These two are so dramatically different. Does anyone know if they’re supposed to have the same solution? or was the first a mistake? If they have the same solution, that’s a good clue in itself.
afaik they have the same solution.
I’m afraid that having read the whole of the other discussion, and realizing that
a) it’s not a riddle or the type of problem where some clever insight will solve it (i.e. running through base64, or noting that it’s referring to Airport abbreviations or genetic sequencing), but rather it’s a cipher,
b) it’s expected to take several hours of trial and error, and
c) it’s already been solved there,
I believe I’ll surrender.
I am still very curious why the similarity to Host Observers company identifier scheme. I played with some random urls there and the expected ones passed or failed appropriately.
well, i at least got the hint figured out, started processing it, and am to the really gory part, sigh- unless i am missing something i need to bang my head against the keyboard for awhile. the hint, fwiw, is straight replacement cipher.
I don’t often revive old threads, so forgive me this time, but did y’all ever get the straightdope on this “riddle”.
Why does the porridge-bird lay its egg in the air?
That’s no belly-pack, that’s a superhero utility belt.–Agh Klemm:p
Never did get the answer. I gave up (
WARNING! TEDIOUS! SPOILERS TOO!
Making use of all available spoilers, I was able to follow most of the progress of the other board. I still don’t have the message, but here is what I know:
First, the hint !@##%&*+^\~?*>=!%=@?+>$=[?+}(?@`really was a “simple” substitution cipher, though there was hardly anything to go on. Somebody actually managed to divine that the last word was CRYPTOGRAM. Then it became clear that it read, in its entirety, BASESIXTYFOURTHENBINARYTHENCRYPTOGRAM.
At this point, I had to go off and learn how Base64 (de)coding works. Someone should correct me if I’m wrong, but apparently each character in the Base64 text stands for a number from 0 to 63, and a chunk of 4 characters in Base64 corresponds to three bytes in binary, and each byte corresponds to an ASCII character. For example, MTEw stands for 12_19_4_48, which in binary becomes 00001100_00010011_00000100_00110000. Next, the first two zeroes of each byte are dropped and the digits regrouped into new bytes, thus our example becomes 001100_010011_000100_110000 and then 00110001_00110001_00110000. These translate into normal decimal numbers as 49_49_48, and in ASCII these correspond to 1_1_0, which finally gets us to the finished binary output as given by the decoder aplet I found.
Next we get to use the information that “The binary needs decoding in the same way as 5.21.” This turns out to mean that the letters of the alphabet are assigned numbers from 1 to 26, in binary (A=1, B=10, C=11…), and you have to figure out where each letter starts in the code, since any letter could have from 1 to 5 binary digits. Making it slightly easier is the promise that the letters will appear in alphabetical order, so that M, for example, will not turn up before we have encountered letters A through L. And, no letter begins with zero. Unfortunately, the binary text decodes to a cipher, so that the letters we find do not even stand for themselves.
At this point it seems sensible to give up, but evidently some were able to solve it anyway, and they dropped some further hints. The first word has three letters; the first sentence has eleven words; there are two common phrases (maybe repeating?); the message contains an instruction that will take the solver to yet another stage in the puzzle. The recommended strategy is to decode the binary and the cryptogram simultaneously.
The first seven letters of the cryptogram are ABCDEFG. I’ve got a few of the others worked out, but I still don’t have even one word of the clear text. If anyone is still interested, I could try to show how the known letters line up with the binary, otherwise I’ll just let you know if I manage to crack it in the next few weeks.
Peregrine
Heh, I got to the same point as you, Peregrine, at which point I decided the puzzle of how to solve it was solved, and the rest would be unpleasant gruntwork, trial and error.i got to the point of figuring out about half of the breaks between letters, and then realized that figuring out both the letter composition and actually decrypting the crypto itself was gonna be more of a pain in the ass than the solution could possibly be worth. Not a well-thought out puzzle, imho-- maybe if it were for a thousand bucks, but not for the sheer edification of solving it.
The riddle! We finished translating it! It’s – it’s – it’s a cookbook!
Zev Steinhardt
LONG! TEDIOUS! AND PROBABLY POINTLESS!
