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Lonelygirl15
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diastres
Lonely Fan
Joined: 28 Nov 2006
Posts: 151
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 Posted: Wed Dec 13, 2006 4:53 am Post subject: [PUZZLE][SOLVED] XX/Xy V.2 (matchmaker) |
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So, we're working on the password for the matchmaker file. I thought I'd post because I'm currently stumped.
To start with, it uses the 13 dominoes. You add the two sides to get a number.
0/0 = 0
0/1 = 1
0/2 = 2
0/5 = 5
0/6 = 6
1/4 = 5
1/5 = 6
1/6 = 7
2/4 = 6
2/5 = 7
3/3 = 6
3/4 = 7
3/6 = 9
And then we can use FCC as well, bringing it to 16 characters. It's hex, so the end result will be eight characters.
| Code: | value | 0 1 2 5 6 7 9 C F
count | 1 1 1 2 4 3 1 2 1 |
Now, the only way F and C could be used to create an alphanumeric character is if we use 4 or 6. We don't have a 4, so...
| Code: | 6C 6C 6F
l l o
value | 0 1 2 5 6 7 9 C F
count | 1 1 1 2 1 3 1 0 0 |
0, 1, 2 and 9 cannot start a pair. (I know 20 is a space, but we've been informed that there are no spaces.)
That means...
| Code: | 6C 6C 6F *0 *1 *2 *9 **
l l o |
(Where * is just to saw we don't know what goes there.)
There are two capital letters, so 5 cannot be a second character in a hex pairing.
Possibilities for each:
*0: 50=P, 70=p
*1: 51=Q, 61=a, 71=q
*2: 52=R, 62=b, 72=r
*9: 59=Y, 69=i, 79=y
**: 56=V, 57=W, 67=g, 76=v, 77=w
Eight characters, three have to be l, l, o and a fourth is some form of p. We don't know if it's upper or lower case.
| Code: | 50 51 52 56 57 59 61 62 67 69 70 71 72 76 77 79
P Q R V W Y a b g i p q r v w y |
Editted because we now know there are two uppercase letters. |
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diastres
Lonely Fan
Joined: 28 Nov 2006
Posts: 151
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| Posted: Wed Dec 13, 2006 6:33 am Post subject: |
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password is WarPylol
I'll write it up fully later.
Filename: 72696e676162656c6c3f
HEX: ringabell?
Add .txt
base64
| Code: | NDkyMDc0NmY2YzY0MjA3OTZmNzUyMDczNjg2NTIwNmM2ZjZm
NmI2NTY0MjA2NjYxNmQ2OTZjNjk2MTcyMjEwZDBhMjg2NTc2NjU2ZTIw
NzQ2ODZmNzU2NzY4MjA0OTIwNjM2MTZlMjc3NDIwNzM3MDY1NmM2
YzJjMjA3OTZmNzUyMDY3NmY3NDIwNmQ3OTIwNmQ2NTYxNmU2OTZl
NjcyOTBkMGEwZDBhNjU2ZDYxNjk2YzIwNmQ2NTNhMjA2YzYxNmU2Nz
c1NjE2NzY1MmU2YzY1NzM3MzZmNmU3MzQwNjc2ZDYxNjk2YzJlNjM2
ZjZkMGQwYTBkMGE2MTZlNjQyMDY4NjE3NjY1MjA3NDY4NjE3NDIwNj
Q2NTYzNmY2NDY1NzIyMDcyNjk2ZTY3MjA3Mzc0NjE2ZTY0Njk2ZTY3Mj
A2Mjc5MjEyMDNhMjk= |
to hex
| Code: | 4920746f6c6420796f7520736865206c6f6f6b65642066616d
696c696172210d0a286576656e2074686f75676820492063616e27
74207370656c6c2c20796f7520676f74206d79206d65616e696e67
290d0a0d0a656d61696c206d653a206c616e67756167652e6c6573
736f6e7340676d61696c2e636f6d0d0a0d0a616e64206861766520
74686174206465636f6465722072696e67207374616e64696e6720
627921203a29 |
to text
| Code: |
I told you she looked familiar!
