Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am preparing for the midterm exam in my Data Security class so I am trying to read and understand the textbook's exercise questions. The sample solution at the end the textbook makes this problem more confusing instead of clarifying it. $\oplus$ denotes XOR operation but what are we trying to XOR with?

$c = \oplus c_i$ and $i \in I(c)$, but $c_i$ is getting XOR with what?

Any help would be appreciated.

Textbook question: Textbook question

Textbook sample solution: Textbook sample solution

share|improve this question
1  
I'm curious, what textbook is this? – d1str0 Feb 29 at 5:00
2  
William Stallings, Cryptography and Network Security: Principles and Practice, 6/E. – Node.JS Feb 29 at 5:49
1  
Thank you. I might want to check this out – d1str0 Feb 29 at 5:50
up vote 15 down vote accepted

The notation $c=\oplus~c_i$ is (terrible) shorthand for $$c=\bigoplus_{i \in I(c)} c_i$$ where the sum sign should be replaced by the big xor sign which could also be written as $$ c=\sum_{i \in I(c)} c_i,$$where $\sum$ denotes vector addition modulo 2.

An example of this decomposition (for length 8 vectors) is $$c=(1,0,1,0,0,0,1,0)=$$ which is nonzero in positions 1,3, and 7, and satisfies $$c=(1,0,0,0,0,0,0,0)\oplus(0,0,1,0,0,0,0,0)\oplus(0,0,0,0,0,0,1,0)$$ $$=c_1\oplus c_3 \oplus c_7=\bigoplus_{i \in I(c)} c_i$$ where $I(c)=\{1,3,7\}.$

Then, since $E$ is assumed to be linear, all the places you marked with an arrow make sense.

share|improve this answer
    
Awesome! Thank you so much for clarification. – Node.JS Feb 29 at 4:42
    
+1 for (terrible) shorthand :D – user32017 Feb 29 at 17:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.