Puzzles from the xkcd book (big puzzle SOLVED!)

[Lukeonia1]

(I really hate to keep bumping a thread, but isn't anyone else interested in what the coded message says?)

I just realized something regarding 101110.

Spoiler:

It turns out I miscounted... The actual sequence is
1516f1410002f1102201010101132011410000000023120a412450a11010002c024,
which is 67 letters and probably not the correct one to look for patterns in. I think the method may be correct though, so I'll continue looking at it.

That's 67 letters. 1402 factors to 21x67.

Spoiler:

Code: Select all

1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024

I'm pretty sure I don't know what I'm doing, but I think our solution lies somewhere in a 21x67 image. Is there another simple way of turning 9x23 into 21x67?

And does anyone have a clue about the puzzle on 111000?

[Adacore]

I'm interested, I just have no ideas on how to solve the remaining unsolved puzzles. Keep posting insights if you have them though - I'm sure more people than just me are reading!

[Carlington]

Well, I don't know if this will be any help - it's 4am and I'm not thinking too clearly.

Spoiler:

I got this by:
--counting the amount of sequential same numbers (e.g. 4 ones, 2 zeroes, 3 ones, etc.)
--Grouping the result into grups of 8 numbers
--Converting this into hexadecimal.

With the result being:
11511161
a1114114
21e11112
21221212
12111131
22111141
19213112
2a141112
14152a11
1121421c
2214

And those four leftover numbers I couldn't figure out what to do with.

EDIT: This is for the red pixel picture puzzle.

[operator[]]

Well, I don't know if this will be any help - it's 4am and I'm not thinking too clearly.

Spoiler:

I got this by:
--counting the amount of sequential same numbers (e.g. 4 ones, 2 zeroes, 3 ones, etc.)
--Grouping the result into grups of 8 numbers
--Converting this into hexadecimal.

With the result being:
11511161
a1114114
21e11112
21221212
12111131
22111141
19213112
2a141112
14152a11
1121421c
2214

And those four leftover numbers I couldn't figure out what to do with.

EDIT: This is for the red pixel picture puzzle.

Spoiler:

First of all, you missed a number (and counted two wrong), the result should be:
11511161
f1114114
21f11112
21221212
12111131
22111141
192131112
2a141112
14152a11
1121421c
2214

A simple sanity check for this is that the picture begins and ends with a red square, so the number of terms should be odd.
Secondly, there is a lot more red than white, which doesn't happen in random bitmaps, so either every other number is treated differently (as in, pairs of two digits - which is not really compatible with an odd length) or the method is wrong. There are also no zeros, which might be an issue.

I just realized something regarding 101110.

Spoiler:

It turns out I miscounted... The actual sequence is
1516f1410002f1102201010101132011410000000023120a412450a11010002c024,
which is 67 letters and probably not the correct one to look for patterns in. I think the method may be correct though, so I'll continue looking at it.

That's 67 letters. 1402 factors to 21x67.

Spoiler:

Code: Select all

1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024
1516f1410002f1102201010101132011410000000023120a412450a11010002c024

I'm pretty sure I don't know what I'm doing, but I think our solution lies somewhere in a 21x67 image. Is there another simple way of turning 9x23 into 21x67?

Spoiler:

I noticed the same thing, but I'm willing to write it off as an coincidence. The reason for that is that an encoding dependent on the factorization of (h*w+1200) would probably limit the number of possible messages.
An idea though, which did unfortunately not work, is to read every 21st number, wrapping at the end. I can't really think of anything else.

And I don't know too much about AES, but wouldn't it be possible to brute-force the key, assuming we solved the puzzle on page 111000? After all, 16^8 is only like 4 billions, so an exhaustive check is not unrealistic.

For reference, here is the remaining puzzle:

19. (page 111000) 123 B *picture of an arrow hitting a wall and bouncing off*
VRDSYLRBYSMRLUVRXGCFHZWXKYNHYKLKWMCLRMFIKOZAIYXJWITOYOVN

[ethan1701]

regarding the puzzle on 111000:

Spoiler:

I'd interpret the 123B as 123 in hex, convert to binary. It converts to 1111011.
as for the symbol under it, this is half of the last symbol in the lower comic on page 1111 (with 3 of 8 on it). Does this mean to half-reverse it? reverse it with whatever the string of letters translates to? I don't know yet.

