Introduction to Algorithms
Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter 4.6, Problem 3E
Program Plan Intro

To show that the case 3 of the master theorem is overstated.

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Match the following sentence to the best suitable answer:          -           A.           B.           C.           D.       for the linear congruence ax=1(mod m), x is the inverse of a, if__________          -           A.           B.           C.           D.       What is -4 mod 9 ?          -           A.           B.           C.           D.       The solution exists for a congruence ax=b(mod m) such that GCD(a,m)=1 and          -           A.           B.           C.           D.       (107+22)mod 10 is equivalent to :_________ A. 5 B. c divides b C. GCD(a,m)=1 D. 9 mod 10
Demonstrate how the following instance of the 3SAT problem can be reduced to its corresponding Sum of Subsets problem. Consequently prove that the Sum of Subsets problem belongs to the class of NPC problems, if 3SAT is in NPC also. = (xyz) (-x-y-z) (xyz) (-x-yz) where x=0, y=0, 2-1, and represents negation
) Show that ∀xP(x) ∧ ∃xQ(x) is logically equivalent to ∀x∃y(P(x) ∧ P(y)) The quantifiers have the same non empty domain. I know that to prove a proposition is logically equivalent to another one, I have to show that ∀xP(x) ∧ ∃xQ(x) ↔ ∀x∃y(P(x) ∧ P(y)) Which means I have to prove that (∀xP(x) ∧ ∃xQ(x)) → ∀x∃y(P(x) ∧ P(y)) ∧ ∀x∃y(P(x) ∧ P(y)) → (∀xP(x) ∧ ∃xQ(x)) I don't know the answer, so I saw the textbook answer. It says (1) Suppose that ∀xP(x) ∧ ∃xQ(x) is true. Then P(x) is true for all x and there is an element y for which Q(y) is true. I get this part. Because P(x) ∧ Q(x) is true for all x and there is a y for which Q(y) is true, ∀x∃y(P(x) ∧ P(y)) is true. Emm... I think ∀x∃y(P(x) ∧ P(y)) is true because ∀x only affects P(x) and ∃y only affects P(y) since their alphabets are different. So, it has the exact same meaning as ∀xP(x) ∧ ∃yQ(y). And since the domains are the same, ∀xP(x) ∧ ∃yQ(y) is actually equal to ∀xP(x) ∧ ∃xQ(x). But the textbook states that "P(x) ∧ Q(x) is…
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