Here we spring a booby trap laid by the “definition by example” policy of the Peter Gordon Guide. It is the Gordonian Polygon, as demonstrated by his Guide to Solving Sudoku. The correct interpretation and application of the strategy are revealed. Also, a counter example shows again why this “Guide” cannot be trusted.
The first sentence of the Guide section on Gordonian Polygons reads: “What works for rectangles also works for figures of more than four sides.” That is true, regarding particular arrangements of identical candidate pairs described in Beyond the Rectangle. But these uniqueness figures are not mentioned, much less identified, in the Gordon Guide, and the reader is left to interpret Gordon’s sentence to mean any polygon (of identical pairs) of more than four sides. Gordon’s first example suggests he means exactly that, with one of the cell vertices of the polygon containing extra candidates along with the bv candidates, and no apparent connection between all vertices. From the “extra candidates” cell, he says, the bv candidates can be removed, because one of the extra candidates must remain to prevent a double solution.
Gordon’s first example, Example 23-1, is a loop of 57 bv and extra cell 357r3c2. Peter’s rationale for removing bv numbers 5 and 7 from r3c2 is that, if 3r3c2 is removed, the polygon loop of six 57 bv can never be resolved.
By this argument, if you can draw in a six- sided polygon, you’re Gordonian, and the removal can be made.
But this just isn’t the case.
Fortunately, one of the Guide’s Gordonian examples demonstrates that this simple rationale is a fantasy. Going back to Example 20-1, the grid for the one sided Gordonian rectangle, a dead remote pair forms four sides, allowing us to complete a Gordonian Polygon in three different ways. Problem is, they place the 1-clue in the three different wrong places.
Do we need a magic rule? Is it because we can add only one side to form the polygon? The Guide doesn’t say.
The Guide does mention another reason why the particular Example 21 case works, but it has nothing to do with polygons. Going back to the diagram, the 3r3c2 removal allows the set of bv to remove all outside 5 and 7 candidates that could resolve whether each cell contained 5 or 7. The puzzle would have at least two solutions. No polygon is necessary.
Gordon’s polygon argument is that, going around the polygon, there would be no way to decide which alternating cells are 5, and which are 7. Again, two solutions.
But what kind of loop is Gordon’s polygon? Why should the bv in r5c3 and r3c5 be considered adjacent cells in a loop? Unfortunately, the Guide has not developed the Sudoku fundamentals to explain it, but the loop is of conjugate pairs. Lines and boxes define most of the pairs, but r5c3 and r3c5 form a conjugate pair as a remote pair, bookending a series three conjugate pairs.
Bottom line, Gordon has stumbled over something that works, but he is unaware of what he is dealing with. There is certainly no reason to name it a Gordonian anything.
BTW, Example 21 is easily solved without assuming a single solution and using any uniqueness argument.
not(blue and red) =>
green or orange,
removing 5r5c2, which proves green, and catches two orange 5’s in c2, proving red. No resistance is left.
To guard against a more complex multiple solution, I tested orange, but it quickly asserted that both blue and green are false.
The Guide follows with the Gordonian Polygon Plus, the One-sided Gordonian Polygon, and Gordonian Extended Rectangles. I have to see if there is actually anything new under these titles, and I invite you to come along on the next post. Have a groovy Christmas.