# A logic puzzle from TES: Arena

Its nice when games have riddles hidden in them. While playing TES:Arena, I came across an unusual logical puzzle: There are 3 cells.

If Cell 3 holds worthless brass, Cell 2 holds the gold key.

If Cell 1 holds the gold key, Cell 3 holds worthless brass.

If Cell 2 holds worthless brass, Cell 1 holds the gold key.

Knowing this brave fool, and knowing that all that is said cannot be true, which cell contains the gold key?

The correct answer is Cell 2 as suggested by the game. I wanted to know how one could logically arrive at the result. Could anyone help me with this?

What I tried: I negated all the above statements. The first implication became Cell 3 holds worthless brass AND Cell 2 does not have gold key. But if this is true, then cell 2 does not have the gold key and the result is incorrect. Hence I had this doubt.

PS: Choosing the wrong door causes man eating spiders to be released.

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Does each cell necessarily contain either worthless brass or a gold key? –  Sp3000 Dec 20 '12 at 12:55
Although not stated in game, there is only ONE cell with a gold key. The other cells may/may not have brass. –  Gautam Shenoy Dec 20 '12 at 12:57

If you label the conditions a-c, and if the gold key exists and is unique, it is enough to show that not in 2 leads to a contradiction. But 'key not in 2' leads to 'key in 2' (so a contradiction), as follows:

By (c), not in 2 implies in 1. By (b) then, not in 3. By (a) then, key is in 2. Contradiction.

To be safe, you can check if the game makers messed up:

As 'key in 1' is a sub chain of the above, it also leads to 'key in 2'. Contradiction.

Similarly, 'key in 3' means by uniqueness that 2 holds brass, which, by (c), implies 1 holds the key. Contradiction.

Finally note that if the key is in 2, it doesn't lead to any contradictions. So game is correct.

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Cell 2 cannot hold worthless brass, and cell 1 cannot have the gold key.

If a cell can have neither brass nor key, then that is the limit of what you can conclude. For example, your conditions are consistent with cell 3 having the key and there being nothing in the other cells.

But if every cell contains either brass or key, then you know cell 2 has the key. You can prove that by assuming cell 2 did not have the key, and hence had worthless brass. That implies that cell 1 has the key, which implies cell 3 has worthless brass which implies that cell 2 has the key. Contradiction.

However, if that additional condition were part of the problem, then your middle condition would be redundant.

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Sorry to be contrary, but I believe you need 2 assumptions for there to be a unique solution. First, that all 3 statements are indeed true. If we allow that one or more of those statements doesn't hold, the whole thing falls apart.

Second, that each cell contains either brass or the key. No empty cells. If you disagree, try looking at these 2 solutions:

Cell 1 = Empty, Cell 2 = Key, Cell 3 = Empty

Cell 1 = Empty, Cell 2 = Empty, Cell 3 = Key

None of the 3 statements apply to either of these, so they're both possible valid solutions and you're left to guess and hope the spiders don't eat your face.

Now, you can brute force your way through by listing all possible solutions and checking which are valid under the given statements. In this case there aren't many possible choices, so that approach is not too bad. However, I'm assuming you're looking for a bit deeper insight, so I'll walk through a technique that can sometimes provide a quicker route to the answer, especially in more complex puzzles.

Since all the implications are one way (they are "if", not "if and only if"), one approach is to start by assuming a condition from the left side of a statement and trace the implications through all the statements to look for inconsistencies.

Just going in order, let's assume first that Cell 3 holds brass. Then by statement 1, Cell 2 holds the key. Since we're assuming key or brass in each and only one key, then Cell 1 would have to be brass. Given that, neither of the last 2 statements apply, so this solution is possible with no contradictions. Let's continue and check the others to be sure we have the only possible valid solution.

Looking at statement 2, let's assume now that Cell 1 holds the key. By statement 2, Cell 3 holds brass. By statement 1, Cell 2 holds the key. Since only 1 cell can hold the key, this is a contradiction. Therefore our original assumption is false, so Cell 1 does not hold the key.

Lastly, look at the third statement. If we assume Cell 2 holds brass, then Cell 1 holds the key. But we already know by our last reasoning that if Cell 1 holds the key we end up with a contradiction. So our assumption here is false, and Cell 2 does not hold brass.

Since Cell 2 cannot hold brass, it must hold the key.

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If you take "not all that is said is true" to mean that at least one statement is false, you can go through negating them and see where it leads. If the first statement is false, you have 3 holds brass and 2 does not hold the gold key. Given that there is a gold key, it must be behind door 1. Then the second and third statements would be true, but we still don't know whether cell 2 holds brass.

If the second statement is false, cell 1 has gold and 3 does not have brass (so is empty). The first and third sentences are then true and we again know nothing about cell 2.

If the third statement is false, 2 holds brass and 1 does not have gold, so it must be cell 3 that has gold. The other two are then true and we don't know what 1 holds.

Since we didn't reach a contradiction from any of these, we can't choose between them. I would downvote the puzzle creator. However, if you ignore the "not all that is said is true" and believe them all, gnometorule has shown how to get there.

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