from:

http://www.cppblog.com/lapcca/archive/2010/09/13/126499.html

Radix sort is the algorithm used by the card-sorting machines you now find only in computer museums. The cards are organized into 80 columns, and in each column a hole can be punched in one of 12 places. The sorter can be mechanically "programmed" to examine a given column of each card in a deck and distribute the card into one of 12 bins depending on which place has been punched. An operator can then gather the cards bin by bin, so that cards with the first place punched are on top of cards with the second place punched, and so on.

For decimal digits, only 10 places are used in each column. (The other two places are used for encoding nonnumeric characters.) A d-digit number would then occupy a field of d columns. Since the card sorter can look at only one column at a time, the problem of sorting n cards on a d-digit number requires a sorting algorithm.

Intuitively, one might want to sort numbers on their most significant digit, sort each of the resulting bins recursively, and then combine the decks in order. Unfortunately, since the cards in 9 of the 10 bins must be put aside to sort each of the bins, this procedure generates many intermediate piles of cards that must be kept track of. (See Exercise 8.3-5.)

Radix sort solves the problem of card sorting counterintuitively by sorting on the least significant digit first. The cards are then combined into a single deck, with the cards in the 0 bin preceding the cards in the 1 bin preceding the cards in the 2 bin, and so on. Then the entire deck is sorted again on the second-least significant digit and recombined in a like manner. The process continues until the cards have been sorted on all d digits. Remarkably, at that point the cards are fully sorted on the d-digit number. Thus, only d passes through the deck are required to sort. Figure 8.3 shows how radix sort operates on a "deck" of seven 3-digit numbers.

 

Figure 8.3: The operation of radix sort on a list of seven 3-digit numbers. The leftmost column is the input. The remaining columns show the list after successive sorts on increasingly significant digit positions. Shading indicates the digit position sorted on to produce each list from the previous one.

It is essential that the digit sorts in this algorithm be stable. The sort performed by a card sorter is stable, but the operator has to be wary about not changing the order of the cards as they come out of a bin, even though all the cards in a bin have the same digit in the chosen column.

In a typical computer, which is a sequential random-access machine, radix sort is sometimes used to sort records of information that are keyed by multiple fields. For example, we might wish to sort dates by three keys: year, month, and day. We could run a sorting algorithm with a comparison function that, given two dates, compares years, and if there is a tie, compares months, and if another tie occurs, compares days. Alternatively, we could sort the information three times with a stable sort: first on day, next on month, and finally on year.

The code for radix sort is straightforward. The following procedure assumes that each element in the n-element array A has d digits, where digit 1 is the lowest-order digit and digit d is the highest-order digit.

RADIX-SORT(A, d)
1  for i ← 1 to d
2     do use a stable sort to sort array A on digit i

Lemma 8.3

Given n d-digit numbers in which each digit can take on up to k possible values, RADIX-SORT correctly sorts these numbers in Θ(d(n + k)) time.

Proof The correctness of radix sort follows by induction on the column being sorted (see Exercise 8.3-3). The analysis of the running time depends on the stable sort used as the intermediate sorting algorithm. When each digit is in the range 0 to k-1 (so that it can take on k possible values), and k is not too large, counting sort is the obvious choice. Each pass over n d-digit numbers then takes time Θ(n + k). There are d passes, so the total time for radix sort is Θ(d(n + k)).

When d is constant and k = O(n), radix sort runs in linear time. More generally, we have some flexibility in how to break each key into digits.

Lemma 8.4

Given n b-bit numbers and any positive integer r ≤ b, RADIX-SORT correctly sorts these numbers in Θ((b/r)(n + 2^r)) time.

Proof For a value r ≤ b, we view each key as having d = ⌈b/r⌉ digits of r bits each. Each digit is an integer in the range 0 to 2^r - 1, so that we can use counting sort with k = 2^r - 1. (For example, we can view a 32-bit word as having 4 8-bit digits, so that b = 32, r = 8, k = 2^r - 1 = 255, and d = b/r = 4.) Each pass of counting sort takes time Θ(n + k) = Θ(n + 2^r) and there are d passes, for a total running time of Θ(d(n + 2^r )) = Θ((b/r)(n + 2^r)).

For given values of n and b, we wish to choose the value of r, with r ≤ b, that minimizes the expression (b/r)(n + 2^r). If b < ⌊lg n⌋, then for any value of r b, we have that (n + 2^r) = Θ(n). Thus, choosing r = b yields a running time of (b/b)(n + 2^b) = Θ(n), which is asymptotically optimal. If b ≥ ⌊lg n⌋, then choosing r = ⌊lg n⌋ gives the best time to within a constant factor, which we can see as follows. Choosing r = ⌊lg n⌋ yields a running time of Θ(bn/ lg n). As we increase r above ⌊lg n⌋, the 2^r term in the numerator increases faster than the r term in the denominator, and so increasing r above ⌊lg n⌋ yields a running time of Θ(bn/ lg n). If instead we were to decrease r below ⌊lg n⌋, then the b/r term increases and the n + 2^r term remains at Θ(n).

Is radix sort preferable to a comparison-based sorting algorithm, such as quick-sort? If b = O(lg n), as is often the case, and we choose r ≈ lg n, then radix sort's running time is Θ(n), which appears to be better than quicksort's average-case time of Θ(n lg n). The constant factors hidden in the Θ-notation differ, however. Although radix sort may make fewer passes than quicksort over the n keys, each pass of radix sort may take significantly longer. Which sorting algorithm is preferable depends on the characteristics of the implementations, of the underlying machine (e.g., quicksort often uses hardware caches more effectively than radix sort), and of the input data. Moreover, the version of radix sort that uses counting sort as the intermediate stable sort does not sort in place, which many of the Θ(n lg n)-time comparison sorts do. Thus, when primary memory storage is at a premium, an in-place algorithm such as quicksort may be preferable

在google的 搜索结果中，PR值越 高的网页排在越前面。

网页权重的算法有很多种，为何我唯独选择了page rank来讨论，不仅因为它是Google搜索引擎采用的搜索结果排名算法之核心，且它把整 个互联网当做一个整体来对待且最终依靠经典的数学模型精准地得到web上 网页的权重。