(even though I can't spell, you got my meaning)
email me: language.lessons@gmail.com
and have that decoder ring standing by! :) |
Time for me to pack. I promise I'll write up how it was completed later. |
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sparkyb
Casual Observer
Joined: 07 Dec 2006
Posts: 35
Location: Somerville, MA
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| Posted: Wed Dec 13, 2006 6:58 am Post subject: |
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| Damn you. I was up all night thinking about this, composing this post and writing my specialized brute force solver. Had I know the info that there were no spaces and 2 capitals that cuts my 21 letter combinations down to just 5 for a total of 544,320 possible passwords. That would probably be crackable in under 6 hours. If I added a restriction that the first letter must be capitalize (seems likely), the 2nd letter isn't (I'm assuming 2 words), and the last isn't (I'm assuming the cap is the beginning of a more than 1 letter word) that would cut it down to just 20 minutes. But then I saw that you'd already figured it out. I can't wait to know how. Also, how did you get confirmation of the adding technique and find out that it had 2 caps and no space? |
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windsong
Casual Observer
Joined: 12 Oct 2006
Posts: 113
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| Posted: Wed Dec 13, 2006 8:18 am Post subject: |
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| Lol, about 3 hours ago I had put together only the word payroll, and then i was like, great now i got this w left. Now what. wpayroll? Like, double your payroll? And then I decided i had been up for waaaaay too long and went to bed. Great job dia. |
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deagol
Thor's Hammer
Joined: 28 Oct 2006
Posts: 1068
Location: No, not here.
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| Posted: Wed Dec 13, 2006 9:27 am Post subject: |
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| sparkyb wrote: | | Damn you. I was up all night thinking about this, composing this post and writing my specialized brute force solver. Had I know the info that there were no spaces and 2 capitals that cuts my 21 letter combinations down to just 5 for a total of 544,320 possible passwords. That would probably be crackable in under 6 hours. If I added a restriction that the first letter must be capitalize (seems likely), the 2nd letter isn't (I'm assuming 2 words), and the last isn't (I'm assuming the cap is the beginning of a more than 1 letter word) that would cut it down to just 20 minutes. But then I saw that you'd already figured it out. I can't wait to know how. Also, how did you get confirmation of the adding technique and find out that it had 2 caps and no space? |
Can you believe that? You know about my puzzle/email from warpylol to Op, and that SupressGemma solution... dang was I channeling Tach or what!
Of course, WarPylol is a member of the YT group ipod opera - a hint? (anagrams to Opertion Aphid). Also, that account was in the latest correspondence list she left us in the last drop. We just didn't know the right capitalization.
That's one of the two accounts of that group that has a public video (the other is jimmyjm):
Title: Add Poe Hinder Vilification (anagrams to D in Aphid location verified)
Description: Erin is in ire. (palindrome sugesting Ireland)
Tags: PAID HIND (D in Aphid)
It's just a one second clip of a street scene with a car (license plate [N2 6941] or [NZ 6941]) and an apartment building, thought to be in europe. http://www.youtube.com/watch?v=6_f7o4RK3oE
Is Tachyon implying that WarPylol is Gemma, or related to her?
Do we have the decoder ring ready? So far I have this:
| Code: | :04 Cx
:08 6t
:12 4v
:16 Fo
:20 Bu
:24 2z
:28 8w
:32 3p
:36 5m
:40 Dr
:44 7n
:48 Ay
:52 9q
:56 1s
:60 El
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The second ring is incomplete, with tentative times (hour:minutes for verification)
| Code: | :06 2x (1:06)
:10 Do (1:10)
:18 8m (1:18)
:20 1z (1:20)
:22 As (1:22)
:34 Ft (0:34)
:38 4v (0:38)
:46 9q (0:46)
:48 3l (0:48)
:56 Bw (0:56)
5
6
7
C
E
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I want some language.lessons! 
