Data Structures and Algorithms | Set 20

Following questions have asked in GATE CS 2006 exam. 1. Let S be an NP-complete problem and Q and R be two other problems not known to be in NP. Q is polynomial time reducible to S and S is polynomial-time reducible to R. Which one of the following statements is true? (A) R is NP-complete (B) R is NP-hard (C) Q is NP-complete (D) Q is NP-hard Answer (B) (A) Incorrect because R is not in NP. A NP Complete problem has to be in both NP and NP-hard. (B) Correct because a NP Complete problem S is polynomial time educable to R. (C) Incorrect because Q is not in NP. (D) Incorrect because there is no NP-complete problem that is polynomial time Turing-reducible to Q.
2) A set X can be represented by an array x[n] as follows: Consider the following algorithm in which x,y and z are Boolean arrays of size n: C

algorithm zzz(x[] , y[], z [])
{
   int i;
   for (i=O; i<n; ++i)
     z[i] = (x[i] ^ ~y[i]) V (~x[i] ^ y[i])
}

The set Z computed by the algorithm is: (A) (X Intersection Y) (B) (X Union Y) (C) (X-Y) Intersection (Y-X) (D) (X-Y) Union (Y-X) Answer (D) The expression x[i] ^ ~y[i]) results the only 1s in x where corresponding entry in y is 0. An array with these set bits represents set X - Y The expression ~x[i] ^ y[i]) results the only 1s in y where corresponding entry in x is 0. An array with these set bits represents set Y - X. The operator "V" results in Union of the above two sets.
3. Consider the following recurrence: Which one of the following is true? (A) T(n) = Θ(loglogn) (B) T(n) = Θ(logn) (C) T(n) = Θ(sqrt(n)) (D) T(n) = Θ(n) Answer (B)

  Let n = 2^m
  T(2^m) = T(2^(m/2)) + 1
  Let T(2^m) =  S(m)
  S(m)  = 2S(m/2) + 1  

Above expression is a binary tree traversal recursion whose time complexity is Θ(m). You can also prove using Master theorem.

S(m)  = Θ(m)  
      = Θ(logn)  /* Since n = 2^m */

Now, let us go back to the original recursive function T(n)

  T(n)  = T(2^m) = S(m)
                 = Θ(Logn)

Please note that the solution of T(n) = T(√n) + 1 is Θ(Log Log n), above recurrence is different, it is T(n) = 2T(√n) + 1
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