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I've looked at the paper Ciphertext-Policy Attribute-Based Encryption by Bethencourt, Sahai, Waters and see this an answer here for the explanation of DecrypNode for a leaf node.

I could understand the derivation of this $f_n$ for a leaf node but not for a non-leaf node in the $s$.

I could not understand the move from step 4 to 5 ( the last move).

I will be very grateful if somebody can help me in understanding the last move here?

How is $q_x(0) = q_x(1) \cdot \Delta_{1, S_x'}(0) + q_x(2) \cdot \Delta_{2, S_x'}(0) + q_x(3) \cdot \Delta_{3, S_x'}(0)\ \dots $ ?

I don't think Lagrange interpolation says this. I think I am missing something here. Please help me in understanding this.

Thanks and Regards

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  • $\begingroup$ Welcome to Cryptography. Your question is not clear. Could you post a link? $\endgroup$
    – kelalaka
    Commented Jan 12, 2019 at 20:17
  • $\begingroup$ It is 11 months ago. Maybe you can delete this and ask again in a better way... $\endgroup$
    – kelalaka
    Commented Jan 6, 2020 at 12:47

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The last two steps is from Lagrange interpolation such that $\sum_{i\in S_x'} q_x(i)\triangle_{i,S_x'}(0)=q_x(0)$. The concept is the same as that of leaf node, but now we are at the parent node. Doing this recursively up to the root node would give the secret exponent.

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