浙江大学离散数学2015期末考试题(英文班)

Department of Information Science and Electronic EngineeringZhejiang UniversityFinal Examination11193011Discrete MathematicsAssoc.Prof.Dr.Thomas HonoldM.Sc.Jingmei AiFall Semester,2014Name:Matric.No.:Question123456Marks ObtainedMaximal Marks151515201520100Time allowed:120minutesInstructions 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 insufficient,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 hasno integer solution with xyz=0and n≥3.Express FLT as a formula ofpredicate calculus with domain Z.Note:You may use the basic operations of ordinary arithmetic,includingpowers‘x n’and the comparison operators’=’,’≤’.No further abbreviationsare 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 thata 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 relationa n=3a n−1−4a n−3+4n.Hint:The recurrence relation has a solution of the form a n=cn+d,wherec,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 subsumif 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 booleanfunction 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 additionsand 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 efficient algorithm for this task?Briefly justify youranswer.Solutions1a)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 twoassigments 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 ) .51=152Write 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|=6,3since 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 intersectionsare all empty.1It follows that the desired number is1680−6·60+6·6−2·6=1344.22=153The number of distinct 3-subsets of {1,2,...,10}is103 =10·9·83·2·1=120.5The 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.53=154a)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 formulae =vr2for 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.HenceS 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}3Since G has cycles of odd length,G cannot be bipartite.24=205a)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 2This 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.3b)The characteristic equation isr 3−3r 2+4=0.2The 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 coefficient 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)thefinal answer isa n=α1·2n+α2·n2n+α3(−1)n+2n+9,whereα1,α2,α3∈C.25=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 listn=len(a)for i in range(n):#range(n)=[0,...,n-1]sum=0for j in range(i,n):#range(n)=[i,...,n-1]sum=sum+a[j]#Initially sum=a[i]if sum==0:return Truereturn False...10 b)During Pass i through the outer for-loop,the inner for-loop is executed at mostn−i times,with1addition and1comparison per iteration.Hence the total number of basic operations is at mostn−1i=02(n−i)=2n+(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,amendedby 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 listfirst,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).56=20Exam=100。