图二所示,则集合A 的最大元为a ,最小元不存在.

错误. (3分) 集合A 的最大元不存在,a 是极大元. (7分) 13.下面的推理是否正确,试予以说明.

(1) (?x )F (x )→G (x ) 前提引入 (2) F (y )→G (y ) US (1).

错误. (3分) (2)应为F (y )→G (x ),换名时,约束变元与自由变元不能混淆. (7分) 14.如图二所示的图G 存在一条欧拉回路.

错误. (3分)

因为图G 为中包含度数为奇数的结点. (7分) 13.如果图G 是无向图,且其结点度数均为偶数,则图G 是欧拉图.

错误. (3分) 当图G 不连通时图G 不为欧拉图. (7分) 14.若偏序集的哈斯图如图二所示,则集合A 的最大元为a ,最小元是f .

图二

错误. (3分) 集合A 的最大元与最小元不存在, a 是极大元,f 是极小元,.

五.计算题(每小题12分,本题共36分)

16.设集合A={1,2,3,4},R={|x, y ∈A ;|x -y|=1或x -y=0},试 (1)写出R 的有序对表示; (2)画出R 的关系图;

(3)说明R 满足自反性,不满足传递性.

(1)R={<1,1>,<2,2>,<3,3>,<4,4>,<1,2>,<2,1>,<2,3>,<3,2>,<3,4>,<4,3>} (3分) (2)关系图为

v 1

v

图二

7

(6分)

(3)因为<1,1>,<2,2>,<3,3>,<4,4>均属于R ,即A 的每个元素构成的有序对均在R 中,故R 在A 上是自反的。 (9分)

因有<2,3>与<3,4>属于R ,但<2,4>不属于R ,所以R 在A 上不是传递的。 17.求P →Q ∨R 的析取范式,合取范式、主析取范式,主合取范式. P →(R ∨Q ) ?┐P ∨(R ∨Q)

? ┐P ∨Q ∨R (析取、合取、主合取范式) (9分) ?(┐P ∧┐Q ∧┐R)∨(┐P ∧┐Q ∧R) ∨(┐P ∧Q ∧R) ∨(P ∧┐Q ∧┐R) ∨(P ∧┐Q ∧R) ∨(P ∧Q ∧┐R) ∨(P ∧Q ∧R) (主析取范式) (12分)

18.设图G=,V={ v1,v2,v3,v4,v5},E={ (v1, v2),(v1, v3),(v2, v3),(v2, v4),(v3, v4),(v3, v5),(v4, v5) },试 画出G 的图形表示; 写出其邻接矩阵; (3) 求出每个结点的度数; (4) 画出图G 的补图的图形.

(1)关系图

(3分) (2)邻接矩阵

???????

?????????01

1

10110110110110

100110 (6分)

(3)deg(v 1)=2

deg(v 2)=3 deg(v 3)=4

deg(v 4)=3 deg(v 5)=2 (9分)

(4)补图

16.设谓词公式)(),())

,,(),((y F z y yR z x y zQ y x P x ??∧?→?,试 (1)写出量词的辖域; (2)指出该公式的自由变元和约束变元.

(1)?x 量词的辖域为)),,(),((z x y zQ y x P ?→, (2分)

?z 量词的辖域为),,(z x y Q , (4分) ?y 量词的辖域为),(z y R . (6分) (2)自由变元为)),,(),((z x y zQ y x P ?→与)(y F 中的y ,以及),(z y R 中的z 约束变元为x 与),,(z x y Q 中的z ,以及),(z y R 中的y . (12分) 17.设A ={{1},{2},1,2},B ={1,2,{1,2}},试计算

(1)(A -B ); (2)(A ∩B ); (3)A ×B .

(1)A -B ={{1},{2}} (4分) (2)A ∩B ={1,2} (8分) (3)A×B={<{1},1>,<{1},2>,<{1},{1,2}>,<{2},1>,<{2},2>, <{2},{1,2}>,<1,1>,<1,2>,<1, {1,2}>,<2,1>,<2,2>, <2, {1,2}>}

18.设G=,V={ v1,v2,v3,v4,v5},E={ (v1,v3),(v2,v3),(v2,v4),(v3,v4),(v3,v5),(v4,v5) },试 (1)给出G 的图形表示; (2)写出其邻接矩阵; (3)求出每个结点的度数; (4)画出其补图的图形. 1)G 的图形表示为:

v 1 v 2 v 3 v 4 v 5 ο ο

ο

ο

v 1

v 2 v 3

v 4 v 5

ο ο

ο

ο ο

8

(3分)

(2)邻接矩阵:

???????