I also enjoy thinking of that symbol as "you've reached the end of the book. Now go back and read it from end to beginning"

-Ethan

[ethan1701]

Still stuck on the last puzzle, but here are some thoughts I had regarding it:

Spoiler:

the string of letters is 56 letters long. that's 7x8.
arranging them in sets of 8 gives us:

Code: Select all

VRDSYLR
BYSMRLU
VRXGCFH
ZWXKYNH
YKLKWMC
LRMFIKO
ZAIYXJW
ITOYOVN


I also checked what the letter frequency was (for the event this is a replacement cipher). here's the letter distribution:
a 1
b 1
c 2
d 1
e 0
f 2
g 1
h 2
i 3
j 1
k 4
l 4
m 3
n 2
o 3
p 0
q 0
r 5
s 2
t 1
u 1
v 3
w 3
x 3
y 6
z 2

Hope this helps, though skeptical,
-Ethan

[Carlington]

Ethan: I'm no expert, but to me those strings don't look like they'd fit with a replacement cipher. Don't hold me to that, I just like to put in my two cents.

[alxndr]

19. (page 111000) 123 B *picture of an arrow hitting a wall and bouncing off*
VRDSYLRBYSMRLUVRXGCFHZWXKYNHYKLKWMCLRMFIKOZAIYXJWITOYOVN

I'm no expert, but to me those strings don't look like they'd fit with a replacement cipher. Don't hold me to that, I just like to put in my two cents.

I agree (also no expert). A tool I found to help with substition cracking http://www.simonsingh.net/The_Black_Chamber/frequencypuzzle.htm I couldn't put together anything that looked like a result.

I also slapped together a Python script that randomly guesses a substitution alphabet, deciphers the text, and looks for common words in the result. Nothing promising, but it only got through just over a million substitution alphabets overnight before I stopped it.

Is it interesting that there are no doubled letters?

The arrow thing sorta looks like the schematic symbol for a transistor, doesn't it? (Don't have the book in front of me...)

regarding the puzzle on 111000:

Spoiler:

I'd interpret the 123B as 123 in hex, convert to binary. It converts to 1111011.

-Ethan

Correction:

Spoiler:

decimal 123 is 1111011 in binary; hex 123 is 000100100011 in binary.

[ethan1701]

alxndr, you are of course, correct.
I meant decimal, not hex.

-Ethan

[Pyris]

Hello. I do believe that there was another difference on page '1111' that was over looked.

Spoiler:

The hair. You guys missed the audience's hair.
The resulting binary ( 0 = bald):

Code: Select all

010010110110010101111
0010011001101101111
0110011000111000

When this gets translated it says "Key3of8" cementing the idea that the hidden code is within the Crypt comic on that page.
Mayhaps we are to use the crypt system on either the codes or the hair to get the key.

[ethan1701]

I can confirm that, Pyris.
I used this for the conversion.

Looks like codes are really thoroughly hidden...

-Ethan

[lira_riu]

I'd like to try and work through the puzzles and not just look up the answers, but I could really use some hints on where to start. So far I've gotten the puzzle in the intro, the braille on 1012, and most of the info for the hex key on 110120 -- I'm just not sure how to xor something in hex, I've only seen it used in binary before.

Thanks!

[skeptical scientist]

I'd like to try and work through the puzzles and not just look up the answers, but I could really use some hints on where to start. So far I've gotten the puzzle in the intro, the braille on 1012, and most of the info for the hex key on 110120 -- I'm just not sure how to xor something in hex, I've only seen it used in binary before.

Thanks!

Convert to binary, xor, then back to hex.

[Xutar]

regarding 101110 (the red and white block)

Spoiler:

We've already established that 207 factors into 9x23 which is the pixel dimensions of the block on 101110.
However, you have to wonder if 1407 is actually relevant to this puzzle. After all, it is just the joke of the comic
(2:07pm becoming 14:07 when switched to 24-hour mode)
I guess in other words, could Randy really have encoded a message, in a 9 by 23 pixel image, which is also related to the prime factors of (9x23)+1200?

Edit: also i dont think wolfram's rule 34 would help, for one thing its one dimensional, secondly its pretty boring, and thirdly Im pretty sure randall just mentioned it because it also happens to be the same number as an internet meme and is somewhat related to life/reproduction

[mwillsey]

I do believe its time for a recap.

here are the parts I have solved so far and after checking with the board these should be right.

