Trying to "decrypt" something.

[Nineshadow]

The technique works like this :

 

There are 3 people who want to communicate secretly. They communicate only at certain times, in so-called "communication sessions". In a single communication session, they can exchange multiple messages. Now, for every message, each person picks an integer x so that x=c(mod3). You could say that x is the message itself,but whatever. For person A, c is 0, for person B, c is 1 and for person C , c is 2 (all the time) . In other words, this means that A,B and C pick numbers of the form 3p,3p+1 and 3p+2 respectively.

At the end of a session, this sum is calculated :

S=Σ(x_i-1)x_i(x_i+1) for i=1 to n (where n is the number of exchanged messages in the respective session)

 

Now, given this sum, I need to find out how many messages were sent, who sent them, and what number each sender chose. If there are multiple solutions, the one with the least number of messages will be shown.

For example, if S=120 then there was 1 message sent by person C who chose the number 5.

If S=1001616, then there were 2 messages sent : 1 by person A, with the number 12, and 1 by person B, with the number 100.

[DXMember]

can't you just use public key encryption?

[Nineshadow]

2 minutes ago, DXMember said:

can't you just use public key encryption?

This is an algorithmic problem I found online. It's not practical. I just need to find a technique inverse to the one they presented. I'm really out of ideas though.

[DXMember]

2 minutes ago, Nineshadow said:

This is an algorithmic problem I found online. It's not practical. I just need to find a technique inverse to the one they presented. I'm really out of ideas though.

that's not how algorithmic problems on the internet work - you don't go around asking for solutions... you solve it or you move on

[Nineshadow]

25 minutes ago, DXMember said:

that's not how algorithmic problems on the internet work - you don't go around asking for solutions... you solve it or you move on

Nevermind, I've got it.

It's really hacky but it somehow works.