Warning: this post contains partial spoilers for the 2009 MIT Mystery Hunt puzzle "Invasion of the Micronauts." You've been warned!

This post is an IPython Notebook, which you can also view online through nbviewer here.

Introduction

I'm a big fan of puzzles of all kinds, especially cryptic crosswords, and I've been participating in the MIT Mystery Hunt for the last few years. Mystery Hunt is a weekend marathon of puzzle solving, consisting of puzzles of every style imaginable, from sudoku to physics to Minecraft. Puzzle solving generally involves some mix of cleverness and persistence, and perhaps a bit of brute force.

This week, I've been going through the Mystery Hunt archives from the 2009 Hunt, and I found a puzzle I'd heard of but never solved: Invasion of the Micronauts. This puzzle consists entirely of an (apparently) blank PDF, with no instructions or hints. I remembered hearing from friends of mine that the PDF actually contains a tiny image or description of some kind, but it's too small to see normally. But after a few minutes of zooming and scrolling, I got bored and decided that this was a perfect opportunity for some brute force. A record of my adventures follows.

In [1]:
%pylab inline
from IPython.display import Image
Populating the interactive namespace from numpy and matplotlib

Breaking Micronauts

If the PDF actually contains an embedded image or text, then we should be able to find it in the PDF source. Opening puzzle.pdf in vim gives the following:

In [2]:
display(Image('pdf_text.png'))

I had never actually looked inside a PDF file before, but there's a lot we can learn from this already. The basic format is a text file, with some blobs of binary data embedded in it. Scrolling through the file, I don't see anything that looked like it might be the text that the image supposedly contained, so the data must be hidden inside those binary blobs. Time to go exploring...

Decoding the binary data

Each binary blob is contained between two readable tags: stream and endstream, but I can't see any obvious patterns in them. This is where we break out the big guns: the Adobe Portable Document Format Reference, third edition, from Addison-Wesley. The PDF reference has pages and pages of data on these Content Stream types. Most of that is going to be irrelevant, so we need something more specific to look for.

Just before the stream is a tag /FlateDecode. Aha! According to the manual, that means:

(PDF 1.2) Decompresses data encoded using the zlib/deflate compression method, reproducing the original text or binary data.

That sounds promising! Let's try it out.

In [3]:
# First, let's get the PDF file contents as a byte stream
data = open('puzzle.pdf', 'rb').read()

Next, I manually searched through the data array to find the chunk between the stream and endstream tags, which turned out to start at index 78 and last for 324 bytes. Decompressing it is as easy as this:

In [4]:
import zlib
decompressed = zlib.decompress(data[78:(78+324)])

And here's the result:

In [5]:
print(''.join(chr(x) for x in decompressed))
q Q q 9 8.489918 594 778.53 re W n /Cs1 cs 0 0 0 sc q 0.2400000 0 0 0.2400000 72.12012 708.54
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 684.54
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 660.3
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 636.3
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 612.3
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 588.3
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 564.0599
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 540.0599
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 516.0599
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 491.82
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 467.82
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 443.82
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 419.82
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 395.58
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 371.58
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 347.58
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 323.34
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 299.34
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 275.34
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 251.34
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 227.1
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 203.1
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 179.1
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 154.86
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 130.86
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 106.86
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 80.21997
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 108.1201 80.21997
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 144.1201 80.21997
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 180.1201 80.21997
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 216.1201 80.21997
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 252.1201 80.21997
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 292.1201 80.21997
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q Q q 290.12 79.97421 1.459839 1.259819
re W n /Perceptual ri q 1.459842 0 0 1.259843 290.12 79.97421 cm /Im1 Do Q
Q

Cool! I have no idea what any of that means, but the last two numbers in every row look a lot like coordinates. Let's try graphing them:

In [6]:
lines = ''.join(chr(x) for x in decompressed).split('\n')
elements = [l.split(' ') for l in lines]
x = []
y = []
for row in elements[:-3]:
    x.append(row[-2])
    y.append(row[-1])
In [7]:
plot(x, y)
axis('equal')
Out[7]:
(50.0, 300.0, 0.0, 800.0)

That looks a lot like a letter 'L'. Maybe the rest of the streams will also come out as letters.

Manually specifying the indices for the streams is a pain, so let's automate that with some regular expression magic.

