I'm currently reading the tutorial on linear/differential crypta here: https://www.engr.mun.ca/~howard/PAPERS/ldc_tutorial.pdf

End of page 14/Beginning of page 15: all the bits from the subkeys are summed up to a sum(k) variable. It is then written as 0 in the equation (5) and the probability stays the same.

I don't understand this step, why is the probability not changing if sum(k) is zero, and why is it 1-p if sum(k) is 1?


1 Answer 1


So after re-reading a few times it becomes clear. The key bits are static, so they won't change the probabilities.

Now suppose that $\sum_K = 0$, then the equation in (5) will be true, with probability $\frac{15}{32}$ since it is the same equation.

Suppose now that $\sum_K = 1$, the equation in (5) is now the inverse, and it will thus happen with probability $1-(15/32).$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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