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I have here a simple binary-like pattern, consisting of crosses (X) and circles (O).

O X X O X O O X O
X O X X X X O O O
X X O O X O O X O
O X O O O O O X X
X O X X O X X X X
O X O O O O X X O
O X O X X O O X O
O X O X O X X O X
O X O O X X O X O

There is, however, a hidden sequence that makes the pattern complete when it is orientated in a different way. So, what is this missing sequence?

RULES:

When the pattern is orientated in this "different way", each sequence is unique. Meaning, there should be no repeats.


The missing sequence should fit snuggly "in the middle" of the new orientation.


The missing sequence is nine (9) characters long.


HINTS (Level 1):

The number 14 should appear a lot… so should the number 42. But 14 is more important.

HINTS (Level 2):

Did I say "long"? I never said the sequence was linear, did I?


Perhaps if we played a different game, the pattern would make some more sense?


HINTS (Level 3):

"Unique" should suggest the current pattern has some overlap.


There are more crosses than circles in the hidden sequence.


This new orientation relies on a diagram we all should have used at least once in our lifetime... I mean, it's quite logical! Or is it plausible?

HINTS (Level 4):

Pips and squares... could we also make them into circles?

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  • $\begingroup$ Been a month and then some... any more hints? $\endgroup$
    – Quintec
    Dec 12, 2019 at 4:02
  • $\begingroup$ … I would place X into 4/5 and 6/5, otherwise I don't see what else should be missing. I would add something like (X O X X O X O O O) or (X O X O X O X O X), but only by random "it fits" my ruling. $\endgroup$
    – Jan Ivan
    Dec 12, 2019 at 11:31
  • $\begingroup$ I assure you randomly fitting an arbitrary sequence or padding random X's or O's will not produce the answer efficiently; I'm well aware of the 512 possible sequences, but if you follow my hints judiciously it should narrow the options down to one. However, if you can convince me (with solid reasoning, mind you) that there could be another viable solution, by all means, feel free to argue your case! $\endgroup$
    – Omicron Zed
    Dec 12, 2019 at 16:56
  • $\begingroup$ At this moment I just find 13 a lot, but can't figure out a 9-long pattern to make it 14. ¯\_(ツ)_/¯ $\endgroup$
    – gustavovelascoh
    Jan 8, 2020 at 12:43
  • $\begingroup$ Well, I never said you had to 'make' 14 (at least, not explicitly), just that it should 'appear'... however, that does not mean it's no less important. Still, maybe it's worth a partial answer; you never know, it might help others... $\endgroup$
    – Omicron Zed
    Jan 8, 2020 at 23:44

3 Answers 3

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If we take this 9-by-9 grid, and

divides it into 9 3-by-3 blocks (similar to a Sudoku), and count the number of Os, we get: 

3   4   7
5   7   2
6   3   5

All columns, and all rows have a

sum of 14. The total number if Os is 42. That's many 14's. (Hint 1.4)

So, the duplicates are (1,1 & 2,3), (3,1 & 2,2) and (2,1 & 3,3). We know from hint 3 that

there should be more X than Os. My guess is that we want to make this into a "sudoku-block", i.e. all digits 1-9 occur exactly once. The number of X's per block is: 

6   5   2
4   2   7
3   6   4

This is one (of many many) solutions to this:

O X X O X O O O O
X X X X X X O O O
X X X O X O O X O
X X X O O O O X X
X X X X O X X X X
X X X O O O X X O
O X O X X O O X O
O X O X O X X O X
O X O O X X O X O
.

And here I'm stuck for now. I don't know the sequence yet, and there's a lateral thinking tag on this, so I might be way way off.

This is my best guess for now:

Rearrange the rows/columns, and align this so that we only need to change the X/O in the inner square. PEr the hints, it's 9 letters long, but not linear, and it snugs in the middle.

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  • $\begingroup$ You're getting quite close, but I think you're either reading too much into my hints or too little. You definitely see a game being played, but my hints should be telling you to use something else at the same time; after all, it seems this game has gotten nowhere, since the 'board' is already filled. $\endgroup$
    – Omicron Zed
    Jan 9, 2020 at 16:43
  • $\begingroup$ Also, I'm not seeing where the new/hidden sequence lies... I believe I noticed 8 changes (unless I miscounted), but that does not satisfy my criteria for a valid answer. Perhaps it's best to remind you that the first level should be sufficient to complete the sequence; ROT13(gur bgure yriryf bs uvagf fubhyq fhccyrzrag/ersvar gur svefg yriry). $\endgroup$
    – Omicron Zed
    Jan 9, 2020 at 17:30
  • $\begingroup$ As I said, I don't know the hidden sequence. The changes I made were just random-ish. $\endgroup$
    – CG.
    Jan 10, 2020 at 10:31
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Partial answer:

At this moment these are my findings, coincidences or not, I don't know:

- Counting the number of X in columns: 3,7,3,4,5,4,3,7,3
- Counting the number of X in rows: 4,5,4,3,7,3,4,5,4
- Column 2 is the same as row 5 (Hint 1A, Hint 2C (?))
- Creating a sudoku-like grid to group the symbols, and adding the number of X in each block generates this square, in which each row and column adds up 13 (coincidence ?):
6 | 5 | 2
4 | 2 | 7
3 | 6 | 4

That's all for now...

Edit 1: This is another pattern I found, Could it be useful or just coincidence?

1 8 2 7 3 7 2 8 1

Could this be the answer? X O X X X O O X X

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  • $\begingroup$ You're not too far off, but I think you got a bit... inverted. There should be no coincidence in the summation, as you've probably read in the hints, so this suggests something is off in your observation; perhaps you need to look at something else? $\endgroup$
    – Omicron Zed
    Jan 9, 2020 at 16:32
  • $\begingroup$ ROT13(Gur flzzrgel vf n erq ureevat); otherwise, I'm interested to see where you got this "other pattern". $\endgroup$
    – Omicron Zed
    Jan 9, 2020 at 16:57
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    $\begingroup$ ROT13(Vg vf sebz pbhagvat gur ahzore bs "k" va gur svefg cbfvgvba bs rnpu guerr-ol-guerr oybpx, gura gur frpbaq cbfvgvba naq fb ba) $\endgroup$
    – gustavovelascoh
    Jan 9, 2020 at 17:09
  • $\begingroup$ I'm not seeing where your "answer" in your edit should be in the pattern; perhaps you should explain why this "answer" is desirable (to save you from the frustration of retrofitting the pattern, it's not the solution... still, maybe explaining your conclusions and reasoning could be fruitful). $\endgroup$
    – Omicron Zed
    Jan 9, 2020 at 17:37
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xo

All rows and all columns are different and in the four coloured corners there are 14 x's or o's.

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  • $\begingroup$ I applaud your efforts and observations at the various patterns and orientations, but I'm not seeing how this is getting you any closer to an answer. Specifically, are you seeing anything that suggests the pattern is incomplete? I believe not... $\endgroup$
    – Omicron Zed
    Jan 9, 2020 at 16:28
  • $\begingroup$ @ Omicron Zed Maybe you meant the following pattern: the eighth sequence, counting from top to bottom, is written 0X0X0XX0X. If you rewrite this as X00X0XX0X then both diagonals of the map are made from zeros. $\endgroup$
    – Vassilis Parassidis
    Jan 9, 2020 at 20:43
  • $\begingroup$ Yes, this does make the diagonals more uniform without altering too much in the pattern --- though, if you read into the other hints, this does not seem enough to prove your answer is the correct one. $\endgroup$
    – Omicron Zed
    Jan 9, 2020 at 21:11

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