Spoiler:

1: EE985118
2: 73CD4542
3: DA101CBC
4: 735696B0 (pretty sure about this one)
5: 9BE32BEO
6: i think this is the pixely thing on 101110
7: CB4D62AE (i checked this it works)
8: code on final page

please join the hunt and help out... all we need is parts 6 and 8!!! and we finally get to unlock the code

[PommyDragon2525]

42.39561 -79.13051 2007 09 23 02 38 00

the changed coordinates from page 10020 (the page with the big code)

has anyone tried going there?
http://maps.google.com/maps?f=q&source= ... 5&t=h&z=17
Google can't get very close so we have no idea what could be there.

The final panel change could be a reference to whatever is there. Who knows.

on another note, any ideas about the puzzle on page 100000?

[alxndr]

any ideas about the puzzle on page 100000?

It's been solved: http://fora.xkcd.com/viewtopic.php?f=3&t=45552#p1826060

[medgno]

Related to the image on 101110:

Spoiler:

I've been bit-twiddling recently, and so decided to try to see if the page number was involved with the solution. I've not had any luck with my various attempts, but I figured I'd post the python code in case anyone else could make use of it. I imagine I'm approaching this the wrong way, but for various factoring ideas, this code might help.

http://pastebin.com/f11c5c0c5

[yehudasa]

111000 solved:

Spoiler:

Code is enigma encoded, using wheels 1, 2 and 3 and reflection-B. Message key is XKC and ring setting is AAA. The 'steckers' are the letter pairs that were scattered along the book (NT, LY, GU, CR, IK, DO, EH).

Using enigma decryption software we get:

$ echo VRDSYLRBYSMRLUVRXGCFHZWXKYNHYKLKWMCLRMFIKOZAIYXJWITOYOVN | ./enigma -M M3 -w 123 -u B -s NTLYGUCRIKDOEH -m XKC

keyparteightofeightiseightsixsixtwothreeeightfourletterf

Can also be tested at:

http://www.enigmaco.de/enigma/enigma.html

So, key #8 is:
8662384f

[Adacore]

Awesome! That leaves only one part... anyone fancy brute forcing the remaining 4 billion combinations?

How on earth did you work that out, btw?

[skeptical scientist]

Very nice. Seven down, one to go.

[mwillsey]

Aright guys. only number six left

Spoiler:

1: EE985118
2: 73CD4542
3: DA101CBC
4: 735696B0
5: 9BE32BEO
6: only puzzle left 101110
7: CB4D62AE
8: 8662384f (thanks yehudasa!!)

ideas so far on number 6:
has something to do with the numbers 207 (9x23) and 1407 (21x67)

lets put all the ideas together and see if we can get the final code

[medgno]

Awesome! That leaves only one part... anyone fancy brute forcing the remaining 4 billion combinations?

Well, I've got more processor than brains, so I'll start looking into giving it a try.

Just to make sure I'm not wasting my time,

Spoiler:

is the text to be decrypted on page 10020?

[skeptical scientist]

Summary of solved puzzles:

A red asterisk means that the code has been deciphered, although the deciphered text may not be entirely explained. A blue asterisk means some non-code puzzle has been solved. A spoiler with no asterisk means we have partial information but not a decipherment and/or complete solution. The solution credit will go to the first person who solved it in this thread.

*1. Page numbers

Spoiler:

The pages are numbered in "skew binary" whereby each digit represents a multiple of 2^k+1-1.

So 10112_skew would be 1*(2⁵-1)+0*(2⁴-1)+1*(2³-1)+1*(2²-1)+2*(2¹-1)=31+0+7+3+2=43.

Skew binary also has the property that all digits are either 0 or 1 except for the rightmost non-zero digit.

Solved by e^iπ+1=0.

*2. (last page of intro, unnumbered) 15 11 1 25 8 5 18 5 23 5 7 15

Spoiler:

These are just letters, with 1=a, 2=b, 3=c, etc., so we have, "okay here we go."

Solved by lu6cifer.

*3. (page 1) CNEG BAR BS RVTUG VA URK: RR AVAR RVTUG SVIR BAR BAR RVTUG.

Spoiler:

It uses ROT13. It translates to "PART ONE OF EIGHT IN HEX: EE NINE EIGHT FIVE ONE ONE EIGHT."

Solved by dkurth.

*4. (page 11)

Code: Select all

1 0 T A E / K E
2 0 1 B S W I Y
0 D E 6 S T N 8
C 0 2 5 . R A P

Spoiler:

If you walk over the letters, starting at the top left, you can get the following:
10020 code 256 bit aes w/ key in 8 parts.

