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3 Pages

Homework 2 Solutions

Course: EECS 203, Spring 2011
School: Michigan
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Word Count: 947

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203 EECS Homework 2 Section 1.3 1. (E) 6bef and 10cde 6a) Every student in school has visited North Dakota 6b) Not every student in the school has visited North Dakota 6c) No student in the school has visited North Dakota 10c) x(C(x) F (x) D(x)) 10d) x(C(x) D(x) F (x)) 10e) (xC(x)) (xD(x)) (xF (x)) 2. (M) 42abc 42a) Let A(x) mean &quot;user x has access to an electronic mailbox&quot; xA(x) 42b) Let...

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203 EECS Homework 2 Section 1.3 1. (E) 6bef and 10cde 6a) Every student in school has visited North Dakota 6b) Not every student in the school has visited North Dakota 6c) No student in the school has visited North Dakota 10c) x(C(x) F (x) D(x)) 10d) x(C(x) D(x) F (x)) 10e) (xC(x)) (xD(x)) (xF (x)) 2. (M) 42abc 42a) Let A(x) mean "user x has access to an electronic mailbox" xA(x) 42b) Let S(x, y) mean "system x is in y state" xS(f ilesystem, locked) = A(x) 42c) Using prepositional function from part 42b S(f irewall, diagnostic) = S(proxyserver, diagnostic) 3. (C) 44 We want P and Q that are sometimes false and sometimes true. Let P(x) mean "x is even" Let Q(x) mean "x is multiple of 3" for two values in domain of integers, x = 4 and x = 9. (x(P (x)) (x(Q(x))) reduces to F alse F alse and hence true. But x(P (x) Q(x)) is false. Hence two are not logically equivalent. Section 1.4 4. (E) 6def 6d) Some student is enrolled in both Math222 and CS252. 6d) There are two distinct students such that second one is enrolled in all the courses that first one is enrolled in. 6d) There are two distinct students that are enrolled in the same courses. 5. (M) 10hij 10h) y((x(F (x, y)) (z(wF (w, z) = z = y))) 10i) xF (x, x) 10j) xy(x = y F (x, y) z((F (x, z) z = x) = z = y)) (We don't assume the sentence is asserting that person can or can't fool him/herself) 6. (E) 16cd 16c) Let P(s, c, m) denote that student s has class standing c and is majoring in m. Domain of s is all students in class, of c all four standings and of m all possible majors. scm(P (s, c, m) c = junior m = mathematics). 1 True because there is a sophomore majoring in computer science. 16d) x(mP (x, sophomore, m) cP (x, c, computer)). False, as there is a freshman mathematics major. 7. (M) 22 pabc(p > 0 p = (a2 + b2 + c2 )) 8. (E) 24bcd 24b) A nonnegetive number minus a negative number is positive. 24c) Difference of two nonpositive numbers is not necassarily nonpositive. 24d) The product of two numbers is non-zero if and only if the two numbers are non zero. 9. (E) 36d 36d) Let S(x, y) mean that student x solved problem y. Let domain of x be all students in this class and of y, all exercises in this book. Then statement means xyS(x, y). Negation is xyS(x, y). In English, for every student there is some exercise that he or she has not solved. Section 1.5 10. (M) 14acd 14a) Let P(x) mean that "x is in the class". Q(x) mean "x has got a convetible". R(x) mean "x has gotten a speeding ticket". 1)x(Q(x) = R(x)) Premise = 2)Q(Linda) R(Linda) Universal instantiation of 1 3)P(Linda) Premise 4)Q(Linda) Premise 5)R(Linda) Modus Ponens using 2 and 4 6)P (Linda) R(Linda) Conjunction using 3 and 5 7)xP (x) R(x) Existential quantification from 6 14c) Let P(x) mean "x was a movie produced by John Sayles"; Q(x) mean "movie x was about Coal Miners"; R(x) mean that "movie x is wonderful". Let domain be all movies and a be the unspecified movie. 1)x(P (x) = R(x)) Hypothesis 2)x(P (x) Q(x)) Hypothesis 3)P (a) Q(a) Existential Instantiation from 2 4)P (a) = R(a) Universal Instantiation from 1 5)P(a) Simplification from 3 6)Q(a) Simplification from 3 7)R(a) Modus ponens with 4 and 5 8)Q(a) R(a) Conjunctions using 6 and 7 9)xQ(x) R(x) Existential generalization from 8 14d) Let P(x) mean "x is in the class"; Q(x) mean "x has been to france"; R(x) mean "x has visited Louvre". Let domain be all people and a be the unspecified person. 