Last edited by deagol on Wed Dec 13, 2006 10:20 am; edited 1 time in total |
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acidfingers
Lonely Fan
Joined: 12 Oct 2006
Posts: 147
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| Posted: Wed Dec 13, 2006 6:36 pm Post subject: |
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Alright, so I'm going to attempt to do a formal write-up of the process Dia and I took to get through this puzzle.
As you all know, in the Petaluma Revisited Drop we got 13 different dominoes along with 3 letter stencils.
The dominoes can be looked at from two different perspectives.
1.) Split each domino. So each domino represents two numbers
2.) Sum up both sides of the dominoes. So each domino represents one number.
Well, case #1 won't work, because we will get 13*2=26 numbers, plus the 3 letters. This gives us 26+3 = 29 characters. By the logic of the last XX/XY puzzle, we have to have an even number of characters to choose from since we need to construct hex pairs. So option 1 is out of the window.
So taking case 2 into consideration, we can now sum up each domino and list them out.
| Quote: | 0/0 = 0+0 = 0
0/1 = 0+1 = 1
0/2 = 0+2 = 2
0/5 = 0+5 = 5
0/6 = 0+6 = 6
1/4 = 1+4 = 5
1/5 = 1+5 = 6
1/6 = 1+6 = 7
2/4 = 2+4 = 6
2/5 = 2+5 = 7
3/3 = 3+3 = 6
3/4 = 3+4 = 7
3/6 = 3+6 = 9
Stencil: C
Stencil: C
Stencil: F |
I will add the letter stencils and put everything in order for clarity
| Quote: | 0
1
2
5
5
6
6
6
6
7
7
7
9
C
C
F |
Now, to get started with the hex->ascii madness. For reference, I will provide the list of hex codes and their corresponding ascii letter.
| Quote: | 41 - A
42 - B
43 - C
44 - D
45 - E
46 - F
47 - G
48 - H
49 - I
4A - J
4B - K
4C - L
4D - M
4E - N
4F - O
50 - P
51 - Q
52 - R
53 - S
54 - T
55 - U
56 - V
57 - W
58 - X
59 - Y
5A - Z
61 - a
62 - b
63 - c
64 - d
65 - e
66 - f
67 - g
68 - h
69 - i
6A - j
6B - k
6C - l
6D - m
6E - n
6F - o
70 - p
71 - q
72 - r
73 - s
74 - t
75 - u
76 - v
77 - w
78 - x
79 - y
7A - z |
Those are all of the alphabetic ascii characters, and their corresponding hex codes.
Before going through all of this, I'd like to get some definitions set up so that we have some consistency.
Plaintext - The desired end result. In this case it is a word.
Hex Pair - two hexidecimal numbers joined together (e.g. 6A, 72, etc..)
Character Mask - A string which dictates which hex pairs go where to form the plaintext. (XXXy -> a number-number hex pair followed by a number-letter hex pair)
Left Digit - The hexidecimal number which appears on the left side of a hex pair.
Right Digit - The hexidecimal number which appears on the right side of a hex pair.
Character Pool - The grouping of hexidecimal numbers which have not yet been used in the plaintext
The first step we took in figuring this out is to identify which characters from the character pool can only be used as right digits. If you look at the hex codes for the alphabetic characters, you will notice that the left digits are exclusively 4, 5, 6, or 7. This means that from our character pool, the characters 0, 1, 2, 9, C, C, F absolutely must be right digits.
Looking further, you'll notice the only hex pairs which have C or F as their right digit are 4C, 4F, 6C, and 6F. Since we do not have any 4s in our character pool, this forces C, C, and F to be paired with 6s.
