Find the smallest number with total number of set bits greater than X












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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.










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    0












    $begingroup$


    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.










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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.










      share|cite|improve this question











      $endgroup$




      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






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      edited Jan 6 at 18:48









      amWhy

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      asked Jan 6 at 17:44









      thebenmanthebenman

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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 $[m+1,b]$.






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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 $[m+1,b]$.






            share|cite|improve this answer









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              0












              $begingroup$

              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 $[m+1,b]$.






              share|cite|improve this 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 $[m+1,b]$.






                share|cite|improve this answer









                $endgroup$



                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 $[m+1,b]$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 18:44









                VincenzoVincenzo

                1916




                1916






























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