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While learning Median of Medians algorithm i came across the following lemma ;

"For any recurrence of the form $T(n)<=T(an)+T(bn)+O(n) $,
if $(a+b)<1$ the reccurence will solve to $O(n)$"

(Which means $T(n)$ is bounded by $O(n)$ for some constant $ $ for any $ n>n_0$ )

So i was trying to prove it (since i could not find any proof) using Substitution Method :
(there might be a proof using Master's theroem but i didnt know how to use it here , and i would also like to see a proof using Substitution Method )

So i tried to prove by induction .
We can assume that for for any $n<N$ , $T(n)<=k_2*n $ for some constant $k_2$.
Assuming the statement is correct from $min(a*N ,b*N)$ since $(a+b)<1$

(i dont feel like the previous sentence is entirley correct , maybe i should assume that for any number smaller than n ? Plus , I think I have to assume that they are positive here ? )

$a*N$ and $b*N$ are less than N and thus , we can use the induction hypothesis ;

$$T(N)<=k_2aN+k_2bN+xN $$ (since were given $O(n)$ i assumed there exists a constant x for which the given function is bounded by)
We want $$k_2aN+k_2bN+xN<=k_2N$$ => $$k_2a+k_2b+x<=k_2$$
since $a+b<1$ then i can continue but i can't see how to finish the proof . I think i have to assume by contradiction that $T(n)$ is not bounded by $O(n)$ but im not entirely sure what my assumptions are . Trying to do so i stated that for any constant $$T(n)<=T(an)+T(bn)+O(n)>O(n) $$ but that dont seem like the right way..

I would appreciate any help , thanks.

Tomer.Ov
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  • Sorry, but i dont understand how this is a duplicate . i did read it though, but i could not answer my own question. I do see that this might fall in the case of Akra-Bazzi (which i did not learn so ill have to read it again) , but as i stated i would like to use the Substitution Method here- if possible of course. I will try to prove by Akra-Bazzi though. – Tomer.Ov Feb 21 '19 at 16:45
  • @Tomer.Ov, this question is covered by this answer to the question mentioned by David. The induction used over there is just "Substitution Method" in disguise. Does it make sense? – John L. Feb 21 '19 at 16:54
  • @Apass.Jack Yes ! this looks just like my solution. can you please explain how can i assume that b=c(1−(∑ai)) ? in my case that means that i have to assume that x=k2/(1-a-b) ? how do i get to this statement in my proof ? – Tomer.Ov Feb 21 '19 at 17:38
  • To follow The Uneven Split Theorem Part 1 of that answer, you let $c$ be large enough and let its $b = \frac{c}{1 - \left(a+(\text{your }b)\right)}$. – John L. Feb 22 '19 at 00:02
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    @Apass.Jack Yes , after some further thought i realized that when i continue to solve my equation i can just see that $x$ is less than $(1-a-b)k_2 $, so i can just substitute $ x$ with $(1-a-b)k_2$ and indeed the given equation would end up being smaller than $k_2n$ , which means$ T(n)<=k_2n$ which is what i wanted to proove . Thank you (and David) very much ! – Tomer.Ov Feb 22 '19 at 18:57

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