Sorting nlog(sqrt(n))

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A researcher claimed that she discovered a comparison-based sorting algorithm that runs in $O(nlog(sqrtn))$. Given the existence of an $Omega(nlog(n))$ lowerbound for sorting, how can this be possible?



Hint: It is possible. Don't waste time trying to disprove it. Just show why it is possible.






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    $n , logsqrtn = frac12n , log(n)= Omega(n ,logn)$. There's only a constant factor difference.
    – Gokul
    3 hours ago















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A researcher claimed that she discovered a comparison-based sorting algorithm that runs in $O(nlog(sqrtn))$. Given the existence of an $Omega(nlog(n))$ lowerbound for sorting, how can this be possible?



Hint: It is possible. Don't waste time trying to disprove it. Just show why it is possible.






algorithms







share|cite|improve this question









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  • 1




    $n , logsqrtn = frac12n , log(n)= Omega(n ,logn)$. There's only a constant factor difference.
    – Gokul
    3 hours ago













up vote
1
down vote

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up vote
1
down vote

favorite











A researcher claimed that she discovered a comparison-based sorting algorithm that runs in $O(nlog(sqrtn))$. Given the existence of an $Omega(nlog(n))$ lowerbound for sorting, how can this be possible?



Hint: It is possible. Don't waste time trying to disprove it. Just show why it is possible.






algorithms







share|cite|improve this question









New contributor




confucius_did_shrooms is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











A researcher claimed that she discovered a comparison-based sorting algorithm that runs in $O(nlog(sqrtn))$. Given the existence of an $Omega(nlog(n))$ lowerbound for sorting, how can this be possible?



Hint: It is possible. Don't waste time trying to disprove it. Just show why it is possible.






algorithms




algorithms






share|cite|improve this question









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confucius_did_shrooms is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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edited 3 hours ago









Thinh D. Nguyen

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  • 1




    $n , logsqrtn = frac12n , log(n)= Omega(n ,logn)$. There's only a constant factor difference.
    – Gokul
    3 hours ago













  • 1




    $n , logsqrtn = frac12n , log(n)= Omega(n ,logn)$. There's only a constant factor difference.
    – Gokul
    3 hours ago








1




1




$n , logsqrtn = frac12n , log(n)= Omega(n ,logn)$. There's only a constant factor difference.
– Gokul
3 hours ago





$n , logsqrtn = frac12n , log(n)= Omega(n ,logn)$. There's only a constant factor difference.
– Gokul
3 hours ago











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As a set of functions $mathbbNlongrightarrowmathbbN$, $O(nlog(sqrtn))=O(nlog(n))$.






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    As a set of functions $mathbbNlongrightarrowmathbbN$, $O(nlog(sqrtn))=O(nlog(n))$.






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      As a set of functions $mathbbNlongrightarrowmathbbN$, $O(nlog(sqrtn))=O(nlog(n))$.






      share|cite|improve this answer






















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        up vote
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        As a set of functions $mathbbNlongrightarrowmathbbN$, $O(nlog(sqrtn))=O(nlog(n))$.






        share|cite|improve this answer












        As a set of functions $mathbbNlongrightarrowmathbbN$, $O(nlog(sqrtn))=O(nlog(n))$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 3 hours ago









        Thinh D. Nguyen

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        3,45111468




















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