UOL CIS Community Forum

william · Joined: 09 Dec 2007 Posts: 28

Please post all questions pertaining to Coursework 2 in this thread.

william · Joined: 09 Dec 2007 Posts: 28

1. Big Hint for CW02:
Why is "Euclid's Extended Algorithm for obtaining the modular inverse of b mod a" provided?
Recall for modular multiplication, if m, s and p are given, then
c = (m*s) mod p
If c, s and p are given and you are to calculate the value of m, then you need to solve:
m = (c/s) mod p
=> = (c * (1/s mod p)) mod p
ie. = (c * (multiplicative inverse of s mod p)) mod p

2. Blocking factor:
For El Gamal, even if you use a blocking factor of 1, it is still less susceptible to statistical analysis attacks compared to RSA because of its polyalphabetic and stream-cipher-like behavior where each block is encrypted using a different key. For greater security strength, a greater blocking factor is recommended.

However, we note the following:
a. the plaintext block m must be smaller than p, the prime modulus;
b. if 7-bit ASCII is used as the message alphabet, then each plaintext character value is < 128.

This implies that the modulus prime must be > 128 for a blocking factor of 1. This is fine for the assignment since 1000 < p < 10000.

If you want a blocking factor of 2, then we note that the plaintext block m can now have a maximum value of 127*128+127 = 16383. So the prime must be at least 16384. Since the assignment limits you to 10,000, you cannot use a blocking factor of more than 1.

But for extra credit, I would encourage you to develop an encoding scheme that supports blocking factors of > 1, and use larger primes to test it. If this assignment has been on RSA, you will definitely need to do this.

3. Here is the sample output for CW02 which uses a blocking factors of 2 and 3. It is useless for you to submit this as your assignment since the assignment requires you to use a prime < 10,000, which limits you to a blocking factor of 1. So just use this output to understand better the requirements of the assignment and nothing more.

william · Joined: 09 Dec 2007 Posts: 28

magustan · Joined: 09 Dec 2007 Posts: 5

Hello Sir, I have completed according to your framework.

at times the program run successfully, but sometimes it doesn't.

Could you please enlighten me or is it normal?

Magus

skyice · Joined: 09 Dec 2007 Posts: 5

Hi William,

Can advice me on how to calculate inverse mod.

8 = (5 * m) mod 19
m = 13

How to get that?

Thanks!

magustan · Joined: 09 Dec 2007 Posts: 5

skyice · Joined: 09 Dec 2007 Posts: 5

magustan · Joined: 09 Dec 2007 Posts: 5

magustan · Joined: 09 Dec 2007 Posts: 5

++++++++++++++++++++++++++++++++++++++++++
> Bob is now sending the following message to Alice:
> Alice allows a maximum blocking factor of 1
> Message will require 54 block/s.
> Bob processes the following...
> PLAINTEXT : Student Name: Tan Hwi Keng, Student Number: U050265703
> ENCODED : 83 116 117 100 101 110 116 32 78 97 109 101 58 32 84 97 110 32 72 119 105 32 75 101 110 103 44 32 83 116 117 100 101 110 116 32 78 117 109 98 101 114 58 32 85 48 53 48 50 54 53 55 48 51
> ENCRYPTED : 839 632 3345 2092 1066 3049 632 819 2230 1431 336 1066 316 819 3552 1431 3049 819 908 1293 701 819 1569 1066 3049 2753 3463 819 839 632 3345 2092 1066 3049 632 819 2230 3345 336 405 1066 2684 316 819 2526 3098 1707 3098 1046 681 1707 3394 3098 20
> PARTIALS : 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713 2713
> Alice processes the following...
> DECRYPTED : 83 116 117 100 101 110 116 32 78 97 109 101 58 32 84 97 110 32 72 119 105 32 75 101 110 103 44 32 83 116 117 100 101 110 116 32 78 117 109 98 101 114 58 32 85 48 53 48 50 54 53 55 48 51
> DECODED : Student Name: Tan Hwi Keng, Student Number: U050265703
> Original message matched received message. Public key crytosystem works.

++++++++++++++++++++++++++++++++++++++++++

lol you see my partial key doesn't change =X

skyice · Joined: 09 Dec 2007 Posts: 5

the partial key is 'k' or 'm'?

skyice · Joined: 09 Dec 2007 Posts: 5

but if you look at william's post, his partial key changes...

magustan · Joined: 09 Dec 2007 Posts: 5

skyice · Joined: 09 Dec 2007 Posts: 5

Hi,

Thanks anyway... i did it. My partial key changes...

Slevin · Joined: 09 Dec 2007 Posts: 3

when submit cw02 ah? i havent start doing them leh... Sad

Slevin · Joined: 09 Dec 2007 Posts: 3

finish liao, so simple. LOL