Pls I need explanation on How i can design a Cryptography that use multiple passwords or passphrase to open a safe(Lock). For example, if i need five people to unlock a secured device whereby all the five need to enter their individual password before the device can be opened.

Please how can this be done? I don't have much knowledge on this, hence i wouldn't expect all the answers right away. I would appreciate a link, or book or somewhere i can read details about it. Thanks.

• Look up 'Secret Sharing' Apr 24, 2014 at 13:24
• Thanks, I feel better hearing that, can u please help with a good source of information, where i can read it up. Thanks once more. Apr 24, 2014 at 14:02
• I've added two tags to your question - the tag wiki should help Apr 24, 2014 at 14:03
• I've seen a lot of replies and answers from the tag you added. Nice. Apr 24, 2014 at 14:35

Abstract. In this paper we show how to divide data D into n pieces in such a way that D is easily reconstructable from any k pieces, but even complete knowledge of k - 1 pieces reveals absolutely no information about D. This technique enables the construction of robust key management schemes for cryptographic systems that can function securely and reliably even when misfortunes destroy half the pieces and security breaches expose all but one of the remaining pieces.

source: http://dl.acm.org

### In a nutshell:

I have a secret $S$. Let's say $S=10$ (or a password turned into the value $10$), and I want to share it between $N=2$ people. I create a polynomial of degree $d=N-1$ (because I care to have a secret for two people): $$f(x) = ax + b$$

We set $b = S$, and let $a$ be a random value larger than $S$ and $N$. For example, let's say $a=13$.

Therefore:

$$f(1) = 13*1+10 = 23 \\ f(2) = 13*2+10 = 36$$ Now, send $f(1)$ to person $1$, and $f(2)$ to person $2$. In order to obtain the secret $S$ and to unlock the door, both person $1$'s value and person $2$'s value are needed.

Using polynomial approximation, we can get the resulting $f(x)$ from the people values: $$f(x) = \sum_{i=1}^N f(i) \cdot \prod_{j\neq i} \frac{x-x_j}{x_i-x_j} = \sum_{i=1}^N f(i) \cdot \prod_{j\neq i} \frac{x-j}{i-j}$$

• WOW!! I feel so excited about this explanation...I think am so much satisfied. The whole concept is very clear to me, and i'll read more on it. Thanks to every one. Apr 24, 2014 at 14:34
• @DrLecter Thanks for mentioning the bad link, I've fixed it with the ACM link (Thanks @Josiah).
– jonm
Apr 28, 2014 at 14:24