Question

1. In a (2, 5) Shamir secret sharing scheme with modulus 19, two of the shares are (0, 11) and (1, 8). Another share is (5, k), but the value of k is unreadable. Find the correct value of k.

Answer 1

Note: Done accordingly. Please comment for any problem. Please Uprate. Thanks.

This scheme is called (k,n) {\displaystyle \left(k,n\right)}()threshold scheme where k is number of pieces and n is number of participants. Now we have (2,5) sharing scheme that means with 2 pieces we can regerate the secret.

Also we have to build the polynomial such that:

Now our k is 2 so equation will go upto only 2 terms i.e.

f(x) = a₀+a₁x

Now a0 is non coefficient part, so it is the key(lets call it s). We have given f(x) = 0 for x = 11 and f(x) = 1 for x =8, putting in equation:

0 = s + 11*a mod 19 ……(i)

1 = s + 8*a mod 19 ……(ii)

Subtracting i and ii

-1 = 3*a mod 19

18 = 3*a mod 19

Or a = 6

Putting in i we get :

0 = s +11*(6) mod 19

0=s + 66 mod 19

S=-66 mod 19

S= 10 mod 19

S= 10

Now

New equation will be:

K= s + 5*a mod 19

K = 10+5*6 mod 19

K= 40 mod 19

K= 2 mod 19

Hence k = 2