虽然今天的搜索引擎的排名系统远远要比这个算法复杂。域名数据，内容质量，用户数据，建站时间等都可能被考虑进去，但是page rank算法仍然是核心的技术之一，使得Google名声大噪。关于page rank的介绍性文章，在Google黑板报里的” 谈 Page Rank – Google 的民主表决式网页排名技术”。关于其更详细的论述，可以参照Google 的两个创始人拉里•佩奇 （Larry Page ）和谢尔盖•布林 (Sergey Brin)的论文 ” The PageRank Citation Ranking: Bringing Order to the Web ”。

首先，page rank基 于以下的假设：”一个 网页被引用 (即反向 链接)的次数越多，则 说明越重要; 一个网 页虽然没有被多次引用，但是被重要的网页引用，则它也可能是很重要的；一个网页的重要性被平均的传递到它所引用的网页。”所以，为了说明问题的方便，就引出了下面这个简化了的Page Rank算法。简化版一：R(u) = cR(1)/N(1) + ……+cR(v)/N(v)。(v∈Bu)。 其中R(v)是网页v的PR值，N(v)是 网页v的正向链接数，B(u)是页面u的反向链接的集合。c是 阻尼系数(Damping Factor)，它的值在0到1之间。因 此，阻尼系数的使用，减少了其它页面对当前页面A的排序贡献。那么这个式子如何用数学的方法解答呢？首先可以认为整个互联网是 一个大的有向图G=(V,E)。V是所有页面的集合，E是有向边的集 合，(i,j)表示页i有指向页j的超链接。由于有向图和矩阵在本质上 是可以互相转换的，下面举例是如何互转的：

此有向连通图模拟网页间的超链接

链接源I D 链接目标 ID

1          2,3 ,4,5, 7

2          1

3          1,2

4          2,3,5

5          1,3,4,6

6          1,5

7          5

如果我们假设Aij＝1 ,if (从页面 i 向页面 j 「有 」 链接的情况) ；Aij＝0, if (从页面 i 向页面 j 「没有」链接的情况) 。 则根据以上我们可以构造如下矩阵

A = [

   0, 1, 1, 1, 1, 0, 1;

   1, 0, 0, 0, 0, 0, 0;

   1, 1, 0, 0, 0, 0, 0;

   0, 1, 1, 0, 1, 0, 0;

   1, 0, 1, 1, 0, 1, 0;

   1, 0, 0, 0, 1, 0, 0;

   0, 0, 0, 0, 1, 0, 0;

]

接下来再来看看公式一中的PR值求法，即

PR(1) = c[1*PR(2) + (1/2)*PR(3) + (1/4)*PR(5) +(1/2)*PR(6)];

PR(2) = c[(1/5)*PR(1) + (1/2)*PR(3) + 0.25*PR(5) + 0.5*PR(5)];

............

PR(7) = c[(1/5)*PR(1) + 0+ 0......];

则，可以得出P^{T =} cMP^T，其中M的方阵如下，

M = [

    0,    1,   1/2,    0,     1/4,    1/2,    0;

    1/5,    0, 1/2,    1/3,    0,     0,     0;

    1/5,    0, 0,     1/3,    1/4,    0,     0;

    1/5,    0, 0,      0,     1/4,    0,     0;

    1/5,    0, 0,     1/3,    0,     1/2,    1;

    0,      0, 0,      0,     1/4,    0,     0;

    1/5,    0, 0,     0,     0,     0,     0;

]

所以，P^T为M的特征根为c的 特征向量。只需求出最大特征根的特征向量，就是网页集对应的最终PageRank值，这可以用迭代方法计算。如何迭代呢？如果 我们给定初始向量P^T1’做第一次迭代，就相当于用初始向量乘以上面的矩阵。第二次迭代就相当于用第一 迭代的结果再乘以上面的矩阵……实际上，在随机过程的理论中，上述矩阵被称为“转移概率矩阵”。这种离散状态按照离散时间的随机转移过程称为马尔可夫链(Markov chain)。设转移概率矩阵为P，若存在正整数N，使得P_N>0(每 一个元素大于0)，这种链被称作正则链，它存在唯一的极限状态概率，并且与初始状态无关。这篇”Google搜索与Inter网中的数学”文章 里，描述了马氏链与page rank的关系。

最后可以看到，从最开始的矩阵A到矩阵M可以 很容易转化得到(将A倒置后将各个数值除以各自的非零要素)。

现在考虑有一个页面（比如是页面7），它不含有任何的超链接，即它的前向 链接或者说出度为0，很显然，方阵M的最后一行为全零，这样，特征向量P^T也 为全零。我们也可以从图论的角度来阐述这个问题。我们可以这样定义一个有向图：图G的顶点集合为V={V1,V2,…,Vn}， 边的集合为E={E_ij}。我们把有向图G的每个顶点都给定一个权值P(Vi)， 即为它的PR值。有向边AB的权值定义为A为B贡 献的PageRank，也即网站A链接到网站B的概率。这样，对于一个 顶点，若它的出度大于0，则从它出去的权值和为1（这一点可以从上述的方阵M中 的列可以看出，满足了全概率）。显然，如果图中有一个顶点X的出度是0，那么经过有限次迭代后所有 顶点的PR值都将变为0。这是因为由于X不对外贡献任何PR， 所以整体的PR总和在不断减少，最终减为0。这个被拉里•佩奇和谢尔盖 •布林称为rank sink。 为了克服这个问题，就有了下面这个公式：Rank算 法简化版二：R(u) = （1-c）+ cR(1)/N(1) + ……+cR(v)/N(v)。(v∈Bu)。 ……公式二

(1-c)实际上为了服从概率分布。这样可以推导出P = (1-c)e + cMP，即

P = [(1-c)ee^T/n + cM]P (e^TP=n)。关于用矩阵方法来推导的更详细的文章，可以参考这篇 "The $25,000,000,000 Eigenvector: The Linear Algebra Behind Google" .