10020 contains the big hexadecimal code, and we have other clues labeled part n/8, so it makes sense. A 256 bit key in 8 parts should have 32 bits per part, so each part should be 8 hex digits. This matches what we saw for part 1.

Jerith points out that the order in which the blocks are traversed is one of the finite approximations to the Hilbert curve, as in the IP block map.

Solved by yehudasa and Vhailor, with additional info provided by Jerith.

*5. (page 111) CY-O CMLROOCXN. YR M.AOGP. NRK.W ABE ,CYDRGY M.AOGP.M.BY YD.P JAB X. BR OJC.BJ.V <D.B CY JRM.O YR NRK.W ,.-P. ANN CB YD. EAPTV [[TCBO.F

Spoiler:

Simple substitution cipher, operating on all characters (letters and punctuation). It deciphers to "It's impossible to measure love, and without measurement there can be no science. When it comes to love, we're all in the dark. --Kinsey" This quote goes with comic #55: "Useless".

Solved by yehudasa.

*6. (page 1012) Dot pattern, not easily typed.

Spoiler:

Braille. It reads, "It takes more time than you expect but less than you fear." This is a reference to the above comic, #128: "dPain over dt".

Solved by FallenNicolae.

*6.5. (page 1101) Pointers in the comic. (I missed this one in my last post, so I had to insert a number between 6 and 7)

Spoiler:

The pointers are different from the original. Pointers 1 and 2, interpreted as ascii, give "Key2of8:" Presumably the key is the third pointer, 73CD4542.

Solved by alxndr

*7. (page 1111) "3 of 8".

Spoiler:

Presumably the third part of the 256 bit AES key referenced in the code on page 11 is concealed on this page. The Penny Arcade non-parody seems to exactly match comic #160, so my guess is nothing is hidden in it. The other comic, by contrast, does NOT match the original comic #153, so I'm guessing the key part is the numbers in the slide, which no longer match the diagrams for shift/invert/reverse as they do in the original.

This gives 1101 1010 0001 0000 0001 1100 1011 1100 = DA101CBC

Solved by skeptical scientist.

8. (page 10020) The comic on this page is #240, incase it's connected.

Code: Select all

772A3A35 DEF88CA7  0BDFD186 20B05684  934721F8 F64762FD
03F8D76B 3FA0CB8C  2756B2D0 A9F00A1B  CFF1603E DB05426C

Spoiler:

This is the 256-bit AES cipher referenced in the code on page 11.

That comic holds a special place in Randall's heart, as he explained in the intro (and elsewhere). Also, this code seems to be the focus of several other codes in the book, so it seems important. I'm guessing that deciphering this code, when deciphered, will lead to another meetup or other real-world happening. That could mean there's a time limit on it, but hopefully Randall will have made it far enough in advance that we have some time.

Two things have changed between this comic and the original. The coordinates went from 42.39561 -71.13051 2007 09 23 14 38 00 to 42.39561 -79.13051 2007 09 23 02 38 00, and the conversation on the last panel changed completely. I'm not sure what either of these changes might mean; possibly deciphering the encrypted message might provide some clues.

Unsolved.

9. (page 10100) 101110

Spoiler:

The comic on page 10100 is #247, but the one in the book is different from the one on the site. In the book, the time is 2:07 and therefore the guy says "207 is 3x3x23" and then in the last panel the time is changed to 14:07.

The number 101110 is a page number. That page has a small red pixel drawing, which is 23 pixels wide and 9=3x3 pixels high, and therefore contains 207 pixels. I'm not sure how this helps, but I suspect it's a clue.

Unsolved.

*10. (page 10101) Sequence of stick figures posed in various ways.

Spoiler:

Simple substitution cipher, but with stick figures, a la The Adventure of the Dancing Men. The plaintext is "started with four letters but only need three what a riddle what a mystery." I'm not certain what it means.

Deciphered by skeptical scientist; still unsolved.

*11. (page 10112) 0011001001011100000101100000110111100100001010100110011000010010110011

Spoiler:

0011001001011100000101100000110111100100001010100110011000010010110011 is 70 binary digits. Breaking this into 14 sequences of 5 digits, and converting binary to decimal, we have 6 9 14 1 12 3 15 4 5 9 19 1 5 19. Making the substitution 1=a, 2=b, etc., we get finalcodeisaes, i.e. "final code is aes". This seems rather redundant (we already knew that from page 11), but it seems to be the solution.

