Give asymptotic upper and lower bounds for the following recurrences with 
the assumption that T(n) is constant for n<=2.

a) T(n)= T(n/3)+n

sol: Assume that n=3^k for some k. Using iterated substitution we 
get:
  T(3^k)= T(3^{k-1}) + 3^k
             = T(3^{k-2}) + 3^{k-1} + 3^k
             ....
             = T(1) + \sum_{i=1}^k (3)^i
             = T(1) + (3^{k+1}-1)/2 - 1

Assuming that T(1)=c and Since k=log_3 n, we guess 
that T(n) = (3n-1)/2 + c - 1 
Now we prove this by induction (assuming that n=3^k again).
(Do the induction!). This implies that T(n) \in \Theta(n)


b) T(n)=2T(n/4)+\sqrt{n}

sol: Using iterative substituion and assuming that n=4^k for some
k we get:

T(4^k)=2*T(4^{k-1}) + \sqrt{4^k}
      =2*[2*T(4^{k-2} + \sqrt{4^{k-1}}] + \sqrt{4^k}
      =2^2 * T(4^{k-2}) + \sqrt{4^k} + \sqrt{4^k}
      ...
      = 2^k T(4^0) + k*\sqrt{4^k}

So with T(1)=c (for some constant c), since 2^k = \sqrt{n} we guess:
  T(n)=\sqrt{n}(k+c)

Now we prove this by induction. This implies 
T(n)\in \Theta(\sqrt{n}\log n)

c) T(n)=T(\sqrt{n})+1

Sol: Assume that n=2^k then we can write the recurrence as:

T(2^k)=T(2^{k/2})+1
      =T(2^{k/4})+1+1 = T(2^{k/2^2})+2
      =T(2^{k/8})+1+2 = T(2^{k/2^3})+3
      ....
      =T(2^{k/2^i})+i
      ...
Assume further that k=2^p. If we continue until i=p we get
T(2^k)=T(2^{k/2^p})+p

Since k/2^p=1 we have T(2^k)=T(2^1)+p = T(2)+p
Now we prove this by inudction on p,  i.e. prove T(2^{2^p})=T(2)+p
for p>=0. 
Base case: k=1 is easy
Ind. Step: assume it is true for k=2^p, i.e. T(2^{2^p})=T(2)+p
prove it for k=2^{p+1}: T(2^{2^{p+1}})=T(2^{2^p})+1 = T(2)+(p+1) as 
wanted.

Because T(2) is a constant T(2^{2^p})\in \Theta(p). 
Since n=2^k and k=2^p, thus T(n)\in \Theta(log log n).

d) T(n)=T(n-1)+lg n

(lg means log_2 here).

sol: we use iterated substitution:
 T(n)=T(n-1) + lg n 
     =T(n-2) + lg (n-1) + lg n
     =T(n-3) + lg (n-2) + lg (n-1) + lg n
     ....
     =T(1) + lg 2 + lg 3 + ... + lg n

We use induction to prove that T(n)=T(1) + \sum_{i=2}^n lg i
(prove this for the student).

Since T(1) is a constant and because lg i <= lg n for every i<=n, this 
implies that T(n) \in O(n lg n) = O(n log n)

On the other hand if we consider the last n/2 terms in the above summation
we have: T(n)>= sum_{i=n/2}^n lg i >= sum_{i=n/2}^n lg (n/2) = (n/2)lg(n/2)
= (n/2)(lg n - 1)= (n/2)lg n - (n/2). Since (n/2)lg n is the dominant term
T(n)\in \Omega (n lg n) = \Omega(n log n).