How does the following evaluate?
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We have a recurrence as follows:
$$T(n) = 2Tleft(sqrtnright) + log n$$
Renaming $m = log n$ yields
$$T(2^m) = 2T(2^fracm2) + m$$
Renaming $S(m) = T(2^m)$ new recurrence becomes:
$$S(m) = 2Sleft(fracm2right) + m$$
How does this $T(2^fracm2)$ become $Sleft(fracm2right)$?
recurrence-relations
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We have a recurrence as follows:
$$T(n) = 2Tleft(sqrtnright) + log n$$
Renaming $m = log n$ yields
$$T(2^m) = 2T(2^fracm2) + m$$
Renaming $S(m) = T(2^m)$ new recurrence becomes:
$$S(m) = 2Sleft(fracm2right) + m$$
How does this $T(2^fracm2)$ become $Sleft(fracm2right)$?
recurrence-relations
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We have a recurrence as follows:
$$T(n) = 2Tleft(sqrtnright) + log n$$
Renaming $m = log n$ yields
$$T(2^m) = 2T(2^fracm2) + m$$
Renaming $S(m) = T(2^m)$ new recurrence becomes:
$$S(m) = 2Sleft(fracm2right) + m$$
How does this $T(2^fracm2)$ become $Sleft(fracm2right)$?
recurrence-relations
We have a recurrence as follows:
$$T(n) = 2Tleft(sqrtnright) + log n$$
Renaming $m = log n$ yields
$$T(2^m) = 2T(2^fracm2) + m$$
Renaming $S(m) = T(2^m)$ new recurrence becomes:
$$S(m) = 2Sleft(fracm2right) + m$$
How does this $T(2^fracm2)$ become $Sleft(fracm2right)$?
recurrence-relations
edited Aug 15 at 12:40
James
2,002619
2,002619
asked Aug 15 at 12:31
Jot Waraich
104
104
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1 Answer
1
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up vote
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accepted
you agree that $$S(m) = T(2^m)$$
now put $fracm2$ inplace of $m$.
$$S(fracm2)=T(2^fracm2)$$
so what is $T(2^fracm2)$ now it is $S(fracm2)$
thus we can replace $T(2^fracm2)$ with $S(fracm2)$ right?
hope it clears your doubt!
Never thought it is going to be this easy : |
â Jot Waraich
Aug 15 at 12:44
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
you agree that $$S(m) = T(2^m)$$
now put $fracm2$ inplace of $m$.
$$S(fracm2)=T(2^fracm2)$$
so what is $T(2^fracm2)$ now it is $S(fracm2)$
thus we can replace $T(2^fracm2)$ with $S(fracm2)$ right?
hope it clears your doubt!
Never thought it is going to be this easy : |
â Jot Waraich
Aug 15 at 12:44
add a comment |Â
up vote
0
down vote
accepted
you agree that $$S(m) = T(2^m)$$
now put $fracm2$ inplace of $m$.
$$S(fracm2)=T(2^fracm2)$$
so what is $T(2^fracm2)$ now it is $S(fracm2)$
thus we can replace $T(2^fracm2)$ with $S(fracm2)$ right?
hope it clears your doubt!
Never thought it is going to be this easy : |
â Jot Waraich
Aug 15 at 12:44
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
you agree that $$S(m) = T(2^m)$$
now put $fracm2$ inplace of $m$.
$$S(fracm2)=T(2^fracm2)$$
so what is $T(2^fracm2)$ now it is $S(fracm2)$
thus we can replace $T(2^fracm2)$ with $S(fracm2)$ right?
hope it clears your doubt!
you agree that $$S(m) = T(2^m)$$
now put $fracm2$ inplace of $m$.
$$S(fracm2)=T(2^fracm2)$$
so what is $T(2^fracm2)$ now it is $S(fracm2)$
thus we can replace $T(2^fracm2)$ with $S(fracm2)$ right?
hope it clears your doubt!
answered Aug 15 at 12:42
James
2,002619
2,002619
Never thought it is going to be this easy : |
â Jot Waraich
Aug 15 at 12:44
add a comment |Â
Never thought it is going to be this easy : |
â Jot Waraich
Aug 15 at 12:44
Never thought it is going to be this easy : |
â Jot Waraich
Aug 15 at 12:44
Never thought it is going to be this easy : |
â Jot Waraich
Aug 15 at 12:44
add a comment |Â
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