Solved by skeptical scientist.

*12. (page 11002) UATAQANUUNG QSICHINGNKUCHAANG CHSICHINGSICHINGNG SICHINGULUUNGUUNG CHAANGATAQANCHATAQANNG SICHINGULUUNGUUNG HSICHINGULUUNGATAQANM-SICHINGULUUNGSICHINGQSICHINGN NUATAQAN

Spoiler:

This one uses Aleut numbers:

uataqanuung qsichingnkuchaang chsichingsichingng sichinguluunguung chaangataqanchataqanng sichinguluunguung hsichinguluuungataqanm-sichinguluungsichingqsichingn nuataqan

Ataqan is the Aleut number 1, for example; siching is 4, uluung is 7 and chaang is 5. We can pick out each occurrence of an aleut number in blue:
uataqanuung qsichingnkuchaang chsichingsichingng sichinguluunguung chaangataqanchataqanng sichinguluunguung hsichinguluuungataqanm-sichinguluungsichingqsichingn nuataqan


Subsituting these gives:
u1uung q4nku5 ch44ng 47uung 51ch1ng 47uung h471m-474q4n nu1

Assuming 1337 speak, we have:
uluung qankus chaang atuung siching hatim-ataqan nul

This is Aleut for 7 3 5 6 4 6 11 nul

Translating to hex gives 735646B0. Presumably this is one of our 8 parts.

Solved by FallenNicolae.

*13. (page 11011) To: *drawing of reddit alien*

Code: Select all

GJNHIYTOTNNNBSFOEVYYVT
NAQGYIUAEIEAIAEURFYV
GULGBIREOUKEGEAEEPFQ
VQTLEDVYSRNVNJULRNAQTVZOY
RVETOHRHEWSWHAGURJNO
RNYYZVZFDESFYIEIOPELJR
ERGUROBEBRLGNVNLSDKETEBI

Spoiler:

I went searching for this one on reddit and of course they already had it worked out.

First, you use ROT13 again to produce the following:

Code: Select all

TWAUVLGBGAAAOFSBRILLIG
ANDTLVHNRVRNVNRHESLI
THYTOVERBHXRTRNRRCSD
IDGYRQILFEAIAWHYEANDGIMBL
EIRGBUEURJFJUNTHEWAB
EALLMIMSQRFSLVRVBCRYWE
RETHEBOROEYTAIAYFQXRGROV

Then, you remove the parts of the code which are part of Lewis Carroll's "Jabberwocky". He actually incorrectly wrote the last word (or the part of it that's there) wrong. It should be borogoves, rather than borogroves.
TWAUVLGBGAAAOFSBRILLIG
ANDTLVHNRVRNVNRHESLI
THYTOVERBHXRTRNRRCSD
IDGYRQILFEAIAWHYEANDGIMBL
EIRGBUEURJFJUNTHEWAB
EALLMIMSQRFSLVRVBCRYWE
RETHEBOROEYTAIAYFQXRGROV

Code: Select all

UVLGBGAAAOF
LVHNRVRNVNR
RBHXRTRNRRC
QILFEAIAWHY
RGBUEURJFJU
QRFSLVRVBCR
EYTAIAYFQXR

Then, you have to use ROT13 again to get this:

Code: Select all

HIYTOTNNNBS
YIUAEIEAIAE
EOUKEGEAEEP
DVYSRNVNJUL
ETOHRHEWSWH
DESFYIEIOPE
RLGNVNLSDKE

Then, rearrange the rows thusly:

Code: Select all

HIYTOTNNNBS
ETOHRHEWSWH
YIUAEIEAIAE
RLGNVNLSDKE
EOUKEGEAEEP
DVYSRNVNJUL
DESFYIEIOPE

And then read down the columns and you get "Hey reddit I love you guys thanks for everything nine eleven was an inside job wake up sheeple"

Solved by redditors (of course).

*14. (page 100000) HALF OF WHAT YOU SEEK IS HERE / AND HALF YOU'VE SEEN BEFORE
SO WHY DON'T YOU WISE UP A BIT / AND DIGG A LITTLE MORE
429C6822BE41C334BE616EFFF8B5A320

Spoiler:

This is a reference to the AACS encryption key controversy. The "half we've seen before" is the 128 bit number 09F91102 9D74E35B D84156C5 635688C0. With the two halves, "09F91102 9D74E35B D84156C5 635688C0" XOR "429C6822 BE41C334 BE616EFF F8B5A320" gives "Key #5 of 8:9be32be0" when interpreted as ASCII.

