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: Select all
value | 0 1 2 5 6 7 9 C F
count | 1 1 1 2 4 3 1 2 1
** ** ** ** ** ** ** **
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
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: Select all
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
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: Select all
*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
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: Select all
*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
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: Select all
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
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: Select all
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
Let's look at the leftover possible combinations, knowing that we no longer have any 6's to use:
Code: Select all
*0: 50=P, 70=p
*2: 52=R, 72=r
*9: 59=Y, 79=y
*7: 57=W, 77=w
Well look at that. We have all the letters, just not necessarily their placement.
Code: Select all
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
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.)