AlgorithmsI Exercises: Analysis of Algorithms
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
- 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 ...
- 6.046 Design and Analysis of Algorithms
课程信息 6.046 Design and Analysis of Algorithms
- "Mathematical Analysis of Algorithms" 阅读心得
"Mathematical Analysis of Algorithms" 阅读心得 "Mathematical Analysis of Algorithms" ...
- 《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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- AlgorithmsI Exercises: UnionFind
Question1 Give the id[] array that results from the following sequence of 6 unionoperations on a set ...
随机推荐
- ARC和非ARC文件混编
在编程过程中,我们会用到很多各种各样的他人封装的第三方代码,但是有很多第三方都是在非ARC情况下运行的,当你使用第三方编译时出现和下图类似的错误,就说明该第三方是非ARC的,需要进行一些配置. 解决方 ...
- php 两个数组是否相同,并且输出全面的数据,相同的加一个字段标示
方法一: $date是数组,数组中有字段id,name; $data1是数组,数组中有字段sort_id,name; 所以要通过$date[$i]['id']==$data1[$j]['sort_id ...
- 10.8 noip模拟试题
1.花 (flower.cpp/c/pas) [问题描述] 商店里出售n种不同品种的花.为了装饰桌面,你打算买m支花回家.你觉得放两支一样的花很难看,因此每种品种的花最多买1支.求总共有几种不同的 ...
- JS cookie 读写操作
/*** ** 功能: cookie操作对象 ***/ var cookies = { /*** ** 功能: 写入cookie操作 ** 参数: name cookie名称 ** value coo ...
- 在DataTable中更新、删除数据
/*在DataTable中选择记录*/ /* 向DataTable中插入记录如上,更新和删除如下: * ----但是在更新和删除前,首先要找出要更新和删除 ...
- WCF,WebAPI,WCFREST和WebService的区别
Web ServiceIt is based on SOAP and return data in XML form.It support only HTTP protocol.It is not o ...
- Get file name without extension.
Ref:How to get file name without the extension? Normally,there are two ways to implements this:use t ...
- 详细查看数据库SQL执行计划
DBCC DROPCLEANBUFFERS 清除数据缓存DBCC FREEPROCCACHE 清除执行计划缓存 SET SHOWPLAN_XML ON 此语句导致 SQL Server 不执行 Tr ...
- Maven搭建Spring+Struts2+Hibernate项目详解
http://www.bubuko.com/infodetail-648898.html
- Deep Learning 学习随记(七)Convolution and Pooling --卷积和池化
图像大小与参数个数: 前面几章都是针对小图像块处理的,这一章则是针对大图像进行处理的.两者在这的区别还是很明显的,小图像(如8*8,MINIST的28*28)可以采用全连接的方式(即输入层和隐含层直接 ...