离散数学试题及答案

Mock ExamNotes___________________________________________________________________________________ 1. There are 38 questions in this mock exam. The real exam will consist of about 25 questions that will be relatively similar to those here... that does not mean “identical” to those...2. If you do these well, you should have no big difficulties in the final exam.3. I encourage you to work these questions first on your own, without help, to see what you do or do not understand. You may seek help after that. Remember that no one will help you or give you hints during the exam. We will clarify the questions if something is not clear but not more than that.#01 - Page 13 #8Let p and q be the propositionsp : I bought a lottery ticket this week.q : I won the million dollar jackpot.Express each of these propositions as an English sentence.a) ￢p I did not buy a lottery ticket this week.b) p ∨q I bought a lottery ticket this week or I won the million dollar jackpot.c) p → q If I bought a lottery ticket this week then I won the million dollar jackpot.d) p ∧q I bought a lottery ticket this week and I won the million dollar jackpot.e) p ↔ q I bought a lottery ticket this week if, and only if, I won the million dollar jackpot.#02 - Page 15 #36Construct a truth table for each of these compound propositions.a) (p ∨q) ∨rp q r p ∨q(p ∨q) ∨rT T T T TT T F T TT F T T TT F F T TF T T T TF T F T TF F T F TF F F F Fc) (p ∧q) ∨r∧(p q)∧∨rp q r p qT T T T TT T F T TT F T F TT F F F FF T T F TF T F F FF F T F TF F F F Fe) (p ∨q)∧￢r∨∧￢rp q r p ∨q￢r(p q)T T T T F FT T F T T TT F T T F FT F F T T TF T T T F FF T F T T TF F T F F FF F F F T F#03 - Page 35 #9Show that each of these conditional statements is a tautology by using truth tables.c) ￢p → (p → q)p q￢p p → q￢p → (p → q)T T F T TT F F F TF T T T TF F T T Td) (p ∧q) → (p → q)∧p → q(p q) → (p → q)∧p q p qT T T T TT F F F TF T T T TF F T T Te) ￢(p → q) → pp q p → q￢(p → q)￢(p → q) → p T T T F TT F F T TF T T F TF F T F T#04 - DNFWrite the following proposition in disjunctive normal form:s = (r → p) → (p∧q)p q r r → p p∧q sT T T T T TT T F T T TT F T T F FT F F T F FF T T F F TF T F T F FF F T F F TF F F T F Fs=(p∧q∧r)∨(p∧q∧¬r)∨(¬p∧q∧r)∨(¬p∧¬q∧r)=(p∧q)∨(¬p∧r)#05 - Page 53 #8Let I (x) be the statement “x has an Internet connection” and C(x, y) be the statement “x and y have chatted over the Internet,” where the domain for the variables x and y consists of all students in your class. Use quantifiers to express each of these statements.b) Rachel has not chatted over the Internet with Chelsea.C(Rachel, Charles)e) Sanjay has chatted with everyone except Joseph.∀x(x ≠ Joseph → C(Sanjay, x))f ) Someone in your class does not have an Internet connection.∃x(￢I(x))i) Everyone except one student in your class has an Internet connection.!∃y[￢I(y) ∧∀x(x ≠ y → I(x))]j) Everyone in your class with an Internet connection has chatted over the Internet with at least one other student in your class.∃∀x yC(x, y)m) There is a student in your class who has chatted with everyone in your class over the Internet.∃x∀yC(x, y)#07 – Page 67 #27Determine the truth value of each of these statements if the domain for all variables consists of all real numbers.a) ∀x∃y(x2 = y) Truec) ∃x∀y(xy = 0)Truee) ∀x(x = 0 → ∃y(xy = 1))False∧x − y = 1)Falsei) ∀x∃y(x + y = 2 2j) ∀x∀y∃z(z = (x + y)/2)TrueUse rules of inference to show that the hypotheses “If it does not rain or if it is not foggy, then the sailing race will be held and the lifesaving demonstration will go on,” “If the sailing race is held, then the trophy will be awarded,” and “The trophy was not awarded” imply the conclusion “It rained.”Define the following literals:r It rainsf It is foggys The sailing race will be heldd The lifesaving demonstration will go ont The trophy will be awardedThe premises are then∧P1(¬r ∨ ¬f) → (s d)P2s → tP3¬tand the conclusion isrThe proof proceeds as follows:1¬t P32s → t P23¬s Modus tollens with 1 and 2'∨Addition to 34¬s ¬d∧De Morgan's law5¬(s d)∨) → (s d)∧P16(¬r ¬f∨)Modus tollens with 5 and 67¬(¬r ¬f∧De Morgan's law8r f9r Simplification of 8#09 – Page 80 #27Use rules of inference to show that if ∀x(P(x) → (Q(x) ∧S(x))) and ∀x(P(x) ∧R(x)) are true, then∀x(R(x) ∧S(x)) is true.