2 1)x(P (x) Q(x)) 2)x(Q(x) = R(x)) 3)P (a) Q(a) 4)Q(a) = R(a) 5)Q(a) 6)R(a) 7)P(a) 8)P (a) R(a) 9)x(P (x) R(x)) Premise Premise Existential instantiation from 1 universalll instantiation from 2 Simplification from 3 Modus ponens with 4 and 5 Simplification from 3 Conjunction of 6 and 7 Existential generalization from 8 11. (E) 18 We know some s exists that makes the existential in premise true but we can't conclude that this s is Max. 12. (E) 24 24) Steps 3 and 5 are wrong as the simplification doesn't apply to disjunctions. 13. (M) 28 1)x(P (x) Q(x)) 2)x((P (x) Q(x)) = R(x)) 3)P (a) Q(a) 4)((P (a) Q(a)) = R(a)) 5)(P (a) Q(a)) R(a) 6)(P (a) Q(a)) R(a) 7)(Q(a) P (a)) R(a) 8)Q(a) (P (a) R(a)) 9)P (a) (P (a) R(a)) 10)(P (a) P (a)) R(a) 11)P (a) R(a) 12)R(a) = P (a) 13)x(R(x) = P (x)) premise premise universal instantiation from 1 universal instantiation from 2 simplification of implification De Morgan's and simplification Commutativity Associativity Resolution on 3 and 8 Associativity Idempotence Implication universal generalization 14. (C) 34bd 34b) Let l mean "logic is difficult"; m mean "mathematics is not easy";s mean "many students like logic". Then: 1)l s Premise 2)m = l Premise To check m = s It is false when m is false and s is true. If s is true then l has to be true for first premise to hold. If l is true then second premise is true. So premises hold even when this statement is false. So we can't conclude it. 34d) This is equivalent to m l. If we take contrapositive of premise 2 and simplify we get the same. So this is true. 3
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Michigan - EECS - 203
EECS 203: Homework 3 SolutionsSection 2.1 1. (E) 8defg d) True e) True f) True g) False, since cfw_, cfw_ = cfw_. 2. (E) 22 a) No, since a power set should contains at least the empty set ; note that P () = cfw_. b) Yes, P (cfw_a) = cfw_, cfw_a. c) No, s
Michigan - EECS - 203
EECS 203 Homework 3 SolutionsSection 1.6 6. and(an odd number) 24. Proof by contradiction: Assume that at most two days are selected from any month and still select as least 25 days. To select as many days as possible, we can select two from each month.
Michigan - EECS - 203
EECS 203: Homework 4 SolutionsSection 2.4 1. (E) 4cd c) a0 = 8, a1 = 11, a2 = 23, a3 = 71 d) a0 = 2, a1 = 0, a2 = 8, a3 = 0 2. (E) 10ef e) The general formula for this sequence is an = (3 an-1 ) + 2, given that a0 = 0 Therefore, this sequence from n=0 to
Michigan - EECS - 203
EECS 203 HOMEWORK 4 SOLUTIONS 1. (M) Prove that the union of all elements of a power set of a set is the set itself.Let S be the original set and P(S) be the power set and UPS be the union of all elements of a power set. To Prove: S = UPS We p
Michigan - EECS - 203
EECS 203: Homework 5 SolutionsSection 3.2 1. (E) 2bcd b) f (x) is O(x2 ) with C = 2 and k =21000.c) f (x) is O(x ) with C = 1 and k = 1. Note that log x &lt; x, so x log x &lt; x2 . d) f (x) is not O(x2 ). Suppose there exist C and k such that x4 /2 Cx2 for
Michigan - EECS - 203
EECS 203: Homework 5 SolutionsSection 2.4 1. (E) 16bc b) 3 j -2 j = 3 j - 2 j 3 -1 2 -1 - 3-1 2-1 =9841-511 =9330 =j=0 9 j=0 9 j=0 8 88c) 23 j32 j j=08=2 3 j3 2 jj=0 j=08823 9-1 329 -1 = 3-1 2-1 =298413511 =212152. (E)n 2i 3 ji=1 j=1 m n
Michigan - EECS - 203
EECS 203: Homework 6 SolutionsSection 4.1 1. (E) 4 a) 13 = (1(1 + 1)/2)2 b) Evaluating both sides 1=1. So True. c) The inductive hypothesis is the statement P (k) : 13 + 23 + . + k 3 =k(k+1) 2 2.(k+1)(k+2) 2 2d) We need to prove that for all k 1, giv
Michigan - EECS - 203
EECS 203: Homework 7 SolutionsSection 5.1 1. (E) 24 a) Since the strings do not contain the same digit, there are 10 ways to choose the first digit, 9 ways to choose the second and so on. Therefore the answer is 10 9 8 7 = 5040. b) There are 10 ways to c
Michigan - EECS - 203
