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.

The standard Pentopia rules apply. However, each clue has been replaced by a letter. Different letters refer to different clues. There are 15 possible clues below the grid, only 9 clues are used.

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For reference, here is the list of pentominoes:

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2 Answers 2


The completed grid:

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Step by step solution:

As a first step,

Note that some of the clues touch the wall. These clues can not contain an arrow pointing towards this wall. In particular, the E can only be the up-clue. Using this, we know that the following cells are empty:
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The H clue in R13C3 cannot contain a down arrow, so H can only consist of up and right arrows. Since E already represents the up-clue, it follows that H definitely contains a right arrow. This yields some shaded cells around the rightmost H. Furthermore, by the clue in R3C3 we see that the F cannot point to the left, and by R3C13 the G cannot point upwards.
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We conclude from the shaded cells from the previous step that H is the right-up-clue. Furthermore, in the lower-left of the grid we have a 2x2 box which needs to be empty. This implies that C is the right-clue. This gives a P-pentomino in the top-right, and therefore F needs to be the up-down clue.
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The top-left 3x2 box is nonempty because of the E below it. R2C3 now needs to be filled to prevent forming another P, this also enables us to complete the F clue below this cell. the H in R13C3 cannot have one of its adjacent cells filled, since this would quickly cause a contradiction with the other nearby H: we would get 6 connected cells (or diagonally touching pentominoes).
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We learn from the top-left that A has either a left or a down arrow (to prevent forming a P-pentomino). In the bottom-left of the grid, this yields that the cell R11C2 needs to be shaded, and therefore A has a down arrow. This information about A can be used in the top-left of the grid again. Now note that R1C5 cannot be shaded, since this would prevent us from properly completing the E clue in R3C1. In particular, A does not have a right arrow.
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The C in the bottom left should have one of the 3 cells to its right filled. This implies that also R14C5 is filled. In particular, the B clue contains a left arrow. This gives a lot of empty cells around the B clue at the top of the grid.
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The E in R4C9 should have some cell above it shaded. This leads to shading the cell directly below the D as well, meaning that the D clue contains a down arrow. This can be applied to the D at the bottom of the grid as well.
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We conclude that the D corresponds to the down-right-clue. Furthermore, the E clue at the bottom of the grid can be completed.
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The pentomino above this E in R15C8 has to cross row 12 (otherwise it would be another P). This implies that one of the first 8 cells to the right of the H in R12C1 is shaded, and therefore also one of the first 8 cells to the top of this H is shaded.
So either R5C1, R7C1, R10C1 or R11C1 needs to be shaded. In the case where either R10C1 or R11C1 is shaded, the H clue in R8C1 forces R7C1 to also be shaded. We can conclude from this that always R5C1 or R7C1 is shaded. In both these cases we see that the cells directly below and above the F in R6C2 need to be shaded as well.
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This leaves only one way for completing both H clues in column 1. Also the E in R7C4 can be completed. The D in R12C7 can furthermore be completed using the fact that the pentomino next to it cannot form a W.
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The B clue cannot contain an arrow to the right, since this would construct another P at the bottom of the grid. This means that the cell directly to the left of the A in the top row is shaded, implying that A corresponds to the down-left-clue, and B corresponds to the left-clue. This yields the following progress:
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The F clue at the right can be completed, doing this also yields that G is the down-clue. Also some progress can be made at the left-hand side of the grid by not creating another F-pentomino.
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Using the last remaining clues, and trying not to get duplicate any pentominoes, get us here:
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All 12 pentominoes have been placed already, so we are done!
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  • $\begingroup$ Nice write-up, Reinier - I have posted a solution which differs to yours in explanation after a certain point, as I realised I didn't quite follow all of your logic - specifically: "Combining this with the H in R8C1 yields that either R5C1 or R7C1 needs to be shaded." Why is this the case? The R8C1 could potentially still be satisfied by shading a square at the top of column 1 above the E clue, so I don't really see how we can deduce (at this point of the logical process) that either of R5C1 or R7C1 must be shaded. Am I being a bit thick?! Happy to be put right :) $\endgroup$
    – Stiv
    Commented Aug 4, 2021 at 18:00
  • $\begingroup$ Wait, I've realised - the shaded square above R8C1 has to be closer than the top ones because of the H in R12C1 needing a shaded square within 8 spaces. If that square isn't in the spaces above R8C1 then it's in the ones below it, which forces the one for R8C1 to be even closer to it, as there is no Down clue. Whatever happens, this forces R6C2's squares in those positions and the rest of your logic follows. I get it now (but it is a hard part of the puzzle to get your head around - phew!). Well done again :) $\endgroup$
    – Stiv
    Commented Aug 4, 2021 at 18:40
  • $\begingroup$ That is indeed the argument which I was thinking about! I have added a slightly more detailed explanation to this step, since I believe it is indeed the trickiest step of this solution. $\endgroup$
    – Reinier
    Commented Aug 4, 2021 at 19:52
  • 1
    $\begingroup$ Impressive explanations from both of you! :D I accept this answer as it still grasps the idea of all of the logics, while also being the first. Shoutout to Stiv for improving the answer especially the tricky step! :) $\endgroup$
    – athin
    Commented Aug 7, 2021 at 3:35

