# How to create (n, t) secret splitting from (n, n) secret splitting?

Given a secret splitting scheme $$(n ,n)$$ that creates $$n$$ shares from secret $$s$$. In this scheme all shares must be combined to create $$s$$.

How do you create a secret splitting scheme $$(n, t)$$? Of $$n$$ parts at least $$t$$ parts must be combined to determine secret $$s$$?

$$n =$$ # of Parts

$$s =$$ Secret

$$t =$$ Threshold of parts needed to create the secrets

$$s_1, s_2, s_3, ... =$$ Shares in a $$(n, n)$$ secret splitting scheme

$$P_2, P_2, P_3, ... =$$ Shares in a $$(n, t)$$ secret splitting scheme

$$l =$$ Intermediary value to determine the n needed in $$(n, n)$$ secret splitting scheme

Example 1 $$(4, 3)$$:

$$l$$ = $$4 \choose 3-1$$ = $$6$$

$$s \rightarrow (6, 6) = [s_1, s_2, s_3, s_4, s_5, s_6]$$

$$s \rightarrow (4, 3) = [P_1= [s_3, s_4, s_5], P_2= [s_1, s_4, s_6], P_3= [s_1, s_2, s_5], P_4= [s_2, s_3, s_6]]$$

Example 2 $$(4, 2)$$:

$$l$$ = $$4 \choose 2-1$$ = $$4$$

$$s \rightarrow (4, 4) = [s_1, s_2, s_3, s_4]$$

$$s \rightarrow (4, 2) = [P_1 = [s_1, s_2, s_3], P_2 = [s_1, s_2, s_4], P_3 = [s_1, s_3, s_4], P_4 = [s_2, s_3, s_4]]$$

What is a methodology to determine an arbitrary $$(n, t)$$ scheme? For example, what would $$(6,3)$$ look like?

• @Reppiz stackexchange did not let me reply to your comment or upvote it. Shamir's Secret Sharing is great, but I don't believe one can implement it with informational security. I like this method because it can be implemented with informational security. I just don't understand how to algorithmically implement it. Thank you for the suggestion. Jul 23, 2021 at 17:25
• What makes you say that you can't implement Shamir's Secret Sharing with information-theoretic security? Jul 27, 2021 at 15:33
• You start the question with "Given a secret splitting scheme (n,n) ... " - this might be impossible to achieve. There are schemes, which are (n,n), which can not be adapted to arbitrary (t,n) secret sharing. For example: The secret is the XOR of all shares. So unless you specify, which secret sharing scheme is given, this can't be answered. Or are you asking for which secret sharing can achieve this?
– tylo
Apr 19, 2022 at 10:26