In CLRS book, in the road cutting example there is a recursion formula
$$ 1+\sum_{j=0}^{n-1}T(j) $$ and it can be proved that the sum is $$ 2^n $$
by simple induction. In 3-rd edition it is a formula 15.3 on a page 364.
As I can understand from the recursion-tree aproach, the recursion tree which can be built for the said problem is $$ 1->(n-1)->(n-2)|each.node->(n-3)|each.node->... $$
and so on, which can be summed as $$ \sum_{j=1}^{n-1}[\Pi_{k=1}^{j}[(n-k)]] $$
How can I come up with the suggestion of $2^n$ from this recursion tree (or why this tree is wrong and what can be the right approach to suggesting $2^n$ running time)?
Asked By : user19668
Answered By : Rick Decker
Minor nit: your first displayed expression should be $$ T(n) = \begin{cases} 1 & \text{if $n=0$,}\\ 1 + \sum_{j=0}^{n-1}T(j) & \text{otherwise} \end{cases} $$ You can indeed solve this by looking at the recurrence tree, but you don't wind up with your sum of products expression (try it for $n=4$ to see that). For this problem there's an easier way, though:
Hint: What is $T(n+1)-T(n)$?
You can take a look at our reference question, on recurrence relations, as well.
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Question Source : http://cs.stackexchange.com/questions/37332
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