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If I have a recurrence equation: $$T(n) = 3T\big(\frac{n}{4}\big) + cn^2$$

It says it splits a problem of size n into 3 subproblems of size n/4. Then it keeps splitting it until the problem size at the leaves is 1 or less. The first part that does not make sense to me is how can you split a problem of size n into 3 problems of size n/4. To me it seems that there are now 3 problems of size n/4, but where is the remaining 1/4 of n? Who is solving that?

Also, the number of leaves is $$n^{\big(\frac{loga}{logb}\big)}$$ Assuming n=1000, it can be calculated that there are 235 leaves. So 235 problems of size 1 are solved, how are the rest of 765 solved?

I have a feeling that I have misunderstood some principle, but I do not know what exactly.

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Here is a very simple example: binary search.

We are looking for an element $x$ inside a sorted array $A$. We compare $x$ to the element at the middle $m$ of $A$. If $m = x$, we're done. If $m > x$, we recurse on the left half, and if $m > x$, we recurse on the right half. The recurrence for the worst-case running time is $$ T(n) = T(n/2) + O(1), $$ whose solution is $T(n) = O(\log n)$.

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  • $\begingroup$ This is great! Thanks, I get the point now! Is it possible to give an example where the problem is split into more parts?(But I accept your answer either way) $\endgroup$ Jan 23 at 16:27
  • $\begingroup$ Linear time selection comes to mind: en.wikipedia.org/wiki/… $\endgroup$ Jan 23 at 16:30

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