   现在可以想象一下整个web中 有250亿不止的网页，收敛速度是至关重要的。所以为什么作者拉里•佩奇 （Larry Page ）和谢尔盖•布林 (Sergey Brin)要取c为0.85,在这篇文章 ”How Google Finds Your Needle in the Web's Haystack ”的最后部分讨论到。

Pagerank算法除了对搜索结果进行排序外，还可 以应用到其它方面，如估算网络流量，向后链接的预测器，为用户导航等。这篇文章 PageRank sculptin g.讲述了当前Google的一些改进.

    后记，2001年9月，PageRank被 授予美国专利，专利被正式颁发给斯坦福大学，佩奇作为发明人列于文件中。最后要说的一点，分析PR算法，用到了离散数学，线性 代数，概率论，数值计算甚至随机过程，所以看来数学确实很好很强大需要认真学习啊。

from:

http://hi.baidu.com/alphahunters/blog/item/e0e8a92d1b2820321f3089a0.html

http://blog.csdn.net/boyplayee/archive/2009/09/13/4548930.aspx

数学表述为：h = H(M) ，其中H( )--单向散列函数，M--任意长度明文，h--固定长度散列值。

在信息安全领域中应用的

第 一当然是单向性(one-way)，从预映射，能够简单迅速的得到散列值，而在计算上不可能构造一个预映射，使其散列结果等于某个特定的散列值，即构造相 应的M=H-1(h)不可行。这样，散列值就能在统计上唯一的表征输入值，因此，密码学上的 Hash 又被称为"消息摘要(message digest)"，就是要求能方便的将"消息"进行"摘要"，但在"摘要"中无法得到比"摘要"本身更多的关于"消息"的信息。

第 二是抗冲突性(collision-resistant)，即在统计上无法产生2个散列值相同的预映射。给定M，计算上无法找到M ，满足H(M)=H(M ) ，此谓弱抗冲突性；计算上也难以寻找一对任意的M和M ，使满足H(M)=H(M ) ，此谓强抗冲突性。要求"强抗冲突性"主要是为了防范所谓"生日攻击(birthday attack)"，在一个10人的团体中，你能找到和你生日相同的人的概率是2.4%，而在同一团体中，有2人生日相同的概率是11.7%。类似的，当预 映射的空间很大的情况下，算法必须有足够的强度来保证不能轻易找到"相同生日"的人。

第 三是映射分布均匀性和差分分布均匀性，散列结果中，为 0 的 bit 和为 1 的 bit ，其总数应该大致相等；输入中一个 bit 的变化，散列结果中将有一半以上的 bit 改变，这又叫做"雪崩效应(avalanche effect)"；要实现使散列结果中出现 1bit 的变化，则输入中至少有一半以上的 bit 必须发生变化。其实质是必须使输入中每一个 bit 的信息，尽量均匀的反映到输出的每一个 bit 上去；输出中的每一个 bit，都是输入中尽可能多 bit 的信息一起作用的结果。

Damgard 和 Merkle 定义了所谓“压缩函数(compression function)”，就是将一个固定长度输入，变换成较短的固定长度的输出，这对密码学实践上 Hash 函数的设计产生了很大的影响。Hash函数就是被设计为基于通过特定压缩函数的不断重复“压缩”输入的分组和前一次压缩处理的结果的过程，直到整个消息都 被压缩完毕，最后的输出作为整个消息的散列值。尽管还缺乏严格的证明，但绝大多数业界的研究者都同意，如果压缩函数是安全的，那么以上述形式散列任意长度 的消息也将是安全的。这就是所谓 Damgard/Merkle 结构：

在 下图中，任意长度的消息被分拆成符合压缩函数输入要求的分组，最后一个分组可能需要在末尾添上特定的填充字节，这些分组将被顺序处理，除了第一个消息分组 将与散列初始化值一起作为压缩函数的输入外，当前分组将和前一个分组的压缩函数输出一起被作为这一次压缩的输入，而其输出又将被作为下一个分组压缩函数输 入的一部分，直到最后一个压缩函数的输出，将被作为整个消息散列的结果。

MD5 和 SHA1 可以说是目前应用最广泛的Hash算法，而它们都是以 MD4 为基础设计的。

1) MD4
MD4(RFC 1320)是 MIT 的 Ronald L. Rivest 在 1990 年设计的，MD 是 Message Digest 的缩写。它适用在32位字长的处理器上用高速软件实现--它是基于 32 位操作数的位操作来实现的。它的安全性不像RSA那样基于数学假设，尽管 Den Boer、Bosselaers 和 Dobbertin 很快就用分析和差分成功的攻击了它3轮变换中的 2 轮，证明了它并不像期望的那样安全，但它的整个算法并没有真正被破解过，Rivest 也很快进行了改进。

下面是一些MD4散列结果的例子：

MD4 ("") = 31d6cfe0d16ae931b73c59d7e0c089c0
MD4 ("a") = bde52cb31de33e46245e05fbdbd6fb24
MD4 ("abc") = a448017aaf21d8525fc10ae87aa6729d
MD4 ("message digest") = d9130a8164549fe818874806e1c7014b
MD4 ("abcdefghijklmnopqrstuvwxyz") = d79e1c308aa5bbcdeea8ed63df412da9
MD4 ("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789") = 043f8582f241db351ce627e153e7f0e4
MD4 ("12345678901234567890123456789012345678901234567890123456789012345678901234567890") = e33b4ddc9c38f2199c3e7b164fcc0536

2) MD5
MD5(RFC 1321)是 Rivest 于1991年对MD4的改进版本。它对输入仍以512位分组，其输出是4个32位字的级联，与 MD4 相同。它较MD4所做的改进是：

1) 加入了第四轮
2) 每一步都有唯一的加法常数；
3) 第二轮中的G函数从((X ∧ Y) ∨ (X ∧ Z) ∨ (Y ∧ Z)) 变为 ((X ∧ Z) ∨ (Y ∧ ～Z))以减小其对称性；
4) 每一步都加入了前一步的结果，以加快"雪崩效应"；
5) 改变了第2轮和第3轮中访问输入子分组的顺序，减小了形式的相似程度；
6) 近似优化了每轮的循环左移位移量，以期加快"雪崩效应"，各轮的循环左移都不同。
尽管MD5比MD4来得复杂，并且速度较之要慢一点，但更安全，在抗分析和抗差分方面表现更好。

