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Home/ Questions/Q 6050539
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T07:41:58+00:00 2026-05-23T07:41:58+00:00

I’m wondering if there is some strong (like AES or so.) encryption function that

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I’m wondering if there is some strong (like AES or so.) encryption function that works like this:

  • symetric
  • 2 keys: plaintext -> 2keys ->ciphered text, however it must not matter order of keys, i.e

Key1 (Key2 (plaintext)) == Key2 (Key1(plaintext))
e.g. “commutative”
(also required for decryption – you need two keys, doesn’t matter order)

thanks

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  1. Editorial Team
    Editorial Team
    2026-05-23T07:41:58+00:00Added an answer on May 23, 2026 at 7:41 am

    It’s not a commutative encryption, but there are well-proven algorithms for secret sharing (note, this is not the same thing as “key agreement.”)

    Two of the best known methods are Shamir’s and Blakley’s. In general, these algorithms take a secret and produce many “shares”. When enough shares are available to reach a threshold, the secret can be recovered. In the simplest case, two shares are required, but the threshold can be higher.

    To explain Shamir’s method in simple terms, think about a line on a graph. If you know any two points on the line, you know everything about the line. Any string of bytes, like the encryption key of a symmetric cipher, is just a large number, in base-256. Shamir’s algorithm treats this secret as the line’s “y-intercept” (the y-coordinate of the line when x=0). Then the line’s slope chosen randomly. The y-coordinates of the line at x=1, x=2, x=3, … are computed, and each point is given to a different share-holder.

    If any two of these share-holders get together, they can draw a line through their two points, back to the y-axis. The y-coordinate at where it crosses the axis is the original secret. However, each share-holder has only one point; by themselves, they can’t guess anything about the original secret.

    The threshold can be increased by increasing the degree of the polynomial. For example, if a parabola is used instead of a line, three shares are needed instead of two.

    There’s more to a real implementation, like the use of modular arithmetic, but this is the concept behind it. Blakley’s approach is similar, but it uses the intersection of planes to encode the secret.

    You can play around with an implementation of Shamir’s method online.

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