solving a recurrence equation $T(n) = T(cn) + T((1-c)n) + 1$ The 2019 Stack Overflow Developer Survey Results Are In$T(n)=T(cn) + T((1-c)n)+1$ while $0<c<1$Recurrence - Master Theorem - Asymptotic Questionrecurrence relation arising from Magic the Gathering scenarioSolving this recurrence relationWhy do we set $u_n=r^n$ to solve recurrence relations?Optimizing an asymptotic recurrence relation with two recursive termsSolving the recurrence $T(n) = sqrt n T(sqrtn) + sqrtn$max-weighted sums of multinomial coefficentFind a solution for the equation $a(4n) = a(n)over 4$Linear recurrence solving for tighest possible big O boundsSolving a recurrence relation with a definite integral?

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solving a recurrence equation $T(n) = T(cn) + T((1-c)n) + 1$



The 2019 Stack Overflow Developer Survey Results Are In$T(n)=T(cn) + T((1-c)n)+1$ while $0<c<1$Recurrence - Master Theorem - Asymptotic Questionrecurrence relation arising from Magic the Gathering scenarioSolving this recurrence relationWhy do we set $u_n=r^n$ to solve recurrence relations?Optimizing an asymptotic recurrence relation with two recursive termsSolving the recurrence $T(n) = sqrt n T(sqrtn) + sqrtn$max-weighted sums of multinomial coefficentFind a solution for the equation $a(4n) = a(n)over 4$Linear recurrence solving for tighest possible big O boundsSolving a recurrence relation with a definite integral?










1












$begingroup$


first, this question has been asked here before, however I have some issues with it that haven't been answered yet :



give asymptotic bounds of $ T(n) $ where




$ T(n) = T(cn)+ T((1-c)n) +1 , 0<c<1 $




this is the original post link
I didn't quite understand why the answer given over there was with a linear solution $S(n) = nS(1)$



and another thing, I've been told by someone that an even tighter bound is $theta(log(n))$ is possible.



so two things, is this true, and why is the linear solution is of the form given before.



thanks ahead.










share|cite|improve this question











$endgroup$











  • $begingroup$
    The title has $+1$ on the rhs, the body has $+n$. The Akra–Bazzi method is applicable to both cases, giving $p = 1$.
    $endgroup$
    – Maxim
    Apr 1 at 13:24










  • $begingroup$
    It’s suppose to be +1
    $endgroup$
    – lidor718
    Apr 1 at 13:45















1












$begingroup$


first, this question has been asked here before, however I have some issues with it that haven't been answered yet :



give asymptotic bounds of $ T(n) $ where




$ T(n) = T(cn)+ T((1-c)n) +1 , 0<c<1 $




this is the original post link
I didn't quite understand why the answer given over there was with a linear solution $S(n) = nS(1)$



and another thing, I've been told by someone that an even tighter bound is $theta(log(n))$ is possible.



so two things, is this true, and why is the linear solution is of the form given before.



thanks ahead.










share|cite|improve this question











$endgroup$











  • $begingroup$
    The title has $+1$ on the rhs, the body has $+n$. The Akra–Bazzi method is applicable to both cases, giving $p = 1$.
    $endgroup$
    – Maxim
    Apr 1 at 13:24










  • $begingroup$
    It’s suppose to be +1
    $endgroup$
    – lidor718
    Apr 1 at 13:45













1












1








1


1



$begingroup$


first, this question has been asked here before, however I have some issues with it that haven't been answered yet :



give asymptotic bounds of $ T(n) $ where




$ T(n) = T(cn)+ T((1-c)n) +1 , 0<c<1 $




this is the original post link
I didn't quite understand why the answer given over there was with a linear solution $S(n) = nS(1)$



and another thing, I've been told by someone that an even tighter bound is $theta(log(n))$ is possible.



so two things, is this true, and why is the linear solution is of the form given before.



thanks ahead.










share|cite|improve this question











$endgroup$




first, this question has been asked here before, however I have some issues with it that haven't been answered yet :



give asymptotic bounds of $ T(n) $ where




$ T(n) = T(cn)+ T((1-c)n) +1 , 0<c<1 $




this is the original post link
I didn't quite understand why the answer given over there was with a linear solution $S(n) = nS(1)$



and another thing, I've been told by someone that an even tighter bound is $theta(log(n))$ is possible.



so two things, is this true, and why is the linear solution is of the form given before.



thanks ahead.







recurrence-relations asymptotics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 1 at 14:48









user

6,48311031




6,48311031










asked Mar 30 at 17:27









lidor718lidor718

175




175











  • $begingroup$
    The title has $+1$ on the rhs, the body has $+n$. The Akra–Bazzi method is applicable to both cases, giving $p = 1$.
    $endgroup$
    – Maxim
    Apr 1 at 13:24










  • $begingroup$
    It’s suppose to be +1
    $endgroup$
    – lidor718
    Apr 1 at 13:45
















  • $begingroup$
    The title has $+1$ on the rhs, the body has $+n$. The Akra–Bazzi method is applicable to both cases, giving $p = 1$.
    $endgroup$
    – Maxim
    Apr 1 at 13:24










  • $begingroup$
    It’s suppose to be +1
    $endgroup$
    – lidor718
    Apr 1 at 13:45















$begingroup$
The title has $+1$ on the rhs, the body has $+n$. The Akra–Bazzi method is applicable to both cases, giving $p = 1$.
$endgroup$
– Maxim
Apr 1 at 13:24




$begingroup$
The title has $+1$ on the rhs, the body has $+n$. The Akra–Bazzi method is applicable to both cases, giving $p = 1$.
$endgroup$
– Maxim
Apr 1 at 13:24












$begingroup$
It’s suppose to be +1
$endgroup$
– lidor718
Apr 1 at 13:45




$begingroup$
It’s suppose to be +1
$endgroup$
– lidor718
Apr 1 at 13:45










0






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