# a few signposts

Assign an integer (1 through 16) to each cell so that:

• 16 is in the cell with the star; and
• each integer's arrow points to its successor (i.e. the next integer), though not necessarily in the adjacent cell.

Assume all arrows are vertical, horizontal, or at a 45° angle.

(This is what Simon Tatham's puzzle collection calls a "Signpost" puzzle. His puzzles, though, always tell you where 1 is, and frequently where other numbers are, too. I decided to try making a puzzle that tells you where only the last cell is. I'm afraid I've therefore made it too easy; please let me know.)

Edited to add: This can also be solved on Mr. Tatham's site.

The first thing to notice is that

there are many cells whose successor is locked to one cell. Let's mark them with variables. a1 is a shorthand for a+1.

Now look at

the cell marked as b1. Its successor cannot be a1 because that would mean b1=a. Therefore its successor is a. Similarly, the successor of f1 is e, and that of c1 is the cell right above it.

Then,

the successors of b3 and f3 are also fixed, since e.g. b1 and b2 are smaller than b3 and therefore cannot be its successor.

The successor of d1 cannot be d or c1, so its successor is f. Then d6 (previously f4) connects to c by the same logic.

Since the only candidate for the successor of b4 is 16, d becomes 1 and d9 connects to b. This finishes the puzzle.

Here is the solution:

Edit:

Here is the improved solution path

The top row because of the ‘b tile’ needs obviously a concatenation of integers like so:

There are only two options for 15 and with one of the options, any way you try to achieve the concatenation will make you have to use the x tile as ‘an exit’. But then we run into a dead end:

So the other option has to be 15 and this leads us to

As you can see a and b needs to be concatenated like so therefore the ‘b’ tile cannot be 8 which then forces:

Previous path

A solution path:

I thought regressing from the star is the best way to start. What I then saw was that there were two possibilities for 15. I tried one of them but it led me to dead ends with every conceivable variation:

For example here is a couple of such cases:

So I tried the only other option that could be 15, 14 being in the square below was then forced and with the first variation I tried with the possibilities for 13 being the ‘variations’, the solution was very straightforward to see using basic deduction! My stategy was if this did lead to a dead end hypothetically, I would have just tried the other options for 13 and other variations based on them until the solution is found (which aren’t too many).

This can be done in three main steps.

First:

All arrows one cell away from the edge that they are pointing towards must lead to that one cell.

Next:

Some arrows point to a line of cells, but all but one of these cells has already been claimed. These arrows must point to the remaining cell.

Lastly:

A cell chain cannot link to itself, because that would create a loop. So the last cell in each chain must lead to the first cell of another chain.