7
$\begingroup$

One day, a colleague from the office hands you the following message without a comment and makes a quick retreat out of your cubicle. Can you make head & tails from this? What is your colleague trying to tell you here?

The code

The solution to this puzzle is an English sentence and this is not a rebus puzzle. Hints will follow if the message is not deciphered in due time.

Hint 1

The first line is part of the code instruction, the second line is a self-verification step: If you can make the number on the left represent a G you have probably found the code algorithm and can apply it to the 3rd line, which is the message.

Hint 2

"G" does not represent anything but the letter G itself.

Hint 3

You did miss it at first, but the paper slip has been very carefully punctuated with a needle on the right edge. It seems that two different needles have been used too...

$\endgroup$
  • 1
    $\begingroup$ Damn, I thought "quire" was a hint before seeing your edit! $\endgroup$ – Rand al'Thor Jun 21 '15 at 12:05
  • $\begingroup$ Na, the puzzle is only contained in the image. (And you were handed a PAPER slip to solve it.) $\endgroup$ – BmyGuest Jun 21 '15 at 12:06
  • $\begingroup$ Is this something related with electric resistances? $\endgroup$ – Cows quack Jun 21 '15 at 12:07
  • $\begingroup$ @KritixiLithos No it isn't. Hints will follow after some cooling-down period. (Some people crack these things so fast, that I don't want to spoil it right away.) $\endgroup$ – BmyGuest Jun 21 '15 at 12:55
  • $\begingroup$ Is the colleague by any chance Japanese? :P $\endgroup$ – Joe Z. Jun 21 '15 at 16:35
10
$\begingroup$

Okay, so first I

translated the binary hole punches on the left. From top to bottom the dots give: 010000100100100101001110010000010101001001011001 which one can translate directly into 8-bit ASCI codes for 6 letters, spelling out: "BINARY".

With this it was obvious

that the next line should be read as binary as well, with 494128555671552 = 1110000010110100001000101000010001100000000000000

Then, I tried out the example with a small pyramid and realized that:

if I "dropped" the binary in, in order from most significant bit to least, I would get this:binary

The rules were:

Drop each bit in from the top. They fall straight down; if the space directly under them is full they go down and right, and if that space is taken they go down and left. If all three are taken, they stay in place. (It helps to visualize it as sand falling down in an hourglass.)

So just using the colors by themselves produced this:

colors.

This is not useful. However, thanks to Quark we already know

that the colour bar could be interpreted as binary string as well. Using RGB flags, i.e. R=(100) G=(010) B=(001) and accordingly for mixed colours. The binary code of the colour stripe is:
100010010110000100101100000110001010100010010100001101000010100000010010110100110101010011001010001011010000010101000101000101010101001000101000101100110100100110100110101001000101010100000100001110101000110000001100001000001011100000010000000010000100000100001000001000010001000000100001000011000000000000000010000000000000

Using this and the general rules give:

Solution

Finally, one can notice that

"I love you ever" is not really a proper English sentence, but looking closely there are 4 pixels between "you" and "ever", so the final message is I love you forever.

$\endgroup$
  • $\begingroup$ Very good work! You might want to edit into your answer that 494128555671552 = 1110000010110100001000101000010001100000000000000b $\endgroup$ – BmyGuest Jun 23 '15 at 18:59
  • $\begingroup$ And I'm awaiting your final solution, of course! $\endgroup$ – BmyGuest Jun 23 '15 at 19:00
  • $\begingroup$ @BmyGuest: I tried, but either I made a transcription error or my colleague is really into puzzly modern art. I can't figure out what the message is supposed to be. $\endgroup$ – Deusovi Jun 23 '15 at 19:02
  • $\begingroup$ ? Let me cross-check. And maybe crosscheck the binary first ( I didn't do that yet ).... $\endgroup$ – BmyGuest Jun 23 '15 at 19:04
  • 1
    $\begingroup$ There must be a transmission error somewhere. When I apply Quark's binary encoding to Deusovi's method, I get a clear message. $\endgroup$ – M Oehm Jun 23 '15 at 20:13
4
$\begingroup$

Partial answer:

Here's the color bar at the bottom converted to a base 2 number (red = 100, green = 010, blue = 001, yellow = 110, cyan = 011, magenta = 101, black = 000). 100010010110000100101100000110001010100010010100001101000010100000010010110100110101010011001010001011010000010101000101000101010101001000101000101100110100100110100110101001000101010100000100001110101000110000001100001000001011100000010000000010000100000100001000001000010001000000100001000011000000000000000010000000000000

Note: the chances of this being necessary are slim because it's near impossible for this sequence to not contain any white bars and yet be a direct translation to something. However as I can't think of any other way this could be decoded, posting it anyway.

$\endgroup$
  • $\begingroup$ +1 for the idea and you might not be so wrong after all... $\endgroup$ – BmyGuest Jun 21 '15 at 15:23
  • $\begingroup$ I think it's something to do with octal/base64 $\endgroup$ – somebody Jun 22 '15 at 10:56
  • $\begingroup$ As you can see, your contribution was quite essential, albeit ahead of it's time ;c) $\endgroup$ – BmyGuest Jun 24 '15 at 11:04
  • $\begingroup$ @BmyGuest I almost didn't post this, glad I did :) $\endgroup$ – Quark Jun 24 '15 at 13:04
1
$\begingroup$

Just a little guess here. The letter "G" in the English Alphabet is number 7. Therefore I thought somehow 7 would be in that equation. I then thought what if you take each of the numbers and plug it into the pyramid. Anyhow this is what I have come up with:

$$\begin{array} &&&&7\\ &&4&9&4\\ &1&2&8&5&5\\ 5&6&7&1&5&5&2 \end{array}$$

This probably means perfectly nothing, but I thought I would throw it out there anyhow. 7 is at the top because the arrow is pointing there, and I thought maybe that is where "G" goes, AKA 7th letter of the alphabet.

$\endgroup$
  • $\begingroup$ Interesting idea: 15 digits + G and 16 slots in the Pyramid... I haven't thought of that. Not the solution, but interesting observation... $\endgroup$ – BmyGuest Jun 23 '15 at 19:02

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.