Solved by operator[] with a little help from skeptical scientist.

*15. (page 100012) Five pixel images.

Spoiler:

These are QR codes. FallenNicolae writes:

I was able to take a picture and decode the top 3 and the bottom, but the 4th wouldn't. I'll take another later and try, but the others are:

1: Toad's Factory 2:00.776
2: Coconut Mall 2:08.147
3: DK Summit 2:03.351
4: DK Mountain (DD): 1:59.814
5: Thanks, David and Ashley.

Which I believe are Mario Kart levels (and presumably times)-- makes sense given the comic on that page.

There are QR code readers for many cellphones, including the iPhone, so if anyone can download one (or already has one) and can try to decipher the fourth code, I would appreciate it. My cellphone, unfortunately, won't work.

Solved by FallenNicolae; 4th code provided by shaav.

16. (page 101110) Pixel image:

Spoiler:

As I said above, I believe this is related to the factoring the time comic on page 10100.

Unsolved.

*17. (various pages) Page 101010, in the upper-right corner, has "LY" in gray text. Other pages have red text in the upper-right corner, always two letters. Page 101120: IK; page 102000: GU; page 110100: EH; page 110102: DO; page 110111: CR. I'm not sure what any of this means, or if the difference between red and grey text is significant.

Spoiler:

These are part of the key to the cipher on page 111000.

Solved by yehudasa.

*18. (page 110120) The start of the tenth favorite word used by Bender
The toon that went south while commanded by Ender
The number of lights that Picard said were on
And the class of the planet where Kirk shouted "Khaaan!"
The rings for the men minus rings for the elves
And the product mod ten of a fivesome of twelves
The end of a code NES gamers know
And the base used to model how quickly things grow
When they're XOR'd together the checksum is "E"
Which will tell you you've got the penultimate key

Spoiler:

The start of the tenth-favorite word used by Bender: C (for Chump)
The toon that went south while commanded by Ender: B (first battle, against Rabbit)
The number of lights that Picard said were on: 4 (Chain of Command)
And the class of the planet where Kirk shouted "Khaaan!": D class planet (Regula)
The rings of the men minus rings for the elves: 9-3=6
And the product mod ten of a fivesome of twelves: 12⁵=2⁵=32=2 (mod 10)
The end of a code NES gamers know: assuming you don't count select+start, A
And the base used to model how quickly things grow: e

So key 7/8 is CB4D62AE.

Solved by skeptical scientist and FallenNicolae.

*19. (page 111000) 123 B *picture of an arrow hitting a wall and bouncing off*
VRDSYLRBYSMRLUVRXGCFHZWXKYNHYKLKWMCLRMFIKOZAIYXJWITOYOVN

Spoiler:

Code is enigma encoded, using wheels 1, 2 and 3 and reflection-B. Message key is XKC and ring setting is AAA. The 'steckers' are the letter pairs that were scattered along the book (NT, LY, GU, CR, IK, DO, EH).

Using Enigma decryption software we get "keyparteightofeightiseightsixsixtwothreeeightfourletterf." (Enigma decryption software can be found at http://www.enigmaco.de/enigma/enigma.html.)

So, key #8 is 8662384F.

Solved by yehudasa.

Progress on the code on page 10020:

Spoiler:

Per page 11, this is encrypted with 256 bit AES, with key in eight parts. So far we have 7 parts.

Part 1: EE985118 (page 1)
Part 2: 73CD4542 (page 1101)
Part 3: DA101CBC (page 1111)
Part 4: 735646B0 (page 11002) (Are we sure this is part 4? All the other parts seemed to be labeled somehow. Also, why does someone else have a 9 where I have a 4?)
Part 5: 9BE32BEO (page 100000)
Part 6: ???????? (presumed to be on page 101110)
Part 7: CB4D62AE (page 110120)
Part 8: 8662384f (page 111000)

[PommyDragon2525]

this is not making sense

[yehudasa]

about 10101 (dancing men):

Spoiler:

'started with 4 letters but only needed 3': probably refers to the 'xkc' key that was used to solve key #8..
what a riddle, what a 'mystery' points at the enigma

[skeptical scientist]

I thought about that. It's possible, but not definite. We can regard it as "provisionally solved" unless someone comes up with a better idea. Our main goal should be finding the last part (and possibly confirming that part 4 is what we think it is), which probably has nothing to do with the dancing men.

