The problem

The problem we've looked at resembles the Sum and Product problem. Like this problem, the problem below can be described with Public Anouncement Logic[1]

Problem description

Three agents A,B and C all have a natural number, x,y and z respectively, pasted on their foreheads. An agent can only see the number on the foreheads of other agents. One of the three numbers is the sum of the two other numbers. All the previous is commonly known to all agents.

The following announcements now take place:

  • A says: "I do not know the numbers"
  • B says: "I do not know the numbers"
  • C says: "I do not know the numbers"

If the numbers are at most 10, a truthful announcement now is:

  • A says: "I know the numbers"

Prove this!

Constraints

There are three common constraints on the group of numbers:

  • There is only one agent that has a number that is the sum of the other two numbers, from now on sum-agent. The other agents are called term-agents.
  • The number range from 0 to 10 (inclusive), and cover only the natural numbers.
  • Numbers can occur more than once.

Given these constraints, easy and hard situations are possible. In the easy situations one of the three agents knows which number he has, before all agents are asked whether they know their number. For example the case where agents A,B and C respectively have numbers 1,9 and 10. Both agents A and B know their number immediately after seeing the other numbers. In this case A sees 9 and 10, and thus knows that since he can't be 19 (which is larger than 10), he must be 1. The same goes for agent B, but C still has doubts between being 8 or 10.

The cases in which there is first a sequence of questions, is far more interesting. Take for instance the case (1,4,3). Although agent A sees the other numbers, he still can be the sum-agent or one of the term-agents. At the same time, he knows that the other agents see him and generate their own options. Based on the fact that they also don't know the number, he can infer which of the options is the true case. This has been evaluated further.