# A maze puzzle for the day

Here's a little maze puzzle I originally built a couple of years ago, that seems apropos to reprise now:

Can you make it from the A in the top left of this grid to the Z in the bottom right, always going either up one letter (for instance, A to B or G to H) or down one letter (for instance, N to M)? The alphabet wraps around, so you can go from Z up to A or A down to Z too. Try as hard as you can (and remember that you can always work backwards if you get stuck forwards), and see where you get!

A small hint for those who are wondering about any possible invalidities in the grid:

Solving the maze is not the same thing as solving the puzzle. Read those instructions carefully!

• Are we moving horizontally/vertically only or diagonally as well? Apr 1, 2015 at 22:53
• @Duncan Only horizontally/vertically (though diagonal moves should be impossible in any case - no two diagonal letters will be alphabetically adjacent, the way the grid is built) Apr 1, 2015 at 22:57
• @StevenStadnicki 'you can go from Z uo to A or A down to Z too' i thought skipping some rows from a - z but there are none in the same row where that would be possible..
– user9543
Apr 1, 2015 at 23:28
• Ah, before reading the comments I was sure that the solution is to go from A to Z in one move - nothing in the instructions says that our moves need to make sense in terms of the grid.
– JiK
Apr 4, 2015 at 13:22

The maze is invalid. However, the point of this puzzle is to find out that it is invalid, and a little something extra. I have made all the connections I can find.

The dead ends of the start path have the letters A P R I L.
The dead ends of the end path have the letters F O O L S.
The puzzle is an April Fools joke!
P.S. The J in [row 6, column 9] could be an L, and it would be a valid maze with none of the paths disconnected from the main route.

• I reached the same conclusion by blacking out squares that did not have two consecutive letters touching, then repeating and ended up blocking off all paths. As another note, the K in (5,5) could be an I to present a solution. Apr 1, 2015 at 23:02
• Your answer is better, and complete, so mine has become obsolete :) Apr 1, 2015 at 23:42
• Lol it's cool. I don't mind if he accepts your answer for being first. You earned it, of course, and if my grid helped you, then it's all the better. :) Apr 1, 2015 at 23:43
• Therefore, the answer to the puzzle is no :) Apr 2, 2015 at 1:49
• Which tool did you use to create that image?
– smci
Apr 5, 2015 at 10:45

Given that (as shown in two other answers), there is no normal path, I extrapolate that this may be the solution to the maze.

Allowing wrap-around in the maze as well. Though if this is a valid solution, it would be nice if the instructions said something about wrap-arounds being allowed. If the below is a solution, then note that you can also make it wrap-around 1 or 2 rows further down (from the Z or the other A).

• Exactly what I was thinking Apr 1, 2015 at 23:07
• also the z ander the y on the left could be attached to the y on the other side than, and the a under that z to the z on the other sid,e
– user9543
Apr 1, 2015 at 23:11
• lol Duncan solved the unsolvable maze! nice job.
– JLee
Apr 2, 2015 at 4:01
• An easier solution would be to go diagonally from the start to the end... ;) Apr 2, 2015 at 7:00
• @Duncan They're not - this is an artifact of the way the maze was built (and the fact that all squares are reachable). You can 'checkerboard color' the grid; then odd letters (A, C, E, G, etc.) show up on one color of the checkerboard and even letters (B, D, F, H, etc.) show up on the other color. Since diagonally adjacent letters will always be an even number of (orthogonal) steps apart, they can never be alphabet-adjacent. Apr 2, 2015 at 21:23

You can't pass the red line,also checked diagonal.

• This is simpler than Bulldogg6404's answer! Dec 22, 2015 at 15:16

While it is intended as unsolvable as explained by Bulldogg6404, it can actually be solved, as long as you look at the puzzle laterally at the right moment.

Specifically...

If you run down the left edge (with a slight detour to get from D to A), you reach a Z in the 8th row. Viewing the problem from the side (that is, laterally), you'll discover that it's also an N, which is convenient, because this allows you to move right onto the O, which is an O irrespective of perspective. From there, it's easy to get to the bottom-right corner by going right six, down one, right two, down three, right two, and down one.