[raldi]

You may have figured out the plaintext of the reddit puzzle, but there's more to it than you realize.

Big hint:

Spoiler:

http://online.wsj.com/article/SB124648494429082661.html

[skeptical scientist]

You may have figured out the plaintext of the reddit puzzle, but there's more to it than you realize.

Big hint:

Spoiler:

http://online.wsj.com/article/SB124648494429082661.html

Are you sure about that? The reddit message seems pretty straightforward; are you sure you aren't reading more into it than is there?

[14_thirty_seven]

Regarding 111, I believe you guys are missing part of it:

Spoiler:

It has been written on a QWERTY keyboard while set on Dvorak.

You got the answer, but thought this may interest some of you...

[Qaanol]

Wow, there is some truly impressive puzzle-solving and code-breaking in this thread.

*3. (page 1) CNEG BAR BS RVTUG VA URK: RR AVAR RVTUG SVIR BAR BAR RVTUG.

Spoiler:

It uses ROT13. It translates to "PART ONE OF EIGHT IN HEX: EE NINE EIGHT FIVE ONE ONE EIGHT."

Solved by dkurth.

First off, I don't have the book (although I did buy one as a present for someone else…) but I'm not entirely satisfied with this.

Spoiler:

Are the spaces in the original? If so, then I'd read "EE" as a whole word, meaning just the letter E once, and that gives only 7 hex characters. I could be wrong though.

*13. (page 11011) To: *drawing of reddit alien*

Code: Select all

GJNHIYTOTNNNBSFOEVYYVT
NAQGYIUAEIEAIAEURFYV
GULGBIREOUKEGEAEEPFQ
VQTLEDVYSRNVNJULRNAQTVZOY
RVETOHRHEWSWHAGURJNO
RNYYZVZFDESFYIEIOPELJR
ERGUROBEBRLGNVNLSDKETEBI

Spoiler:

I went searching for this one on reddit and of course they already had it worked out.

First, you use ROT13 again to produce the following:

Code: Select all

TWAUVLGBGAAAOFSBRILLIG
ANDTLVHNRVRNVNRHESLI
THYTOVERBHXRTRNRRCSD
IDGYRQILFEAIAWHYEANDGIMBL
EIRGBUEURJFJUNTHEWAB
EALLMIMSQRFSLVRVBCRYWE
RETHEBOROEYTAIAYFQXRGROV

Then, you remove the parts of the code which are part of Lewis Carroll's "Jabberwocky". He actually incorrectly wrote the last word (or the part of it that's there) wrong. It should be borogoves, rather than borogroves.
TWAUVLGBGAAAOFSBRILLIG
ANDTLVHNRVRNVNRHESLI
THYTOVERBHXRTRNRRCSD
IDGYRQILFEAIAWHYEANDGIMBL
EIRGBUEURJFJUNTHEWAB
EALLMIMSQRFSLVRVBCRYWE
RETHEBOROEYTAIAYFQXRGROV

Code: Select all

UVLGBGAAAOF
LVHNRVRNVNR
RBHXRTRNRRC
QILFEAIAWHY
RGBUEURJFJU
QRFSLVRVBCR
EYTAIAYFQXR

Then, you have to use ROT13 again to get this:

Code: Select all

HIYTOTNNNBS
YIUAEIEAIAE
EOUKEGEAEEP
DVYSRNVNJUL
ETOHRHEWSWH
DESFYIEIOPE
RLGNVNLSDKE

Then, rearrange the rows thusly:

Code: Select all

HIYTOTNNNBS
ETOHRHEWSWH
YIUAEIEAIAE
RLGNVNLSDKE
EOUKEGEAEEP
DVYSRNVNJUL
DESFYIEIOPE

And then read down the columns and you get "Hey reddit I love you guys thanks for everything nine eleven was an inside job wake up sheeple"

Solved by redditors (of course).

You may have figured out the plaintext of the reddit puzzle, but there's more to it than you realize.

