Question

Determine the positive number c & n0 for the following recurrences (Using Substitution Method):

1. T(n) = T(ceiling(n/2)) + 1 … Guess is Big-Oh(log base 2 of n)

2. T(n) = 3T(floor(n/3)) + n … Guess is Big-Omega (n * log base 3 of n)

3. T(n) = 2T(floor(n/2) + 17) + n … Guess is Big-Oh(n * log base 2 of n).

I am giving my Solution for Problem 1:

Our Guess is: T(n) = O (log_2(n)).
By Induction Hypothesis assume T(k) <= c * log_2(k) for all k < n,here c is a const & c > 0

     T(n) = T(ceiling(n/2)) + 1
<=>  T(n) <= c*log_2(ceiling(n/2)) + 1
<=>   "   <= c*{log_2(n/2) + 1} + 1
<=>   "    = c*log_2(n/2) + c + 1
<=>   "    = c*{log_2(n) - log_2(2)} + c + 1
<=>   "    = c*log_2(n) - c + c + 1
<=>   "    = c*log_2(n) + 1
<=>   T(n) not_<= c*log_2(n)      because c*log_2(n) + 1 not_<= c*log_2(n).

To solve this remedy used a trick a follows:

    T(n) = T(ceiling(n/2)) + 1
<=>  "  <= c*log(ceiling(n/2)) + 1 
<=>  "  <= c*{log_2 (n/2) + b} + 1             where  0 <= b < 1
<=>  "  <= c*{log_2 (n) - log_2(2) + b) + 1  
<=>  "   = c*{log_2(n) - 1 + b} + 1
<=>  "   = c*log_2(n) - c + bc + 1
<=>  "   = c*log_2(n) - (c - bc - 1)      if c - bc -1 >= 0  
                                             c >= 1 / (1 - b)
<=> T(n) <= c*log_2(n) for c >= {1 / (1 - b)}

so T(n) = O(log_2(n)).

This solution is seems to be correct to me ... My Ques is: Is it the proper approach to   do?

Thanks to all of U.

Answer 1

For the first exercise:
We want to show by induction that T(n) <= ceiling(log(n)) + 1.

Let’s assume that T(1) = 1, than T(1) = 1 <= ceiling(log(1)) + 1 = 1 and the base of the induction is proved.

Now, we assume that for every 1 <= i < nhold that T(i) <= ceiling(log(i)) + 1.

For the inductive step we have to distinguish the cases when n is even and when is odd.

If n is even: T(n) = T(ceiling(n/2)) + 1 = T(n/2) + 1 <= ceiling(log(n/2)) + 1 + 1 = ceiling(log(n) - 1) + 1 + 1 = ceiling(log(n)) + 1.

If n is odd: T(n) = T(ceiling(n/2)) + 1 = T((n+1)/2) + 1 <= ceiling(log((n+1)/2)) + 1 + 1 = ceiling(log(n+1) - 1) + 1 + 1 = ceiling(log(n+1)) + 1 = ceiling(log(n)) + 1

The last passage is tricky, but is possibile because n is odd and then it cannot be a power of 2.