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Kerbal Kountdown!!!!

boxley

By boxley
in KSP1 Discussion

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andruszkow

andruszkow

Posted
Posted

Just throwing some thoughts out there.

The sequence of the code:

5566 AA55 66DB

4AA9 AA95 5955

55 66 AA sequence seems repeated. If you go strictly for this you have: (Notice italics)

55 66 AA 55 66 DB 4AA9 AA95 5955

In the end, you have AA95 5955 - Dont know if AA is reversing 95 to 59, or if it's a clue at all.

The AA is a modulator of some kind. Must be. 5566 AA 5566 DB <- End of line?

It COULD also be:

Skip each second, so:

5566 AA55 66DB = 56A56D = Hex colour is definitely green'ish - Does this mean anything?

4AA9 AA95 5955 = 4AA955 = Hex colour Also pretty green

- - - Updated - - -

Also, it's possible to spell out the hex color code #b4d455 with the first line, which is indeed Jebediah's color.

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h4didact

h4didact

Posted
Posted
It's just an illusion though. If you're "obseesed" with a number you'll start seeing it everywhere. Like, you know, I see the number 69 a lot... If you know what I mean... *cough*

Hahahaha. Confirmation Bias is a huge thing in our species' brain processes.

On a side note, I always wanted to become an Austrian citizen. Can you adopt me or something? Or give me a job in Austria?

Seriously though, my dream is to live in Austria. A minute of silence for our dead dreams...

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andruszkow

andruszkow

Posted
Posted
Hi, shouldn't it be "bada55" - also totally valid HEX

Either way.. :)

- - - Updated - - -

I want the riddle to spell out b4d455 Jeb.

I just cant figure out how to get "Jeb" from the second line.

Perhaps something with subtracting. In hex, Jeb is 4a 65 62.

Second line is 4AA9 AA95 5955, so the 4A = J is there in the beginning.

Then I thought about subtracting. If you convert into Base10, so that A = 11, you have A - 9 = 2. It might be a bit far fetched :D

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monophonic

monophonic

Posted
Posted
Either way.. :)

- - - Updated - - -

I want the riddle to spell out b4d455 Jeb.

I just cant figure out how to get "Jeb" from the second line.

Perhaps something with subtracting. In hex, Jeb is 4a 65 62.

Second line is 4AA9 AA95 5955, so the 4A = J is there in the beginning.

Then I thought about subtracting. If you convert into Base10, so that A = 11, you have A - 9 = 2. It might be a bit far fetched :D

Just decide it is a one time pad. Then you can construct a key to get whatever plaintext you want.

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WWEdeadman

WWEdeadman

Posted
Posted
Hahahaha. Confirmation Bias is a huge thing in our species' brain processes.

On a side note, I always wanted to become an Austrian citizen. Can you adopt me or something? Or give me a job in Austria?

Seriously though, my dream is to live in Austria. A minute of silence for our dead dreams...

xD Nah, I don't think I can do either of that for you. RIP dream...

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GoSlash27

GoSlash27

Posted
Posted

I'm gonna give away the first level of the code.

The reason you're seeing so many 5s and As is because they are 0101 and 1010 in binary; alternating bits. Kuzzter has touched on this upstream. 5s and As are often used as test patterns, filler, etc. due to this property.

You can expect to see less randomness if you look at it as a string of binary.

5566 AA55 66DB 
4AA9 AA95 5955

is

010101010110011010101010010101010110011011011011010010101010100110101010100101010101100101010101

This string of binary is *really* not random. There are no groups of 3 or more 1s or 0s in a row and single bits are much more prevalent than pairs. Paired 0s are very rare...

Best,

-Slashy

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Tweety

Tweety

Posted
Posted

I keep messing with a binary translator but I keep getting 3 x P and a J. I am just mashing about with a binary converter while I am at work but this is really bugging me now. Cheers for the puzzle! It is keeping me far more entertained than fitting tyres ever could!

Tweety

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Mister Dilsby

Mister Dilsby

Posted
Posted (edited)
This string of binary is *really* not random. There are no groups of 3 or more 1s or 0s in a row and single bits are much more prevalent than pairs. Paired 0s are very rare...

Thanks Slashy, we really did need this step. Without knowing whether this was a seeded encryption, a substitution cipher, or whatever there was no point in making random manipulations to find out--the string is so short that any or all of those could have been the path.

With the clues of (1) repeating 00 and (2) alternating bits are important, I can see that the string '1001' acts (mostly) as a separator between sections of alternating bits, and I can rearrange the binary to this pattern:

0101010101

1001

10101010

1001

01010101

1001

10 11 01 10 11 01 00

10101010

1001

10101010

1001

01010101

1001

01010101

The string in the middle '10 11 01 10 11 01 00' is the only one that is not either '1001' or alternating bits. (Yes, there is another '1001' string in there, but if I take that out as a separator it breaks the pattern of the repeating strings) Another interpretation is that the message includes the repeating bits in the following line up to the next 1001, making the complete message '10 11 01 10 11 01 00 10 10 10 10'

That's all I've got so far.

