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I'm studying about Additive differential of XOR

I saw two papers that are "H.Lipmaa et al., On the Additive Differential Probability of Exclusive-or" and "V.Velichkov et al., The Additive Differential Probability of ARX"

In this two paper, Limaa et al. and Velichkov et al. compute the probability of additive differential of xor in similar way.

But there is one difference between them.

That is matrix $A_i$ is different.

For example $A_{000}$ in first paper is $A_{000}=\begin{pmatrix} 4& 0& 0& 1& 0& 1& 1& 0 \\0& 0& 0& 1& 0& 1& 0& 0 \\ \vdots & & & \cdots & \cdots & & & \vdots \\ 0& 0& 0& 0& 0& 0& 0& 0 \end{pmatrix}$

and $A_{000}$ in second paper is $A_{000}=\begin{pmatrix} 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\0& 1& 0& 0& 0& 0& 0 & 0 \\ \vdots & & & \cdots & \cdots & & & \vdots \\ 0& 0& 0& 0& 0& 0& 0& 1 \end{pmatrix}$

I hope someone can tell me why there exists a difference.

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  • $\begingroup$ I find the answer... See the table 2. in second paper. $\endgroup$ – jyj Mar 8 at 8:15
  • $\begingroup$ If you found the answer, please be kind enough to share it. $\endgroup$ – DannyNiu Mar 8 at 9:02
  • $\begingroup$ In second paper(V.Velichkov et al., The Additive Differential Probability of ARX), the table 2's order is not same to orginal order. For example, S[i] =0 when (s1[i],s2[i],s3[i])=(0,0,-1). So, there exist a difference $\endgroup$ – jyj Mar 8 at 11:44
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    $\begingroup$ @jyj Please post it as an answer to your question, rather than sharing it in the comments. Comments should not be used to answer questions. $\endgroup$ – Ella Rose Mar 8 at 15:19
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The OP has provided the answer in a comment.

"In the second paper(V.Velichkov et al., The Additive Differential Probability of ARX), the table 2's order is not same as the original order.

For example, $S[i] =0$ when $(s_1[i],s_2[i],s_3[i])=(0,0,-1)$. So, there exists a difference.

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