In [1]:
%matplotlib inline
import pandas as pd

import numpy as np
from __future__ import division
import itertools

import matplotlib.pyplot as plt
import seaborn as sns

import logging
logger = logging.getLogger()

17 Amortized Analysis

In an amortized analysis, we averget the time required to perform a sequence of data-structure operations over all the operations performed.

Amortized analsis differs from average-case analysis in that probability is not involved; an amortized analysis guarantees the _average performance of each operation in the worst case__.

Bear in mind that the charges assigned during an amortized analysis are for analysis purposes only.

When we perform an amortized analysis, we often gain insight into a particular data structure, and this insight can help us optimize the design.

17.1 Aggregate analysis

We show that for all $n$, a sequence of $n$ operations takes worst-case time $T(n)$ in total.

Stack operations

$1 \times O(n) = O(n)$

Incrementing a binary counter
In [2]:
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In general, for $i = 0, 1, \dotsc, k-1$, bit $A[i]$ flips $\lfloor \frac{n}{2^i} \rfloor$ times in a sequence of $n$ INCREMENT operations on an initially zero counter.

\begin{align} \sum_{i=0}^{k-1} \lfloor \frac{n}{2^i} \rfloor &< n \sum_{i=0}^{\infty} \frac{1}{2^i} \\ &= 2n \end{align}

17.2 The accounting method

credit: the cost that an operation's amortized cost $\hat{c_i}$ exceeds its actual cost $c_i$.

requriments: $$\sum_{i=1}^{n} \hat{c_i} \geq \sum_{i=1}^{n} c_i$$

Stack operations
$c_i$ $\hat{c_i}$
PUSH 1 2
POP 1 0
MULTIPOP min(k,s) 0

$2 \times O(n) = O(n)$

Incrementing a binary counter

set a bit to 1: 2
set a bit to 0: 0

The INCREMENT procedure sets at most one bit, $2 \times O(n) = O(n)$

17.3 The potential method

Let $D_i$ be the data structure that results after applying the $i$th operation to data structure $D_{i-1}$.

potential function $\phi$: maps each data structure $D_i$ to a real number $\phi(D_i)$.

$\hat{c_i} = c_i + \phi(D_i) - \phi(D_{i-1})$
hence, the total amortized cost of the $n$ operations is: $$\sum_{i=1}^n \hat{c_i} = \sum_{i=1}^n c_i + \phi(D_n) - \phi(D_0)$$

Different potential functions may yield different amortized costs yet still be upper bounds on the actual costs. The best potential function to use depends on the disired time bounds.

Stack operations

define: $\phi$ to be the number of objects in the stack.

for PUSH: \begin{align} \hat{c_i} &= c_i + \phi(D_i) - \phi(D_{i-1}) \\ &= 1 + (s+1) - s \\ &= 2 \end{align}

for POP: \begin{align} \hat{c_i} &= c_i + \phi(D_i) - \phi(D_{i-1}) \\ &= 1 + (s-1) - s \\ &= 0 \end{align}

for MULTIPOP: \begin{align} \hat{c_i} &= c_i + \phi(D_i) - \phi(D_{i-1}) \\ &= k + (s-k) - s \\ &= 0 \end{align}

Incrementing a binary counter

define: $\phi$ to be $b_i$, the number of 1s in the counter after the $i$th operation.

Suppose: the $i$th INCREMENT operation reset $t_i$ bits.

for INCREMENT: \begin{align} \hat{c_i} &= c_i + \phi(D_i) - \phi(D_{i-1}) \\ &= (t_i + 1) + (b_{i-1} - t_i + 1) - b_{i-1} \\ &= 2 \end{align}

17.4 Dynamic tables

load factor: $$\alpha(T) = \frac{\|\text{items of T}\|}{\|T\|}$$

17.4.1 Table expansion

insert an item into a full table, we expand the table with twice spaces.

The cost of the $i$th operation is: \begin{equation} c_i = \begin{cases} i \quad & \text{expand: if i - 1 is an exact power of 2} \\ 1 \quad & \text{otherwise} \end{cases} \end{equation}

The total cost of $n$ TABLE-INSERT operations is therefore: \begin{align} \sum_{i=1}^{n} c_i &\leq n + \sum_{j=0}^{\lfloor \lg n \rfloor} 2^j \\ &< n + 2n \\ &= 3n \end{align}

17.4.2 Table expansion and contraction

Halve the table size when deleting an item causes the table to become less than 1/4 full, rather than 1/2 full as before(引起振荡).