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Another puzzle in the spirit of the Density™ puzzle. This one requires coordination plus some basic math. Enjoy!

enter image description here

Final answer: (7)

Hint 1

A coordinate consists of two numbers. The words in the intro, "...coordination plus...", is a hint.

Hint 2

Histogram

Hint 3

When hint 1 is done, the diagram is no longer needed.

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It seems to me that the answer is:

TERNANY - although I don't recognise this as a word, it is presumably some kind of pun on the word 'ternary'?

How did I get this? Well, first note that the diagram represents:

A scatter plot, where the 0-axes are alternately black and white, representing units, and the purple pixels represent the points that have been plotted upon it.

If we:

Read off the (x,y)-coordinates of each point (as hinted at by the use of 'coordination' in the flavour text), we find the following list of coordinates:

(0,1), (1,0), (1,2), (3,0), (2,3), (1,5), (5,1), (3,4), (4,3), (6,4), (4,7), (0,12), (10,2), (5,10), (7,9), (7,10), (6,12), (8,10), (10,9), (12,7), (8,12), (11,9), (10,11)

Next, use the second clue provided in the flavour text:

And - cued by 'plus' - add together the numbers in each coordinate pair to get the following list of numbers:

1, 1, 3, 3, 5, 6, 6, 7, 7, 10, 11, 12, 12, 15, 16, 17, 18, 18, 19, 19, 20, 20, 21

Following the second hint, we can now:

Treat these numbers as a dataset and plot each number's frequency of appearance as a histogram (to which there is a subtle graphical hint in the diagram itself, as a curved histogram-like shape can be seen among the purple plotted data points):
enter image description here

We can see from this that:

No number appears more than twice, and the range of the numbers is 21, which is divisible by 3 (and after all, the title tells us '3 is the key'...) - in fact it is 7 times 3, and we are looking for a seven-letter answer... so how about we try dividing this histogram into seven equal parts, and interpret each triple using ternary? So, dividing the histogram like this:

enter image description here

We can now extract the seven ternary triples: 202, 012, 200, 112, 001, 112 and 221.

These convert into decimal as: 20, 5, 18, 14, 1, 14 and 25.

And these then convert via A1Z26 into the final answer of TERNANY.

Disclaimer: This answer was arrived at entirely independently to others. My dataset is the same as @Randal'Thor's (just ordered differently) and was derived in the same way at a similar time, while @Avi's answer (with similar deductions after the publishing of the third hint) only loaded in my browser after pressing the submit button for this answer(!), which contains further steps and deductions not present in the other two :)

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  • $\begingroup$ Well done! The intended answer was rot13(GREANEL) but I somehow messed up with the second R. Bummer! Anyway, super job (as usual)! $\endgroup$ – Jens Feb 20 at 21:59
  • $\begingroup$ @Jens I did wonder if that might have been a slip-up, but I wanted to give you the benefit of the doubt and assume that there was some kind of 'toe tactic' pun going on that I wasn't quite seeing just yet! :) $\endgroup$ – Stiv Feb 20 at 22:01
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I have no idea if I'm on the right track here, but let's give it a try.


Observations:

  • Aside from the green background, every cell is black, white, or purple.
  • The black and white ones are (more or less) alternating along the left and bottom edges.
  • The entire square is 13 cells by 13.

Initial conclusions:

The black and white edges represent coordinate axes, starting from $(0,0)$ in the bottom left corner. There is no real difference between black and white - the two colours are just a handy way to show alternation/parity. (Purple cells on the left and bottom edges are to be seen as purple dots on top of the pre-existing black and white axes.)

The main puzzley content is in the positioning of the purple cells. Somehow from this we should extract an answer.

The size of the grid means we'll probably end up finding some numbers and converting them to letters using A=1, B=2, etc. for the solution.


More detailed data:

There are a total of 23 purple cells. Most columns contain exactly 2, but there are two columns with 3, three columns with 1, and one column with none.

The coordinates of the purple cells are $(0,1),(0,12),(1,0),(1,2),(1,5),(2,3),(3,0),(3,4),(4,3),(4,7),(5,1),(5,10),(6,4),(6,12),(7,9),(7,10),(8,10),(8,12),(10,2),(10,9),(10,11),(11,9),(12,7)$.

Taking sums of these coordinates ("coordination plus") and converting to letters gives $A,L,A,C,F,E,C,G,G,K,F,O,J,R,P,Q,R,T,L,S,U,T,S$. Which doesn't seem to have enough vowels to be an anagram of something meaningful. Anyway the solution is supposed to be just 7 letters.

Probably there is some more meaning in the exact positioning of the purple cells; it's not as simple as just summing coordinates.

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  • $\begingroup$ It's a start! :) Did you see the second hint? Note also that there is no anagram tag. $\endgroup$ – Jens Feb 19 at 18:52
  • $\begingroup$ @Jens I couldn't figure out yet how to use the second hint. The most natural thing would be just to count the top purple dot in each column, but I can't imagine there's that much wasted information in the puzzle. $\endgroup$ – Rand al'Thor Feb 19 at 19:01
  • $\begingroup$ All I'll say is that the order of the hints is not arbitrary. $\endgroup$ – Jens Feb 19 at 19:10
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Partial answer, continuing from where @Rand al'Thor left off:

The points are:

$(0,1),(0,12),(1,0),(1,2),(1,5),(2,3),(3,0),(3,4),(4,3),(4,7),(5,1),(5,10),(6,4),(6,12),(7,9),(7,10),(8,10),(8,12),(10,2),(10,9),(10,11),(11,9),(12,7)$ (Thanks to @Rand al'Thor for doing the busywork here)

After this, the

sums of each pair are: $1, 12, 1, 3, 6, 5, 3, 7, 7, 11, 6, 15, 10, 18, 16, 17, 18, 20, 12, 19, 21, 20, 19$

Making

A histogram of the sums, we get:

 1   * *
 3   * *
 5   *
 6   * *
 7   * *
 10  *
 11  *
 12  * *
 15  *
 16  *
 17  *
 18  * *
 19  * *
 20  * *
 21  *
 

This looks

suspiciously like Morse Code, which I am terrible at decoding, so I'll stop here.

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  • $\begingroup$ To incorporate the possible hint from the title three is the key I binned the histogram using the 3 yielding 7 numbers but I am not sure if this is a dead end. $\endgroup$ – 53RT Feb 20 at 21:05
  • $\begingroup$ @53RT I also tried that but not entirely sure if it would lead anywhere - maybe encoded as ternary 3-bit digits, they mean something? $\endgroup$ – Avi Feb 20 at 21:06
  • $\begingroup$ I think you were pretty close Avi - you just needed to include the whitespace in your graph (see my answer for what I think the intended solution is...). We were clearly working on this at exactly the same time! $\endgroup$ – Stiv Feb 20 at 21:12
  • $\begingroup$ @Stiv I noticed the whitespace at about the same time that 53RT made their comment, but too lazy to actually compute that out :P $\endgroup$ – Avi Feb 20 at 21:21

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