# You could have invented Shamir's secret sharing

Updated 2021-09-25

Tags: cryptography, security, secret-sharing, math, education, python, horcrux

Suppose you have some secret code—a master password or cryptocurrency seed phrase are common examples—and you want to back it up. Is there a clever way to arrange things so that it’s simultaneously secure against the risk of you losing the backup and against the risk of someone else finding it?

We’ll try the naive way, build up to proper secret sharing, and finish with recommendations for how you could actually use it to back up your passwords. Some of this is based on an excellent YouTube video, some on the original 1979 paper by Adi Shamir, and the rest I came up with myself.

# Attempt 1: break it in half

Most people would probably try breaking the secret code in half and storing the two halves in separate locations. For example if you have the 8-character password `!f8sy=06`

, you could write `!f8s`

on one piece of paper and `y=06`

on the other. There are two problems with that though:

- If you lose either piece of paper you lose the password
- If someone tech-savvy finds one piece, they might be able to guess the entire password

The first problem is obvious. You can see the second by using Python to estimate the entropy of each piece:

```
>>> import string
>>> chars = string.ascii_lowercase + string.ascii_uppercase + string.punctuation
>>> chars
'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~'
>>> len(chars)
84
>>> len(chars) ** 8 # possible 8-char passwords
2406758911082496
>>> len(chars) ** 4 # possible 4-char password halves
49067136
```

There are only around 50 million combinations per half. Which sounds like a lot, but a modern computer could probably try all of them in a couple seconds! Another way to look at it is that guessing the missing half is 50 million times easier than guessing the entire password from scratch.

# Attempt 2: add redundancy

Of course, you could compensate for that by making the password longer. But the general problem remains, and it gets exponentially worse if you also try to protect yourself against losing some of the pieces (problem #1 above).

Why the trade-off? Because to protect against losing any given part of the password you have to make more than one copy of it. For example, you could break the password into 3 partially redundant pieces so that any 2 of them can be used to reconstruct it. But then each piece would have to contain more of the password:

`███sy=06`

`!f8███06`

`!f8sy=██`

Now you’re protected from losing any one share, but you’ve sacrificed almost all of the entropy! Someone who finds even one piece can easily guess the missing characters:

```
>>> len(chars) ** 3 # possible missing 3-char chunks
592704
>>> len(chars) ** 2 # possible missing 2-char chunks
7056
```

If you needed to, you could pick a very long password and split it up like that. But it turns out there’s a more elegant way to solve both problems at once…

# “One weird trick”

The trick is to encode your secret as the Y intercept of a curve, then write down coordinates to points on that curve. Here’s how to do the simplest “2 of N” case, where the curve is a straight line and you can reconstruct it from any 2 points:

- Encode secret as Y intercept
- Pick a point randomly to be the first secret share
- Draw a line between them
- Pick more points along the same line to be the other shares

For step 1, there are many ways to encode text as numbers. We’ll use ASCII character codes:

```
>>> def encode(chars):
return [ord(c) for c in chars]
...
...>>> def decode(nums):
return ''.join(chr(n) for n in nums)
...
...>>> encode('!f8sy=06')
33, 102, 56, 115, 121, 61, 48, 54]
[>>>
>>> decode(encode('!f8sy=06'))
'!f8sy=06'
```

To keep it simple, let’s focus on secret sharing just the exclamation point:

Now anyone who has the `(x,y)`

coordinates of at least two of the blue + brown points can draw a line to find the secret number, 33. This is much better than the naive solution above because finding one pair of coordinates leaks *no information at all.* You could draw an infinite number of lines through the point you found, and 128 of them would lead to Y intercepts representing ASCII characters. Which is the same as the total number of characters you would have to guess from anyway. Cool, right?

To split up the whole password we could just repeat that process 8 times.

# Generalize to other curves

Two points are enough to define a straight line. So if we want to require more than 2 shares to reconstruct the secret, we need a more complicated curve. The number of shares needed is called the “threshold”. It’s the same as the number of variables (besides `x`

and `y`

) in the equation for the curve.

Threshold | Equation | Curve |
---|---|---|

2 | y = ax + b |
linear |

3 | y = ax^{2} + bx + c |
quadratic |

4 | y = ax^{3} + bx^{2} + cx + d |
cubic |

… | … | … |

For example, the cubic version (threshold = 4) could be set up like this:

It’s probably counter-intuitive, but even knowing 3 of the blue + brown points on that curve doesn’t get you any closer to finding the intercept. Without 4 you might as well have none at all.

*Note: Shamir’s actual scheme uses number fields defined in terms of large prime numbers; the easier-to-visualize real number curves and equations shown here are just meant to give some basic intuition for how the math works*

# Make it useful!

The final step is to hide the math from regular users who just want to back up their cryptocurrency, passwords, or other data. The `ssss`

(“Shamir’s Secret Sharing Scheme”) package is easy to use. Pick the threshold with `-t`

and the total number of shares with `-n`

:

```
$ ssss-split -t 3 -n 5
Generating shares using a (3,5) scheme with dynamic security level.
Enter the secret, at most 128 ASCII characters: !f8sy=06
Using a 64 bit security level.
1-d8a1c623c3a614a5
2-e38ae6524ad6f239
3-40676cf2da3882d3
4-36b7e761817c962b 5-955a6dc11192e6d3
```

```
$ ssss-combine -t 3 -n 5
Enter 3 shares separated by newlines:
Share [1/3]: 1-d8a1c623c3a614a5
Share [2/3]: 3-40676cf2da3882d3
Share [3/3]: 4-36b7e761817c962b Resulting secret: !f8sy=06
```

Of course there’s some other fanciness going on too. But I hope that it doesn’t look quite like magic anymore, and that you would consider using something similar to back up your actual master password or seed phrase.

One warning though: don’t use the online demo on the `ssss`

site. You would be sending your secrets to the author + anyone snooping on the unsecure HTTP connection. Ideally you should boot into a Linux LiveCD (I like Lubuntu), install it there with `sudo apt install ssss`

, and only save the secret shares on paper.

# Shameless self-promotion

If you do decide to use `ssss`

, consider using it via a program I wrote called Horcrux. It’s only a little more complicated. The main advantage is that it lets you create your secret shares once, hide them, and then encrypt new backups later without gathering enough shares to reconstruct the master password each time. I think that makes it much more likely you’ll keep regular backups, and less likely you’ll have the master password written on a Post-It note or in some similarly insecure location.