3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @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
  • 1
    $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.