?????????01

1

10110110110110

000100 (6分)

(3)v1,v2,v3,v4,v5结点的度数依次为1,2,4,3,2 (9分)

(4)补图如下:

16.试求出(P ∨Q )→R 的析取范式,合取范式,主合取范式.

(P ∨Q )→R ?┐(P ∨Q)∨R ? (┐P ∧┐Q)∨R (析取范式) (3分)

? (┐P ∨R)∧ (┐Q ∨R)(合取范式) (6分) ? ((┐P ∨R)∨(Q ∧┐Q))∧ ((┐Q ∨R)∨(P ∧┐P)) ? (┐P ∨R ∨Q)∧(┐P ∨R ∨┐Q)∧ (┐Q ∨R ∨P) ∧(┐Q ∨R ∨┐P)

? (┐P ∨Q ∨R)∧(┐P ∨┐Q ∨R)∧ (P ∨┐Q ∨R)

(主合取范式) (12分) 17.设A={{a, b}, 1, 2},B={ a, b, {1}, 1},试计算 (1)(A -B ) (2)(A ∪B ) (3)(A ∪B )-(A ∩B ). (1)(A -B )={{a, b}, 2} (4分) (2)(A ∪B )={{a, b}, 1, 2, a, b, {1}} (8分) (3)(A ∪B )-(A ∩B )={{a, b}, 2, a, b, {1}} (12分)

18.图G=,其中V={ a, b, c, d, e},E={ (a, b), (a, c), (a, e), (b, d), (b, e), (c, e), (c, d), (d, e) },对应边的权值依次为2、1、2、3、6、1、4及5,试

(1)画出G 的图形;

(2)写出G 的邻接矩阵;

(3)求出G 权最小的生成树及其权值. (1)G 的图形表示为:

(3分)

(2)邻接矩阵:

???????

?????????01

1

1

1

10110110011100

110110

(3)粗线表示最小的生成树,

(10分)

权为7: (12分)

15.求(P ∨Q )→(R ∨Q )的合取范式. (P ∨Q )→(R ∨Q )

??(P ∨Q )∨(R ∨Q ) (4分) ?(?P ∧?Q)∨(R ∨Q )

?(?P∨R∨Q)∧(?Q∨R∨Q)

?(?P∨R∨Q) ∧R 合取范式(12分)

16.设A={0,1,2,3,4},R={|x∈A,y∈A且x+y<0},S={|x∈A,y∈A且x+y≤3},试求R,S,R?S,R-1,S-1,r(R).

R=?, (2分)

S={<0,0>,<0,1>,<0,2>,<0,3>,<1,0>,<1,1>,<1,2>,<2,0>,<2,1>,<3,0>} (4分)

R?S=?,(6分)

R-1=?,(8分)

S-1= S,(10分)

r(R)=IA.(12分)

17.画一棵带权为1, 2, 2, 3, 4的最优二叉树,计算它们的权.

(10分)

权为1?3+2?3+2?2+3?2+4?2=27 (12分)

15.求(P∨Q)→R的析取范式与合取范式.

(P∨Q)→R ??(P∨Q)∨R (4分)

? (?P∧?Q)∨R (析取范式)(8分)

? (?P∨R)∧(?Q∨R) (合取范式)(12分)

16.设A={0,1,2,3},R={|x∈A,y∈A且x+y<0},S={|x∈A,y∈A且x+y≤2},试求R,S,R?S,S -1,r(R).R=?, S={<0,0>,<0,1>,<0,2>,<1,0>,<1,1>,<2,0>} (3分)

R?S=?,(6分)

S -1= S,(9分)

r(R)=IA={<0,0>,<1,1>,<2,2>,<3,3>}.(12分)

17.画一棵带权为1, 2, 2, 3, 4的最优二叉树,计算它们的权.