消 息首先被拆成若干个512位的分组，其中最后512位一个分组是“消息尾+填充字节(100…0)+64 位消息长度”，以确保对于不同长度的消息，该分组不相同。64位消息长度的限制导致了MD5安全的输入长度必须小于264bit，因为大于64位的长度信 息将被忽略。而4个32位寄存器字初始化为A=0x01234567，B=0x89abcdef，C=0xfedcba98，D=0x76543210， 它们将始终参与运算并形成最终的散列结果。

接着各个512位消息分组以16个32位字的形式进入算法的主循环，512位消息分组的个数据决定了循环的次数。主循环有4轮，每轮分别用到了非线性函数

F(X, Y, Z) = (X ∧ Y) ∨ (～X ∧ Z)
G(X, Y, Z) = (X ∧ Z) ∨ (Y ∧ ～Z)
H(X, Y, Z) =X ⊕ Y ⊕ Z
I(X, Y, Z) = X ⊕ (Y ∨ ～Z)
这 4轮变换是对进入主循环的512位消息分组的16个32位字分别进行如下操作：将A、B、C、D的副本a、b、c、d中的3个经F、G、H、I运算后的结 果与第4个相加，再加上32位字和一个32位字的加法常数，并将所得之值循环左移若干位，最后将所得结果加上a、b、c、d之一，并回送至ABCD，由此 完成一次循环。

所用的加法常数由这样一张表T[i]来定义，其中i为1…64，T[i]是i的正弦绝对值之4294967296次方的整数部分，这样做是为了通过正弦函数和幂函数来进一步消除变换中的线性性。

当所有512位分组都运算完毕后，ABCD的级联将被输出为MD5散列的结果。下面是一些MD5散列结果的例子：

MD5 ("") = d41d8cd98f00b204e9800998ecf8427e
MD5 ("a") = 0cc175b9c0f1b6a831c399e269772661
MD5 ("abc") = 900150983cd24fb0d6963f7d28e17f72
MD5 ("message digest") = f96b697d7cb7938d525a2f31aaf161d0
MD5 ("abcdefghijklmnopqrstuvwxyz") = c3fcd3d76192e4007dfb496cca67e13b
MD5 ("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789") = d174ab98d277d9f5a5611c2c9f419d9f
MD5 ("12345678901234567890123456789012345678901234567890123456789012345678901234567890") = 57edf4a22be3c955ac49da2e2107b67a
参考相应RFC文档可以得到MD4、MD5算法的详细描述和算法的C源代码。

3) SHA1 及其他
SHA1 是由NIST NSA设计为同DSA一起使用的，访问http://www.itl.nist.gov/fipspubs可以得到它的详细规范 --[/url]"FIPS PUB 180-1 SECURE HASH STANDARD"。它对长度小于264的输入，产生长度为160bit的散列值，因此抗穷举(brute-force)性更好。SHA-1 设计时基于和MD4相同原理,并且模仿了该算法。因为它将产生160bit的散列值，因此它有5个参与运算的32位寄存器字，消息分组和填充方式与MD5 相同，主循环也同样是4轮，但每轮进行20次操作，非线性运算、移位和加法运算也与MD5类似，但非线性函数、加法常数和循环左移操作的设计有一些区别， 可以参考上面提到的规范来了解这些细节。下面是一些SHA1散列结果的例子：

SHA1 ("abc") = a9993e36 4706816a ba3e2571 7850c26c 9cd0d89d
SHA1 ("abcdbcdecdefdefgefghfghighijhijkijkljklmklmnlmnomnopnopq") = 84983e44 1c3bd26e baae4aa1 f95129e5 e54670f1
其他一些知名的Hash算法还有MD2、N-Hash、RIPE-MD、HAVAL等等。上面提到的这些都属于"纯"Hash算法。还有另2类Hash算法， 一类就是基于对称分组算法的单向散列算法，典型的例子是基于DES的所谓Davies-Meyer算法，另外还有经IDEA改进的Davies- Meyer算法，它们两者目前都被认为是安全的算法。另一类是基于模运算/离散对数的，也就是基于公开密钥算法的，但因为其运算开销太大，而缺乏很好的应 用前景。

没有通过分析和差分攻击考验的算法，大多都已经夭折在实验室里了，因此，如果目前流行的Hash算法能 完全符合密码学意义上的单向性和抗冲突性，就保证了只有穷举，才是破坏Hash运算安全特性的唯一方法。为了对抗弱抗冲突性，我们可能要穷举个数和散列值 空间长度一样大的输入，即尝试2^128或2^160个不同的输入，目前一台高档个人电脑可能需要10^25年才能完成这一艰巨的工作，即使是最高端的并 行系统，这也不是在几千年里的干得完的事。而因为"生日攻击"有效的降低了需要穷举的空间，将其降低为大约1.2*2^64或1.2*2^80，所以，强 抗冲突性是决定Hash算法安全性的关键。

在NIST新的 Advanced Encryption Standard (AES)中，使用了长度为128、192、256bit 的密钥，因此相应的设计了 SHA256、SHA384、SHA512，它们将提供更好的安全性。

Hash算法在信息安全方面的应用主要体现在以下的3个方面：

1) 文件校验
我们比较熟悉的校验算法有奇偶校验和CRC校验，这2种校验并没有抗数据篡改的能力，它们一定程度上能检测并纠正数据传输中的信道误码，但却不能防止对数据的恶意破坏。

MD5 Hash算法的"数字指纹"特性，使它成为目前应用最广泛的一种文件完整性校验和(Checksum)算法，不少Unix系统有提供计算md5 checksum的命令。它常被用在下面的2种情况下：

