41900 – Fundamentals of Security Project Part-1 (Week 4 – Week 8)
One-Way Property versus Collision-Free Property
In this project you will investigate the difference between hash function’s two distinctive properties: one-way property and collision-free property. You are required to use the brute-force method to see how long it takes to break the hash generated using each of these properties. Instead of using openssl’s command-line tools, you are required to update the C programs provided to invoke the message digest functions in openssl’s crypto library. A sample code can be found at the following link, it would be helpful to familiarize with the sample code:
http://www.openssl.org/docs/crypto/EVP_DigestInit.html.
One-way means that given a function output, you cannot find a matching input, except by trying many potential inputs and getting lucky. A Collision is about finding two distinct inputs which yield the same output, without any predefined constraint on said output.
Since most of the general hash functions are quite strong against the brute-force attack conducted on those two properties, it can take years to break them using brute-force. Hence, to make the task feasible we reduce the length of the hash value to 24 bits. You can use any one- way hash function, but only the first 24 bits of the hash value generated will be used in this task.
There are two skeleton programs i.e. oneway.c and collision.c provided for this assignment. You aim is to complete the code, run it successfully and observe which property is easier to break using the brute-force method.
Required Outcome:
1. You must complete the code and run it to find out the following:
a. How many trials it will take you to break the one-way property using the brute- force method? You should repeat your experiment at least 5 times and report your average number of trials.
b. How many trials it will take you to break the collision-free property using the brute-force method? Similarly, you should report the average.
2. You must demo your code and answer questions based on it.
41900 – Fundamentals of Security Project Part-1 (Week 4 – Week 8)
1. Brute force: One-Way property break
The program’s objective is to hit a “target” hash value by generating the hash for all possible combinations of different strings and matching it with the target hash value.
The approach to generate strings and match them against a target is implemented as follows:
1. Start with string length 1
2. Generate all possible combinations of strings for current length using the character set,
generate its hash and compare it with the target hash.
3. If a match is found, the execution stops and reports the number of iterations used.
4. If no match is found in all possible strings of current length, increment length by ‘1’ and
go to step (2)
The brute force code is used in the one-way property break program oneway.c to generate strings In a systematic manner, exhausting all possibilities for a given length (starting with 1), and incrementing the length if no match was found.
The oneway.c should be executed a minimum of five times to find out the average number of iterations required to match the target hash.
2. Bute force: Collision-Free property break
In the in the
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case of collision free property, there is no restriction on target hash. The approach followed collision.c program to break this property is as follows:
Allocate an array for saving hash values
Generate initial message (call it M)
Find M’s hash
Generate random string S and find S’s hash
Compare S’s hash to M’s hash, if a match is found then exit and report the number of iterations used.
Compare S’s hash to each hash value in array, if any match is found then exit and report the number of iterations used.
If no match is found, save the S’s hash into the array and go to step (4)
To break the collision free property, you just need to find two messages with matching hash values. Matching the random string’s hash value to an initial hash as well as saved hash value (in the array) increases the probability of finding hash value with less number of iterations.
Execute the collision.c program five times to find the average number of iterations required.
41900 – Fundamentals of Security Project Part-1 (Week 4 – Week 8)
Instructions for compiling and running the program:
To compile:
➢ gcc -o oneway oneway.c -lcrypto -ldl -std=c99
➢ gcc -o collision collision.c -lcrypto -ldl -std=c99
To run the code:
➢ ./oneway md5
➢ ./collision md5
Marking Criteria (Total Weightage 20%):
Criteria
Incomplete Work
Partial Work
Full Work
Program Completeness
0 Points
Program Submitted without any changes
3 Points
Program Modified, but missing certain components
5 Points
Program contains all essential components.
Program Functioning
0 Points
Program does not compile / run
3 Points
Program runs but number of iterations are wrong.
5 Points
Program runs and provides number of iterations within acceptable range.
Demo and Questionnaire
0 Points
Unable to answer questions.
5 Points
Able to partially answer the questions.
10 Points
Able to answer all questions.
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