Inference and Resolution Questions and Answers

Q: What is the practical response to a family of formulas a solver cannot handle?

Adding a symmetry breaking constraint, or using a system such as cutting planes that has short proofs where resolution does not, changes the problem the solver faces. Waiting is hopeless against an exponential, other methods may succeed, and shrinking the instance avoids the question. Changing the formulation when the method is provably inadequate, rather than persisting with it, is the same lesson that reformulation teaches in ordinary search.

Q: How does symmetry breaking help a satisfiability solver?

Where permuting objects gives an equivalent assignment, the solver would otherwise explore every permutation, and an ordering constraint keeps only one representative. Pigeonhole formulas are the classic case. Variable counts rise slightly, the clause structure is unchanged, and satisfiability is preserved rather than created. Recognising that a search is repeating equivalent work, and forbidding the repetitions, is a saving available in every symmetric problem.

101.

What is the practical response to a family of formulas a solver cannot handle?

A.Encode the problem differently or use a stronger proof system
B.Increase the time limit until the solver eventually succeeds
C.Accept that the problem cannot be solved by any means
D.Reduce the size of the instance until it becomes tractable

Answer: A

Adding a symmetry breaking constraint, or using a system such as cutting planes that has short proofs where resolution does not, changes the problem the solver faces. Waiting is hopeless against an exponential, other methods may succeed, and shrinking the instance avoids the question. Changing the formulation when the method is provably inadequate, rather than persisting with it, is the same lesson that reformulation teaches in ordinary search.

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102.

How does symmetry breaking help a satisfiability solver?

A.It guarantees that the formula becomes satisfiable
B.It prevents equivalent assignments being explored repeatedly
C.It reduces the number of variables in the encoded problem
D.It converts the formula into a Horn clause set

Answer: B

Where permuting objects gives an equivalent assignment, the solver would otherwise explore every permutation, and an ordering constraint keeps only one representative. Pigeonhole formulas are the classic case. Variable counts rise slightly, the clause structure is unchanged, and satisfiability is preserved rather than created. Recognising that a search is repeating equivalent work, and forbidding the repetitions, is a saving available in every symmetric problem.

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103.

Match the following control strategies of prolog :

List - I	List - II
a. Forward movement	i. Variable can be done with a constant, another variable or a function.
b. Unification	ii. The entire conjunctive goal is executed.
c. Deep backtracking	iii. Previous sub goal to find alternative solutions.
d. Shallow backtracking	iv. Chooses sub goal with possible unifier.

Codes : a b c d

A.iv i ii iii
B.ii iv i iii
C.iii i iv ii
D.ii iii iv i

Exam Year(s)

UGC NET - 2013,

Answer: A

The key pairs a-iv, b-i, c-ii, d-iii. Forward movement chooses a sub goal with a possible unifier (iv). Unification lets a variable be bound to a constant, another variable or a function (i). Deep backtracking is paired with the entire conjunctive goal (ii), and shallow backtracking with returning to the previous sub goal to find alternative solutions (iii).

The other codes assign the unifier choice or the variable binding to different strategies, so they disagree with the key.

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104.

The objective of ________ procedure is to discover at least one ________ that causes two literals to match.

A.unification, validation
B.unification, substitution
C.substitution, unification
D.minimax, maximum

Exam Year(s)

UGC NET - 2013,

Answer: B

The objective of the unification procedure is to discover at least one substitution that causes two literals to match. Unification finds values for the variables, a substitution, that make the two expressions identical.

Validation is not the output of unification. Placing substitution first and unification second reverses the roles, and minimax and maximum belong to game-playing search, not to matching literals.

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105.

If we convert ∃u ∀v ∀x ∃y (P(f(u),v, x, y) → Q(u,v,y)) to ∀v ∀x (P(f(a),v, x, g(v,x)) → Q(a,v,g(v,x))) This process is known as

A.Simplification
B.Unification
C.Skolemization
D.Resolution

Exam Year(s)

UGC NET - 2013,

Answer: C

The process is known as Skolemization. The existentially quantified variable u is replaced by the constant a, and y, which depends on the universally quantified v and x, is replaced by a function g(v, x), after which the existential quantifiers are removed. The answer is (C).

Simplification would reduce an expression by algebra, unification finds a substitution that makes two literals identical, and resolution is the inference rule that is applied to clauses afterwards. None of these removes an existential quantifier.

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106.

The tracing model in Prolog describes program execution in terms of certain events. These events are

A.call and exit
B.call and fail
C.call, exit and redo
D.call, exit, redo and fail

Exam Year(s)

UGC NET - 2013,

Answer: D

The tracing model in Prolog describes execution in terms of four events: call, exit, redo and fail. A call is made when a goal is first entered, exit when it succeeds, redo when backtracking returns to it, and fail when it finally fails.

The options that list two or three of these events are incomplete, since all four make up the model.

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