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Given the following recurrence relation:

$t(1) = 1$

$t(n) = t(\lfloor \frac{n}{2} \rfloor) + t(\lceil \frac{n}{2} \rceil) + n$

How would a proof for the solution $t(n) = n (\lfloor \log n \rfloor + 3) - 2^{\lfloor \log n \rfloor + 1}$ look?

I suppose this should be proven by induction.

The base case would be $t(1) = 1 = 3 - 2 = 1 \cdot (\log 1 + 3) - 2^{\log 1 + 1}$

Next consider two cases. The case $n$ is even and $n$ is odd...

Is that correct? And if yes, how would you choose $n$? $n=2k$ and $n=2k+1$?

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Hint:

If you take the $2k$ and $2k+1$ approach, your recurrence becomes $$t(2k)=2t(k)+2k$$ $$t(2k+1)=t(k)+t(k+1)+2k+1$$

while your inductive hypothesis becomes $$t(2k)=2k(\lfloor\log k\rfloor+4)-2^{\lfloor\log k\rfloor +2}$$ $$t(2k+1)=(2k+1)(\lfloor\log k\rfloor+4)-2^{\lfloor\log k\rfloor +2}$$

and the inductive step looks reasonably simple, with some care needed when $k$ is one less than a power of two. I might start by checking the cases when $k=1$, i.e. $n=2,3$

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  • $\begingroup$ That's plausible thanks. In the end i want $t(2k) = 2k(\lfloor \log 2k \rfloor + 3)-2^{\lfloor \log 2k \rfloor + 1}$ for an even $n$, don't i? $\endgroup$ – upe Nov 17 '18 at 13:49
  • $\begingroup$ @Peter: yes, though for positive real $x$ you have $\log_2(2x)=\log_2(x)+1$ and so for positive integer $k$ you have $\lfloor \log_2(2k)\rfloor=\lfloor \log_2(2k+1)\rfloor = \lfloor \log_2(k)\rfloor +1$ $\endgroup$ – Henry Nov 17 '18 at 20:29

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