Week 1 - Intro to Security
Rishi在课上给了一个取模算法 (Modular Arithmetic) 的例子。
课程大纲
Cryptography
Access Control
Introduction to Networking
Security Protocols
Web Systems and Attacks
Other Common Attacks and Defenses
课程目标
▶Understand basic concepts of cryptography and SQL
▶Understand basic concepts of cloud services, in particular storage
▶Demonstrate an understanding of the threats to data stored on a computer, locally or in the cloud
▶Demonstrate an understanding of the threats to data sent on the network
▶Identify risks and use techniques to eliminate or mitigate them.
什么是网安？
Correctness and Efficient algorithem against an attacker.
我们保护什么？
取决于你的数字资产，具体来说，是Information and Infrastructure.
例如：Sensitive Data, Control System, Hardware devices, etc.
如何保护？
转化一下，就是如何：
Security goal 设立安全目标
estimate impact of attacks 估计攻击影响
design mitigations 设计缓解措施
答：Analyse systems, spot vulnerabilities, build protection.
信息安全目标
Confidentiality: Attacker shoud not retrieve any information. 保密性：攻击者不得获取任何信息。
Integrity and Authenticity: Received data is authentic and the sender is genuine. 完整性和真实性：接收到的数据是真实的，发送者是真实的。
Availability: Data should accessible on demand. 可用性：数据应可按需获取。
潜在攻击者是谁？
所有人。
知名攻击
Ransomware 勒索病毒
Phishing 钓鱼
Unix介绍
file system:
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应知应会command
echo, mkdir, ls, pwd, cat, cp, mv, rmdir, rm, touch, locate(建库索引查找 快), find(实时查找 慢), grep, diff, du(disk usage, 列出目录下磁盘使用情况), head(cat 前十行), diff, tar(打包), top(性能监控), history(显示执行过的命令历史)
Modular Arithmetic
17 mod 5 = 2 32 mod 5 = 2
乘法加法分配率都是双分配，也就是说，括号里面分配之后，括号外还要分配。
- 49 mod 5 = (17 + 32) mod 5 = (17 mod 5 + 32 mod 5) mod 5 = (2+2) mod 5 = 4
- 2^10 mod 5 = 4
- 2^11 mod 5 = (2 * 2^10) mod 5 = (2^10 mod 5) x (2 mod 5) mod 5 = 4 x 2 mod 5 = 3
- 2^10 mod 17 = (2^5 mod 17)(2^5 mod 17) mod 17 = 15x15 mod 17 = 25x9 mod 17 = (25 mod 17)(9 mod 17) mod 17 = 8 x 9 mod 17 = 4
- 2^100 mod 17 = (2^10 mod 17)^10 mod 17 = 4^10 mod 17 = 2^20 mod 17 = (2^10 mod 17)^2 mod 17 = 4^2 mod 7 = 16 mod 17 = 16
Larry Shi