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 29.5, Problem 9E
Program Plan Intro
To state for which values of r, s and t you can assert that
- Both P and D have optimal solutions with finite objective value.
- P is feasible, but D is infeasible.
- D is feasible, but P is infeasible.
- Neither P nor D is feasible.
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Students have asked these similar questions
Let pn(x) be the probability of selling the house to the highest bidder when there are n people, and you adopt the Look-Then-Leap algorithm by rejecting the first x people. For all positive integers x and n with x < n, the probability is equal to p(n(x))= x/n (1/x + 1/(x+1) + 1/(x+2) + … + 1/(n-1))
If n = 100, use the formula above to determine the integer x that maximizes the probability n = 100 that p100(x). For this optimal value of x, calculate the probability p100(x).
Briefly discuss the significance of this result, explaining why the Optimal Stopping algorithm produces a result whose probability is far more than 1/n = 1/100 = 1%.
Which of the following statements are true given A* (admissibility and consistency of heuristics ] ?
The heuristic function h[n] is called admissible if h[n] is never larger than h*[n], namely h[n] is always less or equal to true cheapest cost from n to the goal.
If the heuristic function, h always underestimates the true cost [h[n] is smaller than h*[n]), then A* is guaranteed to find an optimal solution.
When h is inconsistent, it can not be admissible.
If h is consistent and h[goal)=0 then h is admissible
A* is complete and optimal
Suppose we have a heuristic h that over-estimates h* by at most epsilon (i.e., for all n, 0<= h(n) <= h*(n)+epsilon). Show that A* search using h will get a goal whose cost is guaranteed to be at most epsilon more than that of the optimal goal.
Chapter 29 Solutions
Introduction to Algorithms
Ch. 29.1 - Prob. 1ECh. 29.1 - Prob. 2ECh. 29.1 - Prob. 3ECh. 29.1 - Prob. 4ECh. 29.1 - Prob. 5ECh. 29.1 - Prob. 6ECh. 29.1 - Prob. 7ECh. 29.1 - Prob. 8ECh. 29.1 - Prob. 9ECh. 29.2 - Prob. 1E
Ch. 29.2 - Prob. 2ECh. 29.2 - Prob. 3ECh. 29.2 - Prob. 4ECh. 29.2 - Prob. 5ECh. 29.2 - Prob. 6ECh. 29.2 - Prob. 7ECh. 29.3 - Prob. 1ECh. 29.3 - Prob. 2ECh. 29.3 - Prob. 3ECh. 29.3 - Prob. 4ECh. 29.3 - Prob. 5ECh. 29.3 - Prob. 6ECh. 29.3 - Prob. 7ECh. 29.3 - Prob. 8ECh. 29.4 - Prob. 1ECh. 29.4 - Prob. 2ECh. 29.4 - Prob. 3ECh. 29.4 - Prob. 4ECh. 29.4 - Prob. 5ECh. 29.4 - Prob. 6ECh. 29.5 - Prob. 1ECh. 29.5 - Prob. 2ECh. 29.5 - Prob. 3ECh. 29.5 - Prob. 4ECh. 29.5 - Prob. 5ECh. 29.5 - Prob. 6ECh. 29.5 - Prob. 7ECh. 29.5 - Prob. 8ECh. 29.5 - Prob. 9ECh. 29 - Prob. 1PCh. 29 - Prob. 2PCh. 29 - Prob. 3PCh. 29 - Prob. 4PCh. 29 - Prob. 5P
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