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Constraint Satisfaction Problems Questions and Answers

1.

A cryptarithmetic problem of the type SEND + MORE MONEY Can be solved efficiently using

Exam Year(s)

UGC NET - 2012,

Answer: C

A cryptarithmetic problem such as SEND + MORE = MONEY is solved efficiently by constraint satisfaction. Each letter must be given a different digit so that the sum is right, and the constraints on the digits and on the carries cut down the search at once. Depth first, breadth first and bidirectional search try assignments blindly and do not use the constraints in the same way, so they take much longer. The keyed answer is the constraint satisfaction technique.

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2.

The map colouring problem can be solved using which of the following technique ?

Exam Year(s)

UGC NET - 2013,

Answer: B

The map colouring problem can be solved by constraint satisfaction, which is option (B). The regions are the variables, the colours are their values, and the constraint is that two neighbouring regions must have different colours, so a solution is an assignment satisfying all the constraints.

Means-end analysis reduces the difference between a state and the goal, AO* is a search of AND-OR graphs, and breadth first search is a blind search that would not use the constraints. So (A), (C) and (D) are less suited to the problem.

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3.

Assume that each alphabet can have a value between 0 to 9 in a cryptoarithmetic problem

  CROSS
+ROADS
--------
DANGER

Which of the following statement is true ?

(i) No two alphabets can have the same numeric value.
(ii) Any two alphabets may have the same numeric value.
(iii) D = 0
(iv) D = 1

Exam Year(s)

UGC NET - 2013,

Answer: B

The statements true are (i) and (iv): no two letters can share a value, and D = 1. The sum CROSS + ROADS = DANGER has nine different letters (C, R, O, S, A, D, N, G, E), and in a cryptarithmetic puzzle each letter stands for a different digit, so statement (i) holds and (ii) fails. Adding two five-digit numbers gives at most a six-digit number whose leading digit is the carry, which can only be 1, so D = 1 and not 0.

One solution is 96233 + 62513 = 158746, in which D = 1. The options with (iii) or (ii) are therefore wrong.

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Answer: A

A constraint problem gives a set of variables, a domain of permitted values for each and a set of constraints restricting which combinations are allowed, and a solution is an assignment satisfying all of them. The classical search formulation is different, and rules and objectives belong to other frameworks. Adopting a representation in which the structure of a problem is visible, rather than hidden inside a successor function, is what makes the specialised algorithms possible.

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Answer: B

Because the state is described by variables, an algorithm can detect that a partial assignment already violates a constraint and discard every completion of it at once, which an atomic representation cannot express. The space is the same size, search is still required and the problems remain NP hard in general. Recognising that the gain comes from the representation rather than from any cleverness in the algorithm connects this topic directly to the earlier discussion of state representations.

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Answer: C

Drawing a node for each variable and an edge for each binary constraint exposes the structure of the problem, and that structure determines which algorithms will be efficient. A tree shaped graph is solvable in linear time, and a nearly tree shaped one is nearly as easy. Search trees, statistics and domain contents are different objects. Making the structure of a problem visible so that it can be exploited is the practical reason for drawing the graph at all.

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Answer: D

A global constraint such as all different relates any number of variables and has a specialised propagation algorithm far stronger than decomposing it into pairwise inequalities. It need not involve every variable, and it can in principle be decomposed, though at a heavy loss of propagation strength. Providing an efficient algorithm for a commonly occurring pattern, rather than expressing it in primitives, is what makes constraint programming systems practical.

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Answer: A

Three variables each with the same two valued domain are pairwise satisfiable but cannot all differ, and only a global algorithm using matching theory detects that immediately. Readability and storage are genuine but minor benefits. The example is the standard demonstration that decomposition loses information, and it justifies the whole apparatus of specialised global constraints in constraint programming systems.

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Answer: B

Backtracking chooses an unassigned variable, tries each value consistent with the assignment so far and recurses, returning when no value works. It is depth first search specialised to this representation, and its memory is linear. Backward search, restarts and recording failures describe other techniques, the last being no-good learning. A basic algorithm that is transformed by the heuristics layered on it is the pattern of this whole topic.

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Answer: C

Choosing the most constrained variable first means failure is detected as early as possible, pruning the tree near the root where the saving is greatest. It is often called the fail first heuristic for that reason. Constraint counts are the degree heuristic, value selection is a separate decision, and recency is not used. Ordering so that failure comes early, rather than late, is counter-intuitive until one notices that the cost of failure grows with the depth at which it occurs.

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