Note: I initially shelved my write-up of this puzzle when @Reinier posted their solution when I was midway through. However, after reflection - even though the two are similar in their logic up to a point, albeit with some steps in a different order - I figure my write-up may still be useful to people interested in this puzzle as there was one step in @Reinier's that I couldn't quite follow, plus I inserted the confirmed arrow clues into the puzzle graphic as I went, which I think makes it easier to comprehend... You can be the judge of that!

Step 1:

To break into this puzzle, let's use the E clues as an illustration of the general method...

By looking at the E cells positioned around the edges, we can see that the E clue cannot involve a Left arrow (there is an E on the left edge), a Right arrow (E on the right edge) or a Down arrow (E on the bottom edge). We are left to conclude that the E must therefore be the Up arrow clue. Inserting these into the E positions immediately allows us to blank out adjacent Left, Right and Down sides.

Step 1

Of course, we can also blank out all spaces that contain a clue. And we can repeat our edge-based logical deductions to blank out all immediately adjacent spaces above an A, B or C (on top edge), below a C or I (on bottom edge), to the left of an H (on left edge) or right of an F or G (on right edge).

Step 1b

And now we can see that additionally:
- H cannot have a Down arrow (there are no shaded squares below the H in Column 3);
- F cannot have a Left arrow (no shaded squares left of the F in Column 3);
- G cannot have an Up arrow (no shaded squares above the G in R3C13).

Step 1c

Step 2:

No pentomino can fit in the enclosed 2x2 block in the bottom left corner, so blank this out. Looking at the C in the bottom row, this now has no shaded squares to its left, so C cannot have a Left arrow. In the previous step we have already ruled out C having Up and Down arrows too, so C must be the Right arrow clue.

Step 2

Next, note that we've already ascertained that H cannot have a Left or a Down arrow. Now that the Up and Right clues have been identified (E and C, respectively), H must therefore be the the Up-and-Right clue. Some additional shading and blanking out follows.

Step 2b

This in turn fixes F as our Up-and-Down arrow clue.

Step 2c

Step 3:

Consider next the top-left corner of the grid.

If R2C3 is left blank, then a second P pentomino will form in the top left-hand corner area (since there must be a shaded cell above the clue in R3C1). To avoid this illegal situation, R2C3 must be shaded (and thus R4C3 also). Furthermore, at least one of R1C3 and R2C4 must be shaded, again to avoid a P pentomino forming.

