浙江大学离散数学2015期末考试题(英文班)

Department of Information Science and Electronic Engineering

Zhejiang University

Final Examination

11193011Discrete Mathematics

Assoc.Prof.Dr.Thomas Honold

M.Sc.Jingmei Ai

Fall Semester,2014

Name:

Matric.No.:

Question123456

Marks Obtained

Maximal Marks151515201520100

Time allowed:120minutes

Instructions to candidates:

?This examination paper contains six(6)questions.

?Please answer every question and subquestion,and justify your answers.

?For your answers please use the space provided after each question.If this space is insu?cient,please continue on the blank sheets provided.

?This is a CLOSED BOOK examination,

except that you may bring1sheet of A4paper(hand-written only)and a Chinese-English dictionary(paper copy only)to the examination.

Question1(15marks)

a)Is(p1?p2)∨p1a tautology?

b)Are(p1→p2)→p3and p1→(p2→p3)logically equivalent?

c)Fermat’s Last Theorem(FLT)states that the equation x n+y n=z n has

no integer solution with xyz=0and n≥3.Express FLT as a formula of

predicate calculus with domain Z.

Note:You may use the basic operations of ordinary arithmetic,including

powers‘x n’and the comparison operators’=’,’≤’.No further abbreviations

are allowed.The domain(universe of discourse)must be the set of all integers. Question2(15marks)

Three black balls,three red balls and three yellow balls are arranged in a3×3 matrix in such a way that there is no monochromatic row or column(“monochro-matic”refers to balls of the same color).Balls of the same color are considered as indistinguishable.In how many ways can this be done?

Question3(15marks)

Suppose a1,a2,...,a10are integers satisfying1≤a1≤a2≤···≤a10≤40.Show that there exist distinct3-subsets{i,j,k}and{r,s,t}of{1,2,...,10}such that

a i+a j+a k=a r+a s+a t.

Question4(20marks)

Let G be the simple graph whose vertices are the subsets S?{1,2,3,4,5,6,7} with|S|=3,two vertices S and T being adjacent if and only if S∩T=?.

a)How many vertices and edges does G have?

b)Is G regular?If applicable,determine its degree.

c)Is G connected?

d)Is G bipartite?

Question5(15marks)

a)Find a particular solution of the recurrence relation

a n=3a n?1?4a n?3+4n.

Hint:The recurrence relation has a solution of the form a n=cn+d,where

c,d are constants.

b)Determine the general solution of the recurrence relation in a).

Question6(20marks)

A list a=(a0,a1,...,a n?1)of integers is said to have a(contiguous)zero subsum

if there exist0≤i≤j≤n?1such that a i+a i+1+···+a j=0.

a)Write(using a programming language or some form of pseudocode)a boolean

function has zero subsum(a)with parameter a as above,which returns true if a has a zero subsum and false otherwise.

b)Determine the worst-case time complexity of your algorithm,using additions

and comparisons of list elements as basic operations.Both the exact number of basic operations and an asymptotic growth estimate should be given.

c)Does there exist a more e?cient algorithm for this task?Brie?y justify your

answer.

Solutions

1a)No.1This can be proved either by a truth table or,e.g.,as follows:If p 1is false and p 2is true then p 1?p 2is false.Hence the disjunction (p 1?p 2)∨p 1is false as well.4b)No.1For the proof one can use again a truth table or,e.g.,the following shorter argument:p 1→(p 2→p 3)is false for exactly one assigment of truth values to p 1,p 2,p 3,viz.p 1=p 2=T ,p 3=F .On the other hand,(p 1→p 2)→p 3is false whenever p 1=p 3=F (since p 1→p 2=T in this case),and hence for at least two

assigments of truth values (in fact for exactly three).

4c)An appropriate formula is

??x ?y ?z ?n ?(xyz =0)∧n ≥3∧(x n +y n =z n ) .

5

1=15

2Write A for the set of all arrangements of the 9balls (without any restriction)and A i (B j )for the set of arrangements with row i (resp.,column j )monochromatic.Then the desired number is |A \(A 1∪A 2∪A 3∪B 1∪B 2∪B 3)|.2We use the PIE for the computation.The number of unrestricted arrangements is |A |= 93,3,3 = 93 63 =84·20=1680.3Further we have

|A i |=|B j |=3 63

=60,2|A i,j |=|B i,j |=3!=6,

2|A 1,2,3|=|B 1,2,3|=6,3

since 2monochromatic rows force the 3rd row to be monochromatic as well.

