Kerbal Kountdown!!!!

[Mister Dilsby]

There are a lot of 5s and As...

a.k.a five and ten. I also noticed that the sum of digits in the six blocks fits in a relatively narrow range:

5566=22 AA55=30 66DB=36

4AA9=33 AA95=34 5955=24

So if the sums of digits have any meaning, Slashy could have put in a lot of 5s and As to make the sums come out to what he wanted, while keeping the arithmetic easy.

ETA: Note also there are never more than two letters in a block, and only the first and last blocks have no letters.

Edited April 22, 2015 by Kuzzter

[Torham234]

is it 42?

You would be surprised how often that figure comes up in an average working day. It makes me uncomfortably paranoid...

[Wolfos31]

You would be surprised how often that figure comes up in an average working day. It makes me uncomfortably paranoid...

I know that 42 is in the Hitchiker's Guide to the Galaxy, but what other significance is tied to it that would make you paranoid? A failed shuttle launch or something?

[Torham234]

I know that 42 is in the Hitchiker's Guide to the Galaxy, but what other significance is tied to it that would make you paranoid? A failed shuttle launch or something?

If the ultimate answer is 42,based on a sci fi comedy, and you encounter that number at least twice a week in every facet of our world, it makes you question the reality you live in. Either that or its just your brain picking that particular number form all the other numbers you encounter, and trying to find a pattern that does not in fact exist. I think I was joking... or was I ?

[WWEdeadman]

If the ultimate answer is 42,based on a sci fi comedy, and you encounter that number at least twice a week in every facet of our world, it makes you question the reality you live in. Either that or its just your brain picking that particular number form all the other numbers you encounter, and trying to find a pattern that does not in fact exist. I think I was joking... or was I ?

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*

[GoSlash27]

a.k.a five and ten. I also noticed that the sum of digits in the six blocks fits in a relatively narrow range:

5566=22 AA55=30 66DB=36

4AA9=33 AA95=34 5955=24

So if the sums of digits have any meaning, Slashy could have put in a lot of 5s and As to make the sums come out to what he wanted, while keeping the arithmetic easy.

ETA: Note also there are never more than two letters in a block, and only the first and last blocks have no letters.

... except that it's a cryptogram, not a math puzzle

There's no reason to have to guess at the encipherment algorithm; it is possible to tease out the original message logically.

5566 AA55 66DB 
4AA9 AA95 5955

Best,

-Slashy

PS I should also point out that hex characters are grouped as a matter of course, so the fact that they are grouped in fours doesn't necessarily mean anything. It could just as easily be a single 24 digit hex string or a series of 24 hex values. Or anything in between...

Edited April 22, 2015 by GoSlash27

[Starhawk]

There's no reason to have to guess at the encipherment algorithm; it is possible to tease out the original message logically.

My logic skills appear to be failing me. I'm still at a loss. I seem to be missing an understanding of some basic aspect of decryption.

???

[Rocketfeline]

So isnt this now pointless due to the release date already there and stuff

[GoSlash27]

My logic skills appear to be failing me. I'm still at a loss. I seem to be missing an understanding of some basic aspect of decryption.

???

Well... the basic idea is that the sender and recipient had an agreed-upon process for how to get the message back out of the cipher. You were not supposed to be able to read it in plain text, but their method isn't very good. Plus you have a "crib" since you're expecting "5" to be involved with the answer.

You can tell a few things right off the bat about their method.

1) The final enciphered text is obviously hex

2) The final enciphered text is extremely non-random

2a) There are only a few hex digits present out of the 16 possible

2b) Of the digits present, some appear much more often than others.

So you scratch your head and thing "what's significant about these particular digits?" What do they have in common that makes them appear so often? What's the deal with the other digits that makes them not appear?

It's sort of a sleuthing exercise. Logical deduction and google- fu should be sufficient to crack it.

Best,

-Slashy

[Fel]

It's sort of a sleuthing exercise. Logical deduction and google- fu should be sufficient to crack it.

That is a lame way of saying rtfm. It isn't like we're cryptologists.

So, I'm bored.

5566AA5566DB4AA9AA955955 is what we're given... so let's do something random... like take the rot13 of it

5566NN5566QO4NN9NN955955 now let's do some more random, and remove the changing characters

5566556649955955 and more random and remove the repeating characters

565649595

And our answer is VVIY, which is 5 6 Y or 56 years.

