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Could you please help me with the following puzzle:

Consider the following puzzle:

Suppose there are two box makers: Knight and Knave. Knight always writes true statements on his box, while Knave always writes false ones.

(ed: Each box was made by either a knight or a knave, and each one has a note written by its maker -- comments from MJD)

Suppose there are three boxes: A, B, and C. One of the box contains a bomb. The boxes have the following note:

A: There is a bomb in this box.

B: The bomb is not in this box.

C: At most one of these three boxes was made by Knight.

Suppose your task is to avoid choosing a box that contains bomb. Which one should you choose?

My conclusion is that we should choose box C. I derive the conclusion from:

1) Assume that the note in box C is correct.

It means there can only be one box that has correct note i.e. the box C itself. The two other boxes have incorrect notes which mean the bomb will be on box B.

2) Assume that the note in box C is wrong.

This means there will be two (or three boxes) that have correct note. But not all three boxes are correct, because we already assume box C has incorrect note. So, only box A and box B that have the correct note. In this case, it means the bomb is in box A.

So, for both case, the safe choice would be box C. Is this a correct logic reasoning in math?

PS: Additionally, is this a correct way to answer this question? Or is there a more formal way (mathematically)?

Thanks a lot for the help.

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The puzzle is missing something important: it does not state that the notes were written by knights or knaves. Without this assurance, there's no way to be sure that they weren't written by crazy olf Mrs. Dalrymple who lives up the street, and convey no information at all about the contents of the boxes. –  MJD Jul 2 '13 at 6:28
    
@MJD: I assumed the note is written by knights or knaves because it says "Knight always writes true statements on his box". –  user350954 Jul 2 '13 at 6:30
    
Oh, I see. Each box was made by either a knight or a knave, and each one has a note written by its maker. –  MJD Jul 2 '13 at 6:32
1  
Yes, your reasoning is fine. It’s the way that I’d have solved the puzzle. –  Brian M. Scott Jul 2 '13 at 6:35
1  
I wouldn't see any purpose in making it "more formal". –  Jack M Jul 2 '13 at 6:43
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1 Answer

For comparison, here is "a more formal way" to write down your perfectly correct reasoning.

Call the three boxes $\;a,b,c\;$, and one of them has the bomb: call that one $\;d\;$ (for dynamite). Write $\;T(x)\;$ for "box $\;x\;$ was made by a Knight" ($\;T\;$ for Truth).

Using this notation, we are given that \begin{align} T(a) & \;\equiv\; a = d \tag{0} \\ T(b) & \;\equiv\; b \not= d \tag{1} \\ T(c) & \;\equiv\; \text{at most one of }T(a),T(b),T(c)\text{ is true} \tag{2} \\ \end{align}


Looking at the shape of these formulae, $(0)$ and $(1)$ can just be substituted into $(2)$, so we must work on simplifying $(2)$.

Now an expression of the form $\;\text{at most one of }P,Q,R\text{ is true}\;$ is not easy to manipulate. And because $\;T(c)\;$ occurs twice in $(2)$, it seems best to make a case distinction on $\;T(c)\;$.

If $\;T(c)\;$, then $(2)$ is simplifies to \begin{align} & T(c) \;\equiv\; \text{at most one of }T(a),T(b),T(c)\text{ is true} \tag{2} \\ \equiv & \qquad \text{"using assumption $\;T(c)\;$"} \\ & \text{true} \;\equiv\; \text{at most one of }T(a),T(b),\text{true}\text{ is true} \\ \equiv & \qquad \text{"simplify"} \\ & \lnot T(a) \land \lnot T(b) \\ \equiv & \qquad \text{"using $(0)$ and $(1)$"} \\ & a \ne d \land b = d \\ \equiv & \qquad \text{"using $\;a \not= b\;$"} \\ & b = d \tag{3} \\ \end{align}

If $\;\lnot T(c)\;$, then $(2)$ is simplifies to \begin{align} & T(c) \;\equiv\; \text{at most one of }T(a),T(b),T(c)\text{ is true} \tag{2} \\ \equiv & \qquad \text{"using assumption $\;\lnot T(c)\;$"} \\ & \text{false} \;\equiv\; \text{at most one of }T(a),T(b),\text{false}\text{ is true} \\ \equiv & \qquad \text{"simplify"} \\ & T(a) \land T(b) \\ \equiv & \qquad \text{"using $(0)$ and $(1)$"} \\ & a = d \land b \not= d \\ \equiv & \qquad \text{"using $\;b \not= a\;$"} \\ & a = d \tag{4} \\ \end{align} Now we see that both $(3)$ and $(4)$ imply $\;c \not= d\;$ (using $\;b \not= c\;$ and $\;a \not= c\;$, respectively). In other words, in both cases we can be sure that box $\;c\;$ does not contain the bomb.


The advantage of this style of doing proofs, is that often the shape of the formulae help to direct the search for the proof-- and this also helps in presenting the proof, so that it is easy to follow for the readers.

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