# Using a meta-Statement as a clue

June 22, 2010 Leave a comment

Take the self-referential number square puzzle that I posted that I posted the other day. Now ignore the last clue at the bottom, which column 2 has an even number on it. Fill it in as long as far as you can and you get the following:

Filling in the rest with either the green numbers or the yellow numbers gives a valid solution. Hence there are two solutions of this set of clues. (i.e. this set of axioms has two models, where a model is a set of 16 numbers that has the properties.) Really, we want a unique solution, but what if we had a clue which references the set of models that exist without the clue?

Let us call the square that is in column 2 and row 3 ‘J’, to use the convention for the solution on the post that introduced this puzzle.

So for example, our clue could:

Let X be the sum of all the possible values of J without this clue (i.e. X is the sum of the value of J over all models without this clue). Then J = X-3.

This then lets us conclude, upon realizing there are just two models without this clue, that J = 7 and hence we have a unique answer.

Another variation of this is to have a clue which says ‘The solution is unique’, even though it is not, but in fact allows you to deduce a unique solution! I may write about this at a latter date.

The second self-referential square uses another sort of meta-statement, saying that one of the clues is in fact incorrect (and that the meta-statement does not count as one of the clues). This is not interesting from a logic point of view, however.