Write a recurrence relation for this modified Quicksort?
0
In a modified quicksort, you partition randomly on odd levels of recursive calls and on even levels you partition on the largest element. How do you write a recurrence relation for exactly how many comparisons quicksort will do?
algorithm quicksort relation recurrence
asked Nov 11 '18 at 17:52
Michael LinMichael Lin
102
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0
In a modified quicksort, you partition randomly on odd levels of recursive calls and on even levels you partition on the largest element. How do you write a recurrence relation for exactly how many comparisons quicksort will do?
algorithm quicksort relation recurrence
asked Nov 11 '18 at 17:52
Michael LinMichael Lin
102
Exactly sounds a little weird when you consider a random variable ?!
– Yves Daoust
Nov 11 '18 at 18:11
Can you provide the recurrence relation for the case of ordinary Quicksort ?
– Yves Daoust
Nov 11 '18 at 18:12
Yes. T(n) = T(n/4) + T(3n/4) + (n-1) for average case.
– Michael Lin
Nov 12 '18 at 0:36
Then it suffices to add a term n-1 to find the largest element, then another term n-1 to perform the partition with this pivot, and replace n by n-1, as the partition will just leave the largest element apart. T(n)=2(n-1)+T((n-1)/4)+T(3(n-1)/4)+(n-2).
– Yves Daoust
Nov 12 '18 at 13:49
add a comment |
0
0
0
In a modified quicksort, you partition randomly on odd levels of recursive calls and on even levels you partition on the largest element. How do you write a recurrence relation for exactly how many comparisons quicksort will do?
algorithm quicksort relation recurrence
asked Nov 11 '18 at 17:52
Michael LinMichael Lin
102
In a modified quicksort, you partition randomly on odd levels of recursive calls and on even levels you partition on the largest element. How do you write a recurrence relation for exactly how many comparisons quicksort will do?
algorithm quicksort relation recurrence
algorithm quicksort relation recurrence
asked Nov 11 '18 at 17:52
Michael LinMichael Lin
102
asked Nov 11 '18 at 17:52
Michael LinMichael Lin
102
asked Nov 11 '18 at 17:52
Michael LinMichael Lin
102
asked Nov 11 '18 at 17:52
Michael LinMichael Lin
102
asked Nov 11 '18 at 17:52
Michael LinMichael Lin
102
102
Exactly sounds a little weird when you consider a random variable ?!
– Yves Daoust
Nov 11 '18 at 18:11
Can you provide the recurrence relation for the case of ordinary Quicksort ?
– Yves Daoust
Nov 11 '18 at 18:12
Yes. T(n) = T(n/4) + T(3n/4) + (n-1) for average case.
– Michael Lin
Nov 12 '18 at 0:36
Then it suffices to add a term n-1 to find the largest element, then another term n-1 to perform the partition with this pivot, and replace n by n-1, as the partition will just leave the largest element apart. T(n)=2(n-1)+T((n-1)/4)+T(3(n-1)/4)+(n-2).
– Yves Daoust
Nov 12 '18 at 13:49
add a comment |
Exactly sounds a little weird when you consider a random variable ?!
– Yves Daoust
Nov 11 '18 at 18:11
Can you provide the recurrence relation for the case of ordinary Quicksort ?
– Yves Daoust
Nov 11 '18 at 18:12
Yes. T(n) = T(n/4) + T(3n/4) + (n-1) for average case.
– Michael Lin
Nov 12 '18 at 0:36
Then it suffices to add a term n-1 to find the largest element, then another term n-1 to perform the partition with this pivot, and replace n by n-1, as the partition will just leave the largest element apart. T(n)=2(n-1)+T((n-1)/4)+T(3(n-1)/4)+(n-2).
– Yves Daoust
Nov 12 '18 at 13:49
Exactly sounds a little weird when you consider a random variable ?!
– Yves Daoust
Nov 11 '18 at 18:11
Exactly sounds a little weird when you consider a random variable ?!
– Yves Daoust
Nov 11 '18 at 18:11
Can you provide the recurrence relation for the case of ordinary Quicksort ?
– Yves Daoust
Nov 11 '18 at 18:12
Can you provide the recurrence relation for the case of ordinary Quicksort ?
– Yves Daoust
Nov 11 '18 at 18:12
Yes. T(n) = T(n/4) + T(3n/4) + (n-1) for average case.
– Michael Lin
Nov 12 '18 at 0:36
Yes. T(n) = T(n/4) + T(3n/4) + (n-1) for average case.
– Michael Lin
Nov 12 '18 at 0:36
Then it suffices to add a term n-1 to find the largest element, then another term n-1 to perform the partition with this pivot, and replace n by n-1, as the partition will just leave the largest element apart. T(n)=2(n-1)+T((n-1)/4)+T(3(n-1)/4)+(n-2).
– Yves Daoust
Nov 12 '18 at 13:49
Then it suffices to add a term n-1 to find the largest element, then another term n-1 to perform the partition with this pivot, and replace n by n-1, as the partition will just leave the largest element apart. T(n)=2(n-1)+T((n-1)/4)+T(3(n-1)/4)+(n-2).
– Yves Daoust
Nov 12 '18 at 13:49
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Exactly sounds a little weird when you consider a random variable ?!
– Yves Daoust
Nov 11 '18 at 18:11
Can you provide the recurrence relation for the case of ordinary Quicksort ?
– Yves Daoust
Nov 11 '18 at 18:12
Yes. T(n) = T(n/4) + T(3n/4) + (n-1) for average case.
– Michael Lin
Nov 12 '18 at 0:36
Then it suffices to add a term n-1 to find the largest element, then another term n-1 to perform the partition with this pivot, and replace n by n-1, as the partition will just leave the largest element apart. T(n)=2(n-1)+T((n-1)/4)+T(3(n-1)/4)+(n-2).
– Yves Daoust
Nov 12 '18 at 13:49