The following text was found written on the wall of an ancient tomb:


Can you decipher what it means?

Nearby the wall, this piece of parchment was found. Perhaps it is a clue for deciphering the word?


The following instructions were found (seemingly referring only to the grid on the left):

Shade 22 squares in the grid such that:

  • No number appears in a row or column more than once (shade boxes to remove duplicates).
  • No shaded square is adjacent to another shaded square vertically or horizontally.
  • When completed, all un-shaded squares create a single continuous area. (Example)

Hint 1:

1 2 3 4 5 6 7 8 9

Hint 2:

New information reveals that the parchment actually contains notes on how to encode the word written on the wall. I'm sure by working backwards, you can use this note to decipher the text as well.


3 Answers 3


This answer resolves the final message concealed in the puzzle. The initial puzzle was solved by @BeastlyGerbil - go upvote their answer for finding that part of the solution.

Final answer:

The one-word message hidden in the puzzle is meant as feedback from the OP on your efforts in solving it. The OP says:


How to get there...

First, make note of @BeastlyGerbil's solution to the initial puzzle:

enter image description here
For full details on how to solve this step of the puzzle, see BG's answer (and consider leaving a +1).

For our intents and purposes, what is especially crucial to the final answer is overlaying the shaded squares from the left-hand grid onto the right-hand grid.

Note from the second hint that:

the solved puzzle describes how to encode (not decode) the original ciphertext, 'DXFDMKEPU'.

So how to do this? First, note down the shaded 'L' and 'R' sequences in each row (numbered 1-9) of the solved puzzle, gaining us:
LRL / RL / RRR / L / RLR / LLR / LR / RRLR / R

Note then that these letter groups show to us how to create the 9-letter ciphertext. If we treat the alphabet as a letter-line, arranged from A to Z in order, these 'L' and 'R' characters represent LEFT and RIGHT movements through the alphabet in order to reach the letters in 'DXFDMKEPU'.

Immediately, notice that this means one 'L' will cancel out one 'R', and vice versa, since this purely returns us to the spot from which we came. Because of this we can simplify our letter groups to:
L / (no movement) / RRR / L / R / L / (no movement) / RR / R

To find the original letters:

simply take the ciphertext and substitute each 'L' in the instructions for an 'R' and each 'R' for an 'L', their reverse operations. This means we need to apply the following shifts to the letters in the ciphertext:
R / (no movement) / LLL / R / L / R / (no movement) / LL / L

Starting from 'DXFDMKEPU' we then get:
D --> E (R)
X --> X (no movement)
F --> C (LLL)
D --> E (R)
M --> L (L)
K --> L (R)
E --> E (no movement)
P --> N (LL)
U --> T (L)

i.e. it spells out 'EXCELLENT'...

  • 2
    $\begingroup$ If this is the correct answer, in my opinion YOU should receive the checkmark. Solving the enigmatic portion of this puzzle, and most others, is much more challenging than solving the preliminary puzzle, which in this case is rather trivial. I don't feel that anyone on this site, myself included, should receive a checkmark simply for posting an incomplete partial answer in which only the easiest part of the puzzle is solved. $\endgroup$ Commented Apr 24, 2020 at 9:31
  • 1
    $\begingroup$ This is the correct answer. @LannyStrack, both you BG got the idea of how the L's and R's are meant to be used. BG mentioned how it should be solved without giving an answer and you were able to get an answer using an idea similar to mine. Stiv was indeed the first to find my final answer. I don't know who deserves the check most, but I think normally, the first to get my answer should get the check (i.e. Stiv). $\endgroup$ Commented Apr 24, 2020 at 9:38
  • $\begingroup$ @eyl327 - I completely agree. $\endgroup$ Commented Apr 24, 2020 at 9:40
  • 1
    $\begingroup$ @eyl327 Well, it's entirely your shout. I'll edit my answer to include an image of BG's puzzle solve and a reference to his answer. Just didn't want to crash the party... :) $\endgroup$
    – Stiv
    Commented Apr 24, 2020 at 9:54
  • 1
    $\begingroup$ Ahhh very nice!!!! @eyl327 I’d say Stiv deserves it or you could create a community wiki so that there’s a correct answer and everyone gets credit. Stiv got the final answer here though, and the grid wasn’t the hard part to solve so I think he should get it! $\endgroup$ Commented Apr 24, 2020 at 11:27

The final answer is


To see how this is found, check out Stiv’s Answer where he realises what to do with the grid once solved!

