In the Shamir secret sharing scheme, we distribute a secret among a different users as follows. If our secret is a message (m₁, ..., mk) from V (k, q) then, we encode it as a codeword of the Reed-Solomon RSk(q) and give one coordinate to each user. In this problem, we will use q = 7, k = 4 and the parity check matrix H₁ below for RS(7). НА = 1 1 1 1 1 1 1 0 1 2 3 4 5 6 01 4 2 2 4 1 A new secret is selected and user #1 receives share value 0, user #2 receives share value 6 and user #3 receives share value 1 and are collaborating to discover the new secret. They can't recover the secret with only this information Now, suppose users #1, #2 and #3 discover, in addition to the values of their own shares, that users #4 and #5 have identical shares (but they don't necessarily know what the common value is). Using this information, first explain how they can collaborate to recover the secret and then find the secret.
In the Shamir secret sharing scheme, we distribute a secret among a different users as follows. If our secret is a message (m₁, ..., mk) from V (k, q) then, we encode it as a codeword of the Reed-Solomon RSk(q) and give one coordinate to each user. In this problem, we will use q = 7, k = 4 and the parity check matrix H₁ below for RS(7). НА = 1 1 1 1 1 1 1 0 1 2 3 4 5 6 01 4 2 2 4 1 A new secret is selected and user #1 receives share value 0, user #2 receives share value 6 and user #3 receives share value 1 and are collaborating to discover the new secret. They can't recover the secret with only this information Now, suppose users #1, #2 and #3 discover, in addition to the values of their own shares, that users #4 and #5 have identical shares (but they don't necessarily know what the common value is). Using this information, first explain how they can collaborate to recover the secret and then find the secret.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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In Step 2 part, could you please explain why you said:
H4 * [s1, s2, s3, s4, s5] = [0, 6, 1]
As to me, H4 is a 3x7 matrix, while [s1, s2, s3, s4, s5] is a 1x5 matrix. We cannot perform that multiplication
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