Networks exam - Cryptography (S-Boxes)

Shicky

Shicky

Shicky

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This is pretty poorly explained in the notes considering the exam questions we're meant to answer.....basically we've been given one diagram of its implementation.

For example the diagram given basically has some 2-bit inputs:
A |0 0 1 1
B |0 1 0 1

This then goes through an s-box to produce:
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1

Now how are those above being determined? Is it as simple as the inputs, basically you're trying to cover all possibilities? If so surely this doesnt or is there only allowed to be one 1 in each sequence?

In some of the past papers you're told to provide 3-bit inputs. I assume this would be something like:
A 0 0 0 1 1 0 1 1
B 0 0 1 0 1 1 0 1
C 0 1 0 0 0 1 1 1

Futher questions ask about product ciphers (multiple p-boxes and s-boxes.) Such as an 18-bit p-box is used 4 times, a 3-bit s-box is used, how many s-boxes will there be? I assume here s-boxes should be used to 'cover' the p-boxes as such so 6 per p-box in this case? How do you know whether additional s-boxes go before the p-boxes or simply in-between?

What i mean by this is how do you know the product cipher will be like this:

PBox | SBox | PBox | SBox | PBox

rather than:

SBox | PBox | SBox | PBox | SBox | PBox | SBox

Hope that isn't too random and confusing, thanks for any insight you can provide!
 
Shicky said:
For example the diagram given basically has some 2-bit inputs:
A |0 0 1 1
B |0 1 0 1

This then goes through an s-box to produce:
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1

Now how are those above being determined? Is it as simple as the inputs, basically you're trying to cover all possibilities? If so surely this doesnt or is there only allowed to be one 1 in each sequence?
Click to expand...

Only a quick one, I'm sure someone can follow up with a fuller explanation... Using two inputs, there are 4 combinations - each of these maps to an output, hence giving a 4x4 matrix. I think input reads vertically, output horizontally. The example is an identity matrix though, so input = output.

I think. I should really remember more about this from a year ago.
 
gumbald said:
Only a quick one, I'm sure someone can follow up with a fuller explanation... Using two inputs, there are 4 combinations - each of these maps to an output, hence giving a 4x4 matrix. I think input reads vertically, output horizontally. The example is an identity matrix though, so input = output.

I think. I should really remember more about this from a year ago.
Click to expand...

Im with you on how each of those should be read, i.e vertically and horizontally, this is definately correct. So in terms of my predicted inputs for a 3-bit input I was correct? Then the 3 to 6 encoder will produce an identity matrix with 6 1's on the diagonal?

100000
010000
001000
000100
000010
000001
 
any smart people about with any ideas? Would really like to grasp this concept before the exam as it comes up every year. Think I may understand it but have no way to check really as I cant find anything on the net, my books, my notes
 
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