第 一是文件传送后的校验，将得到的目标文件计算 md5 checksum，与源文件的md5 checksum 比对，由两者 md5 checksum 的一致性，可以从统计上保证2个文件的每一个码元也是完全相同的。这可以检验文件传输过程中是否出现错误，更重要的是可以保证文件在传输过程中未被恶意篡 改。一个很典型的应用是ftp服务，用户可以用来保证多次断点续传，特别是从镜像站点下载的文件的正确性。

更 出色的解决方法是所谓的代码签名，文件的提供者在提供文件的同时，提供对文件Hash值用自己的代码签名密钥进行数字签名的值，及自己的代码签名证书。文 件的接受者不仅能验证文件的完整性，还可以依据自己对证书签发者和证书拥有者的信任程度，决定是否接受该文件。浏览器在下载运行插件和java小程序时， 使用的就是这样的模式。

第二是用作保存二进制文件系统的数字指纹，以便检 测文件系统是否未经允许的被修改。不少系统管理/系统安全软件都提供这一文件系统完整性评估的功能，在系统初始安装完毕后，建立对文件系统的基础校验和数 据库，因为散列校验和的长度很小，它们可以方便的被存放在容量很小的存储介质上。此后，可以定期或根据需要，再次计算文件系统的校验和，一旦发现与原来保 存的值有不匹配，说明该文件已经被非法修改，或者是被病毒感染，或者被木马程序替代。TripWire就提供了一个此类应用的典型例子。

更 完美的方法是使用"MAC"。"MAC" 是一个与Hash密切相关的名词，即信息鉴权码(Message Authority Code)。它是与密钥相关的Hash值，必须拥有该密钥才能检验该Hash值。文件系统的数字指纹也许会被保存在不可信任的介质上，只对拥有该密钥者提 供可鉴别性。并且在文件的数字指纹有可能需要被修改的情况下，只有密钥的拥有者可以计算出新的散列值，而企图破坏文件完整性者却不能得逞。

2) 数字签名
Hash 算法也是现代密码体系中的一个重要组成部分。由于非对称算法的运算速度较慢，所以在数字签名协议中，单向散列函数扮演了一个重要的角色。

在这种签名协议中，双方必须事先协商好双方都支持的Hash函数和签名算法。

签名方先对该数据文件进行计算其散列值，然后再对很短的散列值结果--如Md5是16个字节，SHA1是20字节，用非对称算法进行数字签名操作。对方在验证签名时，也是先对该数据文件进行计算其散列值，然后再用非对称算法验证数字签名。

对 Hash 值，又称"数字摘要"进行数字签名，在统计上可以认为与对文件本身进行数字签名是等效的。而且这样的协议还有其他的优点：

首先，数据文件本身可以同它的散列值分开保存，签名验证也可以脱离数据文件本身的存在而进行。

再 者，有些情况下签名密钥可能与解密密钥是同一个，也就是说，如果对一个数据文件签名，与对其进行非对称的解密操作是相同的操作，这是相当危险的，恶意的破 坏者可能将一个试图骗你将其解密的文件，充当一个要求你签名的文件发送给你。因此，在对任何数据文件进行数字签名时，只有对其Hash值进行签名才是安全 的。

3) 鉴权协议
如下的鉴权协议又被称作"挑战--认证模式：在传输信道是可被侦听，但不可被篡改的情况下，这是一种简单而安全的方法。

需 要鉴权的一方，向将被鉴权的一方发送随机串（“挑战”），被鉴权方将该随机串和自己的鉴权口令字一起进行 Hash 运算后，返还鉴权方，鉴权方将收到的Hash值与在己端用该随机串和对方的鉴权口令字进行 Hash 运算的结果相比较（“认证”），如相同，则可在统计上认为对方拥有该口令字，即通过鉴权。

POP3协议中就有这一应用的典型例子：

S: +OK POP3 server ready <1896.697170952@dbc.mtview.ca.us>
C: APOP mrose c4c9334bac560ecc979e58001b3e22fb
S: +OK maildrop has 1 message (369 octets)
在 上面的一段POP3协议会话中，双方都共享的对称密钥（鉴权口令字）是tanstaaf，服务器发出的挑战 是<1896.697170952@dbc.mtview.ca.us>，客户端对挑战的应答是 MD5("<1896.697170952@dbc.mtview.ca.us>tanstaaf") = c4c9334bac560ecc979e58001b3e22fb，这个正确的应答使其通过了认证。

散列算法长期以来一直在计算机科学中大量应用，随着现代密码学的发展，单向散列函数已经成为信息安全领域中一个重要的结构模块，我们有理由深入研究其设计理论和应用方法。

    The algorithm was conceived by Donald Knuth and Vaughan Pratt and independently by J. H. Morris in 1977, but the three published it jointly.

In this article, we will be using zero-based arrays for our strings; thus the 'C' in W (shown below) will be denoted W[2].

To illustrate the algorithm's details, we work through a (relatively artificial) run of the algorithm. At any given time, the algorithm is in a state determined by two integers, m and i, which denote respectively the position within S which is the beginning of a prospective match for W, and the index in W denoting the character currently under consideration. This is depicted, at the start of the run, like

             1         2
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W: ABCDABD
i: 0123456

We proceed by comparing successive characters of W to "parallel" characters of S, moving from one to the next if they match. However, in the fourth step, we get S[3] is a space and W[3] = 'D', a mismatch. Rather than beginning to search again at S[1], we note that no 'A' occurs between positions 0 and 3 in S except at 0; hence, having checked all those characters previously, we know there is no chance of finding the beginning of a match if we check them again. Therefore we move on to the next character, setting m = 4 and i = 0.

             1         2
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:     ABCDABD
i:     0123456

We quickly obtain a nearly complete match "ABCDAB" when, at W[6] (S[10]), we again have a discrepancy. However, just prior to the end of the current partial match, we passed an "AB" which could be the beginning of a new match, so we must take this into consideration. As we already know that these characters match the two characters prior to the current position, we need not check them again; we simply reset m = 8, i = 2 and continue matching the current character. Thus, not only do we omit previously matched characters of S, but also previously matched characters of W.