∀∧Premise1x(P(x) R(x))∧Universal instantiation2P(c) R(c)3P(c)Simplification from 24x(P(x) →∧Premise∀(Q(x) S(x)))∧Universal instantiation5P(c) → (Q(c) S(c))∧Modus ponens with 3 and 56Q(c) S(c)7S(c)Simplification from 68R(c)Simplification from 2∧Conjunction of 7 and 89R(c) S(c)∀∧Universal generalization10x(R(x) S(x))Prove that if n is a positive integer, then n is odd if and only if 5n + 6 is odd.First, assume that n is odd, so that n = 2k+1 for some integer k. Then 5n+6 = 5(2k+1)+6 = 10k + 11 = 2(5k + 5) + 1. Hence, 5n + 6 is odd. To prove the converse, suppose that n is even, so that n = 2k for some integer k. Then 5n + 6 = 10k + 6 = 2(5k + 3), so 5n + 6 is even. Hence, n is odd if and only if 5n + 6 is odd.#11 – Page 126 #19What is the cardinality of each of these sets?a) {a}1b) {{a}}1c) {a, {a}}2d) {a, {a}, {a, {a}}}3#12 – Page 126 #40Explain why (A × B) × (C × D) and A × (B × C) × D are not the same.The tuples in those sets do not have the same composition. The tuplets in (A × B) × (C × D) are pairs of pairs: ((x,y),(u,v)). However, the tuplets in A × (B × C) × D are ordered triplets with two singletons and a pair: (u, (x,y), v).#13 – Page 136 #27Draw the Venn diagrams for each of these combinations of the sets A, B, and C.b) (A ∩ B) ∪(A ∩ C)c) (A ∩ B) ∪(A ∩ C)#14 – Page 153 #22Determine whether each of these functions is a bijection from R to R.a) f (x) = −3x + 4Yesb) f (x) = −3x2 + 7No: elements greater than 7 have no preimages.c) f (x) = (x + 1)/(x + 2)No: -2 has no imaged) f (x) = x5 + 1YesFor each of these sequences find a recurrence relation satisfied by this sequence. (The answers are not unique because there are infinitely many different recurrence relations satisfied by any sequence.)a) a n= 3a n= a n-1c) a n= 2n + 3a n-1= 2(n - 1)+ 3 = 2n + 3 – 2 = a n – 2. This implies that a n= a n-1+ 2.f ) a n= n2 + n(e1)Here we have two independent terms with n. We will need two additional formulas:a n-1= (n-1)2 + n – 1 = n2 – 2n + 1 + n – 1 = n2 – n(e2)a n-2= (n-2)2 + n – 2 = n2 – 4n + 4 + n – 2 = n2 – 3n + 2(e3)From (e1) and (e2), we have a n – a n-1 = 2n(e4)From (e2) and (e3), we have a n-1 – a n-2 = 2n – 2(e5)From (e4) and (e5), we have a n – a n-1 – (a n-1 – a n-2) = a n – 2a n-1+ a n-2 = 2, or a n = 2a n-1– a n-2+ 2 g) a n= n + (−1)n(e1)a n-1 = n – 1 + (−1)n-1 = n – 1 – (−1)n(e2)From (e1) and (e2), we have a n – a n-1 = 1 + 2(−1)nor a n = a n-1 + 1 + 2(−1)nWe can split this into the odd an even n's:a2k = a2k-1 + 1a2k+1 = a2k− 1h) a n= n!a n = n a n-1#16 – Page 583 #30 (+ additional questions)Let R1 = {(1, 2), (2, 3), (3, 4)} and R2 = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4)} be relations from {1, 2, 3, 4} to {1, 2, 3, 4}. Finda) R1∪R2 = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4)} = R2b) R1 ∩ R2 = {(1, 2), (2, 3), (3, 4)} = R1c) R1 − R2 = ∅d) R2 − R1 = {(1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3)}e) R1 ◦ R2 = {(1, 2), (1, 3), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)}f ) R2 ◦ R1 =g) Draw the graph of R1 ◦ R2.h) Find the reflexive, symmetric and transitive closures ofReflexive closure = {(1, 1), (1, 2), (2, 2), (2, 3), (3, 4), (3, 4), (4, 4)}Symmetric