Okay, so I was pretty bored tonight and cooked up a computer program that measured the number of possible occurences of a letter in the text. It does not pay attention to see if it will cause a case where the next letter has to begin 00 or something and it does not allow words to overlap the “end of a sentence” … at least I hope thats what those 0k’s were. If those aren’t sentence breaks, then leave a message and I’ll adjust the program. Sooo, thanks in a large part to Peregrine, I had a hack at it. The source code can be viewed at my high school’s webspace - hopefully they’ll stay up without crashing (not the best tech poeple, but free hosting) for anyone who wants a peek at how I did it. I commented it, but its pretty heavily nested and may be difficult to understand. Sorry. In any case, the results are as follows:
Data Resorted
Letter Usage Occurrence
Probability Letter Usage Occurrence
Diff Probability Binary Digits
A 182 57.59 | C 59.50 18.83 2
C 99 31.33 | B 41.50 13.13 2
B 81 25.63 | G 40.83 12.92 3
G 54 17.09 | E 30.83 9.76 3
F 44 13.92 | F 30.83 9.76 3
E 44 13.92 | A 24.00 7.59 1
D 36 11.39 | N 23.06 7.30 4
N 28 8.86 | D 22.83 7.23 3
O 25 7.91 | O 20.06 6.35 4
M 22 6.96 | I 17.06 5.40 4
K 22 6.96 | J 17.06 5.40 4
J 22 6.96 | K 17.06 5.40 4
I 22 6.96 | M 17.06 5.40 4
L 21 6.65 | L 16.06 5.08 4
W 18 5.70 | W 16.03 5.07 5
U 17 5.38 | U 15.03 4.75 5
Y 15 4.75 | Y 13.03 4.12 5
S 14 4.43 | S 12.03 3.81 5
H 13 4.11 | Q 9.03 2.86 5
Q 11 3.48 | H 8.06 2.55 4
Z 10 3.16 | Z 8.03 2.54 5
R 8 2.53 | R 6.03 1.91 5
X 6 1.90 | X 4.03 1.27 5
T 5 1.58 | T 3.03 0.96 5
V 4 1.27 | V 2.03 0.64 5
P 2 0.63 | P 0.02 0.01 5
Okay, time for some explanation. The left hand column is the actual results. The letter is followed by the number of times it was used (“usage”) and then the probability of its occurrence is in the third column (simply took the usage and divided it by the total number of possible letters (79*4). Not a very exact answer, however, because someone is far more likely to draw a “1” (the letter A) from a random bunch of 1s and 0s than a “11010” (the leter Z). The data reflects this.
So, I tried to correct for that a little bit. That’s the right hand column - I “interpreted” the data and ended up changing it, so I give you both versions. This time there is the letter followed by the “usage diff,” which is a kinda complicated number I mostly made up. Basically, what I did was find the random probability of drawing the number corresponding to the letter (like 11010 for Z) by saying 1/2 chance multiplied by 1/2 chance multiplied by 1/2 chance multiplied by 1/2 chance multiplied by 1/2 chance = 1/32 chance of drawing a Z randomly. Then I took the total number of possible draws (for z it would be 794/5 [because it has a length of 5]) and then multiplied the probability of getting my desired result (1 out of 32 or 1/32) onto that answer, to determine the theoretical number of random Z’s I should get. I then subtracted off these random Z’s (794/5/32 = 1.975) from the actual usage, and got the number of nonrandom Zs (10-1.975=8.025). Lastly, and this is the most important column, is the occurrence probability based on the “nonrandom” usage. This results in an approximately real estimation of the probability of a letter appearing in this cryptogram. Finally, pay attention to the number of binary digits … even though I’ve corrected for basic randomness, other factors cause large digit numbers to come up less often statistically (wall effects, distribution of error).
Use a page like this to compare occurrences in this cryptogram and in the English language. Basically, from all of this data, I would guess that either C, B, or G is the letter E, and I personally would go with G because it has another digit to deal with and still made it to the top. Of course, this still has a significant error, but I personally think the data is about 70% correct (WAG).
Hope I helped some.
Oh, there is one more source for error in the data I didn’t think about (it was getting late last night).
Basically, when you divide the total number of possible characters (79*4) by the length of one letter (5 in the case of Z), you assume that each letter cannot form a part of another … like a 1 in 11010 does not begin the next 11010 (it isn’t possible in this case, but some letters, like G = 111, could theoretically form more times than I’m allowing).
Well, I dunno really how to fix that, so just remember about it…that’s prolly why some of the longer letters are skewed to have fewer occurrences. But the data should still be somewhat right.
ROTFL!
I’ve done it.
I had a look at PoignantSod’s frequency analysis, and then I tried making a comparison with a bit of typical English text rendered into the binary cipher, but it didn’t get me anywhere. The thing that finally cracked it was trying on some double letters where the cryptogram has 10101010. I thought it might be the zz in puzzle, but that didn’t work. Then I thought it could be the dd in riddle. Success! After that it fell together fairly quickly.
Here is what the binary cipher would look like with spaces between the letters:
1 10 11 100 101 110 111 101 11 1000 1 100 101 1001 1010 1010 1000 110
1011 100 1100 111 110 101 1101 11 111 1001 111 1001 1100 1110 111
10 1111 110 101 1 110 111 10000 10 111 10 1011 111 10000 10001 1000 100 1110 110 111
1010 10 111 10010 10 1011 1100 1000 100 1100 10011 101 1001 1010 1010 1000 110
10001 1000 100 1110 110 111 1100 1000 100 1100 10011 110 1000 10100 1001 1110
1010 10 111 10001 10011 10001
That can be converted to the alphabet version of the cryptogram, but it is recommended to decipher as you go. After all the decoding, the resulting message is
You are truly a riddle master but it isn’t over yet. Go to mtgplanet dot com slash riddleplanet slash elfin dot php
So I did. As we were already aware from the discussion on the other board, this only leads to another cryptogram, which I have not yet attempted to solve.
So anyway, Pavlos, there’s your answer.
I must say I am really freaking impressed. Where were you all when we had the Brits had their little Enigma problem?
Wow.
Clap clap clap.
Poignant Sod and Peregrine-- nice work. That’s an impressive show of patience and insight. I can’t imagine having the drive to solve this (much less the ability).
Impressive, Peregrine.
wow thanks Peregrine. 
and Poignant Sod
and everyone else too…
Wow - Extremely impressive job Peregrine. Truly well done.