Question 1

Suppose that you time a program as a function of N and produce
the following table.

N seconds
-------------------
1024 0.000
2048 0.001
4096 0.007
8192 0.029
16384 0.121
32768 0.519
65536 2.156
131072 9.182
262144 38.784
524288 164.585
1048576 698.592
2097152 2958.141

Estimate the order of growth of the running time as a function of N.
Assume that the running time obeys a power law T(N) ~ a N^b. For your
answer, enter the constant b. Your answer will be marked as correct
if it is within 1% of the target answer - we recommend using
two digits after the decimal separator, e.g., 2.34.

Answer: 2.08

Question Explanation

The theoretical order-of-growth is N ^ (25/12) = 2.08
The empirical order-of-growth is N ^ (log_2 ratio)

log_2
N seconds ratio ratio
---------------------------------------
1024 0.000 - -
2048 0.001 - -
4096 0.007 7.00 2.81
8192 0.029 4.14 2.05
16384 0.121 4.17 2.06
32768 0.519 4.29 2.10
65536 2.156 4.15 2.05
131072 9.182 4.26 2.09
262144 38.784 4.22 2.08
524288 164.585 4.24 2.09
1048576 698.592 4.24 2.09
2097152 2958.141 4.23 2.08


Question 2

What is the order of growth of the worst case running time of the following code fragment
as a function of N?

int sum = 0;
for (int i = 0; i < N; i++)
    for (int j = i+1; j < N; j++)
    for (int k = j+1; k < N; k++)
      for (int h = k+1; h < N; h++)
         sum++;
Answer: N^4

int sum = 0;

for (int i = N*N; i > 1; i = i/2)
     sum++;

Answer: logN

The i loops iterates ~ lg (N^2) ~ 2 lg N times.

int sum = 0;
for (int i = 1; i <= N*N; i = i*2)
for (int j = 0; j < i; j++)
sum++;

Answer: N^2

The body of the innermost loop executes 1 + 2 + 4 + 8 + ... + N^2 ~ 2 N^2 times.


Question 3

Given the following definition of a MysteryBox object:

public class MysteryBox {
private final long x0, x1, x2, x3;
private final double y0, y1, y2, y3;
private final boolean z0;
private final int[] a = new int[320];

...
}

Using the 64-bit memory cost model from lecture, how many bytes does
each object of type MysteryBox use? Include all memory allocated when the
client calls new MysteryBox().

Answer: 1400

Question Explanation:

The correct answer is: 1400

public class MysteryBox {                           //   16 (object overhead)
private final long x0, x1, x2, x3; // 32 (4 long)
private final double y0, y1, y2, y3; // 32 (4 double)
private final boolean z0; // 1 (1 boolean)
private final int[] a = new int[320]; // 8 (reference to array)
// 1304 (int array of size 320)
... 7 (padding to round up to a multiple of 8)
} ----
1400

AlgorithmsI Exercises: Analysis of Algorithms的更多相关文章

  1. 算法分析 Analysis of Algorithms -------GeekforGeeker 翻译

    算法分析 Analysis of Algorithms 为什么要做性能分析?Why performance analysis? 在计算机领域有很多重要的因素我们要考虑 比如用户友好度,模块化, 安全性 ...

  2. 6.046 Design and Analysis of Algorithms

    课程信息 6.046 Design and Analysis of Algorithms

  3. "Mathematical Analysis of Algorithms" 阅读心得

    "Mathematical Analysis of Algorithms" 阅读心得 "Mathematical Analysis of Algorithms" ...

  4. 《Mathematical Analysis of Algorithms》中有关“选择第t大的数”的算法分析

    开头废话 这个问题是Donald.E.Knuth在他发表的论文Mathematical Analysis of Algorithms中提到的,这里对他的算法分析过程给出了更详细的解释. 问题描述: 给 ...

  5. 612.1.002 ALGS4 | Analysis of Algorithms

    我们生活在大数的时代 培养数量级的敏感! Tip:见招拆招 作为工程师,你先要能实现出来. 充实基础,没有什么不好意思 哪怕不完美.但是有时候完成比完美更重要. 之后再去想优化 P.S.作者Rober ...

  6. Analysis of Algorithms

    算法分析 Introduction 有各种原因要求我们分析算法,像预测算法性能,比较不同算法优劣等,其中很实际的一条原因是为了避免性能错误,要对自己算法的性能有个概念. 科学方法(scientific ...

  7. Analysis of algorithms: observation

    例子: 3-Sum 给定N个整数,这里面有多少个三元组,使其三个整数相加为0,如上面的例子为有4个三元组. 这个问题是许多问题如计算机几何,图形学等的基础. 用简单粗暴的方式来解决3-Sum问题 通过 ...

  8. Time complexity analysis of algorithms

    时间复杂性的计算一般而言,较小的问题所需要的运行时间通常要比较大的问题所需要的时间少.设一个程序P所占用的时间为T,则 T(P)=编译时间+运行时间. 编译时间与实例特征是无关的,且可假设一个编译过的 ...

  9. AlgorithmsI Exercises: UnionFind

    Question1 Give the id[] array that results from the following sequence of 6 unionoperations on a set ...

随机推荐

  1. Building Tomcat7 source step by step---官方文档

    Table of Contents Introduction Download a Java Development Kit (JDK) version 6 Install Apache Ant 1. ...

  2. spring mvc DispatcherServlet详解之三---request通过ModelAndView中获取View实例的过程

    整个spring mvc的架构如下图所示: 上篇文件讲解了DispatcherServlet第二步:通过request从Controller获取ModelAndView.现在来讲解第三步:reques ...

  3. Java基础知识强化之集合框架笔记47:Set集合之TreeSet保证元素唯一性和比较器排序的原理及代码实现(比较器排序:Comparator)

    1. 比较器排序(定制排序) 前面我们说到的TreeSet的自然排序是根据集合元素的大小,TreeSet将它们以升序排列. 但是如果需要实现定制排序,比如实现降序排序,则要通过比较器排序(定制排序)实 ...

  4. cordova 消息推送,告别,消息推送服务器,和 苹果推送证书

    cordova plugin add org.apache.cordova.vibration cordova plugin add https://github.com/katzer/cordova ...

  5. Linux下长时间ping网络加时间戳并记录到文本(转)

    [root@test ~]# ping 192.168.2.1 -c 10 PING 192.168.2.1 (192.168.2.1) 56(84) bytes of data.64 bytes f ...

  6. Css3渐变实例Demo(一)

    1.指定渐变背景的大小 .div { background: url(../img/1.jpg); /*background-size:contain;*/ width: 500px; height: ...

  7. angularjs-googleMap googleMap api地址解析与反解析

    1.js:根据地址得到经纬度var myplace=$scope.place;//获取输入的地址var geocoder = new google.maps.Geocoder();//创建geocod ...

  8. oracle 10g 恢复dmp文件。

    1. 在winxp下,安装10g,默认选择,一路ok.(安装前自检出现dhcp警告,可直接忽略) 2.命令行,在xp下,输入sqlplus,即可启动,登陆用 sqlplus / as sysdba 用 ...

  9. ORACLE 数据库简单测试

    ORACLE 数据库简单测试 操作系统:Windows 7 – ORACLE:oracle database 10.2.0.4 一.目的 测试 启动监听程序.数据库  非同一个用户的情况,用户是否可以 ...

  10. ajax跨域传值

    <script type="text/javascript"> function xmlpage(){ $.ajax({ url:'http://localhost/3 ...