FIT2093 Assignment 1: Analyzing Encryption & Vulnerabilities at Monash

This document presents a solution to an information security assignment, likely for the FIT2093 course. It covers hybrid encryption methods, including key generation using RSA, and decryption processes. The assignment delves into practical implementation with Java code for key generation and discusses mathematical formulas relevant to ciphertext analysis. Additionally, it addresses vulnerabilities by analyzing corrupted blocks in ciphertext files and attempting plaintext recovery using online converters. The solution provides insights into the practical aspects of encryption and decryption, alongside vulnerability assessment in ciphertext files. Desklib offers more resources for students.

University
Semester
Information Security
Student ID
Student Name
Submission Date
1

Table of Contents
Task 2.........................................................................................................................................................3
a) Hybrid Encryption Method..........................................................................................................3
b) Decryption Process........................................................................................................................5
c) Mathematical Formula..................................................................................................................7
d) Vulnerability..................................................................................................................................8
2

Task 2
a) Hybrid Encryption Method
package com.mkyong.keypair;
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.security.KeyPair;
import java.security.KeyPairGenerator;
import java.security.NoSuchAlgorithmException;
import java.security.NoSuchProviderException;
import java.security.PrivateKey;
import java.security.PublicKey;
public class GenerateKeys {
private KeyPairGenerator keyGen;
private KeyPair pair;
private PrivateKey privateKey;
private PublicKey publicKey;
public GenerateKeys(int keylength)
throws NoSuchAlgorithmException, NoSuchProviderException {
this.keyGen = KeyPairGenerator.getInstance("RSA");
this.keyGen.initialize(keylength);
}
public void createKeys() {
this.pair = this.keyGen.generateKeyPair();
this.privateKey = pair.getPrivate();
this.publicKey = pair.getPublic();
}
3

public PrivateKey getPrivateKey() {
return this.privateKey;
}
public PublicKey getPublicKey() {
return this.publicKey;
}
public void writeToFile(String path, byte[] key) throws IOException {
File f = new File(path);
f.getParentFile().mkdirs();
FileOutputStream fos = new FileOutputStream(f);
fos.write(key);
fos.flush();
fos.close();
}
public static void main(String[] args)
throws NoSuchAlgorithmException, NoSuchProviderException, IOException {
GenerateKeys gk_Alice;
GenerateKeys gk_Bob;
gk_Alice = new GenerateKeys(1024);
gk_Alice.createKeys();
gk_Alice.writeToFile("KeyPair/publicKey_Alice", gk_Alice.getPublicKey().getEncoded());
gk_Alice.writeToFile("KeyPair/privateKey_Alice", gk_Alice.getPrivateKey().getEncoded());
gk_Bob = new GenerateKeys(1024);
gk_Bob.createKeys();
gk_Bob.writeToFile("KeyPair/publicKey_Bob", gk_Bob.getPublicKey().getEncoded());
gk_Bob.writeToFile("KeyPair/privateKey_Bob", gk_Bob.getPrivateKey().getEncoded());
}
}
4

b) Decryption Process
-----BEGIN PGP PRIVATE KEY BLOCK-----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sI/zCRA0rh/fwhw2vu8LOacGJiLBaTKILWGPMvhm4ZHrcj7d3GwTbrb7v69ehOs+
R9IYyj4p0YiDWecgcTQM1I7JGBIkHVMF9WKMjYSycoim/x1CiSw88IXMX5maCj2w
5

RQzc2PpxI4ePa0OxSCyxLI9glTeCjswGKGe31k4AO4U+J1ydA8YEXLgOkAEIAK00
myChnYLpMCnBbjCYnUJhAmuYNpFAQGmRMwNeet1BuOGzFx+wPqKHDBLW+kGPeIRi
Qf4y+OFw3iPk0HUlW5uNGI7kXQ0aoCOk3HoYqLyX7G3d8anQmcKZgu9HUguKy/v8
WcWZeU4SfyvfU/igkhE4PT20+U42FRQN+ojSJbI9qI82qJ/pfq+rzLFge69OKGAZ
++/zxxjI/SVEDX+/4oEW5v3/n4BK+gIuiuGRfgM6fbw3prkgDGI6F8HVIsdkODVn
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2mAHPi5mI+jWWjLTgY2Q66YKob4=
=Dojk
6

-----END PGP PRIVATE KEY BLOCK-----
Hex Decimal value is illustrated as below.
c) Mathematical Formula
In this task, we are compute the how many blocks are corrupted on tow cipher text file
such as c_aes1_qd.bin and c_aes2_qd.bin by using the mathematical formula.
Decrypting with the incorrect IV causes the first block of plaintext to be corrupt but
subsequent plaintext blocks will be correct. However there is also the case where your IV is
partially right (say the first n < 16 bytes are correct) then you will see the correct first n bytes in
the decrypted file and the remaining (16-n) bytes incorrectly. This means that even if you try an
IV of all zeroes you might still get the first n characters correct. The computer blocks
information is presented as below.
 data offset: 0x13=19
 hex8 - 0x41
 int8 - 65
 int16 - 16724
 int32 - 1096041760
 int64 - 4707463515513117000
 uint8 - 65
 uint16 - 16724
7

 uint32 - 1096041760
 uint64 - 4707463515513117000
d) Vulnerability
In this task, we are compute the plain text on two cipher text file such as c_aes1_qd.bin
and c_aes2_qd.bin by using the converter in online which is illustrated as below.
Plain Text is shown below.
c_aes1_qd
`....I...% .GA$
..AH..-....?....
.a....E......p..
c_aes1_qd
.n2`.GC8..e#..a.
G..2af....]...!.
....."...:}..Y..
8