Recurrence formality question Announcing the arrival of Valued Associate #679: Cesar Manara ...
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Recurrence formality question
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Solving recurrence relation with unrolling techniqueSolving recurrence relation in 2 variablesSolving recurrence relation of algorithm complexity?Solving this recurrence relationNonlinear recurrence relationSolving recurrence relation, no clue how to approachSolve the recurrence $T(n) = T(lfloor n/2 rfloor)+ T(lfloor n/3 rfloor) + lfloor n log_2 nrfloor$.Solve the recurrence relation with no initial conditionsSolving a recurrence relation with floor functionWhy can this recurrence relation be rewritten like this?
$begingroup$
I have tried to solve this recurrence relation using induction. $$T(n) = T(lfloor log_2 n rfloor) +1$$
It is clear that I should get something similar to $log *n$, but I don't know how to formalize this kind of questions. Thank you for your kind help.
recurrence-relations
asked Mar 24 at 19:32
ga asga as
84
$endgroup$
add a comment |
$begingroup$
I have tried to solve this recurrence relation using induction. $$T(n) = T(lfloor log_2 n rfloor) +1$$
It is clear that I should get something similar to $log *n$, but I don't know how to formalize this kind of questions. Thank you for your kind help.
recurrence-relations
asked Mar 24 at 19:32
ga asga as
84
$endgroup$
add a comment |
$begingroup$
I have tried to solve this recurrence relation using induction. $$T(n) = T(lfloor log_2 n rfloor) +1$$
It is clear that I should get something similar to $log *n$, but I don't know how to formalize this kind of questions. Thank you for your kind help.
recurrence-relations
asked Mar 24 at 19:32
ga asga as
84
$endgroup$
I have tried to solve this recurrence relation using induction. $$T(n) = T(lfloor log_2 n rfloor) +1$$
It is clear that I should get something similar to $log *n$, but I don't know how to formalize this kind of questions. Thank you for your kind help.
recurrence-relations
recurrence-relations
asked Mar 24 at 19:32
ga asga as
84
asked Mar 24 at 19:32
ga asga as
84
edited Mar 24 at 19:40
ga as
asked Mar 24 at 19:32
ga asga as
84
asked Mar 24 at 19:32
ga asga as
84
asked Mar 24 at 19:32
ga asga as
84
84
add a comment |
add a comment |
1 Answer
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$begingroup$
That $lfloor log_2 n rfloor$ suggests you should plug in powers of $2$ and see if something interesting happens: (Assume $T(1)$ is given)
begin{align}
T(2) &= T(lfloor log_2 2 rfloor) +1 \ &= T(1) +1; \
T(4) &= T(lfloor log_2 4 rfloor) +1 \ &= T(2) +1 \ &= T(1) +2; \
T(8) &= T(lfloor log_2 8 rfloor) +1 \ &= T(3) +1 \ &= T(1) +3; \
T(16) &= T(lfloor log_2 16 rfloor) +1 \ &= T(4) +1 \ &= T(1) +4;
end{align}
and so on. Then, you notice that $T(2^m) = T(1) +m$; you may prove it by induction (the inductive step will make use of the recurrence relation.)
answered Mar 24 at 19:48
RócherzRócherz
3,0263823
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1 Answer
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1 Answer
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$begingroup$
That $lfloor log_2 n rfloor$ suggests you should plug in powers of $2$ and see if something interesting happens: (Assume $T(1)$ is given)
begin{align}
T(2) &= T(lfloor log_2 2 rfloor) +1 \ &= T(1) +1; \
T(4) &= T(lfloor log_2 4 rfloor) +1 \ &= T(2) +1 \ &= T(1) +2; \
T(8) &= T(lfloor log_2 8 rfloor) +1 \ &= T(3) +1 \ &= T(1) +3; \
T(16) &= T(lfloor log_2 16 rfloor) +1 \ &= T(4) +1 \ &= T(1) +4;
end{align}
and so on. Then, you notice that $T(2^m) = T(1) +m$; you may prove it by induction (the inductive step will make use of the recurrence relation.)
answered Mar 24 at 19:48
RócherzRócherz
3,0263823
$endgroup$
add a comment |
$begingroup$
That $lfloor log_2 n rfloor$ suggests you should plug in powers of $2$ and see if something interesting happens: (Assume $T(1)$ is given)
begin{align}
T(2) &= T(lfloor log_2 2 rfloor) +1 \ &= T(1) +1; \
T(4) &= T(lfloor log_2 4 rfloor) +1 \ &= T(2) +1 \ &= T(1) +2; \
T(8) &= T(lfloor log_2 8 rfloor) +1 \ &= T(3) +1 \ &= T(1) +3; \
T(16) &= T(lfloor log_2 16 rfloor) +1 \ &= T(4) +1 \ &= T(1) +4;
end{align}
and so on. Then, you notice that $T(2^m) = T(1) +m$; you may prove it by induction (the inductive step will make use of the recurrence relation.)
answered Mar 24 at 19:48
RócherzRócherz
3,0263823
$endgroup$
add a comment |
$begingroup$
That $lfloor log_2 n rfloor$ suggests you should plug in powers of $2$ and see if something interesting happens: (Assume $T(1)$ is given)
begin{align}
T(2) &= T(lfloor log_2 2 rfloor) +1 \ &= T(1) +1; \
T(4) &= T(lfloor log_2 4 rfloor) +1 \ &= T(2) +1 \ &= T(1) +2; \
T(8) &= T(lfloor log_2 8 rfloor) +1 \ &= T(3) +1 \ &= T(1) +3; \
T(16) &= T(lfloor log_2 16 rfloor) +1 \ &= T(4) +1 \ &= T(1) +4;
end{align}
and so on. Then, you notice that $T(2^m) = T(1) +m$; you may prove it by induction (the inductive step will make use of the recurrence relation.)
answered Mar 24 at 19:48
RócherzRócherz
3,0263823
$endgroup$
That $lfloor log_2 n rfloor$ suggests you should plug in powers of $2$ and see if something interesting happens: (Assume $T(1)$ is given)
begin{align}
T(2) &= T(lfloor log_2 2 rfloor) +1 \ &= T(1) +1; \
T(4) &= T(lfloor log_2 4 rfloor) +1 \ &= T(2) +1 \ &= T(1) +2; \
T(8) &= T(lfloor log_2 8 rfloor) +1 \ &= T(3) +1 \ &= T(1) +3; \
T(16) &= T(lfloor log_2 16 rfloor) +1 \ &= T(4) +1 \ &= T(1) +4;
end{align}
and so on. Then, you notice that $T(2^m) = T(1) +m$; you may prove it by induction (the inductive step will make use of the recurrence relation.)
answered Mar 24 at 19:48
RócherzRócherz
3,0263823
answered Mar 24 at 19:48
RócherzRócherz
3,0263823
answered Mar 24 at 19:48
RócherzRócherz
3,0263823
answered Mar 24 at 19:48
RócherzRócherz
3,0263823
3,0263823
add a comment |
add a comment |
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