Using 6 as the left digits and 2 Cs and an F as the right digits, we get 6C, 6C, 6F
If you look at the hex table above, this gives us l, l, and o.
Now we have at least 3 characters from the plaintext, and we can pull three 6s, two Cs, and one F:
| Quote: | Plaintext: l l o ? ? ? ? ?
100% For sure Right digits: 0 1 2 9
Remaining characters: 5 5 6 7 7 7
Hex Pairs Needed: _0 _1 _2 _9 __
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This is where we hit a wall. We were then informed that the solution contains 2 capitalized letter though. If you look at the list of hex values for capital letters, the left digit must be 4 or 5, and we only have two 5s (and no 4s) so those 5s must be left digits.
This is the point where it loses any kind of continuity. At this point, we have the following contraints:
| Quote: | Plaintext: l l o _ _ _ _ _
Character Pool: 0 1 2 5 5 6 7 7 7 9
Left Digits : 5 5 7 7 6/7
Right Digits : 0 1 2 9 6/7
Possible Hex Pairs:
50 51 52 56 57 59
61 62 67 69
70 71 72 76 77 79
=========
P Q R V W Y
a b g i
p q r v w y
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Since there's no other completely constrained part of the puzzle at this point, the only way to reach the solution now is to begin guessing.
One important thing to note is that once you begin choosing hex pairs, your character pool decreases in size, which gets rid of some of the possible hex pairs from before.
For example, lets say we chose the hex pair 57 (W)
our character pool now loses a 5 and a 7, which leaves us with
| Quote: | Plaintext: W l o l _ _ _ _
Character Pool: 0 1 2 5 6 7 7 9
Left Digits: 5 6 7 7
Right Digits: 0 1 2 9
Possible Hex Pairs:
50 51 52 59
61 62 69
70 71 72 79
=========
P Q R Y
a b i
p q r y |
Notice how the number of potential hex pairs decreases since the character pool has decreased in size when we took two characters from it to make the hex pair for W.
So basically at this point, Dia and I chose different capital letters to use, and saw how the character pool would change after that and got a list of letters. There really isn't any elegant way to do this aside from just going through all of the possible options. (This is essentially brute forcing, which isn't very elegant, but whatever)
At this point, Tachyon updated her profile with a character mask.
| Quote: | | XX XX XX XX XX Xy Xy Xy |
Using the logic from last XX/XY puzzle, we noticed that the last 3 hex pairs must contain a letter as their right digit. The hex pairs we have for sure are 6C, 6C and 6F so we put those at the end. The word now for sure ends with either "lol" "oll" or "llo"
So, time for the very unelegant part of the solution. After staring at the list of possible hex pairs and their corresponding letters, someone got reminded of the user that is in that ipod opera group on youtube, warpylol. it ends with "lol" which looks promising, and the list of possible hex pairs certainly seems like it would fit. So after choosing a capital 'W' for the first letter, we worked out the rest of the hex pairs, and surely enough the letters all fit. It left us with a capital W and capital P, so we concluded that "WarPylol" was plaintext and, indeed, to the puzzle.
Surely enough, this password opened the password-protected rar file, and much celebration took place
_________________
[13:09:39] <marlasinger> acid's like a pussy sore, if you keep playing with it it just gets worse. |
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diastres
Lonely Fan
Joined: 28 Nov 2006
Posts: 151
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| Posted: Wed Dec 13, 2006 8:15 pm Post subject: |
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Now that I've woken up from passing out on a sudoku puzzle while a plane was taking off... (Don't ask how I do it. I don't know. Possibly from walking the airport twice. P.S. Acid! I figured out why it didn't go in a circle!)
I don't have a single serve friend. The information is secure. Though, a small child did try to get me to sit with him and his mother. He just wanted to steal my white tiger plushie. 
This is XX/Xy Version 2.0. Acidfingers and I spent many hours in Tachyon - er - #tachyon, discussing this. So much <3 to him.
