GoSlash27

This looks like fun. I'll give it a go. Best, -Slashy *edit* PM sent

Well, I don't know that I'd call it a pun, but the important part is that it is solvable without any guessing or speculation (though google-fu would be handy), which is what I was going for. So excellent puzzle and well done! You can tell what needs to be done simply by studying the encrypted form. *edit* it would also count as a cypher, since it could be used to transmit walls of encoded text. Best, -Slashy

Sad part is, there will probably be people who have beaten the game before I even get it downloaded and installed... Best, -Slashy

Ooo! Hype hype hype!

10 30 50 90 F8 10 10 00 Wait a second! I see what you did there. Merely stating that it's based on your old school programming experience was enough to tip me off. So as not to spoil the fun for everybody else, I'll post what tomorrow's solution would look like and see if you agree. F8 10 30 08 08 88 70 00 Is that about right? -Slashy

I'm still scratching my head over Renegrade's puzzle. It's extremely non-random, so there should be something to give us a toehold there...

Well... stressing that it was a *guess*, I would multiply 3 by the first 3 prime numbers, then convert it to hex. 3*2*3*5= 90, which would be 0x5A. Best, -Slashy

4x the first 4 prime numbers? Best, -Slashy - - - Updated - - - Ooo! This looks interesting.... -Slashy

It's 1101001000 in binary and 840 in decimal. I'm not seeing anything that stands out that I'd consider non-random in such a small sample. Best, -Slashy - - - Updated - - - That's what I said. Turns out not to be the case... - - - Updated - - - Hmm... Speculating (which I'll have to do with such a small sample size), these 3 hex characters are supposed to be 2 hex characters. The only 2 possible hex values that would hold these bits are 69(divide by 8) or D2(divide by 4). It happens that D2 would trip 52 in an ASCII conversion, which would be "4". What are the odds that this would just happen to have ASCII 4 sitting in it? 7 bits in the correct sequence? 1 in 128. So if it's a coincidence, it's a huge one. This suggests that the algorithm goes something like this: Plain text -> ascii left shift twice Put a "0x1" tag on the front and pretend it's a number. ^ That's a big conclusion to draw from just a single character, though.

I'll take a wild stab at it based on strictly the info presented. 0x348 implies hex (with "x" being "don't care"). Writing out these values as binary shows one bit high in each column, which is non-random. the x implies that the solution doesn't care whether these bits are high, so... all of these are added "or"ed, and the result will be "F" regardless of what "X" is. F is 4 high bits. The cryptographic method is to or together the hex values and count the high bits. Am I close? -Slashy

Fel, I think you may have misinterpret my comments regarding the puzzle. I was merely describing the way the puzzle was constructed; intended to be solved without resorting to trial- and-error, guessing, or assuming I muddied it up in order to hide the solution. Nothing I said was intended to be a reflection on the person solving it. It was definitely tougher than I thought it would be, but that's my failure, not yours. I'm sure you would've noticed all the important details and made the correct logical conclusions given enough time. Apologies, -Slashy

Kasper, Knowing you, you probably spotted it in the raw hex without even bothering converting it to binary You're just like "Hmm.. 5 5 6 6 a a 5... wait a minute... is this Morse code??" /we all understand you're a busy guy Best, -Slashy

In anticipation of an upcoming official riddle, I'll post the answer as a spoiler. So we are at the point where our code is whether out high bits are one bit long or 2. What kind of code would be based on that where you wouldn't have to guess at my alphabet? Morse code. Reading it as morse yields "46495645"; all valid characters. This part is probably something you folks can read in your sleep; ASCII characters F I V and E. This was clearly more difficult than I expected it to be, so iFail. But as you can see, there was absolutely nothing about this encryption that couldn't be reasoned out simply by looking at it. Best, -Slashy

You're on the trail! Best, -Slashy

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

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

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

That was my thought as well when I saw the video; "or on second thought..." but as 5thHorseman pointed out, the spaceplane parts seem to be surviving. Best, -Slashy

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

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

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

I'm thinkin' SQUAD will just make spaceplane parts impervious to heat effects. Best, -Slashy

^ This. She will go into the rotation when available. I imagine she'll be busy in the early stages helping to sort out the new aerodynamics. Best, -Slashy

SQUAD still has a few surprises to unveil during the countdown. Can't have the video folks spoiling it. -Slashy

... 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...