最优二叉树如图三所示

(10分)

图三

权为1?3+2?3+2?2+3?2+4?2=27 (12分)

15.设谓词公式

))

,

,

(

)

(

)

,

(

)(

(z

x

y

B

z

y

x

A

x?

?

,试

(1)写出量词的辖域;(2)指出该公式的自由变元和约束变元.

(1)?x量词的辖域为

))

,

,

(

)

(

)

,

(

(z

x

y

B

z

y

x

A?

,(3分)

?z量词的辖域为

)

,

,

(z

x

y

B

, (6分)

(2)自由变元为

))

,

,

(

)

(

)

,

(

(z

x

y

B

z

y

x

A?

中的y,(9分)

约束变元为x与z.(12分)

16.设集合A={{1},1,2},B={1,{1,2}},试计算

(1)(A-B);(2)(A∩B);(3)A×B.

(1)A-B ={{1},2} (4分)

(2)A∩B ={1} (8分)

(3)A×B={<{1},1>,<{1},{1,2}>,<1,1>,<1, {1,2}>,<2,1>,<2, {1,2}>} (12分)

17.设G=,V={ v1,v2,v3,v4 },E={ (v1,v3),(v2,v3),(v2,v4),(v3,v4) },试

(1)给出G的图形表示;(2)写出其邻接矩阵;

(3)求出每个结点的度数;(4)画出其补图的图形.

(1)G的图形表示为(如图三):

(3分)

(2)邻接矩阵:

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

1

1

1

1

1

1

1

1

0(6分)

ο

οο

οο

ο

οο

1 2

2 3

3

4

7 5

12

ο

οο

οοο

ο

οο

1 2

2 3

3

4

7 5

12

9

10

(3)v1,v2,v3,v4结点的度数依次为1,2,3,2 (9分)

(4)补图如图四所示:

21.化简下列集合表示式:

)()())(()(C B A C B A C B A C B A ---- = )()~()~()~~(C B A C B A C B A C B A = ))(~)(())(~)~((C C B A C C B A ?

= ))(())~((E B A E B A 设E 为全集 = )()~(B A B A = )(~B B A = E A = A

22.设},21|{R x x x A ∈≤≤=,},0|{R y y y B ∈≥=,求B A ?,A B ?,并画出其图像. ⑴ B A ?=},0|{},21|{R y y y R x x x ∈≥?∈≤≤ =},,0,21|,{R y x y x y x ∈≥≤≤>< B A ?的图像如下图1所示的阴影部分.

图1 图2 ⑵A B ?=},21|{},0|{R x x x R y y y ∈≤≤?∈≥ =},,0,21|,{R y x y x x y ∈≥≤≤>< A B ?的图像如上图2所示的阴影部分.

23.设G=,V={ v1,v2,v3,v4,v5},E={ (v1,v3),(v2,v3),(v2,v4),(v3,v4),(v3,v5),(v4,v5) },试: ⑴ 给出G 的图形表示; ⑵ 画出其补图的图形. .⑴ G 的图形表示见图3;⑵ G 的补图的图形,见图4

图3 图4 24.构造权为2,3,4,4,5,5,7的最优树。 最优树如下图5所示.

11

21.设A ,B 和C 是全集E 的子集,化简下列集合表示式: )(~)~()(C B A C B A C B A )(~)~()(C B A C B A C B A

= )(~)()~()(C B A C B A C B A C B A = ))()~(())~()((C B A A B B C A = ))(())((C B E E C A = )()(C B C A = C B A )(

22.设A ={1,2,3},用列举法给出A 上的恒等关系I A ,全关系E A ,A 上的小于关系 },,{y x A y x y x L A <∧∈><= 及其逆关系和关系矩阵.

}3,3,2,2,1,1{><><><=A I (2分)

}3,3,2,3,1,3,3,2,2,2,1,2,3,1,2,1,1,1{><><><><><><><><><=A E (2分) }3,2,3,1,2,1{><><><=A L (2分)

L A 的逆关系}2,3,1,3,1,2{1

><><><=-A L (2分)

????? ??=000100110A

L M ???

?

?

??=-0110010001A

L M . (2分)

23.图G =,其图形如右图1所示。

⑴ 写出G 的邻接矩阵;

⑵ 画出G 的权最小的生成树以及计算出其权值.