[GoSlash27]

Fel,

You shouldn't need to RTFM or even have access to it. And while you don't have to be a cryptologist to solve it, the thought process by which it would be cracked is the same, but you don't need any special tools to do it.

Finally, it's not a code that requires trying random stuff to break.

If nobody solves it by tomorrow morning, I'll post a walkthrough to the solution and somebody else can have a go if SQUAD doesn't post a riddle.

Best,

-Slashy

*edit* New page repost

5566 AA55 66DB 
4AA9 AA95 5955

Edited April 22, 2015 by GoSlash27

[Rocketfeline]

So the real update will be released 56 years?!?!

[Starhawk]

Well... ...sleuthing exercise. Logical deduction and google- fu should be sufficient to crack it.

Everything you say describes the road I've been walking up and down. Except... google-fu. I have some small skill in this area, but I can't for the life of me even see what to search.

I'll take another crack at it later.

Happy landings!

[Mister Dilsby]

Ok, Slashy saying google-fu is useful is a very important clue, it means that the encryption method must be something we can look up. I'm guessing this was encoded with the XOR method because that's what comes up highest in the google results. I know nothing about this, but through google have learned that the key to breaking it is knowing the length of the key. Maybe the length is 5 numbers, or the key is 55AA, or 5566AA, dunno. Can't do any more now as I must get the kids in bed. http://reverseengineering.stackexchange.com/questions/2062/what-is-the-most-efficient-way-to-detect-and-to-break-xor-encryption

[Fel]

Fel,

You shouldn't need to RTFM or even have access to it. And while you don't have to be a cryptologist to solve it, the thought process by which it would be cracked is the same, but you don't need any special tools to do it.

Finally, it's not a code that requires trying random stuff to break.

If nobody solves it by tomorrow morning, I'll post a walkthrough to the solution and somebody else can have a go if SQUAD doesn't post a riddle.

Best,

-Slashy

I guess you, having inside knowledge, don't quite get it. You've developed a thought process that assists in decryption; or know the outcome of this particular decryption. You keep teasing the answer, but no one is stepping forward. Your last post indicated that this should be easy and we could solve it if we just used google to figure it out.

Like many lateral thinking problems, the solution probably is infinitely simple; but also deceptive. Puzzles like "who shaves the barber" or "where do they bury the survivors" test your ability to process language in a manner different from how we know people process language. Implying that not only does the barber get shaved, but someone shaves him, raises the assumption about the barber's gender more than the profession. When you realize the trick, you can "solve" similar problems fairly quickly; but it is dependent on you knowing to look for the trick to begin with.

[ThatGuyWithALongUsername]

NEW VIDEO:

[GoSlash27]

Everything you say describes the road I've been walking up and down. Except... google-fu. I have some small skill in this area, but I can't for the life of me even see what to search.

I'll take another crack at it later.

Happy landings!

As always, the guy who composed a riddle is a bit like Alex Trebek; I have the answer sitting in front of me, so there's a really good chance that it *seems* easier than it really is...

Kuzzter,

The method itself isn't necessarily something you can look up, but google will probably prove useful during the process.

As a freebie, I would not include a key in a riddle I expect to be solved in less than a day. That would require trial and error (too much time) or specialized tools I would not expect you to have.

Best,

-Slashy

- - - Updated - - -

Fel,

That is all true, but the fact remains that the cipher contains all the clues necessary to deduce how it was encoded and reverse the process.

No specialized knowledge should be required to unscramble the egg... or at least nothing that can't be easily googled.

I set it up to be that way. The thought process should be logical rather than lateral.

But as I said above, it *is* entirely possible that it's much more difficult than I would assume, since I already know how it was done.

Best,

-Slashy

[Mister Dilsby]

And hey, we interrupt this cryptography lesson to bring you a new video!

So it's simpler than we make it out to be, and 'the cipher contains all the clues'. All right. So it starts 5566AA, then 5566DB -- a clear pointer that the DB (which appear nowhere else) is significant. Then 4, which also appears nowhere else, then another repeat: AA9, AA9 again, then our friend 55, 9, 55.

