MIT算法导论课程：Lec08 全域哈希和完全哈希，对应书上的章节：Section 11.5
1. A weakness of hashing
Problem: For any hash function h, a set of keys exists that can cause the average access time of a hash table to skyrocket.存在一组keys插入同一个slot，影响速度。
IDEA：random
2. Universal hashing 全域哈希
定义： Let \(U\) be a universe of keys, and let \(H\) be a finite collection of hash functions, each mapping \(U\) to \(\{0, 1, …, m–1\}\). We say H is universal if for all \(x, y ∈ U\), where \(x ≠ y\), we have \(|{h\in H:h(x)=h(y)}|=|H|/m\) .
随机的从H中选取h，x和y发生collision的概率是1/m.
3. Universality theorem
Theorem: Let h be a hash function chosen (uniformly) at random from a universal set H of hash functions. Suppose h is used to hash n arbitrary keys into the m slots of a table T. Then, for a given key x, we have\(E[collisions\ with x]<n/m\). 其中 \(n/m\)：load factor.
为什么是小于？
Proof
Let \(C_x\) be the random variable denoting the total number of collisions of keys in T with x and let \(c_{x y}=\left\{\begin{array}{ll} 1 & \text { if } h(x)=h(y) \\ 0 & \text { otherwise. } \end{array}\right.\) ，\(C_x\)表示和\(x\)碰撞的key的总数。
可知：\(E[c_xy]=1/m\)，\(C_{x}=\sum_{y \in T-\{x\}} c_{x y}\)
可证： \[ \begin{aligned} E\left[C_{x}\right] &=E\left[\sum_{y \in T-\{x\}} c_{x y}\right] \\ &=\sum_{y \in T-\{x\}} E\left[c_{x y}\right] \\ &=\sum_{y \in T-\{x\}} 1 / m \\ &=\frac{n-1}{m} . \end{aligned} \]
4. Constructing a set of universal hash functions
Let m be prime.Decompose key k into r + 1 digits, each with value in the set {0, 1, …, m–1}. That is, let \(k = 〈 k_0, k_1, …, k_r 〉\), where \(0 ≤ ki < m\).将k进行“m进制表示”。
Randomized strategy
Pick \(a = 〈 a_0, a_1, …, a_r 〉\) where each \(a_i\) is chosen randomly from {0, 1, …, m–1}.
Define \(h_{a}(k)=\sum_{i=0}^{r} a_{i} k_{i} \bmod m\) 对k和a做点积再对m取余。
How big is \(H=\{h_a\}\), \(|H|=m^{r+1}\)
Theorem. The set \(H = {h_a}\) is universal.
Proof. Suppose that \(x = 〈x_0, x_1, …, x_r 〉\) and$ y = 〈y_0, y_1, …, y_r 〉$ be distinct keys. Thus, they differ in at least one digit position, wlog position 0.（假设为0位置）For how many \(h_a ∈ H\) do x and y collide. We must have\(h_a(x)=h_a(y)\). which implies that:\(\sum_{i=0}^{r} a_{i} x_{i} \equiv \sum_{i=0}^{r} a_{i} y_{i} \quad(\bmod m)\), \[ \sum_{i=0}^{r} a_{i}\left(x_{i}-y_{i}\right) \equiv 0 \quad(\bmod m) \]
\[ a_{0}\left(x_{0}-y_{0}\right)+\sum_{i=1}^{r} a_{i}\left(x_{i}-y_{i}\right) \equiv 0 \quad(\bmod m) \]
\[ a_{0}\left(x_{0}-y_{0}\right) \equiv-\sum_{i=1}^{r} a_{i}\left(x_{i}-y_{i}\right) \quad(\bmod m) \]
因为假设了\(x_0\neq y_0\)，所以存在\((x_0-y_0)^{-1}\)，所以： \[ a_{0} \equiv\left(-\sum_{i=1}^{r} a_{i}\left(x_{i}-y_{i}\right)\right) \cdot\left(x_{0}-y_{0}\right)^{-1} \quad(\bmod m) \] 可以看出\(a_0\)是由其他\(a_i\)决定的，其他\(a_i\)确定使得collide的\(a_0\)就确定了。
如果发生碰撞，the number of h为\(m^r*1=|H|/m\).证明为全域哈希。
Fact from number theory
Theorem. Let m be prime. For any \(z ∈ Z_m\) such that \(z ≠ 0\), there exists a unique \(z^{–1} ∈ Z_m\) such that:\(z \cdot z^{-1} \equiv 1 \quad(\bmod m)\)
如果不是质数就不成立。
5. Perfect hashing
Given a set of n keys, construct a static hash table of size \(m = O(n)\) such that SEARCH takes \(\Theta(1)\) time in the worst case.固定keys静态表\(O(1)\)时间查找。
IDEA: Two level scheme with universal hashing at both levels.No collisions at level 2.
如果有\(n_i\)个项被同时哈希到一级表的槽i，那么我们将有\(m_i=n_i^2\)个槽在二级表。
Theorem. Let H be a class of universal hash functions for a table of size \(m = n^2\). Then, if we use a random \(h ∈ H\) to hash n keys into the table, the expected number of collisions is at most 1/2. 如果使用全域哈希\(m = n^2\)，如果随机采样\(h ∈ H\) 期望碰撞数为1/2.
Proof
马尔可夫不等式（Markov’s inequality）
for any nonnegative random variable X, we have \(Pr\{X>=t\}<=E[X]/t\)
Corollary. The probability of no collisions is at least 1/2.
应用马尔可夫不等式，设置t=1， the probability of 1 or more collisions is at most 1/2。
Thus, just by testing random hash functions in H, we’ll quickly find one that works.
目标是创建静态表，所以这样证明很容易找到合适的哈希函数。
Analysis of storage
For the level-1 hash table T, choose m = n, and let ni be random variable for the number of keys that hash to slot i in T.By using \(n_i^2\) slots for the level-2 hash table Si, the expected total storage required for the two-level scheme is therefore \[ E\left[\sum_{i=0}^{m-1} \Theta\left(n_{i}^{2}\right)\right]=\Theta(n) \] since the analysis is identical to the analysis from recitation of the expected running time of bucket sort.