Homework 7 SolutionsSection 4.2 4. a)b) Assume for all c) d) for some non-negative i, j e) We have shown that the statement is true for several basis cases ( ) and that in the inductive steps all other cases are reducible to smaller cases, which are the
Michigan - EECS - 203
EECS 203: Homework 8 SolutionsSection 5.4 1. (E) 8 By binomial theorem, the coefficient of the term x8 y 9 in the expansion of (x + y)17 is 17 = 24310. 8 When considering the term x8 y 9 in the expansion of (3x + 2y)17 , let x = 3x and y = 2y, meaning th
Michigan - EECS - 203
Homework 8 Solutions Section 5.3 30 a). There are C(16,5) ways to choose a committee if there are no restrictions. There are C(9,5) ways to select a committee from just the 9 men. Therefore there are C(16,5) C(9,5) = 4242 committees with at least one woma
Michigan - EECS - 203
EECS 203: Homework 9 SolutionsSection 8.1 1. (E) 4 a) antisymmetric, transitive b) reflexive, symmetric, transitive c) reflexive, symmetric, transitive d) reflexive, symmetric 2. (E) 10 Only the relation in part (a) is irreflexive. 3. (E) 20 An asymmetri
Michigan - EECS - 203
EECS 203: Homework 9 SolutionsSection 3.8 1. (E) 4b[4 -1 -7 6 -7 -5 8 5 4 0 7 3]2. (E) 12a Let A and B be m x k matrices and C be a k x n matrix. By the definition of matrix sum, A+B is an m x k matrix such that each (i,j) entry is equal to aij + bij
Michigan - EECS - 203
EECS 203: Homework 10 SolutionsSection 8.4 1. (E) 2 When we add the pairs (x, x) to the given relation we have all of Z Z; in other words, we have the relation that always holds. 2. (E) 6 We form the reflexive closure by taking the given directed graph a
Michigan - EECS - 203
Homework 10Section 8.11.(M) 48a) Since R and S are both having all pairs like (x,x) where x is an element of set A the union will also have it. Thus R S is also reflexive. 48b) R S will be reflexive as both R and S have pairs like (x,x). 48c) R S will b
Michigan - EECS - 203
EECS 203: Homework 11 SolutionsSection 8.5 1. (E) 14 In order to show that R is an equivalence relation, we need to show that it is reflexive, symmetric and transitive. It is easy to see that R is reflexive since x has the same nth character as x for any
Michigan - EECS - 203
Homework 11 SolutionsSection 8.4 Initial Matrix Using cfw_a as an interior vertex Using cfw_a,b as interior vertices Using cfw_a,b,c as interior vertices Using cfw_a,b,c,d as interior verticesUsing cfw_a,b,c,d,e as interior verticesFinal AnswerSection
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EECS 203: Homework 12 SolutionsSection 9.3 1. (E) 6,8 (6) a 0 1 0 1 0 b 1 0 0 1 1 c 0 0 0 1 1 d 1 1 1 0 0 e 0 1 1 0 0a b c d e(8) a 0 1 0 1 0 b 1 0 1 0 0 c 0 1 1 0 1 d 1 1 0 0 0 e 0 1 0 1 1a b c d e2. (E) 14 a 0 3 0 1 b 3 0 1 0 c 0 1 0 3 d 1 0 3 0a
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EECS 203: Homework 13 SolutionsSection 4.3 1. (E) 6cd c) This is not valid because, for example, f (2) can have many different values and still be consistent with the recursive definition. d) This is not valid for the opposite reason. There are no possib
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Homework 6Section 1.31. (E) 10. If x &gt; 1 then x3 &lt; x4 . So clearly x3 is O(x4 ) for C=1 and k=1. But if x4 is O(x3 ) then x4 &lt; C.x3 for some C and k. Dividing both sides by x3 we get x &lt; C. Clearly this condition will not hold for all large x. 2. (E) (n
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EECS 203: Homework 7Instructions: The questions below come from the course textbook and are rated as (E)asy, (M)edium or (C)hallenging. Answer them as completely, concisely, and legibly as possible and your answers will be scored and weighted according t
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Mail : Inbox: EECS 203 001 LEC Teaching Questionnaires Submission C.https:/web.mail.umich.edu/blue/imp/message.php?actionID=print_messa.Date: From: To: Subject:Mon, 13 Dec 2010 00:10:40 -0500 [12:10:40 AM EST] um_evaluations@umich.edu Undisclosed Recip
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