Step 3

This means that the A clue must contain a Left and/or a Down arrow (an additional Right arrow is still a possibility too). With this in mind, we know that one of R10C1, R11C2 and R12C2 must be shaded. Importantly, whichever of these is shaded, we would definitely end up needing to shade R11C2 for continuity. This means that the A clue definitely includes a Down arrow and we can shade R2C4 for definite.

This has a further knock-on effect in the top-left corner - since at least one of R1C1 and R2C1 must be shaded (due to the Up clue immediately below) and connected to the shaded cells in R2C3-4, it is impossible for R1C5 to be shaded - the A clue is therefore either Down or Left-and-Down (it cannot include a Right arrow - some blankings out follow, including the now isolated A1C7).

Step 3b

In fact, R2C5 must also be blanked out too, since if the A clue is Left-and-Down, shading this space would result in two separate pentominoes touching.

Note also at this point that it is impossible for R12C3 to be shaded - we cannot use a pentomino to satisfy the Right-Up clue in R12C1 whilst also ensuring that R12C3 and R13C4 are part of the same pentomino. There are a couple of knock-on shadings and blankings out for continuity.

Step 3c

Step 4:

Because we have now shaded R11C3, there must be at least one shaded square in R15C4-6, to satisfy the Right clue in R15C3. This means that R14C5 must be shaded as a result, which means that the B clue must include a Left arrow. This has a big knock-on effect at the top of the grid, where the B clue in row 1 has a long line of blanks to its left - equivalent blanks must now also be entered on its other available flanks.

Note that at this stage we can narrow B down to one of the Left clue, the Left-and-Down clue, or the Left-and-Right clue - it cannot be Left-Right-and-Down, or two separate pentominoes will touch illegally at the bottom of the grid.

Step 4

These additional blankings-out mean we now know R3C9-11 at a minimum must all be shaded to satisfy the Up clue in row 4. This means that the D clue must include a Down arrow. We can apply this information to the D clue in row 14 and then note that D must also include a Right arrow and D cannot include a Left arrow, since R14C9 must be blank to avoid separate pentominoes touching. This means D must be the Right-and-Down clue (we know it cannot include an Up arrow due to the blank in R1C10).

We can also resolve the Up clue in the bottom row now, since we know R14C8 must be shaded (so that the shaded square upwards is closer than the one to its right).

Step 4b

Step 5:

Consider now the Up-and-Down clue in column 14. Can we shade the spaces directly above and below it? If we do then the Z pentomino is immediately placed, the B clue in row 9 then becomes a Left-and-Right arrow and the G clue in row 3 is forced to be a Left arrow. Knock-on deductions for the G in R13C13 and the B in row 14 then create two W pentominoes at the bottom - a contradiction!

Step 5

Instead, this means the Up-and-Down arrow in column 14 can only be satisfied by the already-shaded square in R5C14 and its counterpart in the bottom row, which immediately places the U pentomino in that corner and forces the pentomino to the left of I to be the L shape. This then means that I is the Left-and-Right arrow clue and G is the Down arrow clue.

Step 5b

We can now place the Z pentomino on the right, and also conclude that the A clue must be Down-and-Left, as this is its only option remaining. In turn, the B clue can now only be the Left arrow clue and we have identified all of the clues!

This immediately allows us to resolve many more squares:

Step 5c

Step 6:

Now focus on the unresolved Up-Right clue in column 1. Because there is a shaded square two spaces below it, this must be resolved by shading its two immediately adjacent squares. This instantly gives us the W pentomino.

Step 6

Next note that R5C4 cannot be shaded, or a second F pentomino will be formed. This forces the placement of the N pentomino, which in turn forces the top left pentomino to be the Y.

Step 6b

The T, X and V pentominoes are now all forced.

Step 6c

And finally, we note that only the I pentomino remains to be placed, and there is only one way to do that while satisfying the Right arrow clue in the bottom row. All other spaces are blank and the puzzle is complete at last!

Final answer


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