Partial answer: Solved grid

enter image description here

And if it correlates to the second grid then:

enter image description here

And some notes:

- The encrypted word is 9 letters, so probably corresponds to rows/columns
- The Ls and Rs are obviously standing for left and right, so perhaps this is a caesar shift on each letter
- Perhps the row number is how much we shift the letter? Just not sure how the word correlates to the grid

Step by step for the grid:


enter image description here

The fact there are three 2s in a row means this is the only way for one 2 to be left in the column using the adjacency rule. The 2s then rule out the adjacent 7s from being greyed out, so it must be the other 2


enter image description here

The three ones down the left can now only be greyed out as so. This aslo greys out a one on the right.


enter image description here

The 5 at the bottom means the 5 top left must be greyed out. Now if we try greying out the 4s and 5s at the top like so we get a contradiction with 2 5s in one column which cannot be greyed out. So it must be like this instead:

enter image description here


enter image description here

The 6 bottom right means the 5 and 6 on the right must be like so. The three above the greyed out 5 means the 3 lower down must be greyed out.


enter image description here

The 8s next to the 6 and 2 that are greyed out means the other 8s must be greyed out. This leads to a 2 also being greyed out.


enter image description here

An 8 under the greyed out 5 top left means a different 8 gets greyed out. Finally we have to use the 'continuous area' rule to finish this off, so no area is isolated. And voila! 22 greyed out squares and no repeating numbers

Now whats the next step....

  • $\begingroup$ I've found what you're missing to get the final answer - feel free to incorporate it into your solution for completeness; this time I didn't get to the initial puzzle in time to solve it myself, so it would be unfair of me to snipe you! :) $\endgroup$
    – Stiv
    Commented Apr 24, 2020 at 9:14
  • $\begingroup$ Okay, I ended up editing my answer to point towards yours for the first part after encouragement from the OP. Hope you don't mind! :) $\endgroup$
    – Stiv
    Commented Apr 24, 2020 at 10:21
  • 1
    $\begingroup$ @Stiv not at all!! I was thinking in the complete opposite direction, you deserve the credit! I’ll edit in the answer and link your solution but you still solved half if not more, and the grid wasn’t too difficult so IMO you deserve the tick too :) $\endgroup$ Commented Apr 24, 2020 at 11:31

Is the answer...


I arrived at this answer as follows:

In the lefthand grid, after the 22 squares are shaded, combine the remaining numbers with the corresponding 'L' or 'R' indicators on the right grid, to get the image below.


Step through each row, summing the total or L's and R's in the row to arrive at a single number.
For example, in the first row, we have 7L 6R 4L 5L 1R 8R. This adds up to 1L.

Doing this for each row, we get 1L 9R 3R 11L 19R 8R 23R 26R 14L. Now apply each of those as a Caesar shift to the corresponding ciphertext letter. For example, the first letter in the ciphertext (which, as we were told, applies to the first row) is D. If enciphering, we shift 1L to C, but if deciphering, we shift 1R to E.

Using this method for each character in DXFDMKEPU, we get EOCSTCHPI as the deciphered text. This anagrams to SCOTCH PIE.

The solution to the lefthand grid is as follows. Note that BeastlyGerbil was the first to complete this grid.

First, note the column with three adjacent 2's. If the middle of these were shaded, the other two could not be, and there would be redundant 2's in the column. So the middle 2 CANNOT be shaded, and other two MUST be shaded, to avoid redundancy. Also, because the middle 2 is not shaded, the other 2 in its row MUST be shaded.
No square adjacent to a shaded square can, itself, be shaded, as that would violate the puzzle rules. So none of the 7s adjacent to the upper shaded 2, the 1 adjacent to the lower shaded 2, or the 4 adjacent to the righthand shaded 2 can themselves be shaded. Therefore, we can shade all of the identical numbers in the same columns and rows as these unshaded numbers.
Using the same reasoning, additional protected (cannot be shaded) squares are: the 5 directly above the rightmost shaded 2 and the 6 directly below the rightmost shaded 1. The other identical numbers in the same columns must be shaded. Once this is done, we can also shade a 4 at the top and 5 on the right, using the same reasoning.
Now we can go through the whole grid, row-by-row and column-by-column, looking for two or more of the same digit. If found, and if one of these digits is immediately adjacent to a shaded square, we know that it cannot be shaded, but the other must be shaded. Doing that, the grid now looks like this:
Two repeated numbers are still in the grid: two 7s in the second row, and two 3s in the 9th column. We can use another rule to determine which to shade. In each case, shading a particular one of the 7s and one of the 3s would result in a single isolated white square, which does not connect with the rest of the grid. This violates the puzzle rules, so the OTHER of these two squares must be shaded. Doing so, we get the final grid:

  • $\begingroup$ Great job! I feel like this should have been the actual answer since I like it better than what I had in mind. I checked your work and your answer definitely fits. $\endgroup$ Commented Apr 24, 2020 at 9:29
  • $\begingroup$ @Stiv got the answer I was thinking of (which uses the shaded squares only and not the unshaded ones) $\endgroup$ Commented Apr 24, 2020 at 9:31

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