## A Problem of Logic

### The Problem

Recently, I stumbled across a very interesting problem. I closed the site and started to think about the solution. Therefore, I neither have a link nor the solution given on the page. Let us try our luck with it. (I found a similar problem here. The solution is similar to the one I found. But here, I try to say more about the problem.)

Problem: Two prisoners A and B are given two numbers, each between 0 and 10. A is given the number a=6 and B is given the number b=4. They are told that the numbers add to 9 or 10. As usual, the prisoners cannot communicate in any way. Each day, A is asked if the sum is 9 or 10, and then B is asked the same question. Either prisoner can pass or claim to know the answer. If the answer is correct both regain their freedom, if not both are executed.

Can they find a secure solution without ever communicating between each other? If they both are clever enough they can. Now, you can leave this page for thinking if you wish. But mark it so that you can find back in case you cannot solve the problem.

The logic gets very involved if you start with the given values a=6 and b=4. A knows from the start that either b=4 or b=3. This he knows that B knows that he has a number between 5 and 7 etc. If you think that way, you are in for a problematic approach.

### The Solution

It took me several attempts to change my thinking. Let us call „shared knowledge“ the facts that both prisoners know and that both prisoners know the other prisoner knows. In fact, we think of what an observer would know. We ignore the specific numbers a=6 and b=4 for a moment.

Then, after A passes, everyone knows (including the observer) that A cannot have a=10. So he has any number between 0 and 9. Likewise, B cannot have b=10 when he passes. But notice that B can not have the number b=0. Because A has less than 10 he would immediately know the answer a+b=9. Thus, after the first day, we write the shared knowledge as

\(D_1 : \quad 0 \le a \le 9, \quad 1 \le b \le 9\)

Once, A passes the second day we know that he has neither a=0 or a=9. Otherwise, he would have known the sum as you will easily check. So a is between 1 and 8. If B passes too we know that b=1 and b=9 are impossible. So after the second day everyone knows

\(D_2 : \quad 1 \le a \le 8, \quad 2 \le b \le 8\)

Continuing like that we get

\(D_3 : \quad 2 \le a \le 7, \quad 3 \le b \le 7\)

\(D_4 : \quad 3 \le a \le 6, \quad 4 \le b \le 6\)

Now, A knows that a=6. Thus either b=3 or b=4. But b=3 no longer is possible. So when A is asked at Day 5, A knows for certain that a=6 and b=4.

Note that B has to decide between a=5 and a=6 which he cannot do earlier with this reasoning. But assume the correct numbers are a=6 and b=3. Then B has to decide between a=6 and a=7. We can do so when he is asked at Day 4. He will then already know A:3-6.

This solved the problem in our specific case. The logic is rather simple, but weird.

### Deceptive Ideas

Let us try to shorten the process and put our logic to a test. We ignored the specific knowledge of A and B about the number they have. A knows a=6, and B knows b=4. So A knows that b=3 or b=4. Thus A knows that B knows that A has a number between 5 and 7. Likewise, B knows that A knows that B has a number between 3 and 5. So, you might deduce that the shared knowledge between A and B at Day 0 is

\(D_0 : \quad 5 \le a \le 7, \quad 3 \le b \le 5 \tag{1}\)

With the logic from above, we deduce that A would know that answer with a=7. So, after a pass, B knows a=5 or a=6. With either b=3 or b=5 he could immediately deduce a+b=9 or a+b=10. Thus, when he passes he must have b=4. A could then declare the result after only two passes when asked at Day 2.

This is a false and deceptive logic. It cannot be correct, because if a=6 and b=3 then A will declare a+b=9 with the same logic on Day 2. In fact, the „shared knowledge“ becomes

\(D_0 : \quad 5 \le a \le 7, \quad 2 \le b \le 4 \tag{2}\)

From that B will exclude a=5 after a pass from A, which changes the outcome completely.

What goes wrong here? While both prisoners will agree to the facts in (1) and(2), and while both are true, they cannot come up with these facts out of their own knowledge. So they have no common agreement about the situation. They do not know what the other prisoner knows. Thus they cannot conclude anything form the assumed knowledge of the other prisoner. To be more precise, B cannot derive the answer from b=3, because from his viewpoint A can still have a=6 or a=7.

### Simulation

It is quite an interesting question if we can simulate the procedure. In fact, we have just given an algorithm to deduce a+b from the number of passes and a or b. Extending from our special case N=10 to a general N, our algorithm goes as follows. The shared knowledge after Day d is

\(D_d : \quad d-1 \le a \le N-d, \quad d \le b \le N-d\)

Now we take into account the true values of a and b, known privately to A and B, respectively. E.g., A knows that b=N-1-a or N-a. Thus, after Day d-1, at Day d, he knows the value of b if d-1=N-a or N-(d-1)=N-a-1. Using similar arguments for B, we get

- B knows the sum at Day d if d=N+1-b (sum=N) or d=b+1 (sum=N-1), after the pass from A at Day d.
- A knows the sum at Day d if d=N+1-a (sum=N) or d=a+2 (sum=N-1), after the pass from B at Day d-1.

Thus the result is known at day

\(d = \min \{N+1-a,a+2,N+1-b,b+1\}.\)

If N=10, a=6, b=4, then d=5. If N=10, a=6, b=3, then d=4. We can prove that this yields the correct answer for all a, b, and N. Or we can simulate, e.g. in EMT.

>function map check (a,b,N) ... $ v=[N+1-b,b+1,N+1-a,a+2]; $ d=min(v); $ if d==N+1-a then $ s = N; $ elseif d==a+2 then $ s = N-1; $ elseif d==N+1-b then $ s = N; $ elseif d==b+1 then $ s = N-1; $ endif $ return s; $ endfunction >a=0:10; check(a,10-a,10) [10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10] >a=0:9; check(a,9-a,10) [9, 9, 9, 9, 9, 9, 9, 9, 9, 9]

But there is a logical problem. Does the algorithm proof that the two prisoners can escape? The problem is that the prisoners are not allowed to communicate and agree on a specific algorithm. Are there alternatives to this algorithm? Is this the fastest algorithm? These questions are not so easy to answer.

But starting from the observation that A cannot declare anything on Day 1 unless a=10, one can work the way down to the solution as written above. It becomes obvious that the given path is the only possible algorithm

### Generalizations

On the page linked at the start the problem is to decide between a+b=18 and a+b=20. Of course, we could set any other set of possible sums. The logic remains the same. Just don’t let yourself trapped into shortcuts taking into account the knowledge of his own number by any prisoner.