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# Shamir's Secret Sharing
<img src="https://arnaucube.com/img/logoArnauCubeTransparent.png" style="max-width:20%; float:right;" />
- https://arnaucube.com- https://github.com/arnaucube- https://twitter.com/arnaucube<br><br><br><div style="float:right;font-size:80%;"> <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/"><img src="https://licensebuttons.net/l/by-nc-sa/4.0/88x31.png" /></a> <br> 2019-07-05</div>
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# Intro
- I'm not an expert on the field, neither a mathematician. Just an engineer with interest for cryptography- Short talk (15 min), with the objective to make a practical introduction to the Shamir's Secret Sharing algorithm- Is not a talk about mathematical demostrations, is a talk with the objective to get the basic notions to be able to do a practical implementation of the algorithm- After the talk, we will do a practical workshop to implement the concepts. We can offer support for Go, Rust, Python and Nodejs (you can choose any other language, but we will not be able to help)
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- Cryptographic algorithm- Created by Adi Shamir, in 1979 - also known by the $RSA$ cryptosystem - explained in few months ago in a similar talk: https://github.com/arnaucube/slides/rsa
---## What's this about?
- imagine having a password that you want to share with 5 persons, in a way that they need to join their parts to get the original password- take the password, split it in 5 parts, and give one part to each one- when they need to recover it, they just need to get together, put all the pieces and recover the password (the `secret`)- this, has the problem that if a person looses its part, the secret will not be recovered anymore.. luckly we have a solution here:
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- Shamir's Secret Sharing: - from a secret to be shared, we generate 5 parts, but we can specify a number of parts that are needed to recover the secret - so for example, we generate 5 parts, where we will need only 3 of that 5 parts to recover the secret, and the order doesn't matter - we have the ability to define the thresholds of $M$ parts to be created, and $N$ parts to be able the recover
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- 2 points are sufficient to define a line- 3 points are sufficient to define a parabola- 4 points are sufficient to define a cubic curve- $K$ points are suficient to define a polynomial of degree $k-1$
We can create infinity of polynomials of degree 2, that goes through 2 points, but with 3 points, we can define a polynomial of degree 2 unique.
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## Naming
- `s`: secret- `m`: number of parts to be created- `n`: number of minimum parts necessary to recover the secret- `p`: random prime number, the Finite Field will be over that value
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## Secret generation
- we want that are necessary $n$ parts of $m$ to recover $s$ - where $n<m$- need to create a polynomial of degree $n-1$$f(x) = \alpha_0 + \alpha_1 x + \alpha_2 x^2 + \alpha_3 x^3 + ... + + \alpha_{n-1} x^{n-1}$- where $\alpha_0$ is the secret $s$- $\alpha_i$ are random values that build the polynomial*where $\alpha_0$ is the secret to share, and $\alpha_i$ are the random values inside the $Finite Field$
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$f(x) = \alpha_0 + \alpha_1 x + \alpha_2 x^2 + \alpha_3 x^3 + ... + + \alpha_{n-1} x^{n-1}$
- the packets that we will generate are $P = (x, f(x))$ - where $x$ is each one of the values between $1$ and $m$ - $P_1=(1, f(1))$ - $P_2=(2, f(2))$ - $P_3=(3, f(3))$ - ... - $P_m=(m, f(m))$
---## Secret recovery
- in order to recover the secret $s$, we will need a minimum of $n$ points of the polynomial - the order doesn't matter- with that $n$ parts, we do Lagrange Interpolation/Polynomial Interpolation
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## Polynomial Interpolation / Lagrange Interpolation
- for a group of points, we can find the smallest degree polynomial that goees through all that points - this polynomial is unique for each group of points
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$L(x) = \sum_{j=0}^{n} y_j l_j(x)$
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## Wikipedia example
*example over real numbers, in the practical world, we use the algorithm in the Finite Field over $p$<span style="font-size:70%;float:right;">(more details: https://en.wikipedia.org/wiki/Shamir's_Secret_Sharing#Problem)</span><br>- $s=1234$- $m=6$- $n=3$- $f(x) = \alpha_0 + \alpha_1 x + \alpha_2 x^2$ - $\alpha_0 = s = 1234$ - $\alpha_1 = 166$ *(random)* - $\alpha_2 = 94$ *(random)*- $f(x) = 1234 + 166 x + 94 x^2$
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- $f(x) = 1234 + 166 x + 94 x^2$- we calculate the points $P = (x, f(x))$ - where $x$ is each one of the values between $1$ and $m$ - $P_1=(1, f(1)) = (1, 1494)$ - $P_2=(2, f(2)) = (2, 1942)$ - $P_3=(3, f(3)) = (3, 2578)$ - $P_4=(4, f(4)) = (4, 3402)$ - $P_5=(5, f(5)) = (5, 4414)$ - $P_6=(6, f(6)) = (6, 5614)$---
- to recover the secret, let's imagine that we take the packets 2, 4, 5 - $(x_0, y_0) = (2, 1942)$ - $(x_0, y_0) = (4, 3402)$ - $(x_0, y_0) = (5, 4414)$
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- let's calculate the Lagrange Interpolation -  -  -  - - obtaining $f(x) = \alpha_0 + \alpha_1 x + \alpha_2 x^2$, where $\alpha_0$ is the secret $s$ recovered - where we eavluate the polynomial at $f(0)$, obtaining $\alpha_0 = s$- *we are not going into details now, but if you want in the practical workshop we can analyze the 'mathematical' part of all of this
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# And now... practical implementation
- full night long- big ints are your friends- $L(x) = \sum_{j=0}^{n} y_j l_j(x)$
# About
<img src="https://arnaucube.com/img/logoArnauCubeTransparent.png" style="max-width:20%; float:right;" />
- https://arnaucube.com- https://github.com/arnaucube- https://twitter.com/arnaucube<br><div style="float:right;font-size:80%;"> <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/"><img src="https://licensebuttons.net/l/by-nc-sa/4.0/88x31.png" /></a> <br> 2019-07-05</div>
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