             1         2
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:         ABCDABD
i:         0123456

This search fails immediately, however, as the pattern still does not contain a space, so as in the first trial, we return to the beginning of W and begin searching at the next character of S: m = 11, reset i = 0.

             1         2
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:            ABCDABD
i:            0123456

Once again we immediately hit upon a match "ABCDAB" but the next character, 'C', does not match the final character 'D' of the word W. Reasoning as before, we set m = 15, to start at the two-character string "AB" leading up to the current position, set i = 2, and continue matching from the current position.

             1         2
m: 01234567890123456789012
S: ABC ABCDAB ABCDABCDABDE
W:                ABCDABD
i:                0123456

This time we are able to complete the match, whose first character is S[15].

The above example contains all the elements of the algorithm. For the moment, we assume the existence of a "partial match" table T, described below, which indicates where we need to look for the start of a new match in the event that the current one ends in a mismatch. The entries of T are constructed so that if we have a match starting at S[m] that fails when comparing S[m + i] to W[i], then the next possible match will start at index m + i - T[i] in S (that is, T[i] is the amount of "backtracking" we need to do after a mismatch). This has two implications: first, T[0] = -1, which indicates that if W[0] is a mismatch, we cannot backtrack and must simply check the next character; and second, although the next possible match will begin at index m + i - T[i], as in the example above, we need not actually check any of the T[i] characters after that, so that we continue searching from W[T[i]]. The following is a sample pseudocode implementation of the KMP search algorithm.

algorithm kmp_search:
input:
an array of characters, S (the text to be searched)
an array of characters, W (the word sought)
output:
an integer (the zero-based position in S at which W is found)
define variables:
an integer, m ← 0 (the beginning of the current match in S)
an integer, i ← 0 (the position of the current character in W)
an array of integers, T (the table, computed elsewhere)
while m + i is less than the length of S, 
do
    if W[i] = S[m + i],
       let i ← i + 1
       if i equals the length of W,
       return m
    otherwise{
        let m ← m + i - T[i],
       // this if else statement is utilized to reposition i
       if T[i] is greater than -1,
          let i ← T[i]
       else
          let i ← 0
}
done (if we reach here, we have searched all of S unsuccessfully)
return the length of S, that is m+i. 

Assuming the prior existence of the table T, the search portion of the Knuth–Morris–Pratt algorithm has complexity O(k), where k is the length of S and the O is big-O notation. As except for the fixed overhead incurred in entering and exiting the function, all the computations are performed in the while loop, we will calculate a bound on the number of iterations of this loop; in order to do this we first make a key observation about the nature of T. By definition it is constructed so that if a match which had begun at S[m] fails while comparing S[m + i] to W[i], then the next possible match must begin at S[m + (i - T[i])]. In particular the next possible match must occur at a higher index than m, so that T[i] < i.

Using this fact, we will show that the loop can execute at most 2k times. For in each iteration, it executes one of the two branches in the loop. The first branch invariably increases i and does not change m, so that the index m + i of the currently scrutinized character of S is increased. The second branch adds i - T[i] to m, and as we have seen, this is always a positive number. Thus the location m of the beginning of the current potential match is increased. Now, the loop ends if m + i = k; therefore each branch of the loop can be reached at most k times, since they respectively increase either m + i or m, and m ≤ m + i: if m = k, then certainly m + i ≥ k, so that since it increases by unit increments at most, we must have had m + i = k at some point in the past, and therefore either way we would be done.

Thus the loop executes at most 2k times, showing that the time complexity of the search algorithm is O(k).

The goal of the table is to allow the algorithm not to match any character of S more than once. The key observation about the nature of a linear search that allows this to happen is that in having checked some segment of the main string against an initial segment of the pattern, we know exactly at which places a new potential match which could continue to the current position could begin prior to the current position. In other words, we "pre-search" the pattern itself and compile a list of all possible fallback positions that bypass a maximum of hopeless characters while not sacrificing any potential matches in doing so.

We want to be able to look up, for each position in W, the length of the longest possible initial segment of W leading up to (but not including) that position, other than the full segment starting at W[0] that just failed to match; this is how far we have to backtrack in finding the next match. Hence T[i] is exactly the length of the longest possible proper initial segment of W which is also a segment of the substring ending at W[i - 1]. We use the convention that the empty string has length 0. Since a mismatch at the very start of the pattern is a special case (there is no possibility of backtracking), we set T[0] = -1, as discussed above.

We consider the example of W = "ABCDABD" first. We will see that it follows much the same pattern as the main search, and is efficient for similar reasons. We set T[0] = -1. To find T[1], we must discover a proper suffix of "A" which is also a prefix of W. But there are no proper suffixes of "A", so we set T[1] = 0. Likewise, T[2] = 0.

Continuing to T[3], we note that there is a shortcut to checking all suffixes: let us say that we discovered a proper suffix which is a proper prefix and ending at W[2] with length 2 (the maximum possible); then its first character is a proper prefix of a proper prefix of W, hence a proper prefix itself, and it ends at W[1], which we already determined cannot occur in case T[2]. Hence at each stage, the shortcut rule is that one needs to consider checking suffixes of a given size m>1 only if a valid suffix of size m-1 was found at the previous stage (e.g. T[x]=m-1).

Therefore we need not even concern ourselves with substrings having length 2, and as in the previous case the sole one with length 1 fails, so T[3] = 0.

We pass to the subsequent W[4], 'A'. The same logic shows that the longest substring we need consider has length 1, and although in this case 'A' does work, recall that we are looking for segments ending before the current character; hence T[4] = 0 as well.

Considering now the next character, W[5], which is 'B', we exercise the following logic: if we were to find a subpattern beginning before the previous character W[4], yet continuing to the current one W[5], then in particular it would itself have a proper initial segment ending at W[4] yet beginning before it, which contradicts the fact that we already found that 'A' itself is the earliest occurrence of a proper segment ending at W[4]. Therefore we need not look before W[4] to find a terminal string for W[5]. Therefore T[5] = 1.

