Algorithm

I have taken some time to rewrite memchr(3) to utilize similiar algorithm as my strlen(3) did. Two different formular, however, would have a roughly 1/3 difference.

Currently I have taken an approach that when memchr(3) is trying to match a pattern of 8-bit, then consider the string as 8-bit, which uses the first formular which is expected to have uniform O(n) time complexity by reducing unnecessary branches.

This approach, however, might be bogus since if you find 7-bit character within a 8-bit string, in that case, memchr(3) would perform poorly, if there are a lot of characters having their MSB as 1.

dengxf同学提到的一个问题，想了一下，想到的一个算法：

令$x$为已知量，$y$为目标量，$k$为迭代轮数，

1. 令 $y=x-1$，$k=0$
2. 对于 $y$ 右移 $2^k$ 位 $> 0$的情形，执行下列操作，直到此条件不符合为止：
   y与前述结果相与；$k$累进$1$。
3. $y$累进$1$。

原理：第(2)步的循环将除最高位的所有0均置位。最后一步(3)修正这一值。

求一系列字符串的最长公共子串的问题大家都很了解，今天在琢磨垃圾信息的时候突然想到另一个问题。

好比说，5个字符串：

www.domain.com
domain.com
domain.net
www.domain.net
spammer.domain.net

I have an idea: let we have “a” is the number to determine.

return ( (0 == (a & (a-1))) && (a > 0))

Despite the style of the code, is there some better solution? Note that a is a native integer type (say, int, long, long long, etc).

The “Distribution Set” is an important idea on GridDistribute’s vendor distribution. Here comes different ideas on implementing it, which will influence the search approach we will choose. Personally, I think we must have a final decision before next Wednesday.

It’s a common practise to choose a prime number as the hast table size. This article has analyzed Java’s implementation.