Find the smallest number with total number of set bits greater than X
$begingroup$
Find the smallest number $n$ such that the sum of set bits(1's in binary representation) for all number from $1$ to $n$ is greater than a given integer $X$ where $1 lt X lt 10^{18}$
Example for $X = 5$, the answer is $4$,since sum of $1$ to $4$ is $5$
$0001 = 1$
$0010 = 1$
$0011 = 2$
$0100 = 1$
From this answer
$F(n) = F(m) + F(n - m - 1) + (n - m)$
where $m = 2^k - 1$ and $m < n$
I can only think of looping though all possible integers in the range and find which is the smallest one that satisfies this condition.
How do I solve this without looping though the entire range of possible answers.
linear-algebra exponential-function
edited Jan 6 at 18:48
amWhy
1
asked Jan 6 at 17:44
thebenmanthebenman
1254
$endgroup$
add a comment |
$begingroup$
Find the smallest number $n$ such that the sum of set bits(1's in binary representation) for all number from $1$ to $n$ is greater than a given integer $X$ where $1 lt X lt 10^{18}$
Example for $X = 5$, the answer is $4$,since sum of $1$ to $4$ is $5$
$0001 = 1$
$0010 = 1$
$0011 = 2$
$0100 = 1$
From this answer
$F(n) = F(m) + F(n - m - 1) + (n - m)$
where $m = 2^k - 1$ and $m < n$
I can only think of looping though all possible integers in the range and find which is the smallest one that satisfies this condition.
How do I solve this without looping though the entire range of possible answers.
linear-algebra exponential-function
edited Jan 6 at 18:48
amWhy
1
asked Jan 6 at 17:44
thebenmanthebenman
1254
$endgroup$
add a comment |
$begingroup$
Find the smallest number $n$ such that the sum of set bits(1's in binary representation) for all number from $1$ to $n$ is greater than a given integer $X$ where $1 lt X lt 10^{18}$
Example for $X = 5$, the answer is $4$,since sum of $1$ to $4$ is $5$
$0001 = 1$
$0010 = 1$
$0011 = 2$
$0100 = 1$
From this answer
$F(n) = F(m) + F(n - m - 1) + (n - m)$
where $m = 2^k - 1$ and $m < n$
I can only think of looping though all possible integers in the range and find which is the smallest one that satisfies this condition.
How do I solve this without looping though the entire range of possible answers.
linear-algebra exponential-function
edited Jan 6 at 18:48
amWhy
1
asked Jan 6 at 17:44
thebenmanthebenman
1254
$endgroup$
Find the smallest number $n$ such that the sum of set bits(1's in binary representation) for all number from $1$ to $n$ is greater than a given integer $X$ where $1 lt X lt 10^{18}$
Example for $X = 5$, the answer is $4$,since sum of $1$ to $4$ is $5$
$0001 = 1$
$0010 = 1$
$0011 = 2$
$0100 = 1$
From this answer
$F(n) = F(m) + F(n - m - 1) + (n - m)$
where $m = 2^k - 1$ and $m < n$
I can only think of looping though all possible integers in the range and find which is the smallest one that satisfies this condition.
How do I solve this without looping though the entire range of possible answers.
linear-algebra exponential-function
linear-algebra exponential-function
edited Jan 6 at 18:48
amWhy
1
asked Jan 6 at 17:44
thebenmanthebenman
1254
edited Jan 6 at 18:48
amWhy
1
asked Jan 6 at 17:44
thebenmanthebenman
1254
edited Jan 6 at 18:48
amWhy
1
edited Jan 6 at 18:48
amWhy
1
edited Jan 6 at 18:48
amWhy
1
1
asked Jan 6 at 17:44
thebenmanthebenman
1254
asked Jan 6 at 17:44
thebenmanthebenman
1254
asked Jan 6 at 17:44
thebenmanthebenman
1254
1254
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $P_X(n)$ be true if the sum of set bits of the numbers from 1 to $n$ is greater or equal to $X$. Then $P_X(n)$ true implies $P_X(n')$ true for any $n' > n$.
Therefore you can apply binary search: suppose you are searching the interval $[a,b]$. Let $m=lfloor(a+b)/2rfloor$. If $P_X(m)$ is true, then you can reduce your interval to $[a,m]$; if it is false, you can reduce your interval to $[m+1,b]$.
answered Jan 6 at 18:44
VincenzoVincenzo
1916
$endgroup$
add a comment |
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$begingroup$
Let $P_X(n)$ be true if the sum of set bits of the numbers from 1 to $n$ is greater or equal to $X$. Then $P_X(n)$ true implies $P_X(n')$ true for any $n' > n$.
Therefore you can apply binary search: suppose you are searching the interval $[a,b]$. Let $m=lfloor(a+b)/2rfloor$. If $P_X(m)$ is true, then you can reduce your interval to $[a,m]$; if it is false, you can reduce your interval to $[m+1,b]$.
answered Jan 6 at 18:44
VincenzoVincenzo
1916
$endgroup$
add a comment |
$begingroup$
Let $P_X(n)$ be true if the sum of set bits of the numbers from 1 to $n$ is greater or equal to $X$. Then $P_X(n)$ true implies $P_X(n')$ true for any $n' > n$.
Therefore you can apply binary search: suppose you are searching the interval $[a,b]$. Let $m=lfloor(a+b)/2rfloor$. If $P_X(m)$ is true, then you can reduce your interval to $[a,m]$; if it is false, you can reduce your interval to $[m+1,b]$.
answered Jan 6 at 18:44
VincenzoVincenzo
1916
$endgroup$
add a comment |
$begingroup$
Let $P_X(n)$ be true if the sum of set bits of the numbers from 1 to $n$ is greater or equal to $X$. Then $P_X(n)$ true implies $P_X(n')$ true for any $n' > n$.
Therefore you can apply binary search: suppose you are searching the interval $[a,b]$. Let $m=lfloor(a+b)/2rfloor$. If $P_X(m)$ is true, then you can reduce your interval to $[a,m]$; if it is false, you can reduce your interval to $[m+1,b]$.
answered Jan 6 at 18:44
VincenzoVincenzo
1916
$endgroup$
Let $P_X(n)$ be true if the sum of set bits of the numbers from 1 to $n$ is greater or equal to $X$. Then $P_X(n)$ true implies $P_X(n')$ true for any $n' > n$.
Therefore you can apply binary search: suppose you are searching the interval $[a,b]$. Let $m=lfloor(a+b)/2rfloor$. If $P_X(m)$ is true, then you can reduce your interval to $[a,m]$; if it is false, you can reduce your interval to $[m+1,b]$.
answered Jan 6 at 18:44
VincenzoVincenzo
1916
answered Jan 6 at 18:44
VincenzoVincenzo
1916
answered Jan 6 at 18:44
VincenzoVincenzo
1916
answered Jan 6 at 18:44
VincenzoVincenzo
1916
1916
add a comment |
add a comment |
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