11
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{0,1,1}-{1,1,1}-{0,0,1}-{1,1,1}-{0,0,0}-{1,1,0}-{1,1,1}!

Welcome! You have entered the REALM OF CUBIC REPETITION.

Messages passed through here undergo a drastic transformation. All their letters are converted by means of a so-called FPR-substitution and are thereby split up into three dimensions. In all three dimensions they can be either highlighted (1) or not-highlighted (0). Eight possibilities in total of which only the eight(h) is unique.

In case you don't understand how this substitution is carried out: take two steps back and be the first to take a gamble.

Your only way out is to find the number hidden in the grid below. In order to do so, decrypt the message within and follow the instructions. START doesn't count, the other cells are numbered 1 to 36, going from left to right, top to bottom. As the substitution is not unique, the numbers next to and below the grid could be helpful in determining the correct decryption.

For the final part: to find the right cells you need to carry out an FPR-substitution of your own. The input for this will be the corresponding letters of the alphabet of the cell numbers (after 26 start with A again). If you've performed it correctly you will observe that only two cells match the description.

                                                    Grid

After you've decrypted all 36 cells in this grid and found the message, your answer should fit in the following square precisely:
Square

By now, you may realize something is wrong, i.e., the number you've just found doesn't actually exist in that form. Please provide the number in its correct form here: ( _ _ _ _ ); that will be your passcode to exit the realm of cubic repetition.

Good luck!


N.B.: Below you can find a text version of the grid:

            START
47-{0,0,0}-{0,0,1}-{0,0,1}
33-{0,0,1}-{0,0,1}-{1,1,1}
24-{1,0,0}-{0,0,1}-{0,1,1}
29-{1,0,0}-{0,0,1}-{0,0,1}
35-{0,1,1}-{1,0,0}-{0,0,0}
27-{1,1,1}-{0,0,0}-{0,0,1}
29-{0,0,0}-{1,1,0}-{1,0,1}
42-{0,0,1}-{0,1,1}-{0,0,1}
47-{0,1,1}-{0,0,0}-{1,0,0}
39-{1,1,1}-{0,1,1}-{0,0,0}
53-{0,0,1}-{1,0,1}-{0,0,0}
38-{0,0,1}-{0,0,0}-{0,0,1}
     138  -  170  -  135

Hint:

A good place to start is to try and figure what the name of this substitution (it's an acronym) could stand for. An anagram of it can be found somewhere in this puzzle.

Hint 2:

What word could the first line ("{0,1,1}-{1,1,1}-{0,0,1}-{1,1,1}-{0,0,0}-{1,1,0}-{1,1,1}!") spell?

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  • $\begingroup$ Can you add some hints? It has been over 24 hours with no answers or comments. $\endgroup$ – FrodCube Jan 19 '17 at 14:12
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    $\begingroup$ @FrodCube: I added a hint. Hopefully this will get you started. Good luck! $\endgroup$ – Levieux Jan 19 '17 at 14:34
11
+150
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The phrase

REALM OF CUBIC REPETITION

is an anagram of

FIBONACCI PRIME ROULETTE

which surely means that our three bits are

Fibonacci or not?, prime or not?, red or black in roulette?.

Presumably these are applied to

letter numbers, A=1, Z=26; this fits for the ones I have checked so far if indeed the initial exclamation is WELCOME.

This means that our letter-to-bits table looks like this:

A 1 1 0 1 B 2 1 1 0 C 3 1 1 1 D 4 0 0 0 E 5 1 1 1 F 6 0 0 0 G 7 0 1 1 H 8 1 0 0 I 9 0 0 1 J 10 0 0 0 K 11 0 1 0 L 12 0 0 1 M 13 1 1 0 N 14 0 0 1 O 15 0 0 0 P 16 0 0 1 Q 17 0 1 0 R 18 0 0 1 S 19 0 1 1 T 20 0 0 0 U 21 1 0 1 V 22 0 0 0 W 23 0 1 1 X 24 0 0 0 Y 25 0 0 1 Z 26 0 0 0

so the bits-to-letters table looks like

000 DFJOTVXZ 001 ILNPRY 010 KQ 011 GSW 100 H 101 AU 110 BM 111 CE

and the reason for the "eight(h)" is that

the only "unique letter" is H, which is the eighth letter of the alphabet.

Now let's attack the cipher. It's an obvious guess that

the row and column "marginals" are sums of letter numbers

and in fact they turn out to be. By a combination of inference from these sums and plausible-word spotting (you can find the details in an older revision of this answer, if you care; credit to @elias for finding the first word) the message turns out to be

TRIPLE HIGHLIGHTED ROMANS IS THE SOLUTION.

There are several roughly equally plausible ways to interpret this. I think the intended one (which took me a while to think of; the edit history contains my other attempts) is this:

The numbers (letter indices) that get three highlights according to the FPR rule are 3 and 5. The cells numbered 3 and 5 contain, respectively, I and L. Interpreting this as a Roman numeral yields 49 (a square number, which I think is all that was meant by fitting precisely into the square). But the Romans didn't actually write 49 as IL. They wrote it as XLIX, which is therefore the passcode we need.

