Solving a recurrence relation for a recursive function - algorithm

Question

I need to create and solve a recurrence relation for the worst-case analysis for the following psuedocode. I am counting the number additions (not including the for loop counter) as my basic operation.
I am assuming n=2^k.
Here is the progress I have made...
Base Case:
T(n<=4) = 1
W(n)=W(2^k)=additions to calculate answer+additions in next recursion+addition in for loop
W(2^k) = 2 + W(2^(k-2)) + (2^k) - 2 = W(2^(k-2)) + (2^k)
I use back substitution and get the following recurrence relation...
for the jth recursive call
W(2^k) = W(2^(k-2j)) + (2^k) + sum(t=1,j,2^(k-2(t-1)))
I know that I can simplify this because I take W(2^(k-2j)) = W(4) and solve for j to see how many recursive steps the code takes.
In this case, j=(k/2) - 1. Reducing the recurrence gives me...
W(2^k) = 1 + (2^k) + sum(t=1,j,2^(k-2(t-1))).
Reducing the summation gives me...
W(2^k) = 1 + (2^k) + (2^k)*(2^2)*sum(t=1,j,2^(-2t)) or
W(n) = 1 + n + 4n*sum(t=1,j,2^(-2t))
What I cannot simplify is the summation. In lectures, we may have a summation of sum(i=1,n,2^i), which would be 2^(n+1)-1, but this one is different.
int function calc(int n) {
int num,answer;
if(n<=4) {return n+10;}
else {
num=calc(n/4);
answer=(num+num+10);
for(int i=2;i<=n-1;i++) {
answer=answer+answer;
}
return answer;
}
}
Any help would be appreciated. This assignment is due tonight. Thanks

Answer 1

The time complexity of the problem is T(n) = T(n/4) + n. The term n could mean \Theta(n). Hence, T(n) = n + n/4 + n/4^2 + ... + n/(4^log_4(n)) = n(1 + 1/4 + ... + 1/n) = \Theta(n). Notice that lim_{n\to \infty} 1 + 1/4 + ... + 1/4^log_4(n) = 4/3 which is a constant number.

Answer 2

T(n) = T(2^k) // by assumption
T(n) = T(n/4) + n // is the recurrence relation
T(2^k) = T(2^(k-2)) + (2^k) // rewriting
T(2^k) = T(2^(k-2j)) + (2^k)*SUM(i=0;j-1;(1/4)^i) // is the relation for iteration j
T(4) = T(2^(k-2j)) = 1 // when the recursion ends, base case reached, only 1 addition
2^2 = 2^(k-2j) // rewrite and remove T
2=k-2j // remove common base
j=(k/2)-1 // solve for jth iteration
Notice: cannot have decimal for iteration, so j=CEILING((k/2)-1)
SUM(i=0;j-1;(1/4)^i) = (1-(1/4)^j)/(1-(1/4))
For geometric sums and proof of transformation above, see wiki link below
Substituting j with CEILING((k/2)-1), the summation is...
SUM = (1-(1/4)^(CEILING((k/2)-1)))/(1-(1/4))
Finally, T(2^k) = 1 + (2^k)*SUM
In terms of the input n, T(n) = 1 + (n*SUM)
This formula is good for all k>=1
Testing several values...
k=1, T(2) = 1 + (2*0) = 1 // we know this is true because T(2) is a base case
k=2, T(4) = 1 + (4*0) = 1 // we know this is true because T(4) is a base case
k=3, T(8) = 1 + (8*1) = 9 // by recurrence formula, T(8) = T(2) + 8 = 9, true
k=4, T(16) = 1 + (16*1) = 17 // by recurrence formula, T(16) = T(4) + 16 = 17, true
k=5, T(32) = 1 + (32*1.25) = 41 // by recurrence formula, T(32) = T(8) + 32, T(8) = 9, 32+9=41 true
For geometric sums
https://en.wikipedia.org/wiki/Geometric_series