Self-referential Puzzle 2 Solution

The steps of the solution are as follows. This solution is from Kris Hodge, and so far as I recall is the intended method.

• • If clue 10 were false, then at least one of clue 1, 2 and 3 are false. However, only one clue is false and hence 10 must be true. Thus 1, 2 and 3 are true.

  • The only unique numbers between 1 and 5 with A+B = CD are 1+5 = 2*3, 2+3 = 1*5 and 3+5 = 4*2. The constraint given by Clue 2 means that the numbers A and B are 1 and 5 in some order, and C and D are 2 and 3 in some order.

  • Hence the cell in Row 1, Column 3 must be 4.

  • Clue 3 says that 1 is on an edge only twice. Given that 1 must be in the two side columns and the two side rows, if it is to touch the edge only twice it means that it must be in a corner both times. Since we already know A or B is 1, that means A=1 and the opposite corner must also be a 1. It follows that B=5. Thus we have so far:

• • On the third two, the two elements can sum to either 3, 4 or 5. But as the result goes in the centre square and there is already a '4' in the column, it must be either 3 or 5. If it were 3, then clue 8 and clue 9 would be false, and so it cannot be 3. Hence it is 5. (To observe that Clue 9 would be false, note that the first column is the only row, column or diagonal which could possibly fulfill this property).

  • Suppose for the moment that Clue 9 is true. Then the first column would be in order, and so the first cell in the sum box on row three would be a 3. Thus the second cell in the sum box would be 2. Completing this line it must read "3,2,5,1,4", as if the final cell on this row (labelled G) cannot be 1, as it is already in the same column as a G. If Clue 6 were true too, then as G=4 there would be 4 1's in the downward diagonal. However, this cannot occur as it would result in multiple 1's in the same column. Thus either Clue 9 or Clue 6 is false.

  • This means that Clues 4, 5, 7 and 8 are true. In particular, F = 3.

  • 3 is on the edge four times, and so is never in a corner. This means that C = 3 and D = 2.

  • Considering row five, and the fact that 3 is never in a corner, it must be concluded that the number in Row 5, Column 2 is 3.

  • By the Latin Square property, the number on Row 2, Column 3 must be equal to 1.

  • Thus E = 2. Hence all numbers except for 2 are diagonally adjacent to themselves. Consider where 2 should go on Row 4. Clearly it must be in the first column. The square so far should look as follows:

• • Whence Clue 9 is false, and so Clue 6 is true. This implies there must be another 1 in the downward diagonal. It can only go in one position: Row 4.

  • The rest can now be filled in using the Latin Square property. It should look as follows:

History

Below shows the first drawing of it. I redrew it July 2010.