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  • $\begingroup$ Brilliant! +1 Are we supposed to find a number from it? Also, I am wondering if we need a letter from each set there or all the letters from each set? $\endgroup$ – Techidiot Jan 23 '17 at 16:57
  • $\begingroup$ rows 3-4-5 stand definitely for 'highlight' $\endgroup$ – elias Jan 23 '17 at 16:58
  • $\begingroup$ @elias yes (I was figuring that out at the same time as you :-)). $\endgroup$ – Gareth McCaughan Jan 23 '17 at 17:00
  • $\begingroup$ first two rows say 'triple' $\endgroup$ – elias Jan 23 '17 at 17:07
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    $\begingroup$ It's a cryptic hint at what FPR stands for. "take two steps back" is meant to remind you of the Fibonacci sequence; "be the first" alludes to "prime"; "take a gamble" to roulette. $\endgroup$ – Gareth McCaughan Jan 24 '17 at 18:13
4
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Looks arbitrary, but since there is no attempt being made yet, here is what I assume...

If we consider the first set of binaries = WELCOME! we get -
{0,1,1} = W
{1,1,1} = E
{0,0,1} = L
{1,1,1} = C
{0,0,0} = O
{1,1,0} = M
{1,1,1} = E

So,

The table should look like
enter image description here

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  • $\begingroup$ Your first step is indeed correct, but simply plugging these letters into the table below doesn't work, because the substitution is, as mentioned, not unique. Your next step should be to figure out what the substitution does precisely. Having this first word deciphered might prove helpful there. $\endgroup$ – Levieux Jan 23 '17 at 14:48
  • $\begingroup$ I think the {} were meant to acknowledge that those aren't necessarily the letters involved. $\endgroup$ – Gareth McCaughan Jan 23 '17 at 16:35
1
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Attacking based on hint 1:

the acronym mentioned might be an anagram of 'REALM OF CUBIC REPETITION'. For me the word starting with P cries out to be PERMUTATION, but I cannot make too much sense of the rest. 'FIBO PERMUTATION RECICLE' seems to be too much of a stretch with its typo.

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  • $\begingroup$ The most obvious candidate phrases in the puzzle for anagramming are REALM OF CUBIC REPETITION, as you say; FOLLOWING SQUARE PRECISELY; and maybe FOR THE FINAL PART. (Because each contains F, P, and R, and each reads a little oddly.) I've fed all of these to the Internet Anagram Server and not found any plausible-sounding anagrams. (Which doesn't prove much.) $\endgroup$ – Gareth McCaughan Jan 23 '17 at 16:25
  • $\begingroup$ @GarethMcCaughan, how about 'take two steps back and be the first to take a gamble'? $\endgroup$ – elias Jan 23 '17 at 16:32
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    $\begingroup$ Well, "two steps back" suggests Fibonacci (a thought I guess you had too from your FIBO) and you can actually get all of FIBONACCI out of r.o.c.r. You can in fact get FIBONACCI PERMUTE but what's left doesn't form a word starting with R. $\endgroup$ – Gareth McCaughan Jan 23 '17 at 16:32
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    $\begingroup$ The first mathematical treatments of gambling were, I think, by Fermat and Pascal, who have nice friendly initial letters :-), but I suspect this is a red herring. $\endgroup$ – Gareth McCaughan Jan 23 '17 at 16:33
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    $\begingroup$ aha! FIBONACCI PRIME ROULETTE must be it $\endgroup$ – Gareth McCaughan Jan 23 '17 at 16:35
0
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Is the puzzle accurate or have I not finished? I get:

TRIPLE HIGHLIGHTED RFBUNS IS THE SOLUTION

for the cells

(with S = 0,1,1. I think it should be 0,1,0). If I use the vertical >!addition instead of horizontal, ignoring the cells I don't understand, I >!can finagle DEMONS instead of RFBUNS, but this gives me no insight.

I have some ideas how to proceed from there, but they aren't giving me anything meaningful (to me), so I wanted to check my progress so far.

Thanks for the intricate puzzle!

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  • $\begingroup$ S is 0,1,1. See my answer for (I'm 99% sure) the correct message. $\endgroup$ – Gareth McCaughan Jan 23 '17 at 22:08
  • $\begingroup$ In any case, at this point you have not finished. (Neither have I.) $\endgroup$ – Gareth McCaughan Jan 23 '17 at 22:09
  • $\begingroup$ @GarethMcCaughan, Thanks -- your answer corrected my alphabet in a couple places. Still working... $\endgroup$ – alr3000 Jan 23 '17 at 22:23
  • $\begingroup$ (I now have what I believe to be a complete answer, FYI.) $\endgroup$ – Gareth McCaughan Jan 23 '17 at 22:33
  • $\begingroup$ (Except that ffao has raised an objection, which is not without merit.) $\endgroup$ – Gareth McCaughan Jan 23 '17 at 23:00

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