I have been given this puzzle by my friend (who is a qualified military cryptographer). I am completely stuck as for what to do.

Edit: Just got given a big hint!!!!!!!

   "Don't take it so literal, think in all three dimensions."

Im thinking, what if it isnt a grid we are looking at, what if its a cube?

Edit 2:

"The puzzle was made on physical paper, I also solved it on physical paper Even when I made the solution to show Llama, I had to print out my digital spreadsheet for physical paper(edited) because only physical paper has three dimensions"

I have no idea where im going with this

enter image description here

  • $\begingroup$ @dcfyj not a whole lot, to be honest, cos im so stumped. i have found that akseli is finnish for aksis, but I have no idea what "Y-Ehe" is. $\endgroup$ – user36160 Apr 21 '17 at 14:40
  • $\begingroup$ The only hint I have been given are "dont take the numbers too literally" $\endgroup$ – user36160 Apr 21 '17 at 14:41
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    $\begingroup$ The same puzzle can be found here warosu.org/sci/thread/8842230 without anything about "qualified military cryptographers". I think that page may be a mirror of something 4chan-related. $\endgroup$ – Gareth McCaughan Apr 21 '17 at 15:15
  • $\begingroup$ "Akseli" appears to be Finnish for both "axis" and "axle". $\endgroup$ – Gareth McCaughan Apr 21 '17 at 15:16
  • $\begingroup$ oh, I take it back: there is something in the discussion there about the source of the puzzle being a military cryptographer. user36160, are you also the person who posted the puzzle there? $\endgroup$ – Gareth McCaughan Apr 21 '17 at 15:18

The first thing to notice is that the the values of Y-ehe are multiples of the corresponding row. 45 is a multiple of 9, 40 a multiple of 8 and 14 a multiple of 7, etc.

The other thing is that the X-akseli and Y-ehe values sum up to the same value, 194.

If you look at the matrix, you see that the multiples in Y-ehe almost match the number of blank cells. There are 2 empty cells in row 7 and Y-ehe(7) is 2x7. So if you fill the blank cells with the row number, the cells in a row nicely sum up to the Y-ehe value.

Naturally you will want to compute the column sums, and you discover the X-akseli values.

enter image description here

Note however that you have to remove the initial 6 to make it work. Probably a typo.

All this teaches us one thing: the value of a cell is the row number.

Now I will boldly assume, out of nowhere, that the goal is to go from the top-left to the bottom-right corner by minimizing the total cell count. Here are 2 options:

enter image description here and enter image description here

The number at the bottom-right is the sum of all cells. As you can see, the second path, with a score of 80, is better than the first one, with a score of 82. It is longer but uses smaller cell values. There are other paths, but they result still larger scores. The solution on the right shows the optimal path.

I have no explanation about the mismatched sums or where the hint "Think in all 3 dimensions" comes into play.

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  • $\begingroup$ I AM CLUELESS!!!!!!!! i really really apreciate the work, but i asked her and she said its incorrect. love you for doing this, though! (Spread this around, this is hurting me now! i need an answer) $\endgroup$ – user36160 Apr 24 '17 at 2:21
  • $\begingroup$ ok, dude, so i sent her your answe, she said that the "6" was in fact a typo, but it was intentional, and it is one of only 2 typos!!!!!! you con do this, bro! $\endgroup$ – user36160 Apr 24 '17 at 2:37
  • $\begingroup$ That first image is correct! you filled it in correctly! $\endgroup$ – user36160 Apr 24 '17 at 3:07
  • $\begingroup$ It seems my guess about the final goal was incorrect after all. $\endgroup$ – Florian F Apr 24 '17 at 18:37
  • $\begingroup$ Sorry to tell you this, but it starts to look like a "guess what I think" puzzle. If you guess correctly what the riddler had in mind, you might get it, but else you are just shooting in the dark. $\endgroup$ – Florian F Apr 24 '17 at 18:47

I've tried a few things, but haven't got that much, maybe someone can develop on this or find something I missed.

First off, using the numbers as $x$ and $y$ coordinates we can plot a scatter graph:

enter image description here

But as you can see there is very little correlation. Indeed using a spearman's rank calculator, we get a value of $-0.109244$:

enter image description here

So very little negative correlation.

All I can think off for the table, is that perhaps the values in each cell is $\sqrt{a+b}$ with $a$ being the corresponding horizontal value and $b$ the corresponding vertical value.

However this really doesn't give a nice table.

The last thing I tried was superimposing the graph on to the table:

enter image description here

Doesn't really give anything...

What I find really suspicious is that there are 9 x values and 9 y values.

However, despite these attempts I was unable to find anything. The only thing I can think of which I am sure about is the values for the diagonal cells ($-x$ means that the cell is a path and has an integer $x$.

Here is a copyable mathjax table:

$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} & \text{7} & \text{8} & \text{9} \\ \hline \text{1} & 1 & - & & - & - & - & - \\ \hline \text{2} & & -2& & - & & & - & - & - \\ \hline \text{3} & & - &-3 & - & & & & & - \\ \hline \text{4} & & & & 4 & - & - & & & - \\ \hline \text{5} & & & - & - & 5 & - & & - & - \\ \hline \text{6} & - & - & - & & & 6 & & - & \\ \hline \text{7} & - & & - & - & - & - & -7& - & \\ \hline \text{8} & - & & & - & & & & -8& - \\ \hline \text{9} & - & - & - & - & & & & & 9 \\ \hline \end{array}$$

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  • $\begingroup$ I really, really appreciate the work mate. I really hope someone else can build on it. you got further than I ever did $\endgroup$ – user36160 Apr 22 '17 at 1:40
  • $\begingroup$ Please do check the edit, though. i feel if anyone can solve it, its you $\endgroup$ – user36160 Apr 22 '17 at 2:09
  • $\begingroup$ @user36160 hmmm, 3d? In that case I imagine that the grid will give us a set of numbers we can use on the $z$ axis. I'll see if I can think of anything... $\endgroup$ – Beastly Gerbil Apr 22 '17 at 11:14
  • $\begingroup$ cheers, mate. i really think you can do this! $\endgroup$ – user36160 Apr 23 '17 at 1:35

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