⑴ G 的邻接矩阵为: (4分)

?????????

?

??????????01

1

101111110010010001011001010110 ⑵ G 的权最小的生成树如右上图1所示. (4分) 最小的生成树的权为:1+1+5+2+3=12. (2分) 六、证明题(本题共8分)

19.试证明(?x )(P (x )∧R (x ))? (?x )P (x )∧(?x )R (x ).

证明: (1)(?x )(P (x )∧R (x )) P

(2)P (a )∧R (a ) ES(1) (2分) (3)P (a ) T(2)I (4)(?x )P (x ) EG(3) (4分) (5)R (a ) T(2)I (6)(?x )R (x ) EG(5) (6分) (7)(?x )P (x )∧(?x )R (x ) T(5)(6)I (2分)

19.试证明集合等式A ? (B ?C)=(A ?B) ? (A ?C) .

证明:设S= A ? (B ?C),T=(A ?B) ? (A ?C),若x ∈S ,则x ∈A 或x ∈B ?C ,即 x ∈A 或x ∈B 且 x ∈A 或x ∈C . 也即x ∈A ?B 且 x ∈A ?C ,即 x ∈T ,所以S ?T . (4分) 反之,若x ∈T ,则x ∈A ?B 且 x ∈A ?C , 即x ∈A 或x ∈B 且 x ∈A 或x ∈C ,

也即x ∈A 或x ∈B ?C ,即x ∈S ,所以T ?S .

因此T=S . 19.试证明集合等式A ? (B ?C)=(A ?B) ? (A ?C).

证明:设S= A ? (B ?C),T=(A ?B) ? (A ?C),若x ∈S ,则x ∈A 或x ∈B ?C ,即 x ∈A 或x ∈B 且 x ∈A 或x ∈C . 也即x ∈A ?B 且 x ∈A ?C ,即 x ∈T ,所以S ?T . (4分) 反之,若x ∈T ,则x ∈A ?B 且 x ∈A ?C , 即x ∈A 或x ∈B 且 x ∈A 或x ∈C ,

也即x ∈A 或x ∈B ?C ,即x ∈S ,所以T ?S .

18.设G 是一个n 阶无向简单图,n 是大于等于2的奇数.证明G 与G 中的奇数度顶点个数相等(G 是G 的补图). 证明:因为n 是奇数,所以n 阶完全图每个顶点度数为偶数, (3分)

因此,若G 中顶点v 的度数为奇数,则在G 中v 的度数一定也是奇数, (6分)

图1

12

所以G 与G 中的奇数度顶点个数相等. (8分) 18.试证明集合等式A ? (B ?C)=(A ?B) ? (A ?C) .

证明:设S= A ? (B ?C),T=(A ?B) ? (A ?C),若x ∈S ,则x ∈A 或x ∈B ?C ,即 x ∈A 或x ∈B 且 x ∈A 或x ∈C . 也即x ∈A ?B 且 x ∈A ?C ,即 x ∈T ,所以S ?T . (4分) 反之,若x ∈T ,则x ∈A ?B 且 x ∈A ?C , 即x ∈A 或x ∈B 且 x ∈A 或x ∈C , 也即x ∈A 或x ∈B ?C ,即x ∈S ,所以T ?S . 因此T=S .

18.设A ,B 是任意集合,试证明:若A ?A=B ?B ,则A=B .

证明:设x ∈A ,则∈A ?A , (1分)

因为A ?A=B ?B ,故∈B ?B ,则有x ∈B , (3分) 所以A ?B . (5分) 设x ∈B ,则∈B ?B , (6分) 因为A ?A=B ?B ,故∈A ?A ,则有x ∈A ,所以B ?A . (7分)

故得A=B . (8分)

试证明(?x )(P (x )∧R (x ))? (?x )P (x )∧(?x )R (x ). 证明:

⑴(?x )(P (x )∧R (x )) P

⑵ P (a )∧R (a ) ES(1) ⑶ P (a ) T(2)I ⑷(?x )P (x ) EG(3) ⑸ R (a ) T(2)I ⑹(?x )R (x ) EG(5) ⑺(?x )P (x )∧(?x )R (x ) T(5)(6)I 26.试证明 )()()()(x P x x P x ????? 成立。

证明:设公式中的个体变元为a 1,a 2,…,a n ,即个体域E ={a 1,a 2,…,a n },则有: ))()()(()()(21n a P a P a P x P x ∧∧∧???? )()()(21n a P a P a P ?∨∨?∨?? )()(x P x ???

