1. Secret Sharing CPS 290 - Computer Security Nisarg Raval Sep 24, 2014
Material is adapted from CS513 lecture notes (Cornell)

2. Why share a secret?

3. Goal Given a secret s and n parties All n parties together recover s
Less than n parties can not recover s

4. Naive Scheme S=10011 S1 = 100 S2 = 11 High Order Low Order
Concat shares to reveal secret - S = (S1)(S2) = (100)(11) = 10011
What is the problem? - Think of a salary or password

5. Partial Disclosure Given a secret s and n parties
All n parties together recover s
Less than n can not recover any information about s

6. Generate Shares using XOR
S1 = Rand
S2 = S XOR S1
10100
00111
10011
S = S1 XOR S2

7. General Scheme Given a secret s and n parties
Generate n-1 random strings as first n-1 shares
Last share is the bitwise XORing of s with all the other n-1 shares

8. General Scheme Given a secret s and n parties
Generate n-1 random strings as first n-1 shares
Last share is the bitwise XORing of s with all the other n-1 shares
Security Check
Can n parties generate s?

9. General Scheme Given a secret s and n parties
Generate n-1 random strings as first n-1 shares
Last share is the bitwise XORing of s with all the other n-1 shares
Security Check
Can n parties generate s?
Can any n-1 parties generate s?

10. Example
S=10011 S1 S2 S3 S

11. Problem? S=10011 ? S1 S2 S3 S can be constructed by 2 or more generals
Less than 2 generals can not construct s

12. (n,t) Secret Sharing Given a secret s and n parties
Any t or more parties can recover s
Less than t parties have not information about s
(3,2) secret sharing
S=10011 S1 S2 S3 S

13. (n,2) Secret Sharing y
(0,S) x

14. (n,2) Secret Sharing
(xn-1,yn-1)
(xn,yn)
(x1,y1) y
(x2,y2)
(0,S) x

15. (n,2) Secret Sharing y Shares x (xn-1,yn-1) (xn,yn) (x1,y1) (x2,y2)

16. (n,2) Secret Sharing
(xn-1,yn-1)
(x1,y1) y
(0,S) x

17. (n,2) Secret Sharing
Exist a line for every S
(x1,y1) y
(0,S) x

18. (n,3) Secret Sharing
(0,S)
(x1,y1)
(x2,y2)
(xn-1,yn-1)
(xn,yn)

19. Shamir’s Secret Sharing
It takes t points to define a polynomial of degree t-1
Create a t-1 degree polynomial with secret as the first coefficient and the remaining coefficient picked at random
Find n points on the curve and give one to each of the parties.
At least t points are required to fit the polynomial and hence to recover secret
y = at-1 * xt-1 + at-2 * xt-2 + … + a1 * x + a0
Shamir, Adi (1979), "How to share a secret", Communications of the ACM

20. Use Case S1
(3,2)
Secret Sharing Scheme S2 S3
Private Key

21. Problem? S1
S1 compromised
S2 compromised S2
S1 + S2 = Secret S3
Time

22. Refresh Shares S’’1 S’’3 S’’2 S’1 S’3 S’2 S1 S2 S3 Time
Trusted Third Party
S’’1
S’’3
S’’2
S’1
S’3
S’2 S1 S2 S3
Time

23. Refresh Shares S’1 S’’1 S1 S’2 S’’2 S2 S’3 S’’3 can not
Trusted Third Party
S’1
S’’1 S1
S1 compromised
S’2
S’’2 S2
S’2 compromised
S’3
S’’3
can not
construct secret S3
Time

24. Proactive Secret Sharing
Server 1
Server 2 S1 S2

25. Proactive Secret Sharing
Server 1
Server 2 S1 S2
S11
S12
S21
S22

26. Proactive Secret Sharing
Server 1
Server 2 S1 S2
Exchange
Partial Shares
S11
S12
S21
S22
S21
S12

27. Proactive Secret Sharing
Server 1
Server 2 S1 S2
Exchange
Partial Shares
S11
S12
S21
S22
S21
S12
S’2
S’1

28. Proactive Secret Sharing
Server 1
Server 2 S1 S2
Exchange
Partial Shares
S11
S12
S21
S22
S21
S12
S’2
S’1
Recover S
(S11 + S21) + (S12 + S22) S

29. Summary Useful technique to distribute secret Confidentiality
Reliability
Each share must be as long as the secret itself
Require random bits of length proportional to the number of parties as well as length of the secret