平摊分析、表的扩增、势能方法

平摊分析、表的扩增、势能方法

hash表需要多大？

越大越好(time)

尽可能小(space )

$\Theta $(n) for n items

Problem :don't know n?

Solution: dynamic tables

overflow => grow it

• 1.Reallocate (molloc ot new) 2*n ,double it

• 2.move to new

• 3.free old table

Analytics

Seq of insert ops

Worst cost of 1 insert \(\Theta\)(n)

=>Worst cost of n inserts = n* \(\Theta\)(n)=\(\Theta\)(\(n^{2}\))

Wrong :Worst cost of n inserts = \(\Theta\)(n)

Let $c_i $ = i if i- is exaotct power of 2 , = 1 otherwise

i. 1. 2. 3 4 5 6 7. 8 9 10

size. 1. 2. 4. 4 8 8 8 8 16 16

c_i. 1. 2. 3. 1 5 1 1 1 9 1

​ 1 1 1 1 1 1 1 1. 1. 1

​ 1 2. 4 8

= \(\sum _{i = 1}^{n} =n + \sum_{j = 0}^{log(n-1)}2^{j}\\<=3n=\Theta(n)\)

Thus ,average cost of n insert = \(\Theta (n)/n=\Theta(1)\)

Amortized analysis

Analyze a seq of ops to shwo that ave cost per op is small, even though 1 op may be expensive

no probability - ave perf in worst case

Types of amortized arguments

• Aggregate (just -saw) 聚集分析

• Accounting 记账方法. }

• potential 势能分析 }. 更精确，为每个操作分配了特定的平摊代价(不能轻松分析时)

Accounting method

• Charage i -th op a 虚构的平摊代价 cost \(\hat{c_i}\) ($1 pays for 1unit of work)

• 费用是支付给每个操作的，

• 剩余的钱将会存放在银行里，为了后续的操作

• bank balance not go neg，Must have \(\sum c_i <= \sum\hat{c_i}\) ,真实代价小于等于平摊代价

Dynamic table

Charge \(\hat{c_i}\) = $3 for i-th insert

$1 for an immediate insert

$2 ,用于将表翻倍的存储费用

When table double

$1 moves recent item

$1 moves old item

0	0	0	0

再插入一个剩下$2

0	0	0	0	2

再插入一个剩下$2

0	0	0	0	2	2

再插入一个..

0	0	0	0	2	2	2	2

加倍后，正好$8

0	0	0	0	0	0	0	0	2

0	0	0	0	0	0	0	0	2	2	2

这样总是够，银行总不负

所以\(\hat{c_i}\) 平摊代价

\(\sum c_i <= \sum \hat{c_i}\)

​ <= 3n

i.-----1-2-3--4-5-6-7--8--9---10

size--1-2-4--4-8-8-8--8--16---16

c_i---1-2-3--1-5-1-1--1---9----1

------1-1-1--1-1-1-1--1- -1- ---1

------1-2------4------8

\(\hat{c_i}\) --- 2 3. 3 3 3. 3 3 3. 3. 3

Bnk. 1 2 2 4 2. 4 6 8. 2 4

第一个可以收3，也可以

并且尝试全是2，不够

删除时要注意反复

Potential method

"Bank account " viewed as potential

energy of dynamic set

Framework

• Start with data struct \(D_0\)

• Op i transforms \(D_{i-1} => D_{i}\)

• Cost of operate \(c_i\)

=>Define potential function

找到函数\(\Phi - \{ D_i \}->R\) 将数据结构映射到实数

\(\Phi(D_0) = 0\quad and\quad \Phi(D_i) >=0\quad \forall i\)

(可能D_0不等于0，这只是一种)

定义平摊代价 \(\hat{c_i}\) for $\Phi $ is

\(\hat{c_i} = c_i + \Phi(D_i)-\Phi(D_{i-1})\)

​ potential difference,$ \Delta \Phi_i$

If $ \Delta \Phi_i > 0$ then \(\hat{c_i}\) > c_i

stores work in data structure for later

说明收取的费用多余实际费用，多余的费用存进银行变为势能

If $ \Delta \Phi_i < 0$ then $\hat{c_i} $> c_i

Structure delivers up work to help pay for op i

Total amor cost of n ops is

\(\sum_{i=1}^{n}\hat{c_i} = \sum_{i=1}^{n}( c_i + \Phi(D_i)-\Phi(D_{i-1}))\\ =\sum_{i=1}^{n}c_i+\Phi(D_n)-\Phi(D_0)\\ >=\sum_{i=1}^{n}c_i\)

For \(\Phi(D_n)>=0,\Phi(D_0) = 0\)

Table doubling

Define \(\Phi(D_i) = 2i-2^{\lceil log(i) \rceil}\)

Assume \(2^{\lceil log(0) \rceil} = 0\)

Note: \(\Phi(D_0) = 0,\Phi(D_i) >= 0\quad \forall i,\lceil log(i) \rceil\)<=log(i)+1

0	0	0	0	0	0

\(\Phi = 2*6-2^2=4\)

Counting : 0 0 0 0 2 2

Amortizations Cost of i -th insert

\(\hat{c_i} = c_i + \Phi(D_i)-\Phi(D_{i-1})\\ =i\quad if \ i-1 \ is\ power\ of\ 2,\\ \quad1\quad otherwise\\ + (2i-2^{\lceil log(i) \rceil})-( 2(i-1)-2^{\lceil log(i-1) \rceil})\\ =i\quad if \ i-1 \ is\ power\ of\ 2,\\ \quad1\quad otherwise\\ +2-2^{\lceil log(i) \rceil}+2^{\lceil log(i-1) \rceil}\)

Case 1, i -1 is exact power of 2

\(\hat{c_i} = i +2-2(i-1)+(i-1)\\ =3\)

Case 2,i -1 is not an exact power of 2

\(\hat{c_i} = 1 +2\\ =3\)

=>n inserts costs \(\Theta(n)\) in the worst case

(bug：第一个其实是2)

\(\hat{c_i}=1+2-1+0\\ =2\)

sum

• 平摊分析为数据结构的性能提供了一个简洁的抽象概念。

• 通常情况下每种都能用，每种都有某种情况是最简单或精确的。

• 不同的势能函数或不能的记账代价可能产生不同的上限