recurrence

I am trying to find the big O bound for the following recurrence relation: T(n) = T(n-1) + n^c, where c >= 1 is a constant So I've decided to solve this by using iteration: T(n) = T(n-1) + n^c T(n-1) = T(n-2) + (n-1)^c T(n) = T(n-2) + n^c + (n-1)^c T(n-2) = T(n-3) + (n-2)^c T(n) = T(n-3) + n^c + (n-1)^c + (n-2)^c T(n) = T(n-k) + n^c + (n-1)^c + .... + (n-k+1)^c Suppose k = n-1, then: T(n) = T(1) + n^c + (n-1)^c + (n-n+1+1)^c T(n) = n^c + (n-1)^c + 2^c + 1 I'm not sure if this is correct however, plus I would really appreciate some guidance as to how to derive Big O from this. Thanks a lot! Yes

The question is : UNBALANCED MERGE SORT is a sorting algorithm, which is a modified version of the standard MERGE SORT algorithm. The only difference is that instead of dividing the input into 2 equal parts in each stage, we divide it into two unequal parts – the first 2/5 of the input, and the other 3/5. a. Write the recurrence relation for the worst case time complexity of the UNBALANCED MERGE SORT algorithm. b. What is the worst case time complexity of the UNBALANCEDMERGESORT algorithm? Solve the recurrence relation from the previous section. So i'm thinkin the recurrence relation is : T(n)

So I am pretty sure it is O(n) (but it might not be?), but how do you solve it with substitution? If you assume T(n) <= c * n, what is the induction steps? First of all,I'd like to assume clearly that Θ(1)=k,some constant. Next,proceeding using substitution method, we get T(n)=2T(n/2)+Θ(1) =2T(n/2)+k =2{2T(n/4)+k)+k =4T(n/4)+3k =... =n.T(1)+(n-1)k =n.k+(n-1)k =2nk-k =O(n). If you assume it as T(n) <= c * n ,you should start with T(1) and assume for T(n) to be correct,and then proceed to show that it is correct for T(n+1) by using the assumption for T(n). BTW,your assumption was right! For

I am solving this recurrence using a recursion tree. The total cost of each level is n, and the depth of the tree is between log (n) base 4 and log (n) base 4/3 . Intuitively, I expect the solution to be at most the number of levels times the cost at each level. O(cn log (n) base 4/3) = O(n log n) . I was wondering if my approach towards the problem, and my solution is correct? Think of it this way: for the first log 4 n layers of the recursion tree, the sum of the work across those layers will be cn, because if you sum up the total sizes of all the subproblems, it should total n so the total

Consider this example : T(n) = T(7n/8) + 2n I assumed T(1) = 0 and tried to solve it in the following way T(n) = T(7n/8) + 2n = T(49n/64) + 2.(7n/8) + 2n = T(343n/512) + 2.(7n/8).(7n/8)+ 2.(7n/8) + 2n = T(1) + 2n ( (7n/8)^i + ..... + 1) but I could not come to any conclusion about this. I am confused about what should I do in the next step. Your approach is sound, but you'll see what to do if you rewrite it slightly differently: T(n) = T((7/8)^1 * n) + 2 * (7/8)^0 * n = T((7/8)^2 * n) + 2 * (7/8)^1 * n + 2 * (7/8)^0 * n = T((7/8)^3 * n) + 2 * (7/8)^2 * n + 2 * (7/8)^1 * n + 2 * (7/8)^0 * n . .

I got stuck in this recurrence: T(n) = T(n − 1) + lg(1 + 1/n), T(1) = 1? for a while and it seems the master method cannot be applied on this one. We have: lg(1 + 1/n) = lg((n + 1) / n) = lg(n+1) - lg(n) Hence: T(n) - T(n - 1) = lg(n + 1) - lg(n) T(n-1) - T(n - 2) = lg(n) - lg(n - 1) ... T(3) - T(2) = lg(3) - lg(2) T(2) - T(1) = lg(2) - lg(1) Adding and eliminating, we get: T(n) - T(1) = lg(n + 1) - lg(1) = lg(n + 1) or T(n) = 1 + lg(n + 1) Hence T(n) = O(lg(n)) Same answer as the other correct answer here, just proved differently. All the following equations are created from the given