Finally, we see that the next character in the ongoing segment starting at W[4] = 'A' would be 'B', and indeed this is also W[5]. Furthermore, the same argument as above shows that we need not look before W[4] to find a segment for W[6], so that this is it, and we take T[6] = 2.

Therefore we compile the following table:

The example above illustrates the general technique for assembling the table with a minimum of fuss. The principle is that of the overall search: most of the work was already done in getting to the current position, so very little needs to be done in leaving it. The only minor complication is that the logic which is correct late in the string erroneously gives non-proper substrings at the beginning. This necessitates some initialization code.

algorithm kmp_table:
input:
an array of characters, W (the word to be analyzed)
an array of integers, T (the table to be filled)
output:
nothing (but during operation, it populates the table)
define variables:
an integer, pos ← 2 (the current position we are computing in T)
an integer, cnd ← 0 (the zero-based index in W of the next 
character of the current candidate substring)
(the first few values are fixed but different from what the algorithm 
might suggest)
let T[0] ← -1, T[1] ← 0
while pos is less than the length of W, do:
(first case: the substring continues)
if W[pos - 1] = W[cnd], 
          let T[pos] ← cnd + 1, pos ← pos + 1, cnd ← cnd + 1
(second case: it doesn't, but we can fall back)
otherwise, if cnd > 0, let cnd ← T[cnd]
(third case: we have run out of candidates.  Note cnd = 0)
otherwise, let T[pos] ← 0, pos ← pos + 1

The complexity of the table algorithm is O(n), where n is the length of W. As except for some initialization all the work is done in the while loop, it is sufficient to show that this loop executes in O(n) time, which will be done by simultaneously examining the quantities pos and pos - cnd. In the first branch, pos - cnd is preserved, as both pos and cnd are incremented simultaneously, but naturally, pos is increased. In the second branch, cnd is replaced by T[cnd], which we saw above is always strictly less than cnd, thus increasing pos - cnd. In the third branch, pos is incremented and cnd is not, so both pos and pos - cnd increase. Since pos ≥ pos - cnd, this means that at each stage either pos or a lower bound for pos increases; therefore since the algorithm terminates once pos = n, it must terminate after at most 2n iterations of the loop, since pos - cnd begins at 1. Therefore the complexity of the table algorithm is O(n).

Since the two portions of the algorithm have, respectively, complexities of O(k) and O(n), the complexity of the overall algorithm is O(n + k).

As is apparent from the worked example, the algorithm makes its greatest gains over a naive string-searching algorithm when it can skip the most characters at a time. That is, the less it has to backtrack, the faster it goes, which is reflected in the table T by the presence of zeroes. A word such as "ABCDEFG" works well with this algorithm because it has no repetitions of its beginning, so that its table is simply all zeroes with a -1 at the beginning. On the opposite end of the spectrum, W = "AAAAAAA" works terribly, because its table is

This is the worst possible pattern for T, and it can be exploited by a word such as S = "AAAAAABAAAAAABAAAAAAA", in which the algorithm tries to match each 'A' against each 'B' before giving up; this results in the maximum possible number of iterations, verging on twice the number of characters of S as the number of repetitions of "AAAAAAB" increases. Although the table-building routine is rapid for this word (since no backtracking ever occurs), this routine is run only once for a given W, while the search routine is potentially run many times. If each time, this W is searched for in strings like S, the overall efficiency will be as poor as possible. By way of comparison, this particular combination is a best case for the Boyer-Moore string search algorithm.

Note that in practice, the KMP algorithm is not good at searching in natural language texts because it can only skip characters when the first part of the pattern actually matches a part of the text. In practice this only happens occasionally in natural language texts. For instance, consider how many times a prefix of the pattern "text" matches something in this paragraph.

    1.哈希函数的构造方法

    构造哈希函数的方法有很多，这里介绍几种常用的。

直接定址法:H(k)=k 或H(k)=a*k+b(线形函数)

如:人口数字统计表

数字分析法:取关键字的若干数位组成哈希地址

如：关键字如下：若哈希表长为100则可取中间两位10进制数作为哈希地址。  

平方取中法：关键字平方后取中间几位数组成哈希地址

折叠法：将关键数字分割成位数相同的几部分（最后一部分的位数可以不同）然后取几部分的叠加和（舍去进位）作为哈希地址。

除留余数法：取关键字被某个不大于表长m的数p除后所得的余数为哈希地址。

           H(k)=k mod p  p<=m

随机数法：H(k)=rondom(k)。

    2.处理冲突的方法

    假设地址集为0..n-1,由关键字得到的哈希地址为j(0<=j<=n-1)的位置已存有记录,处理冲突就是为该关键字的记录找到另一个" 空"的哈希地址。在处理中可能得到一个地址序列Hi i=1，2，...k 0<=Hi<=n-1),即在处理冲突时若得到的另一个哈希地址H1仍发生冲突,再求下一地址H2,若仍冲突,再求H3...。怎样得到Hi 呢？

开放定址法：Hi=(H(k)+di) mod m  （H(k)为哈希函数；m为哈希表长；di为增量序列）

当di=1，2，3，... m-1 时叫线性探测再散列。

当di=1²，-1²，2²，-2²，3²，-3²，...,k²,-k²时叫二次探测再散列。

当di=random（m）时叫伪随机探测序列。

例：长度为11的哈希表关键字分别为17，60，29，哈希函数为H(k)=k mod 11,第四个记录的关键字为38，分别按上述方法添入哈希表的地址为8，4，3（随机数=9）。---为什么不是6，5，7呢

再哈希法：Hi=RHi(key) i=1,2,...,k,其中RHi均为不同的哈希函数。

链地址法：这种方法很象基数排序，相同的地址的关键字值均链入对应的链表中。

建立公益区法：另设一个溢出表，不管得到的哈希地址如何，一旦发生冲突，都填入溢出表。

    3.哈希表的查找

例:如下一组关键字按哈希函数H(k)=k mod 13和线性探测处理冲突所得的哈希表a[0..15]:

当给定值k=84,则首先和a[6]比，再依次和a[7],a[8]比，结果a[8]=84查找成功。

当给定值k=38,则首先和a[12]比,再和a[13]比,由于a[13]没有,查找不成功,表中不存在关键字等于38的记录。

其基本性质可以归纳为：
1. 根节点不包含字符，除根节点外每一个节点都只包含一个字符。
2. 从根节点到某一节点，路径上经过的字符连接起来，为该节点对应的字符串。
3. 每个节点的所有子节点包含的字符都不相同。

其基本操作有:查找 插入和删除,当然删除操作比较少见.我在这里只是实现了对整个树的删除操作,至于单个word的删除操作也很简单.