And then--yeah, nothing. I can't make anything out of DB4: not as hex to ascii or as the numbers 13 and 11, or anything else. I think I'm done. Problem is that we don't know what kind of answer we're looking for other than it might be 'five'. Sorry Slashy, I'm out. Looking forward to learning how this sort of thing is done.

[Fel]

?

Not that I know how to get there even if... but brute forcing tends to do things like that.

[Merinsan]

I've tried to solve this puzzle, but I am now stumped.

I went really simple and just assumed FIVE was in the answer (because it should be), and got this:

55 66 AA 55 66 DB

4A A9 AA 95 59 55

E _ F E _ _

_ _ F I V E

The problem is, I have no idea what would fit in the blanks, and realistically I can't think of a word which would match that pattern, where the same letter is repeated in position 2 and 5. Throwing this up here in case it's on the right track and inspires someone else.

[Fel]

octave:10> isprime([0x5566 0xaa55 0x66db 0x4aa9 0xaa95 0x5955])

ans =

Columns 1 through 5:

0.0000e+00 0.0000e+00 0.0000e+00 0.0000e+00 1.0000e+00

Column 6:

0.0000e+00

http://braillebug.afb.org/braille_deciphering.asp

Braille dot 5

Edited April 23, 2015 by Fel

[khyron42]

He's referred to this variously as a "cryptogram" and a "cipher." From the wikipedia page on cryptograms:

A cryptogram is a type of puzzle that consists of a short piece of encrypted text. Frequently used are substitution ciphers where each letter is replaced by a different letter or number. To solve the puzzle, one must recover the original lettering.

So, I'm thinking to stop looking for encryption keys and such. Just figure out the character encoding and solve it like the cryptograms that used to be in the daily comic puzzles.

Two possiblities come to mind from that - either it's a 24 letter message where there are only 7 unique letters used, or a 12 character message where only 8 unique letters are used. The 24-letter version would have an awful lot of weird letter constructs. Breaking it into two-number sets gives:

55 66 AA 55 66 DB 

4A A9 AA 95 59 55

I'll be trying to find simple number-to-letter substitutions that give a useful message (AKA 55 = A) within that. Of course, it's possible that I'm taking way TOO simple of an approach...

[dyson]

Hello everybody. I also tried to solve Slashy's riddle.

My small input is: It does not help to substitute the Hex-Codes with the help of the ASCII-Table, because "AA" would already fall into "Extended ASCII" and therefore a little useless (because not standardized).

If we assume the answer is "FIVE" or "Five" or "five" only 4 letters are needed!

That would make:

0x5566AA = f/F

0x5566DB = i/I

0x4AA9AA = v/V

0x955955 = e/E

But how and why?

[MiniMatt]

Well... the basic idea is that the sender and recipient had an agreed-upon process for how to get the message back out of the cipher. You were not supposed to be able to read it in plain text, but their method isn't very good. Plus you have a "crib" since you're expecting "5" to be involved with the answer.

You can tell a few things right off the bat about their method.

1) The final enciphered text is obviously hex

2) The final enciphered text is extremely non-random

2a) There are only a few hex digits present out of the 16 possible

2b) Of the digits present, some appear much more often than others.

So you scratch your head and thing "what's significant about these particular digits?" What do they have in common that makes them appear so often? What's the deal with the other digits that makes them not appear?

It's sort of a sleuthing exercise. Logical deduction and google- fu should be sufficient to crack it.

Best,

-Slashy

This would lead me to suspect coincidence counting will be useful. Ie. if the letter "E" is the most common letter in the English language it may be reasonable to suspect the most common character in the encrypted text is also an "E". Other characters don't appear in the encrypted text because the letters X,Z,Q etc don't appear with high frequency in the English language.

Now it's not going to be quite so simple in this example, as the most frequent characters in the encrypted text are often found in pairs - and there aren't that many English words with double consecutive vowels in them.

Now, on the plus side, I have finally cracked Regzz's puzzle of yesterday:

Ok... I'll try to make a riddle/puzzle for people like me.... here goes:

1234?6789

Good luck!

The answer to ? is clearly "Geoff Hurst in the 1966 World Cup"

[WWEdeadman]

NEW VIDEO:

So, as it seems you need to go into the atmosphere pretty steep to have explosions happen. So much about people worrying about their spaceplanes.