Passcode Walkthrough for JoJo’s Word of the Day 2014-05-20
The current standard passcode format is:
Long Passcode Format 3 (Mostly for Niantic Project document posts) [2-9][a-z][a-z][a-z][2-9]keyword[a-z][2-9][a-z][2-9][a-z]
Difficulty Rating: (out of 8. 1-5 are regular difficulty levels, 6-8 are very challenging):
Source of document:
How/where to find the source of the passcode:
For JoJo’s Word of the Day posts, most often you have to view the source of the linked HTML document to find the actual code. Looking at the source code, you’ll find an HTML comment with the following digits:
Many, many (many) times, encoded passcodes use characters that have been replaced by their ASCII encoding. This can be represented as normal decimal, base 10, number (i.e. “k” becomes “107”); or as a hexadecimal, base 16, number (“k” becomes “6b”); or less commonly octal, base 8 (“k” becomes “153”).
Seeing a long string of numbers means this is a possible option for this passcode. However, seeing that “A” stuck in the middle should immediately point to hexadecimal being the primary suspect, since hexadecimal encoding can use both digits 0-9 as well as letters A-F. If it is hexadecimal, there are a few things we can look for. Let’s consider the hexadecimal ranges for some character types.
|Character types||Characters||Hex range|
Examining the string, it looks like there are a lot of sixes and a lot of threes, which means hexadecimal numbers and lowercase letters are very likely. A single character in hexadecimal ASCII is made up of two hex digits, so let’s break the string up that way.
06 78 32 68 39 63 A2 93 86 36 66 33
Some potential with this, as we see a very good candidate for the suffix. However, it’s near the start of the string instead of the end. Remembering that the passcode format should most likely end
[a-z][2-9][a-z][2-9][a-z], we can look for a pattern
6x 3x 6x 3x 6x for that suffix. (The sixes could also be sevens.) That pattern shows up here:
78 32 68 39 63
However, we have an extra 06 at the start, and still need a prefix and keyword. Let’s look for the prefix first. That should have the pattern
[2-9][a-z][a-z][a-z][2-9], which would appear
3x 6x 6x 6x 3x. (Again, the sixes could also be sevens.) That doesn’t appear to show up in our pairs at first glance, but if you look a little closer, you might notice that the pattern does sneak in slightly modified. Look at the last five pairs, and you’ll see
x3 x6 x6 x6 x3. So the second half of the string has been reversed! There are still a couple of bytes that aren’t quite accounted for, but let’s go ahead and reverse the second half of the string, and then move it to the front, since it looks like the prefix of the code.
33 66 63 68 39 2A 06 78 32 68 39 63
So, we now have good hex values for the prefix and suffix, and just need to figure out what to do for a keyword. We’ve been given two bytes, but they fall well outside the standard ASCII hex values for anything we might normally expect. However, it turns out that ASCII isn’t the only way to encode characters! There’s another character encoding called Unicode, which supports way more characters than ASCII could ever dream of, covering multiple languages, mathematical symbols, and nearly anything you could ever want to represent. So, let’s go look up what Unicode character
Apparently, this is the N-ary Square Union Operator. That may not seem helpful, but take a closer look at the character itself: ⨆ (This may not render right in your browser. Imagine a squared-off U.) Weird shaped characters? Better look at the list of glyphs! Sure enough, there’s a match.
So, let’s go back to our adjusted string and treat those middle bytes as one unit.
33 66 63 68 39 2A06 78 32 68 39 63
Converting from hex ASCII and Unicode, we get:
And now, replacing the Unicode character with the glyph that it strongly resembles, we get the passcode (which is already fully redeemed, sorry!):
Passcode items gained:
L4 Resonator (2)
L7 Xmp Burster (2)
Portal Shield (2)