# Pentopia: One-way Crossing

This is an entry for Fortnightly Topic Challenge #44: Introduce a new grid deduction genre to the community.

Here is a standard Pentopia puzzle.

Rules of Pentopia:

1. Place some pentominoes into the grid (not necessarily all of them) so that they do not touch each other, not even diagonally.
2. Pentominoes cannot repeat in the grid; rotations and reflections of a pentomino are considered the same shape.
3. The arrow denotes all of the 4 directions where the nearest pentominoes are located when looking from that square.
4. The arrow cannot be covered by any pentomino.

• When "all" denotes the all-directions-arrow, is this allowed? i.sstatic.net/JD5F1.png Commented Dec 2, 2020 at 7:18
• @OmegaKrypton do you mean not pointing to any pentomino? in that case, it's not allowed as arrows should be pointing to some pentominoes. Commented Dec 2, 2020 at 7:22
• i mean the diagonal "distance" is 1. the pentonimo is distanced more than 1 from the "all". are diagonal counted in the "directions"? Commented Dec 2, 2020 at 7:24
• @OmegaKrypton "4 directions" imply cardinal directions, i.e. up, down, left, and right. So the diagonal distance does not count. Commented Dec 2, 2020 at 7:25
• Yep, as mentioned by @Bubbler, the 4 directions are up, down, left, and right; also take a look at the second arrow in the example. Commented Dec 2, 2020 at 7:28

### Part 1: finding the starters

First thing to note is that all diagonally touching shaded cells must be part of the same pentomino. Given this, we can see that the cells adjacent to the 4-way arrows cannot be shaded, since the 4 shaded cells cannot form one pentomino.

R9C5 must be seeing shaded cells at 4 or fewer cells away (by the left side). On its right side, the only cell that can be shaded at such distance is the one 2 cells away.

R9C7 must escape upwards, and R10C8 must be unshaded. By the 4-way arrow at R10C6, we know that the cells up to distance 3 from it are empty.

The domino at the center cannot escape by turning left, so it must join with R7C9 to form either a V or a W. This gives the "distance 2" information to R5C8 (the top 4-way arrow). Meanwhile, R13C10 (the bottom 4-way arrow) has only one choice downwards.

### Part 2: placing our first piece(s)

If the R10C6 has distance 4, the two pieces shaded in red are forced to be both W's. So its distance must be 5.

Quick straightforward deduction using (dis)connectivity and distance gives this:

If the red cell is shaded, the arrow above it will force a hexomino. Therefore that cell is unshaded.

### Part 3: major breakthrough

Now, the four isolated shaded cells at the bottom cannot form four isolated pentominoes. To see this, start with the bottom middle one which can only be an N. Then it forces the right bottom one to escape to the right, forming an I, then the rightmost one does not have enough space. Therefore, at least two of them must be in the same pentomino. This can only be done with the two on the left (other horizontally close pairs are forced to be V/W, and the bottom two forming an I gives too small space for the one over them). The two can only be a Z other than V/W. Straightforward deduction places I and L on the right side of it.

Then the bottom leftmost one is a Y, upper-left-middle is an N. Apply distance argument to R4C4. Leftmost middle one must be a P, since it can be neither Y nor N.
Pieces left: F T U X

### Part 4: finishing off

The bottom left-middle-ish one can only be a U. Out of F T X, the top middle shaded cell must be an F, and it has only one way to satisfy the R1C7 arrow.

Finally, given an unused shaded cell (R5C10) and two unsatisfied arrows, both T and X must be in the upper right area. With a bit of case bashing, I believe the answer is the following. Using either T or X to cover R5C10 requires two more cells shaded to the right. There's no way X does not touch the segment, so the segment itself must be X. Then T must cover R3C13 to satisfy the arrow above it, and extend downwards to satisfy the last arrow.

• woah that was fast! nicely done, enjoy a +1 Commented Dec 2, 2020 at 7:38
• To avoid case bashing at the end: ROT13(Nf bayl G naq K yrsg, E5P10 zhfg or rkgraqrq gb E5P11 naq E5P12. E2P13 vzcyvrf ng yrnfg bar bs gur svefg guerr pryyf orybj gurz zhfg or funqrq. Guvf gura svanyvmr gur K, gura yngre gur G ol E8P14.) Other than that, very well done! :) Commented Dec 2, 2020 at 9:56
• @athin Yeah, I realized it too. Thanks for great puzzle as always :) Commented Dec 2, 2020 at 22:51