A New Gap Sequence for Shellsort: RL-Driven Algorithm Discovery Beyond $N^{4/3}$
https://arxiv.org/abs/2609.29881
A New Gap Sequence for Shellsort: RL-Driven Algorithm Discovery Beyond $N^{4/3}$
https://arxiv.org/abs/2609.29881
Improved lower bounds of the time complexity of shellsort
https://arxiv.org/abs/2607.08997
Improved lower bounds of the time complexity of shellsort
https://arxiv.org/abs/2607.08997
It is not the world’s fastest sort, But it was the best sort known until quicksort came around.
It is not the world’s fastest sort, But it was the best sort known until quicksort came around.
Description: You will write a program which implements the following sorts and compares the performance for operations on arrays of integers of growing sizes 10, 100, 1000, 5000, 10000, 25000, etc.... You will graph the performance of…
Description: You will write a program which implements the following sorts and compares the performance for operations on arrays of integers of growing sizes 10, 100, 1000, 5000, 10000, 25000, etc.... You will graph the performance of…
Java Advent Calendar 2024 series 2、十七日目のエントリーです。ご参加ありがとうございます。
qiita.com/_mi/items/35...
Java Advent Calendar 2024 series 2、十七日目のエントリーです。ご参加ありがとうございます。
qiita.com/_mi/items/35...
Using several incomplete #pixelSorting techniques.
#HeapSort #ShellSort #QuickSort #MergeSort #Coding #Processing #pixelSort #pixel
Using several incomplete #pixelSorting techniques.
#HeapSort #ShellSort #QuickSort #MergeSort #Coding #Processing #pixelSort #pixel
If you are able to improve the Shellsort implementation above, please, drop me a line telling me how!
If you are able to improve the Shellsort implementation above, please, drop me a line telling me how!
I know people are usually afraid of using Shellsort, but give it a try to this one before dismissing it!
I know people are usually afraid of using Shellsort, but give it a try to this one before dismissing it!
size_t i, j, k;
int t;
for (k = 3; k <= n/9; k = k*9/4 + 1);
for (k = k*4/9; k; k = k*4/9) {
for (i = k; i < n; ++i) {
t = A[i];
for (j = i; k <= j && t < A[j-k]); j -= k) { A[j] = A[j-k]; }
A[j] = t;
}
}
}
size_t i, j, k;
int t;
for (k = 3; k <= n/9; k = k*9/4 + 1);
for (k = k*4/9; k; k = k*4/9) {
for (i = k; i < n; ++i) {
t = A[i];
for (j = i; k <= j && t < A[j-k]); j -= k) { A[j] = A[j-k]; }
A[j] = t;
}
}
}