Machine-translated by Google. Questions are authored in English.
Top

Informed and Heuristic Search Questions and Answers

Answer: C

Running with a large weight first gives a quick solution, and the weight is then reduced and the search resumed, so a usable answer exists at every moment and improves with time. That suits an agent under a deadline it may not know in advance. Start points, fixed running times and temporal weighting are different notions. Making the quality of an answer a function of the time available, rather than requiring completion, is exactly what a real time system needs.

Please log in or register to report this question.

Answer: D

An agent that may be interrupted at an unpredictable moment needs an answer available throughout, and one that improves monotonically lets it use whatever time it turns out to have. Memory profiles vary, improvement is by construction rather than a separate guarantee, and a heuristic is still needed. Designing for an unknown deadline, rather than assuming one, is what distinguishes an algorithm intended for an agent from one intended for a batch computation.

Please log in or register to report this question.

Answer: A

By convention g is the cost so far, h is the estimate of the cost remaining and f is their sum. Keeping the three letters straight is essential to reading any description of A star or its variants. Tree height, frontier size and accumulated cost are denoted differently. Notation that is consistent across the literature is worth learning exactly, since a misread symbol inverts the meaning of an algorithm.

Please log in or register to report this question.

Answer: B

g is the exact cost accumulated from the initial state to the node in question, known rather than estimated, and it is what uniform cost search orders on. The estimate of the remainder is h and the goal itself is not usually given a letter. Distinguishing an exact quantity from an estimated one in the notation is what makes the properties of the algorithms provable, since the guarantees rest on which of the two may be wrong.

Please log in or register to report this question.

Answer: C

Greedy best first orders the frontier by the estimated remaining cost alone, ignoring what has already been spent, which makes it fast and suboptimal. Uniform cost uses g alone, A star uses both, and breadth first uses depth. Placing each algorithm by which terms of the evaluation function it uses turns a list of methods into a single family with a two dimensional structure, which is much easier to remember.

Please log in or register to report this question.

76.

Which algorithm minimises the sum of both cost terms?

Answer: D

A star adds the cost so far to the estimate of the cost remaining, so it minimises an estimate of the total rather than either half, and that is what gives it optimality with an admissible heuristic. Greedy uses h alone, uniform cost g alone, and depth first neither. The family is completely described by which terms each member uses, which is the most compact way of holding it in mind.

Please log in or register to report this question.

Answer: A

Local search holds one complete configuration and moves to a neighbouring one, so its memory is a single state whatever the size of the space. That is quite unlike the systematic methods, which build partial paths and remember frontiers. The contrast between searching among partial solutions and among complete ones is the deepest division in this subject and it determines both the memory profile and the guarantees available.

Please log in or register to report this question.

Answer: B

The neighbourhood is defined by the move operator, such as changing one variable or swapping two elements, and choosing it well is as important as choosing the algorithm. A larger neighbourhood escapes more local optima but costs more to evaluate. Similar values, graph adjacency in the systematic sense and visited states are different notions. The move set defines the landscape, so redesigning it is often the most effective response to a search that gets stuck.

Please log in or register to report this question.

Answer: C

Enlarging the move set flattens the landscape by connecting states that were previously far apart, so fewer configurations are locally optimal, but every step requires more evaluations. Compound moves and variable neighbourhood search exploit the trade deliberately. Smaller neighbourhoods are cheaper rather than better, the effect is large rather than absent, and termination depends on the acceptance rule. Reshaping the landscape by changing the moves is a lever unavailable in systematic search.

Please log in or register to report this question.

Answer: D

Local search evaluates each complete state by an objective and moves towards better values, so the objective plays the role that both the path cost and the heuristic play in systematic search. Constraint counts are one common choice of objective in satisfaction problems. Path cost is meaningless when no path is kept. Recognising that the objective does all the work, and that its shape is the landscape, focuses attention where the difficulty actually lies.

Please log in or register to report this question.

Loading…

Other Subjects

Tags: Informed and Heuristic Search Questions and Answers || Informed and Heuristic Search MCQ Questions and Answers || Informed and Heuristic Search GK Questions and Answers || Informed and Heuristic Search GK MCQ Questions || Informed and Heuristic Search Multiple Choice Questions and Answers || Informed and Heuristic Search GK || GK on Informed and Heuristic Search || Artificial Intelligence Questions and Answers || Artificial Intelligence MCQ Questions and Answers || Artificial Intelligence GK Questions and Answers || GK on Artificial Intelligence