Edited by Kuzzter
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GoSlash27

GoSlash27

Posted
Posted

Kuzzter,

We're running short of time , so I'll help it along a bit more.

Look at the occurrence of double zeroes in the string and how they relate to single zeroes.

010101010110011010101010010101010110011011011011010010101010100110101010100101010101100101010101

The pairs occur every 5th single digit zero, which is way to regular to be coincidence.

Also note that because of this there are always 5 ones (or pairs of ones) between zero pairs.

Best,

-Slashy

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Trentendegreth

Trentendegreth

Posted
Posted
Kuzzter,

We're running short of time , so I'll help it along a bit more.

Look at the occurrence of double zeroes in the string and how they relate to single zeroes.

010101010110011010101010010101010110011011011011010010101010100110101010100101010101100101010101

The pairs occur every 5th single digit zero, which is way to regular to be coincidence.

Also note that because of this there are always 5 ones (or pairs of ones) between zero pairs.

Best,

-Slashy

My stupid is showing *grey matter leaks out of head...

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Starhawk

Starhawk

Posted
Posted
010101010110011010101010010101010110011011011011010010101010100110101010100101010101100101010101

The number of digits between the 00's varies from 9 to 13. The position of the excess ones in each segment varies. There are eight segments.

???

Happy landings!

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GoSlash27

GoSlash27

Posted
Posted
The number of digits between the 00's varies from 9 to 13. The position of the excess ones in each segment varies. There are eight segments.

???

Happy landings!

Aye.

The zero pairs are perfectly regular compared to single zeroes. Therefore there is no intelligence there; it must be framing.

The single zeroes are the same; framing.

The fact that there are exactly 5 ones (or pairs of ones) per segment means that there is no intelligence to be found there either. Also framing.

The only thing that could possibly carry information is whether or not a particular "high state" is one bit long or two.

ignoring the framing stuff, it looks like this: 11112/21111/11112/22221/11111/21111/11112/11111

Best,

-Slashy

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Mister Dilsby

Mister Dilsby

Posted
Posted (edited)
The pairs occur every 5th single digit zero, which is way to regular to be coincidence.

Actually that's not entirely true. There are five single digit zeros in the first string before the double, then four between each succeeding set. But there are indeed five ones or pairs of ones in each set, like this:

single single single single double

double single single single single

single single single single double

double double double double single

single single single single single

double single single single single

single single single single double

single single single single single

What does that mean? I don't know. Third base.

ETA: Ninja'd by the puzzle master, and oh good I seem to be on the trail...

Edited by Kuzzter
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GoSlash27

GoSlash27

Posted
Posted
Actually that's not entirely true. There are five single digit zeros in the first string before the double, then four between each succeeding set. But there are indeed five ones or pairs of ones in each set, like this:

single single single single double

double single single single single

single single single single double

double double double double single

single single single single single

double single single single single

single single single single double

single single single single single

What does that mean? I don't know. Third base.

ETA: Ninja'd by the puzzle master, and oh good I seem to be on the trail...

You're on the trail!

Best,

-Slashy

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Fel

Fel

Posted
Posted

....

...

..

Are you serious slashy? This is getting highly into specialized knowledge and knowing the solution before hand.

/me still likes my original solution of it being a hidden Braille Code.

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h4didact

h4didact

Posted
Posted
....

...

..

Are you serious slashy? This is getting highly into specialized knowledge and knowing the solution before hand.

/me still likes my original solution of it being a hidden Braille Code.

The Brasille Code you discovered was genius. With all respect to Slashy, Fel's way of solving the puzzle seems to be more interesting than the real result.

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Mister Dilsby

Mister Dilsby

Posted
Posted (edited)
You're on the trail!

Yayyyyyy.... so if there is only one piece of information, whether a high state is one or two bits long, then we have:

11112

21111

11112

22221

11111

21111

11112

11111

Which could also be written:

00001

10000

00001

11110

00000

10000

00001

00000

We note that there are a lot of repeats here as well, and that a five-character string doesn't translate well to most text encoding--assuming we are looking for text encoding, which maybe is something that always gets assumed in cryptography puzzles (which none of the rest of us seem to do, Slashy, if you haven't figured that out by now :) ) or maybe we're looking for a numeral.

Sigh. In any case, it does seem a pattern that in each of these five 'bytes' there are no alternating bits at all. That makes me wonder if I want to compress the repeats, and get:

01

10

01

10

00

10

01

00

This makes me happy because now I have some eight bit binary strings and I can convert those to characters. I am even happier because I know that 01100110 = 'f'! But 01100110 00100100= f$ so perhaps I've gone astray again.

Edited by Kuzzter
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Mister Dilsby

Mister Dilsby

Posted
Posted
I guess I just lack some background understanding of cryptography (specifically cryptanalysis). It doesn't even look like there's enough info left to make a message.

Unless my decoding is correct, and the message is indeed 'f$', and the $ stands for 'Slashy' and the 'f' stands for... well... :blush:

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