Master Theorem for homework outsourcing
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John is looking to outsource his homework. If he gets homework with one task he solves it in a constant number of minutes. Every bigger homework ($2$ or more tasks) he will split among $dgeq 3$ of his colleagues and asks each one of them to solve approximately one third of the tasks. To split $n$ tasks among colleagues it takes him $theta(n^2)$ minutes. Let's assume that each student has the same strategy as John. We need to asymptotically approximate the number of time spent in minutes needed to solve homework with $n$ tasks depending on the natural parameter $dgeq 3$.
I believe this should be solved with Master Theorem. The equation would be as following: $t(n) = d cdot (n/3) + theta(n^2)$. I think we can therefore split it into three cases:
a) $d < 9$ where the solution would be: $theta(n^{log_3(d)})$
b) $d > 9$ with solution $theta(n^2)$
c) $d = 9$ with solution $theta(n^{log3(d)(log(n)})$.
Is this correct?
discrete-mathematics asymptotics
edited Jan 18 at 13:52
idriskameni
641319
asked Jan 18 at 12:44
ponikoliponikoli
416
$endgroup$
add a comment |
$begingroup$
John is looking to outsource his homework. If he gets homework with one task he solves it in a constant number of minutes. Every bigger homework ($2$ or more tasks) he will split among $dgeq 3$ of his colleagues and asks each one of them to solve approximately one third of the tasks. To split $n$ tasks among colleagues it takes him $theta(n^2)$ minutes. Let's assume that each student has the same strategy as John. We need to asymptotically approximate the number of time spent in minutes needed to solve homework with $n$ tasks depending on the natural parameter $dgeq 3$.
I believe this should be solved with Master Theorem. The equation would be as following: $t(n) = d cdot (n/3) + theta(n^2)$. I think we can therefore split it into three cases:
a) $d < 9$ where the solution would be: $theta(n^{log_3(d)})$
b) $d > 9$ with solution $theta(n^2)$
c) $d = 9$ with solution $theta(n^{log3(d)(log(n)})$.
Is this correct?
discrete-mathematics asymptotics
edited Jan 18 at 13:52
idriskameni
641319
asked Jan 18 at 12:44
ponikoliponikoli
416
$endgroup$
$begingroup$
It's directly n**2 according to your recurrence
$endgroup$
– Aditya
Jan 18 at 14:05
add a comment |
$begingroup$
John is looking to outsource his homework. If he gets homework with one task he solves it in a constant number of minutes. Every bigger homework ($2$ or more tasks) he will split among $dgeq 3$ of his colleagues and asks each one of them to solve approximately one third of the tasks. To split $n$ tasks among colleagues it takes him $theta(n^2)$ minutes. Let's assume that each student has the same strategy as John. We need to asymptotically approximate the number of time spent in minutes needed to solve homework with $n$ tasks depending on the natural parameter $dgeq 3$.
I believe this should be solved with Master Theorem. The equation would be as following: $t(n) = d cdot (n/3) + theta(n^2)$. I think we can therefore split it into three cases:
a) $d < 9$ where the solution would be: $theta(n^{log_3(d)})$
b) $d > 9$ with solution $theta(n^2)$
c) $d = 9$ with solution $theta(n^{log3(d)(log(n)})$.
Is this correct?
discrete-mathematics asymptotics
edited Jan 18 at 13:52
idriskameni
641319
asked Jan 18 at 12:44
ponikoliponikoli
416
$endgroup$
John is looking to outsource his homework. If he gets homework with one task he solves it in a constant number of minutes. Every bigger homework ($2$ or more tasks) he will split among $dgeq 3$ of his colleagues and asks each one of them to solve approximately one third of the tasks. To split $n$ tasks among colleagues it takes him $theta(n^2)$ minutes. Let's assume that each student has the same strategy as John. We need to asymptotically approximate the number of time spent in minutes needed to solve homework with $n$ tasks depending on the natural parameter $dgeq 3$.
I believe this should be solved with Master Theorem. The equation would be as following: $t(n) = d cdot (n/3) + theta(n^2)$. I think we can therefore split it into three cases:
a) $d < 9$ where the solution would be: $theta(n^{log_3(d)})$
b) $d > 9$ with solution $theta(n^2)$
c) $d = 9$ with solution $theta(n^{log3(d)(log(n)})$.
Is this correct?
discrete-mathematics asymptotics
discrete-mathematics asymptotics
edited Jan 18 at 13:52
idriskameni
641319
asked Jan 18 at 12:44
ponikoliponikoli
416
edited Jan 18 at 13:52
idriskameni
641319
asked Jan 18 at 12:44
ponikoliponikoli
416
edited Jan 18 at 13:52
idriskameni
641319
edited Jan 18 at 13:52
idriskameni
641319
edited Jan 18 at 13:52
idriskameni
641319
641319
asked Jan 18 at 12:44
ponikoliponikoli
416
asked Jan 18 at 12:44
ponikoliponikoli
416
asked Jan 18 at 12:44
ponikoliponikoli
416
416
$begingroup$
It's directly n**2 according to your recurrence
$endgroup$
– Aditya
Jan 18 at 14:05
add a comment |
$begingroup$
It's directly n**2 according to your recurrence
$endgroup$
– Aditya
Jan 18 at 14:05
$begingroup$
It's directly n**2 according to your recurrence
$endgroup$
– Aditya
Jan 18 at 14:05
$begingroup$
It's directly n**2 according to your recurrence
$endgroup$
– Aditya
Jan 18 at 14:05
add a comment |
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$begingroup$
It's directly n**2 according to your recurrence
$endgroup$
– Aditya
Jan 18 at 14:05