closure = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 3), (4, 3)}Transitive closure = {(1, 2), (1, 3), (2, 3), (2, 4), (3, 1), (3, 4), (4, 1), (4, 2)}Let R be the relation consisting of all pairs (x, y) such that x and y are strings of uppercase and lowercase English letters with the property that for every positive integer n, the n th characters in x and y are the same letter, either uppercase or lowercase. Show that R is an equivalence relation.That definition basically means that two strings are equivalent if, and only if, they have the same length and every corresponding characters x i and y i are the same letter, either lower or upper case.Let c and C stand for the lower and upper cases of a same letter in the English alphabet.Clearly k, x∀k = x k. So (x, x) ∈ R.So R is reflexive.If (x,y)∈ R, then k, x∀k∈ {c, C} and y k∈ {c, C}. So (y, x) ∈ R also. So R is symmetric.If (x,y)∈ R and (y,z)∈ R, then [x k∈ {c, C}, y k∈ {c, C}] and [y k∈ {c, C}, z k∈ {c, C}], which implies that [x k∈ {c, C},z k∈ {c, C}]. So (x, z) ∈ R. So R is transitive.Therefore R is an equivalence relation.#18 – Page 631 #34Answer these questions for the poset ({2, 4, 6, 9, 12, 18, 27, 36, 48, 60, 72}, |).a) Find the maximal elements.27, 48, 60, 72b) Find the minimal elements.2, 9c) Is there a greatest element?Nod) Is there a least element?Noe) Find all upper bounds of {2, 9}.18,36 72f ) Find the least upper bound of {2, 9}, if it exists.18g) Find all lower bounds of {60, 72}.2, 4, 6, 12h) Find the greatest lower bound of {60, 72},if it exists.12#19 – Group TheoryConsider the set G={a1+a2√3 | a1,a2∈ℚ∧a1a2≠0}with the usual multiplication operation.a) Show that G is a group by verifying the axioms of closure, associativity, existence of an identity element, and existence of an inverse element for every element. Specify what the identity element is and the form an an inverse element.1. Closure: Let a,b∈G. Then ab=(a1+a2√3)(b1+b2√3)=(a1b1+3a2b2)+(a1b2+a2b1)√(3)∈G2 Associativity: Let a ,b ,c ∈G . Then (ab )c =[(a 1+a 2√3)(b 1+b 2√3)]c=[(a 1b 1+3a 2b 2)+(a 1b 2+a 2b 1)√(3)](c 1+c 2√3)=[(a 1b 1+3a 2b 2)c 1+3(a 1b 2+a 2b 1)c 2]+[(a 1b 1+3a 2b 2)c 2+(a 1b 2+a 2b 1)c 1]√(3)=a 1b 1c 1+3a 2b 2c 1+3a 1b 2c 2+3a 2b 1c 2+[a 1b 1c 2+3a 2b 2c 2+a 1b 2c 1+a 2b 1c 1]√(3)=[a 1(b 1c 1+3b 2c 2)+3a 2(b 1c 2+b 2c 1)]+[a 2(b 1c 1+3b 2c 2)+a 1(b 1c 2+b 2c 1)]√(3)=(a 1+a 2√3)[(b 1c 1+3b 2c 2)+(b 1c 2+b 2c 1)√(3)]=a [(b 1+b 2√3)(c 1+c 2√3)]=a (bc )3. Identity element. Let this element be e =e 1+e 2√. Thenea =(e 1+e 2√3)(a 1+a 2√3)=(e 1a 1+3e 2a 2)+(e 1a 2+e 2a 1)√(3)=(a 1+a 2√3).This implies that for every a 1 and a 2:e 1a 1+3e 2a 2=a 1e 1a 2+e 2a 1=a 2That implies e 1=1and e 2=0. Thus e =1.4. Inverse element. Consider a =a 1+a 2√3∈G and let its inverse be a −1=x 1+x 2√3if it exists. Then we must havea −1a =(x 1+x 2√3)(a 1+a 2√3)=(x 1a 1+3x 2a 2)+(x 1a 2+x 2a 1)√(3)=1=e This impliesx 1a 1+3x 2a 2=1x 1a 2+x 2a 1=0The solution isa −1=a 1−a 2√3a 12−3a 22.Thus G forms a group.b) Is G Abelian?Yes: ab =(a 1+a 2√3)(b 1+b 2√3)=(a 1b 1+3a 2b 2)+(a 1b 2+a 2b 1)√(3)=ba because this expression is symmetric.#20 – Page 665 #9Determine the number of vertices and edges and find the in-degree and out-degree of each vertex for the shown directed multigraph:5 vertices 13 edgesdeg+(a) = 1, deg+(b) = 1, deg+(c) = 5, deg+(d) = 4, deg+(e) = 0deg−(a) = 6, deg−(b) = 5, deg−(c) = 2, deg−(d) = 2, deg−(2) = 0#21 – Page 666 #28Suppose that a newcompany has five employees: Zamora, Agraharam, Smith, Chou, and Macintyre. Each employee will assume one of six responsiblities: planning, publicity, sales, marketing,development, and industry relations. Each employee is capable of doing one or more of these jobs: Zamora could do planning, sales, marketing, or industry relations; Agraharam could do planning or development; Smith could do publicity, sales, or industry relations; Chou could do planning, sales, or industry relations; and Macintyre could do planning, publicity, sales, or industry relations.a) Model the capabilities of these employees using a bipartite graph.b) Find an assignment of responsibilites such that each employee is assigned one responsibility.Note: the assignment is not unique. The only forced choices are (Z, ma) and (A, de). There is a variety of possibilities for the other 3.c) Is the matching of responsibilities you found in part (b) a complete matching? Is it a maximum matching?The matching (from {Z, A, S, C, M} to {ma, de, sa, pl, pu, ir}) is complete because every employee is matched with a job. It is a maximum because |M| = 5 = |{Z, A, S, C, M}|#22 – Page 676 #21 (+ additional questions)Consider the following grapha) Find the adjacency matrix A of the graph A =(1110100220111210)b) Find how many paths of length 3 there are from c to b A 3=A (1110100220111210)(1110100220111210)=(1110100220111210)(4123353054415125)=(12109414361318710121512124)So there are 6 paths from c to b.#23 – Page 676 #38Determine whether the following two graphs are isomorphic. If so, construct an isomorphism.Notice the second graph can be deformed like this (by moving v 2 all the way down and rotating the other vertices by about a quarter of a turn):It has 2 circuits of length 4 whereas the graph on the left has only 1. That immediately implies that these graphs are not isomorphic.#24 – Page 692 #31-32Consider the following graphs#31#31a) List the cut vertices c c, db) List the cut edges none(c,d)c) What is the vertex connectivity κ(G)?11d) What is the edge connectivity λ(G)?21#25 – Page 704 #22Determine whether the directed graph shown has an Euler circuit. Construct an Euler circuit if one exists. If no Euler circuit exists, determine whether the directed graph has an Euler path. Construct an Euler path if one exists.The vertices' total degrees are all even except for vertices b and c. So it has no Euler circuit but there might be an Euler path, although this is not guaranteed because the graph is directed. However every vertex with an even total degree has equal in and out degrees. Beccause the out-degree of c is larger than its in-degree, then the starting point has to be c. In fact, we o find an Euler path:c → e → b → c → b → f → a → f → e → f →d →e → a → b → d → c → b#26 – Page 716 #8Find a shortest path (in mileage) between each of the following pairs of cities in the airline system shown in Figure 1.Note: You must show every steps of the algorithmB N M ACD S L-0------N 191/N-1090/N760/N722/N-2534/N2451/N B --1090/N760/N722/N-2534/N2451/N C --1090/N760/N-1630/C2534/N2451/N A --1090/N--1630/C2534/N2451/N D --1090/N---2534/N2451/N M ------2534/N2451/N LPath = N → L Distance = 2451b) Boston and San FranciscoB N M ACD S L0-------B -191/B--860/B---N --1281/N951N860/B-2725/N2642/N C --1281/N951N-1768/C2715/C2642/N A --1281/N--1768/C2715/C2642/N M -----1768/C2715/C2642/N D ------2715/C2602/D L ------2715/C-SPath = B → C → S Distance = 2715c) Miami and DenverB N M ACD S L--0-----M -1090/M-595/M----A -1090/M--1201/A---N 1281/N---1201/A-3624/N3541/N C 1281/N----2109/C3056/C3541/N B -----2109/C3056/C3541/N DPath = M → A → C → D Distance = 2109B N M ACD S L--0-----M-1090/M-595/M----A-1090/M--1201/A---N1281/N---1201/A-3624/N3541/N C1281/N----2109/C3056/C3541/N B-----2109/C3056/C3541/N D------3056/C2943/D LPath = M → A → C → D → L Distance = 2943#27 – Page 726 #12Suppose that a connected planar graph has eight vertices, each of degree three. Into how many regions is the plane divided by a planar representation of this graph?We have V = 8. Each node has a degree equal to 3. The sum of all the degrees is therefore 24 and we know it is equal to twice the number of edges; thus E = 12. Recall Euler's formula: V – E + F = 2. So we have 8 – 12 + F = 2, which implies that F = 6.