搜索字典项目的方法为：

(1) 从根结点开始一次搜索；

(2) 取得要查找关键词的第一个字母，并根据该字母选择对应的子树并转到该子树继续进行检索；

/**//*
Name: Trie树的基本实现 
Author: MaiK 
Description: Trie树的基本实现 ,包括查找 插入和删除操作*/
#include<algorithm>
#include<iostream>
using namespace std;

const int sonnum=26,base='a';
struct Trie
{
    int num;//to remember how many word can reach here,that is to say,prefix
    bool terminal;//If terminal==true ,the current point has no following point
    struct Trie *son[sonnum];//the following point
};
Trie *NewTrie()// create a new node
{
    Trie *temp=new Trie;
    temp->num=1;temp->terminal=false;
    for(int i=0;i<sonnum;++i)temp->son[i]=NULL;
    return temp;
}
void Insert(Trie *pnt,char *s,int len)// insert a new word to Trie tree
{
    Trie *temp=pnt;
    for(int i=0;i<len;++i)
    {
        if(temp->son[s[i]-base]==NULL)temp->son[s[i]-base]=NewTrie();
        else temp->son[s[i]-base]->num++;
        temp=temp->son[s[i]-base];
    }
    temp->terminal=true;
}
void Delete(Trie *pnt)// delete the whole tree
{
    if(pnt!=NULL)
    {
        for(int i=0;i<sonnum;++i)if(pnt->son[i]!=NULL)Delete(pnt->son[i]);
        delete pnt; 
        pnt=NULL;
    }
}
Trie* Find(Trie *pnt,char *s,int len)//trie to find the current word
{
    Trie *temp=pnt;
    for(int i=0;i<len;++i)
        if(temp->son[s[i]-base]!=NULL)temp=temp->son[s[i]-base];
        else return NULL;
    return temp;
} 

厂长分配任务给这1万个工人干，按工人编号一个一个来，到最后一个工人就又从头开始，任务完成时间各不相同，
可能一个工人在分配任务的时候手里还有任务， 就得换下一个。

但是这1万个工人都很懒，领到了任务先不做，需要监工1个1个去问，如果工人有任务，就做，如果工人没任务，则不做。 
厂长只管分任务，1个1个来，可能几天也没新任务，不累； 
但是监工很累，监工每天都要看所有工人的情况，即使这些工人都没有任务, 实际上每天工人（80%左右）是没任务的，
请问，怎么让监工的工作轻松下来. 比如说每天只问1小半工人.

欧几里德算法
欧几里德算法又称辗转相除法，用于计算两个整数a,b的最大公约数。其计算原理依赖于下面的定理：

定理：gcd(a,b) = gcd(b,a mod b)

证明：a可以表示成a = kb + r，则r = a mod b
假设d是a,b的一个公约数，则有
d|a, d|b，而r = a - kb，因此d|r
因此d是(b,a mod b)的公约数

假设d 是(b,a mod b)的公约数，则
d | b , d |r ，但是a = kb +r
因此d也是(a,b)的公约数

因此(a,b)和(b,a mod b)的公约数是一样的，其最大公约数也必然相等，得证

欧几里德算法就是根据这个原理来做的，其算法用C++语言描述为：

int Gcd(int a, int b)
{
    if(b == 0)
        return a;
    return Gcd(b, a % b);
}

int Gcd(int a, int b)
{
    while(b != 0)
    {
        int r = b;
        b = a % b;
        a = r;
    }
    return a;
}

int exGcd(int a, int b, int &x, int &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int r = exGcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;

    return r;
}

有了上述规律就可以给出Stein算法如下：

如果A=0，B是最大公约数，算法结束
如果B=0，A是最大公约数，算法结束
设置A1 = A、B1=B和C1 = 1
如果An和Bn都是偶数，则An+1 =An /2，Bn+1 =Bn /2，Cn+1 =Cn *2(注意，乘2只要把整数左移一位即可，除2只要把整数右移一位即可)
如果An是偶数，Bn不是偶数，则An+1 =An /2，Bn+1 =Bn ，Cn+1 =Cn (很显然啦，2不是奇数的约数)
如果Bn是偶数，An不是偶数，则Bn+1 =Bn /2，An+1 =An ，Cn+1 =Cn (很显然啦，2不是奇数的约数)
如果An和Bn都不是偶数，则An+1 =|An -Bn|，Bn+1 =min(An,Bn)，Cn+1 =Cn
n++，转4
这个算法的原理很显然，所以就不再证明了。现在考察一下该算法和欧几里德方法效率上的差别。

给出一个C++的实现:

int Gcd(int a, int b)
{
    if(a == 0) return b;
    if(b == 0) return a;
    if(a % 2 == 0 && b % 2 == 0) return 2 * gcd(a >> 1, b >> 1);
    else if(a % 2 == 0)  return gcd(a >> 1, b);
    else if(b % 2 == 0) return gcd(a, b >> 1);
    else return gcd(abs(a - b), Min(a, b));
}

int SumYe(BiTree T)
{//求二叉树叶结点数之和
 if(!T) return 0;
 if(!T->lchild && !T->rchild ) return 1;
 return SumYe(T->lchild)+SumYe(T->rchild);
}

int HightTree(BiTree T)
{//求二叉树高
 int hl = 0;//记录左子树高
 int hr = 0;//记录右子树高
 if(!T)  return 0;
 hl = HightTree(T->lchild);
 hr = HightTree(T->rchild);
 return (hl>hr) ? hl+1 : hr+1 ;
}