First up – I’ll post this week’s puzzle – which is a fairly gentle nurikabe. It’s been a while since I had a nice and easy puzzle so I hope this pleases the more casual of my readers!
But let’s move on. I want to discuss the numberlink puzzles from last week, and in particular make a few remarks about uniqueness of puzzle. As regular readers know, my approach to numberlink uniqueness is to find a solution to given array of clues and then try and establish its rigidity.
The first puzzle I’ll discuss is #106. Here is the solution:
The first intuition I have about numberlink rigidity is the idea of “unnecessary wrapping”. The general idea is to suppose you have a pair of clues A, and another pair of clues B – and draw the straight line between these two clues. If these two straight lines intersect, then it is clear that one of the solution lines joining the clue must wrap around the other solution line. Conversely, alarm bells should be ringing with any candidate solution if the straight lines are disjoint and yet one solution line wraps round the other.
It is a quick check with this puzzle to see that there is no such unnecessary wrapping. I haven’t rigorously proved to myself that this means the solution must be rigid, but I’d be very surprised if this wasn’t the case (with a few highly symmetric exceptions). Ok, fine, I haven’t got anything more to add to this at the moment so let’s move on.
Here’s what I had in mind (in black) with the original #105 I posted, together with (in red) where things go wrong:
So there are a couple of points to consider here. Firstly, look at that pair of 2’s, and the amount of what I’ve just defined as “unnecessary wrapping” it does! There’s something more than meets the eye going on here. The initial idea for the break-in to this puzzle was with the 1’s and the 5’s. A little spatial reasoning quickly suggests that one of the solution lines must go via the top-left of the grid, and the other along the bottom edge – so that the cluster of 3/4/6/7 is avoided.
A little more inspection reveals that the 1’s would be causing all sorts of difficulty with the 6’s if it went out top left, so the 5 goes round the top left. It was here that I made the mistake. I thought there was only one way for the 5 to get past the 4 and the 6, but completely overlooked the obvious route as marked in red. This was how I fixed the puzzle – by closing up the gap (and adding a 13th pair of clues):
This alteration made the puzzle a little easier than intended. The 3/4/6/7/13 are now essentially forced. A by product of this is that it forces the 2 to squeeze the “wrong” side of the 1. This is a bit of “logic” that only came in to play later in the original – but essentially reveals the origin of the “unnecessary wrapping” of the 2. The given clues in the puzzle are acting as blocks to the paths, which must channel in between them, with the obvious restriction that there are only so many paths that can fit into a small space. So once the 2 goes round the back of the 1, it also has to go round the back of the 8 and the 9.
The rest of the puzzle is then fairly trivial.
What I find interesting with this puzzle is that actually it wasn’t too hard to find a sketch framework to logically prove the solution is the only one. Ok, I’ll hold my hands up with the stupid oversight with the 5’s, but the fix was essentially minimal and still retains most of the features I wanted from the original puzzle.
Going back to #106, I think this one is much harder to make a start to the puzzle and go step by step in trying to establish forced paths. Instead, it isn’t too tricky to use a little metalogic (i.e. assuming uniqueness of the solution) to get the solution out – and then to do a quick check to see that this solution turns out to be rigid. Which is a completely different approach to things!
Anyhow, for all those numberlink fans out there, I hope this has provided a little food for thought.