3

How would I go about solving this 3 Chests problem?

I need to either solve or prove that the following riddle is unsolvable.

  1. Chest A contains gold if Chest B contains gold or if Chest C contains gold.

  2. Chest B contains gold if Chest C contains gold and Chest A contains silver.

  3. Chest C contains gold if Chest A contains silver.

I've created truth tables based on the 3 statements. How do I go about either solving it or proving that it can't be solved?

Brad Goodall
  • 33
  • 1
    Too many "if"s in $1$. Can a chest have silver AND gold? And what is "chest" anyway? – markvs Nov 19 '21 at 01:55
  • Treat it like a system of equations: 3 variables and 3 equations. You can immediately reduce this to two variables and 2 equations by substitution. What you then have should simplify and the answer should be clear. I am assuming from your tables that each chest must contain silver or gold, but not both. – tkf Nov 19 '21 at 01:58
  • Assume $A$, then by 3. $!C$ and by 1. $B\ ||\ C$. But since $!C$ already $B\ ||\ C$ yields just $B$. But then by 2. $C\ &&\ !A$ ... contradiction.

    Now assume $!A$, and continue the argument ...

     – balddraz Nov 19 '21 at 02:07
  • question should specify that each chest must contain either silver or gold. otherwise $A$ can have anything other than silver or gold and $B$ and $C$ anything other than gold. – okzoomer Nov 19 '21 at 02:32

3 Answers3

2

Assume your translation is correct, and you can just build a truth table of the conjunction of all your above 3 clauses to see if it's satisfiable or not.

$$A~~|B~~|C~~|~~(A↔(B∨C))∧(B↔(C∧(¬A)))∧(C↔(¬A))$$

And use some online tool like this you can quickly see its negation is a tautology, thus this puzzle as interpreted by your above logic translation is unsolvable since it's a negation of a tautology (aka a contradiction)...

cinch
  • 1,648
  • 1
  • 4
  • 12
  • Thanks! How do we know to translate "p if q" into $p \leftrightarrow q$ instead of, for example, $p \leftarrow q$? – joseville Dec 13 '21 at 04:56
  • 1
    @joseville Thanks for your further checking! This was from a while ago and I remember I interpreted "p if q" here as a definitional logical equivalence of "p", and no one pointed out to me so far... Yes you have a point here and luckily I just checked same after changing ↔ to ←, and same conclusion still holds. And the gist of my answer here is the object language formation from meta. – cinch Dec 13 '21 at 05:10
1

There is no solution to this problem.

If $A$ contains gold then either chest $B$ or $C$ contain gold. $C$ contains gold if and only if $A$ contains silver. Therefore $C$ does not contain gold and so, $B$ must contain gold which means $C$ contains gold which is a contradiction.

If $A$ contains silver then $C$ contains gold which means $A$ contains gold. This is also a contradiction.

Therefore, there is no solution.

John Douma
  • 11,426
  • 2
  • 23
  • 24
1

I'm assuming each chest contains either silver or gold.

If C contains gold, then A contains silver (Statement $3$), so B contains gold (Statement $2$). This contradicts Statement $1$.

If C does not contain gold, then A does not contain silver (Statement $3$), so B does not contain gold (Statement $2$). Again, (if A contains either silver or gold, this forces A to contain gold), this contradicts Statement $1$.

The combination therefore has no solution.

Robert Shore
  • 23,332