# Re: Median: Oasis and Fast Sorting Algorithm

• From: Leonard Mada <discoleo gmx net>
• To: gnumeric-list gnome org
• Subject: Re: Median: Oasis and Fast Sorting Algorithm
• Date: Sat, 10 Feb 2007 20:27:13 +0200

```Sorry, I noticed, I made 2 small errors:

1. search inside array: log[(n/2)!] << log[(n/4)^(n/2)] = n/2 * log(n/4) !!!
2. move array: log[(n/4)!] << log[(n/8)^(n/4)] = n/4 * log(n/8) !!!

This is because (n-k) * k is always < (n/2)^2 and is equal only and only when k = n/2!!! (where k = 0::n)

In this way:
O(...) << 5/4 * n * log(n) -5/4 * n -1

Please note, that the approximations presented at 1 and 2 are very conservative:
10! / 5^10 = 0.4
12! / 6^12 = 0.2
14! / 7^14 = 0.13

So basically, the terms: log[(n/2)!] are always much less than n/2 * log(n/4); [<< means much less]

For really big numbers, even my previous formula was correct, with n/4 instead of n/2.

Andreas J. Guelzow wrote:
```
```On Sat, 2007-10-02 at 19:49 +0200, Leonard Mada wrote:

```
```- move array (with worst algorithm): log[(n/2)!] << log[(n/4)^(n/4)] =
n/4 * log(n/4)
```
```
```
for n=100 log[(n/2)!] > 140 but n/4 * log(n/4) < 81
```
What exactly is << supposed to mean?

```
so, even under the worst assumption, it will be less than n * log(n),
```
assuming this is true

```
and because log[(n/2-1)!] << log[(n/4)^(n/4)], there will indeed be a significant difference. I presume, that more realistically, this algorithm will be 2 times faster than the corresponding sort and uses only half the memory.
```
"2 times faster" means nothing. You are counting artifical actions
assuming that a comparison is the same as moving a memory block. When
you write the real code a small constant factor can easily disappear.
And typically with complex algorithms it will disappear.

Andreas
```
```

```