We received thirteen dominoes and three stencils at the Petaluma: Revisited (Sixth) Drop.
The stencils were:
C, C, F
The dominoes were:
0/0, 0/1, 0/2, 0/5, 0/6, 1/4, 1/5, 1/6, 2/4, 2/5, 3/3, 3/4, 3/6
If we treated each number on the dominoes seperately, we'd get 26 values, plus the three letters, making 29. This would not give us valid hex, as we'd need an even number of values to get all the pairs.
Instead, we have to add together the two sides of the dominoes to get new values:
0+0 = 0
0+1 = 1
0+2 = 2
0+5 = 5
0+6 = 6
1+4 = 5
1+5 = 6
1+6 = 7
2+4 = 6
2+5 = 7
3+3 = 6
3+4 = 7
3+6 = 9
This way, the dominoes give us thirteen values. Add the three letters and we have sixteen. This is an even number, so we could go ahead, knowing that the end result will have eight characters. We know this because they pair off and 16/2=8.
Now, we'll take a look at my normal count.
| Code: |
value | 0 1 2 5 6 7 9 C F
count | 1 1 1 2 4 3 1 2 1
** ** ** ** ** ** ** **
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The following values must be in the second half of a hex pair, because if they are first we do not receive an alpha-numeric character:
0 1 2 9 C C F
| Code: |
*0 *1 *2 *9 *C *C *F **
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NOTE: I do not remove them from the count until we figure out the pair they're in. It just made more sense to me mentally to do it that way.
Let's take a look at the letters. The only way they become alphanumeric characters is if they are fronted by a 4 or 6. 4 is unavailable to us, so that means we must have 6C 6C 6F.
| Code: |
value | 0 1 2 5 6 7 9 C F
count | 1 1 1 2 1 3 1 0 0
*0 *1 *2 *9 6C 6C 6F **
l l o
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So we have llo. Or lol. (I mentioned this to acidfingers right away... I wish we'd kept it in mind longer...)
Possible combinations to fit the rest:
| Code: |
*0: 50=P, 70=p
*1: 51=Q, 61=a, 71=q
*2: 52=R, 62=b, 72=r
*9: 59=Y, 69=i, 79=y
**: 55=U, 56=V, 57=W, 65=e, 67=g, 75=u, 76=v, 77=w
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We'd done a dictionary attack (both with a program and manually [go us!]) long before this so we knew it wasn't one word. For that reason, several of us believed it was two, and would have two capital letters. (We kept it only having one in mind, as well as it not being two, but... you know how things go...) Later, it was confirmed by SparkleBot that there were two uppercase letters. I'll apologize now for not posting it. Once we got that hint, things started moving quickly for a while, and I don't think either of us were thinking about it.
The only possible uppercase letters are if 5 is the front of the hex pair. That means we can remove 55=U, 65=e and 75=u.
| Code: |
*0: 50=P, 70=p
*1: 51=Q, 61=a, 71=q
*2: 52=R, 62=b, 72=r
*9: 59=Y, 69=i, 79=y
**: 56=V, 57=W, 67=g, 76=v, 77=w
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Now, I looked to *1 immediately. The option of a q without a u? No way. So I took that out, meaning *1 must be 61.
| Code: |
value | 0 1 2 5 6 7 9 C F
count | 1 0 1 2 0 3 1 0 0
*0 61 *2 *9 6C 6C 6F **
a l l o
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The lack of sixes means that the second spot in ** must be a 7, as there aren't enough front spaces for all of the fives and sevens and the 5 takes priority, as it is required to make capitalized letters.
| Code: |
value | 0 1 2 5 6 7 9 C F
count | 1 0 1 2 0 3 1 0 0
*0 61 *2 *9 6C 6C 6F *7
a l l o
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Let's look at the leftover possible combinations, knowing that we no longer have any 6's to use:
| Code: |
*0: 50=P, 70=p
*2: 52=R, 72=r
*9: 59=Y, 79=y
*7: 57=W, 77=w
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Well look at that. We have all the letters, just not necessarily their placement.
| Code: |
value | 0 1 2 5 6 7 9 C F
count | 1 0 1 2 0 3 1 0 0
*0 61 *2 *9 6C 6C 6F *7
a l l o
Pp Rr Yy Ww
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paryllow, huh?
