Dining-Cryptographer Net (DC-Net) Scheme that Handles Collisions?

Question

[Question edited at the request of Mods]

I recently became fascinated with the elegant and simple solution that Chaum proposed for the Dining Cryptographers problem. If you are unfamiliar, please checkout Wikipedia for a summary and solution.

What is interesting is that DC-Nets allow for participants in a group to share messages to the rest of the group anonymously. Wow! But there is a clear issue - two participants in a DC-Net cannot publish messages at the same time (what we call a collision).

So now I am trying to figure out a protocol that alleges to solve this problem by doing the following alterations to the original DC-net:

• Let Alice, Bob and Carol be three participants in this DC-net A,B,C
• Suppose that each participant has a shared secret to the right and to the left, such that |shared secret| = |message|
• Alice hides her message by XORing with her two shared secrets. Call this encrypted message m_a. Bob and Carol do this same with m_b and m_c
• All participants publish their respective messages in three slots. In the first slot they publish m, in the second $m^2$, and so forth (in the third $m^3$). These messages are interpreted as elements in a finite field and thus multiplication is done in this field.
• These (encrypted) messages are then interpreted as finite field elements (not an extension field, but a prime field, to be specific) and they are added together according to the arithmetic operators defined by the finite field.
• Once the three rounds are done, we are left with three power sums $S_1,S_2,S_3$. ($\sum_{i\in{1,2,3}}{m_i} = S_1, \sum_{i\in{1,2,3}}{m_i^2}= S_2, \sum_{i\in{1,2,3}}{m^3_i}= S_3, in \space \mathbb{F}$)

Here is where I get completely lost - the author claims that given these power sums and newtons identities, we can construct a polynomial and we are able to extract the un-encrypted messages! This is really baffling and would be incredible if true.

We construct a polynomial of the form: $a_3x^3+a_2x^2+a_1x^1+a_0$

$a_3 = 1 $

$a_2 = S_1$

$a_1 = \frac{(a_2S_1 - S_2)}{2}$

$a_0 = \frac{(a_1S_1 - a_2S_2 +S_3)}{2}$

Is anyone able to shed light on how powersum of encrypted messages over a finite field can be decrypted using polynomials and newtons identities, I am all ears. If this is too broad of a question, it can be closed.

(Solution is described on page 4) Paper