Big hint:

Spoiler:

http://online.wsj.com/article/SB124648494429082661.html

If you're right, then

Spoiler:

Using the method from the article you linked

The key to the code consisted of a series of two-digit pairs. The first digit indicated the line number within a section, while the second was the number of letters added to the beginning of that row. For instance, if the key was 58, 71, 33, that meant that Mr. Patterson moved row five to the first line of a section and added eight random letters; then moved row seven to the second line and added one letter, and then moved row three to the third line and added three random letters.

with one section of 7 lines, gives a code of 1b 39 59 6e 2b 7b 4d. Or put together and capitalized, 1B39596E2B7B4D, which might be l33t that I cannot decode, or might have another use.

[MoonBuggy]

This is quite possibly a really stupid question, but how does one go about decrypting a message like this, once the last part of the key is found?

I was going through the documentation for OpenSSL in anticipation of getting it finished and I found it opaque enough to make me realise just how little I know about cryptography!

[Lukeonia1]

This is quite possibly a really stupid question, but how does one go about decrypting a message like this, once the last part of the key is found?

I should think you can download utilities that can encrypt/decrypt text in 256-bit AES, given the keys. And the information on how to implement AES is available online, so an ambitious programmer would probably be able to whip up a little program if we can't find one to use.

[danm]

Cool puzzles. I solved some of them and searched the plaintext and found this forum.

Page 10101

Spoiler:

It really seems likely that this refers to the Enigma cipher on page 111000, since "riddle" and "mystery" are synonyms for "enigma."

Page 11011

Spoiler:

Good job finding the Jefferson code. Qaanol found the key and had the right order, but the prefix is wrong. The key is really:

13 37 52 68 24 79 45

It's very tempting as leetspeak with it starting with 1337! (b|le)ets[rz][bg]b[rz]at[gp]as
13375268247945 = C2A2C1CD989 which is too long for a part of the key.

[danm]

This is quite possibly a really stupid question, but how does one go about decrypting a message like this, once the last part of the key is found?

I was going through the documentation for OpenSSL in anticipation of getting it finished and I found it opaque enough to make me realise just how little I know about cryptography!

OpenSSL is probably the easiest way to do it:

openssl aes-256-ecb -d -K <insert key here> -iv <insert Initialization Vector here> -in <source file>

We'd need the key, the IV, and the source file in binary or ascii-armored encoded in BASE64 (add -a).

Spoiler:

My theory is that the changed coordinates in the comic may be the IV. We need 16 bytes for the IV, which could be 00000000000000000000000000000000, or it could be something like 00004239561791305120070923023800 (coordinates, zero-padded). It could even be the first 16 bytes of the code - 772A3A35DEF88CA70BDFD18620B05684.

Note there's a typo above for the transcription of page 10020. It should start with 77, not 66.

[Elryc]

Ok, looking at all this makes me feel dumb. But I figured for the image puzzle let me look at it simply...

Spoiler:

So I started looking at the image upright, and picked out two hex characters, rotated clockwise (due to the reference with the 207 pixels and the clock comic) picked out two more, repeat for the other two directions, and came up with...



...which could potentially make the key 527AC8FC.

Probably grasping, especially on the last rotation, but... ^,^;;

[jestingrabbit]

Seeing what you did makes me think braille. But I haven't really been engaging with this.

[alxndr]

Seeing what you did makes me think braille. But I haven't really been engaging with this.

I don't see a way to slice the pixels up so you end up with sets of 2x3 with nothing left over.

[ethan1701]

23 being a prime number, you won't. this image can only be broken into 3x3x23.

I think you're going at this wrong.

Has anyone tried turning this into a QR code?

-Ethan

[yehudasa]

23 being a prime number, you won't. this image can only be broken into 3x3x23.

I think you're going at this wrong.

Has anyone tried turning this into a QR code?

-Ethan

Yes.. didn't get very far. Was thinking that it's some kind of a reed-solomon encoding, but wasn't very successful in deciphering it that way. Also, other barcodes types didn't yield much, but I'm not an expert on the subject. Tried to use code from the zebra-crossing project in order to get it done. Note that the QR code encoded in the books have (probably) been encoded using the zebra crossing online tool.

Also tried braille as well as morse code, but nothing there.

The direction I'm currently looking into (or would have if I had the time) is some kind of binary coded decimal or some other packed bcd. If you split the columns into 3 bits nibbles and go from left to right, there's some pattern that goes every 4 nibbles. So I would expect to have 1 nibble that is a header and 3 nibbles that are the data. Each bit in the header corresponds to one of the nibbles that follows. For example, if it's 1 it could mean that the numbers should be shifted by 5, or something like it.
However I wasn't able to decipher it this way so I could be wrong..