25.设T 是正则二叉树,有t 片树叶,证明T 的阶数n =2t -1. 证明:根据正则二叉树的概念和握手定理得 ⑴ n =t +i ,i 为分支点数 ⑵ n =m +1 ,m 为T 的边数

⑶ m =2i (正则二叉树的定义) 由⑵和⑶可解得 i =21-n

代入⑴,解出 n =2t -1.

26.试证明 ))()(())()((B x A x B x A x →??→? 成立。 B x A x B x A x ∨???→?)()())()((

B x A x ∨???))()(( ))()((B x A x ∨??? ))()((B x A x →??

13

请您删除一下内容,O(∩_∩)O 谢谢!!!2015年中央电大期末复习考试小抄大全,电大期末考试必备小抄,电大考试必过小抄Shanghai’s Suzhou Creek has witnessed much of the city’s history. Zhou Wenting travels this storied body of water and finds its most fascinating spot s. Some lucky cities can boast a great body of water, like London with the river Thames and Paris with the river Seine. Shanghai is privileged enough to have two great bodies of water: Huangpu River and Suzhou

Creek.Huangpu River became famous when colonists established clusters of grand buildings on its banks on what became known as the bund. Today, the bund overlooks the breathtaking skyline of Lujiazui financial district. Shanghai’s other body of water, however, Suzhou Creek, has been somewhat overshadowed. Suzhou Creek links the inland cities of Jiangsu province with Shanghai. When the British colonists, who arrived in the city after it was opened as a commercial port in 1843 found they could reach Suzhou, Jiangsu province, via the creek, they named it Suzhou Creek. Thanks to its location, a large amount of cargo and travelers were transported via the creek before rail links were established. But after a century of being utilized as a waterway to transport goods and labor, the creek grew dark and smelly. Industrial factories were established along the banks. In the 1990s it became a key task of the city government to clean the creek. Suzhou Creek, which snakes 17 km from the iconic Waibaidu Bridge downtown to the outer ring road in west Shanghai, maps the changing periods of the city’s history, including the imprints of the concessions, the beginning of industrialization and the improvement in people’s living conditions. Where the Bund began In-between the shopping street of East Nanjing Road and the Bund, are a cluster of streets that give me the illusion that I am no longer in modern Shanghai. The streets are narrow and old and criss-cross each other. Any old residential house may turn out to be a former office of the British, constructed in the 1880s. Pawnshops and hardware stores that are hard to find elsewhere, are plentiful here. This area, at the confluence of Huangpu River and Suzhou Creek, is called the Bund Origin. Countless tour buses stop at the site every day and visitors from around the world get off to see this place, the starting point of the concessions in the city. It all started in 1872, when the former British Consulate General was constructed and the Bund began its transformation into an the financial street of the East. Now the site of the former consulate is called “No 1 Waitanyuan”, which translates to “the Bu nd Origin”, to honor its beginnings. The entire complex of this historical site comprises of five buildings, the former British Consulate General, the official residence of the consul, the former Union Church, the church apartments and the former Shanghai Rowing Club. The size of the courtyard is equivalent to that of four standard soccer fields. The building of the former consulate is a two-storey masonry building on an H-shaped plan in typical English renaissance style. The building is designed with a five-arch verandah on the ground floor with a raised terrace facing the garden, while the facade features an entry portico beneath a colonnaded loggia. It has been turned into a café where dinner and afternoon tea are available. Visitors can choose to sit indoors or outdoors to enjoy the magnificent gardens with nearly 30 ancient trees.