Two will be capitalized and two will not, which means that no matter the combination, the count will drop to 0.
Tachyon updated her profile to include a character map, much like the one we got in the last XX/Xy problem:
XX XX XX XX XX Xy Xy Xy
Xs are numbers, ys are letters. The last three must be the llo. So what combinations are there?
llo
oll
lol
Then we have the rest: paryw. This can be anagrammed to warpy.
warpylol.
Acidfingers is actually the one who figured out what should be capitalized.
The password is WarPylol.
It makes the most sense to me, but I didn't know that it would be warpylol before he gave it. War is a word in itself, not to mention it was a tag in one of OpAphid's latest vids (I can't say exactly which one, as I am not connected to the internet at the moment.)
warpylol is part of ipod opera and is OpAphid's only friend. Refer to deagol's post for background information, as I really don't know anything about it myself.
If there are any further questions, please feel free to post and I will do my best to recall the answer.
Much <3 to acidfingers for staying up with me all night and figuring this out. I probably wouldn't have gotten it without him for a lot longer.
I'm not sure how much longer until the plane lands, but it's 5pm in Arizona. I'll post as soon as I get there. Maybe I'll take another nap. I've got 38% battery remaining, so I can't futz around on the computer. Hopefully I'll find an outlet in Chicago. FallingIntoSin tells me the airport I'm going to has free wifi. Let's hope!
Hello from Chicago! You guys are insane, I'm going to freeze. New Hampshire looks bleaker by the moment. I have commandeered an outlet (thank goodness). I didn't sleep afterall, and it might show. Did you know that David Hume says that you have no good reason to believe a cause you have witnessed many times with the same result will have the same result next time?
I know acid posted, but I had typed this up already, and #operationaphid told me to. Bring it up with them.
(I still <3 you acid.) |
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FallingIntoSin
Lonely Fan
Joined: 26 Nov 2006
Posts: 212
Location: Chicago
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| Posted: Thu Dec 14, 2006 10:30 am Post subject: |
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Dia,
I love you. Yesterday was a WARM chicago day  Hahaha.
You (and Acid) are so smart, it's amazing to see how you came up with this. I can't imagine where we'd be without you I'm glad you joined us Dia! ///HUG HUG///
love,
Fall
_________________
I fall down, a lot.
~*~*~*~*~*~*~*~*~*~*
OpAphid Angels Unite!
~*~*~*~*~*~*~*~*~*~*
"Men think if women are grabbing and clawing at each other there's a chance they might somehow you know... kiss." --Jerry Seinfeld, on the appeal of Catfights |
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Ziola
The Order of Denderah
Joined: 17 Oct 2006
Posts: 5774
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| Posted: Thu Dec 14, 2006 10:42 am Post subject: |
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I would like to thank Dia and Acid for all their hard work.
Now if only I knew what the hell she was talking about  |
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butterfly
Lonely Fan
Joined: 05 Oct 2006
Posts: 148
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| Posted: Thu Dec 14, 2006 12:30 pm Post subject: |
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Wow you people are smart!
I cannot understand any of this!  |
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askewcutie
Lonely Fan
Joined: 14 Sep 2006
Posts: 198
Location: Philly
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| Posted: Thu Dec 14, 2006 1:36 pm Post subject: |
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I'm with you butterfly...
I'm glad though that we have such smart people to take care of this for us. I just hope soon I can contribute something soon
_________________
There used to be a real me, but I had it surgically removed.
- Peter Sellers |
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