#28 – Page 732 #4Construct the dual graph for the map shown. Then find the number of colors needed to color the map so that no two adjacent regions have the same color.#29 – Page 733 #17Schedule the final exams for Math 115, Math 116, Math 185, Math 195, CS 101, CS 102, CS 273, and CS 473, using the fewest number of different time slots, if there are no students taking both Math 115and CS 473, both Math 116 and CS 473, both Math 195 and CS 101, both Math 195 and CS 102, both Math 115 and Math 116, both Math 115 and Math 185, and both Math 185 and Math 195, but there are students in every other pair of courses.The best way to obtain a graph for this is to draw a complete graph and then remove edges according to the description in the above paragraph.{MAT115, MAT116, CS473}{MAT185, MAT195}{CS101}{CS102}{CS273}The scheduling is not unique.#30 – Page 755 #4Consider the following rooted tree:a) Which vertex is the root?ab) Which vertices are internal?b, d, e, g, h, i, oc) Which vertices are leaves?c, f, j, k, l, m, n, p, q, r, sd) Which vertices are children of n?nonee) Which vertex is the parent of g?bf ) Which vertices are siblings of k?jg) Which vertices are ancestors of o?a, d, ih) Which vertices are descendants of d?h, i, n, o, p, q, r, s#31 – Page 756 #20How many leaves does a full 3-ary tree with 100 vertices have?L=(m−1)n+1n =(3−1)×100+13=2013=67MATMATMAT185MAT195CS473CS273CS101CS102#32 – Page 769 #2Build a binary search tree for the words oenology, phrenology, campanology, ornithology, ichthyology , limnology, alchemy , and astrology using alphabetical order.#33 – Page 770 #24Use Huffman coding to encode these symbols with given frequencies: A: 0.10, B: 0.25, C: 0.05, D: 0.15, E: 0.30, F: 0.07, G: 0.08. What is the average number of bits required to encode a symbol?0.050.070.080.100.150.250.30 C F G A D B E0.080.100.120.150.250.30 GADBE0.120.150.180.250.30 DB E0.180.250.270.30 BE0.270.300.43Eoenologyphrenologycampanology ichthyology alchemy astrologylimnologyornithology0.430.571.00Codes:A = 110B = 10C = 0111D = 010E = 00F = 0110G = 111Average nuber of bits = (3 + 2 + 4 + 3 + 2 + 4 + 3) / 7 = 21/7 = 3#34 – Page 783 #10-11Consider the following rooted tree:In which order are the vertices visited using a preorder traversal?a, b, d, e, i, j, m, n, o, c, f, g, h, k, p, l#35 – Page 784 #23What is the value of the following prefix expression?a) − 2 / 8 4 3∗− 2 ∗/ 8 4 3=− 2 2∗ 3=− 4 3=1GACFCFb) ↑ − 3 3 4 2 5∗∗∗ 5=↑ − 3 3∗ 8 5∗ 4 2↑ − 3 3=↑ − 9 8 5=↑ 1 5=1c) + − ↑ 3 2 ↑ 2 3 / 6 − 4 2+ − ↑ 3 2 ↑ 2 3 / 6 − 4 2=+ − ↑ 3 2 ↑ 2 3 / 6 2=+ − ↑ 3 2 ↑ 2 3 3=+ − ↑ 3 2 8 3=+ − 9 8 3=+ 1 3=4∗d) + 3 + 3 ↑ 3 + 3 3 3+ 3 + 3 ↑ 3∗↑ 3 6 3∗+ 3 3 3= + 3 + 3∗+ 3 729 3= + 3=∗+ 3 732 3∗= 735 3=2205#36 – Page 795 #13Use depth-first search to produce a spanning tree for the following simple graph. Choose vertex 'a' as the root of this spanning tree and assume that the vertices are ordered alphabetically.a →b →c →d →e →f →g →h → Ig → j#37 – Page 802 #3Use Prim's algorithm to find a minimum spanning tree (and its total weight) for the following weighted graph:(ef)1(cf)3(eh)3(hi)2(gh)4(bc)4(bd)3(ad)2Total weight = 22#38 – Page 802 #8Use Kruskal’s algorithm to find a minimum spanning tree for the weighted graph in Exercise 4 (#37). (ef)1(ad)2(hi)2(bd)3(cf)3(eh)3(bc)4(gh)4Total weight = 22The spanning tree is identical to that in Exercise 4 (#37).。