All other intersections between the sets A i ,B j involve at least one A i and at least one B j .Since monochromatic rows and columns exclude each other,these intersections

are all empty.

1It follows that the desired number is

1680?6·60+6·6?2·6=1344.2

2=15

3The number of distinct 3-subsets of {1,2,...,10}is

103 =10·9·83·2·1=120.5

The 120sums a i +a j +a k are integers in the range [3,120].There are 118such integers.5

=?By the pigeonhole principle,since 120>118,the sums cannot be distinct.Hence there exist at least two 3-subsets with the same sum.5

3

=15

4a)The number of vertices of G is 73 =35.3The number of edges of G is 73 43 ·12=70.Alternatively,use b)and the formula

e =vr

2

for the number of edges of a regular graph.)3b)Yes.The degree of a vertex {a,b,c }is the number of 3-subsets of {1,2,...,7}\{a,b,c },which is a 4-set.Hence G is regular of degree 43 =4.4c)Yes.Consider two non-adjacent vertices S and T .Then |S ∩T |∈{1,2}.1If |S ∩T |=2then |S ∪T |=4and S ,T have a common neighbor,viz.{1,2,...,7}\(S ∪T ).Hence S and T are connected by a path of length 2.2If |S ∩T |=1,let w.l.o.g.S ={1,2,3},T ={1,4,5}.Then,e.g.,M ={4,5,6}is adjacent to S ,N ={2,3,7}is adjacent to T ,and M ,N are adjacent as well.Hence

S and T are connected by a path of length 3.

2In all G is connected and has diameter 3.

d)No.G has cycles of length 7,viz.

{1,2,3}→{4,5,6}→{7,1,2}→{3,4,5}→{6,7,1}→{2,3,4}→{5,6,7}→{1,2,3}

3

Since G has cycles of odd length,G cannot be bipartite.2

4

=20

5a)The sequence a n =cn +d is a solution if and only if,for every n ,

cn +d =3 c (n ?1)+d ?4 c (n ?3)+d +4n

=3cn ?3c +3d ?4cn +12c ?4d +4n

=(4?c )n +9c ?d 2

This is in turn equivalent to the system c =4?c ∧d =9c ?d ,which has the (unique)

solution c =2,d =9.Hence a (p )n =2n +9is a particular solution.3

b)The characteristic equation is

r 3?3r 2+4=0.2

The roots of this equation are 2(of multiplicity 2)and ?1(of multiplicity 1).3This can be proved,for example,by checking that 2and ?1are solutions,and using the fact that the coe?cient of r 2is the negative sum of all three roots.

Hence the general solution of the associated homogeneous linear recurrence relation is a(h)n=α1·2n+α2·n2n+α3(?1)n(α1,α2,α3∈C).3 From this and a)the?nal answer is

a n=α1·2n+α2·n2n+α3(?1)n+2n+9,whereα1,α2,α3∈C.2

5

=15 6a)

def has_subset_sum(a):

#a=[a(0),...,a(n-1)]list of integers

#We use Python’s convention that indexing starts with0

#n denotes the length of the list

n=len(a)

for i in range(n):#range(n)=[0,...,n-1]

sum=0

for j in range(i,n):#range(n)=[i,...,n-1]

sum=sum+a[j]#Initially sum=a[i]

if sum==0:

return True

return False

...10 b)During Pass i through the outer for-loop,the inner for-loop is executed at most

n?i times,with1addition and1comparison per iteration.Hence the total number of basic operations is at most

n?1

i=02(n?i)=2

n+(n?1)+···+1

=n(n+1)=Θ(n2).5

(The worst case,n(n+1)operations,occurs if and only if a has no zero subsum except possibly a n?1=0.)

c)Yes.Consider the list of partial sums s i=a0+a1+···+a i,0≤i≤n?1,amended

by s?1=0.Then a i+a i+1+···+a j=s j?s i?1,and hence a has a zero subset sum if and only if two elements of the list s=(s?1,s0,s1,...,s n?1)are equal.

The list s can be computed in time O(n).Further we can test distinctness of its members in time O(n log n)(by sorting the list?rst,and then checking whether s i< s i+1by one further pass through the list).The total time complexity of this algorithm is only O(n log n).5

6

=20

Exam

=100