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 34.4, Problem 3E
Program Plan Intro
To show that the strategy of Professor Jagger does not yield a polynomial time reduction.
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Check out a sample textbook solutionStudents have asked these similar questions
) 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…
Let R=ABCDEGHK and F= {ABK→C, A→DG, B→K, K→ADH, H→GE} . Is it in BCNF? Prove your answer.
Using the method of full or partial truth tables, say whetherthe following sentences of L1 are tautologies or not. If not,give a counterexample.
(a) ¬(P → Q) → (P → ¬P)
Chapter 34 Solutions
Introduction to Algorithms
Ch. 34.1 - Prob. 1ECh. 34.1 - Prob. 2ECh. 34.1 - Prob. 3ECh. 34.1 - Prob. 4ECh. 34.1 - Prob. 5ECh. 34.1 - Prob. 6ECh. 34.2 - Prob. 1ECh. 34.2 - Prob. 2ECh. 34.2 - Prob. 3ECh. 34.2 - Prob. 4E
Ch. 34.2 - Prob. 5ECh. 34.2 - Prob. 6ECh. 34.2 - Prob. 7ECh. 34.2 - Prob. 8ECh. 34.2 - Prob. 9ECh. 34.2 - Prob. 10ECh. 34.2 - Prob. 11ECh. 34.3 - Prob. 1ECh. 34.3 - Prob. 2ECh. 34.3 - Prob. 3ECh. 34.3 - Prob. 4ECh. 34.3 - Prob. 5ECh. 34.3 - Prob. 6ECh. 34.3 - Prob. 7ECh. 34.3 - Prob. 8ECh. 34.4 - Prob. 1ECh. 34.4 - Prob. 2ECh. 34.4 - Prob. 3ECh. 34.4 - Prob. 4ECh. 34.4 - Prob. 5ECh. 34.4 - Prob. 6ECh. 34.4 - Prob. 7ECh. 34.5 - Prob. 1ECh. 34.5 - Prob. 2ECh. 34.5 - Prob. 3ECh. 34.5 - Prob. 4ECh. 34.5 - Prob. 5ECh. 34.5 - Prob. 6ECh. 34.5 - Prob. 7ECh. 34.5 - Prob. 8ECh. 34 - Prob. 1PCh. 34 - Prob. 2PCh. 34 - Prob. 3PCh. 34 - Prob. 4P
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