Yuanmingyuan Road behind the complex is also a historical site. The road has been revamped as a pedestrian shopping street and high-end brands have Church, stands at the intersection of Yuanmingyuan Road and Suzhou Creek. The church, designed in the style of the English countryside, has a capacity of 500 people. It was very popular during the concession period but was converted into factory offices after 1949. The church we see today is a replica, the original burned down in 2007. There used to be an outdoor swimming pool, the first of its kind in Shanghai, beside the church but has been filled-in and is now a small garden. Bridge of romance There is perhaps no other place that’s more repr esentative of Shanghai than this bridge, which appears in quite a lot of movies about the city. Dozens of couples visit every day to pose for their pre-wedding photos on the bridge where Suzhou Creek begins and interconnects with Huangpu River. This is Waibaidu Bridge, or the Garden Bridge. The soon-to-be-wed couples pose in splendid attire on the bridge, leaning against the railing or sitting on the wooden floor. Some even risk walking into the middle of the road to get the perfect shot.Colorful lights illuminate the bridge throughout the night, making it a picturesque place for pre-wedding

portraits and lovers to meet. Constructed in 1873 and designed by a British company, the 106-meter-long bridge was the first-ever major bridge in Shanghai. In 1856, the first large wooden bridge, Wells Bridge, was built over Suzhou Creek but the bridge toll led to complaints from citizens. So 17 years later, another wooden bridge, which did not require tolls, was built. People called it Waibaidu, which means “going across for free”. The bridge was renovated as a steel truss structure in 1907. Because nearly 40 bridges have now been built over Suzhou Creek, the bridge is no longer a traffic artery but is more of an observation deck for tourists. It is a tradition in Shanghai for a grandmother to walk across a bridge with their grandchild when he or she reaches one month. This represents that the newborn has overcome all the twists and turns and its journey will be safe and smooth throughout his or her life. "Waibaidu Bridge is always the best option because it’s the icon of Shanghai. The picture of my daughter when she was a baby held by her grandmother was also taken here. It’s l ike a family tradition," says Wang Xuefen, a Shanghai native who has a newborn grandson. Changning Riverside There is a 5-km stretch of waterfront by Suzhou Creek in

Changning district on Changning Road from the intersection of Hami Road to Jiangsu Road. It has become a popular place to take a walk and sunbathe on the lawn. There is an overpass at the intersection of Changning Road and Gubei Road for people to enjoy the view of the creek and a 3-km plastic runway on both sides of Changning Road, which attracts people of all ages, Chinese and expat. "Jogging on the two sides gives a different feeling because the north side is next to the creek, and the south side is adjacent to the residential highrises, which is like jogging in the jungle," says Xiao Xu, a

27-year-old woman who lives nearby. The riverside used to be completely different. Dozens of textile mills, chemical plants and machine manufacturing factories were set up along the creek in the 1920s. They brought industrialization but also pollution. From the 1930s the creek could no longer be used as a source for tap water, and no living fish or shrimp could be found. "Suzhou Creek in my memory is dark and smelly. I used to go to the riverbank to watch the sewage disposal running out from the chemical plants when I was a little girl. We didn’t know it was pollution. We thought it was a red waterfall," says Huang Qi, a 57-year-old Shanghai resident. "So the residential houses along the creek were unpopular, and only migrants with low

incomes would live in that area," she says. However, things have changed. The plants were closed and turned into riverside parks and the apartments in the new highrises, especially those facing the creek, are much sought after. East China University of Political Science and Law This is the famous former Saint John’s University, China’s first -ever modern institution of higher education established by missionaries from the United States in 1879. The buildings combine Chinese and Western elements. Address: 1575 Wanhangdu Road, Changning district The old residential area After you leave the university from its east gate you will enter a shabby neighborhood that retains its original look. The alleys are narrow and the houses are overcrowded. Some things have not changed for many generations, such as raising chickens at home. Address: West Guangfu Road Moganshan Road This is an artsy street that has become very popular among artists and fashionistas in recent years. Graffiti covers the walls on the winding street, where you can find a cluster of art galleries and creative industry offices. Sihang Warehouse

Four banks jointly funded the construction of this warehouse, so it is named

sihang, or four banks. The warehouse, built in 1931, was used for the storage of food, first-aid supplies and ammunition during the years of war. The building, which is also a masterpiece left by the Hungarian architect Laszlo Hudec in the 1930s, has been recently transformed into a center of creative industry workshops. Address: 1 Guangfu Road, Zhabei district