Compare two algorithms with each other
I am fairly new algorithms and the math behind it.
I am very sorry if this is the wrong place to ask if so please let me know and I can delete this question.
I would need help with the following algorithms.
The task is to choose α so that B is asymptotically faster than A.
A: $ T_a$(n) = 7$T_a$($frac{n}{2}$)+n²
B: $ T_b$(n) = α$T_b$($frac{n}{4}$)+n²
I have tried to figure out what the asymptotically upper bound would be and I have come up with this formula for A:
A: T(n) = n² + $sum_{k=1}^k 7^{k-1} frac{(n^2)}{2^k}+7^kT(frac{n}{2^k})$
k is for n < 2 = ld(n)
But I do not understand how to go on from here.
Any help would be very apricated.
sequences-and-series algorithms asymptotics
edited Nov 21 '18 at 20:23
Scientifica
6,37641335
asked Nov 21 '18 at 20:01
Mark
111
add a comment |
I am fairly new algorithms and the math behind it.
I am very sorry if this is the wrong place to ask if so please let me know and I can delete this question.
I would need help with the following algorithms.
The task is to choose α so that B is asymptotically faster than A.
A: $ T_a$(n) = 7$T_a$($frac{n}{2}$)+n²
B: $ T_b$(n) = α$T_b$($frac{n}{4}$)+n²
I have tried to figure out what the asymptotically upper bound would be and I have come up with this formula for A:
A: T(n) = n² + $sum_{k=1}^k 7^{k-1} frac{(n^2)}{2^k}+7^kT(frac{n}{2^k})$
k is for n < 2 = ld(n)
But I do not understand how to go on from here.
Any help would be very apricated.
sequences-and-series algorithms asymptotics
edited Nov 21 '18 at 20:23
Scientifica
6,37641335
asked Nov 21 '18 at 20:01
Mark
111
add a comment |
I am fairly new algorithms and the math behind it.
I am very sorry if this is the wrong place to ask if so please let me know and I can delete this question.
I would need help with the following algorithms.
The task is to choose α so that B is asymptotically faster than A.
A: $ T_a$(n) = 7$T_a$($frac{n}{2}$)+n²
B: $ T_b$(n) = α$T_b$($frac{n}{4}$)+n²
I have tried to figure out what the asymptotically upper bound would be and I have come up with this formula for A:
A: T(n) = n² + $sum_{k=1}^k 7^{k-1} frac{(n^2)}{2^k}+7^kT(frac{n}{2^k})$
k is for n < 2 = ld(n)
But I do not understand how to go on from here.
Any help would be very apricated.
sequences-and-series algorithms asymptotics
edited Nov 21 '18 at 20:23
Scientifica
6,37641335
asked Nov 21 '18 at 20:01
Mark
111
I am fairly new algorithms and the math behind it.
I am very sorry if this is the wrong place to ask if so please let me know and I can delete this question.
I would need help with the following algorithms.
The task is to choose α so that B is asymptotically faster than A.
A: $ T_a$(n) = 7$T_a$($frac{n}{2}$)+n²
B: $ T_b$(n) = α$T_b$($frac{n}{4}$)+n²
I have tried to figure out what the asymptotically upper bound would be and I have come up with this formula for A:
A: T(n) = n² + $sum_{k=1}^k 7^{k-1} frac{(n^2)}{2^k}+7^kT(frac{n}{2^k})$
k is for n < 2 = ld(n)
But I do not understand how to go on from here.
Any help would be very apricated.
sequences-and-series algorithms asymptotics
sequences-and-series algorithms asymptotics
edited Nov 21 '18 at 20:23
Scientifica
6,37641335
asked Nov 21 '18 at 20:01
Mark
111
edited Nov 21 '18 at 20:23
Scientifica
6,37641335
asked Nov 21 '18 at 20:01
Mark
111
edited Nov 21 '18 at 20:23
Scientifica
6,37641335
edited Nov 21 '18 at 20:23
Scientifica
6,37641335
edited Nov 21 '18 at 20:23
Scientifica
6,37641335
6,37641335
asked Nov 21 '18 at 20:01
Mark
111
asked Nov 21 '18 at 20:01
Mark
111
asked Nov 21 '18 at 20:01
Mark
111
111
add a comment |
add a comment |
1 Answer
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If you are familiar with Master theorem, you can solve it easily. From the master theorem as $c_{crit} = log_27 sim 2.81$ we know that $A = O(n^{2.81}) = Theta(n^{log_27})$. Hence, if $a = 4^2 = 16$, from master thorem $B = Theta(n^2log(n))$ and asymptotically faster than $A$. Also, for all $a < 16$ this would be true.
answered Nov 22 '18 at 11:10
OmG
2,200620
add a comment |
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1 Answer
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1 Answer
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active
oldest
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If you are familiar with Master theorem, you can solve it easily. From the master theorem as $c_{crit} = log_27 sim 2.81$ we know that $A = O(n^{2.81}) = Theta(n^{log_27})$. Hence, if $a = 4^2 = 16$, from master thorem $B = Theta(n^2log(n))$ and asymptotically faster than $A$. Also, for all $a < 16$ this would be true.
answered Nov 22 '18 at 11:10
OmG
2,200620
add a comment |
If you are familiar with Master theorem, you can solve it easily. From the master theorem as $c_{crit} = log_27 sim 2.81$ we know that $A = O(n^{2.81}) = Theta(n^{log_27})$. Hence, if $a = 4^2 = 16$, from master thorem $B = Theta(n^2log(n))$ and asymptotically faster than $A$. Also, for all $a < 16$ this would be true.
answered Nov 22 '18 at 11:10
OmG
2,200620
add a comment |
If you are familiar with Master theorem, you can solve it easily. From the master theorem as $c_{crit} = log_27 sim 2.81$ we know that $A = O(n^{2.81}) = Theta(n^{log_27})$. Hence, if $a = 4^2 = 16$, from master thorem $B = Theta(n^2log(n))$ and asymptotically faster than $A$. Also, for all $a < 16$ this would be true.
answered Nov 22 '18 at 11:10
OmG
2,200620
If you are familiar with Master theorem, you can solve it easily. From the master theorem as $c_{crit} = log_27 sim 2.81$ we know that $A = O(n^{2.81}) = Theta(n^{log_27})$. Hence, if $a = 4^2 = 16$, from master thorem $B = Theta(n^2log(n))$ and asymptotically faster than $A$. Also, for all $a < 16$ this would be true.
answered Nov 22 '18 at 11:10
OmG
2,200620
answered Nov 22 '18 at 11:10
OmG
2,200620
answered Nov 22 '18 at 11:10
OmG
2,200620
answered Nov 22 '18 at 11:10
OmG
2,200620
2,200620
add a comment |
add a comment |
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