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Kerbal Kountdown!!!!


boxley

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

Well, just as a note

I noticed there were exactly EIGHT of those

Eight as in 1 byte

as did you... XD

i don't know what it is... trying to think of something smart-alecy that would be a "puzzle" but

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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

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Kasper,

Knowing you, you probably spotted it in the raw hex without even bothering converting it to binary :confused:

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

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In anticipation of an upcoming official riddle, I'll post the answer as a spoiler.

Argh. I'm sticking with 'f$' :D

Let's no one get bent out of shape over this. We didn't solve it, but if we were not entertained we were at least 'diverted' from not having an official puzzle to solve--and Slashy was quite good at responding to guesses and giving hints along the way.

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But as you can see, there was absolutely nothing about this encryption that couldn't be reasoned out simply by "knowing the answer"

Best,

-Slashy

Fixed....

I suspected it was morse code in some form, but don't currently have the time to sit down and solve it :)

I guess it usually comes down to the first level common replacement.

Morse, Braille, ASCII, even telephone-digits

And that's without mixing math into it, primes especially being common, but we also have things like roots or squares pop up (they're all squares of primes + 1! It's obvious.)

Then we also have shape transforms.

But finding the exact combination is a headache... which is why slashy's "You can just tease it out" comments tend to be kind of, yeah.

Edited by Fel
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Fixed....

I guess it usually comes down to the first level common replacement.

Morse, Braille, ASCII, even telephone-digits

And that's without mixing math into it, primes especially being common, but we also have things like roots or squares pop up (they're all squares of primes + 1! It's obvious.)

Then we also have shape transforms.

But finding the exact combination is a headache... which is why slashy's "You can just tease it out" comments tend to be kind of, yeah.

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

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That's the brilliant thing about cryptography. If you invent a cryptographic method, it invariably takes other people longer to discover what the method is than it does for you to invent it.

So, here's today's cryptographic challenge (ignore this one if an official challenge comes out)

The answer is "4"

Your Clue: "0x348"

Now, how did I arrive at that number? That is, what was my cryptographic method?

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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

I know little of cryptography or information theory. I've heard of Claude Shannon before.

I'm afraid that it seems to me that solving this requires a lot of specialized background knowledge and techniques. As I said, I thought I was being silly when I originally posted the binary. I spent almost all the time looking at the hex representation. I tried compressing all the pairs into singles and then seeing if it made any sense in ASCII. I tried leaving out all pairs and looking at what was left.

When you mentioned the clue about 5's and A's I googled cryptography A 5 and didn't get any useful hits.

5s and As are often used as test patterns, filler, etc.

But I had already tried leaving out the 5's and A's and compressing the pairs without success, so I didn't know how to use this info.

I was surprised that you posted the binary representation, but I took a look it again. I didn't see anything that was obvious enough to jump out at me after looking for a bit.

I thought to myself, if he wants us to look at the binary, there must be something obvious there. I couldn't see it.

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

Wait, what?? How would I guess to do that? That's some awfully low-level pattern recognition.

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.

This seems very unobvious to me. I can't imagine how long one would have to look at that string of ones and zeros to tease this info out of it.

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.

This idea of 'framing' must be a common concept in cryptography, perhaps related to the 'salt the message' idea that Kuzzter referred to early on.

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

I have to say that it seems to me that only a person who is experienced with cryptograms could get that out of the original message.

So we are at the point where our code is whether out high bits are one bit long or 2.

At this point, I was looking at Kuzzter's 1's and 0's block rather than the block of 1's and 2's.

What kind of code would be based on that where you wouldn't have to guess at my alphabet? Morse code.

Wow! That seems like serious lateral thinking to start from 1's and 2's combined with alphabet and get to Morse code.

I was not able to successfully complete even one of the steps required to solve this puzzle. The real irony here is that I successfully cracked an encryption scheme for an employer of mine several years ago.

Thanks anyway Slashy, I had fun trying.

Happy landings!

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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

Edited by GoSlash27
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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

For the fun of it, I will respond with affirmations of only those pieces you have correctly identified:

You have correctly identified that as Hexadecimal.

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I'll take a wild stab at it based on strictly the info presented.

0x348 implies hex (with "x" being "don't care").

-Slashy

Actually 0x is the lead-in that indicates that the following digits are hex.

As such 0x348 translates to '348 in base 16'

It seems a little odd that it is an odd number of digits long however, so I suspect it is really 0348 base 16 (every digit is 4 bits, and 8 bits makes a byte, so Hex is usually in pairs)

In binary:

0000 0011 0100 1000

One thing that jumps out at me is that there are 4 1's, and if we add them together we get 4.

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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 - - -

Actually 0x is the lead-in that indicates that the following digits are hex.

As such 0x348 translates to '348 in base 16'

It seems a little odd that it is an odd number of digits long however, so I suspect it is really 0348 base 16 (every digit is 4 bits, and 8 bits makes a byte, so Hex is usually in pairs)

In binary:

0000 0011 0100 1000

One thing that jumps out at me is that there are 4 1's, and if we add them together we get 4.

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.

Edited by GoSlash27
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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.

looking at your representation, I noticed something:

1 10 100 1000

next in the sequence would be 10000 with 4 zeroes.

So, perhaps the puzzle is: in binary, how many zeroes follow the next one?

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Actually 0x is the lead-in that indicates that the following digits are hex.

As such 0x348 translates to '348 in base 16'

It seems a little odd that it is an odd number of digits long however, so I suspect it is really 0348 base 16 (every digit is 4 bits, and 8 bits makes a byte, so Hex is usually in pairs)

In binary:

0000 0011 0100 1000

One thing that jumps out at me is that there are 4 1's, and if we add them together we get 4.

The (full) hex number has been correctly identified as 0348 (base 16)

- - - Updated - - -

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

And the decimal Number of 840 is correct!

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And the decimal Number of 840 is correct!

4x the first 4 prime numbers?

Best,

-Slashy

- - - Updated - - -

Here's my version~

10 30 50 90

F8 10 10 00

It's hexadecimal and the clue is four.

Spacing is significant. The line breaks though are not.

Ooo! This looks interesting....

-Slashy

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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 just layered too much in there for us merely mortal professional software developers to understand.

We had to go through:

  1. Realize it is hexadecimal numbers
  2. Convert it to binary
  3. Realize it is morse code
  4. Convert it to text
  5. Realize it is hexadecimal numbers (again)
  6. Realize they are ASCII codepoints
  7. Convert them to text

That's a lot of steps to take with many wrong directions to take at any one of them. And we only know if we went wrong at the end. Some of the steps weren't really that obvious either. E.g. in morse, lines are strictly required to be exactly three times the duration of the dots. You gave us two bits to one. For me that immediately screamed "DCC signal encoding." Just goes to show how different things can seem when you don't already know what you are looking at.

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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

Well I think it was as someone else pointed out too many steps without any indication of whether you were on the right track. That being said for a forum puzzle that was not intended to be solved in 10 minutes by half of the forumgoers it was a fair challenge and I applaud and thank you for the effort and the fun.

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