In [8]:
import re
matches = list(re.finditer(b'stream\n((.|\n)*?)\nendstream', data))

We now have a list matches containing each binary stream blob. Let's look at a few (printing just the first 200 characters from each match):

In [9]:
for m in matches:
    decompressed = zlib.decompress(m.group(1))
    print('============ MATCH =============')
    print(''.join(chr(x) for x in decompressed[:2000]))
============ MATCH =============
q Q q 9 8.489918 594 778.53 re W n /Cs1 cs 0 0 0 sc q 0.2400000 0 0 0.2400000 72.12012 708.54
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 684.54
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 660.3
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 636.3
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 612.3
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 588.3
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 564.0599
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 540.0599
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 516.0599
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 491.82
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 467.82
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 443.82
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 419.82
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 395.58
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 371.58
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 347.58
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 323.34
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 299.34
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 275.34
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 251.34
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 227.1
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 0 0 0.2400000 72.12012 203.1
cm BT 50 0 0 50 0 0 Tm /F1.1 1 Tf (!) Tj ET Q q 0.2400000 
============ MATCH =============
ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ
============ MATCH =============
ÜapplmntrRGB XYZ Ø07acspAPPLöÖÓ-applž3OxWl.틂È)Ò6rXYZ,gXYZ@bXYZTwtpthchad|,rTRC¨gTRC¸bTRCÈvcgtØndinì>desc,ddscmþmmod(cprt¸$XYZ [|4dzXYZ sųDôXYZ '”­~XYZ óRÏsf32BÞÿÿó&’ý‘ÿÿû¢ÿÿý£ÜÀlcurvÍcurvÍcurvÍvcgt…¦1¹	F
0z1å3O4³67w8Ô:/;…<Ú>,[email protected]
vw'x2y;zC{J|O}Q~S[email protected] ‡ˆˆò‰ßŠË‹¶ŒŸ‡ŽnS7‘‘ü’Þ“¾”•{–Y—5˜˜ì™Çš¡›zœS*žžÙŸ¯ …¡Z¢.££Ö¤©¥|¦N§ §ò¨Ã©”ªc«3¬¬Ò­¡®o¯=°[email protected]´Øµ¤¶p·;¸¸Ò¹ºg»2»ü¼Ç½‘¾Z¿#¿íÀ¶Á~ÂGÃÃ×ğÅgÆ.ÆõǼȂÉHÊÊÓ˘Ì]Í!ÍåΩÏlÐ/ÐòѵÒwÓ8ÓúÔ»Õ|Ö<Öý×¼Ø|Ù;ÙûÚºÛxÜ7ÜöÝ´Þsß2ßñà±áqâ2âóãµäxå=ææÉç’è\é'éõêÅë—ìlíCîî÷ïÖð·ñ›ò‚ókôWõEö5÷(øùúûüüûýúþúÿÿ[ólíiã
KÈ>·-©“óeÕ I!µ#$$%ø'`(Æ*++Š,é.F/0ó2F3—4ã6.7t8·9õ;2<l=¡>Ó@A0BZCD£EÅFãGÿIJ2KGL[MlN|OŒP™Q¦R±S½TÆUÎV×WßXæYîZô[ú]^_`abc"d'e,f2g6h:i?jCkFlImKnNoOpPqPrOsNtLuHvDw?x9y1z({|}}û~íހ΁½‚«ƒ—„ƒ…m†V‡?ˆ&‰‰òŠ×‹ºŒž€ŽaB!‘‘Þ’¼“™”u•Q–,——à˜¹™’šj›BœœðÇžŸs I¡¡ô¢É£ž¤r¥G¦¦ð§Å¨™©nªB««ë¬À­”®h¯=°°æ±»²³e´:µµå¶º·¸f¹=ººê»À¼—½o¾G¿¿øÀÒÁ¬Â†ÃaÄ=ÅÅöÆÔDzȒÉrÊSË5ÌÌüÍáÎÇϮЖÑ[email protected]××ûØìÙßÚÒÛÆܺݰަߜà“áŠâãxäoåfæ\çQèFé:ê,ëììüíèîÒïºðŸñƒòdóBôôùõÏö¤÷vøFùùàú©ûpü7üýýÁþƒÿCÿÿ,?UmŠ¡¹	Ï
(29 A!I"K#N$P%N&J'E(=)4*'+,	,ô-ß.Ç/­0‘1s2S304
ÍŒ‘L’’Ë“Š”J•
•Ë–Œ—M˜˜Ð™“šV››Ýœ¡gž-žôŸ» ƒ¡L¢¢á£¬¤y¥G¦¦æ§·¨Š©^ª4««ä¬¿­œ®|¯_°C±+²³³ø´ïµë¶ì·ó¹º»0¼S½~¾²¿ñÁ:ÃìÅXÆÎÈUÉæˈÍ9Î÷ÐÄҢԏ֋ؗڳÜÛß
============ MATCH =============
rXYZ gXYZ4bXYZHwtpt\chadp,rTRCœgTRCœbTRCœdesc@ocprt8vcgt¬0ndinÜ8dscmLòXYZ tK>ËXYZ Zs¬¦&XYZ (W¸3XYZ óRÏsf32BÞÿÿó&’ý‘ÿÿû¢ÿÿý£ÜÀlcurvÍvcgt¸R¸R¸Rndin8¡HW
T9€€€textCopyright 2007 Apple Inc., all rights reserved.mlucenUS&~esES&‚daDK.êdeDE,¨fiFI(ÜfrFU(*itIT(VnlNL(nbNO&ptBR&‚svSE&jaJPRkoKR@zhTWlzhCNÔruRU"¤plPL,ÆYleinen RGB-profiiliGenerisk RGB-profilProfil Générique RVBN‚, RGB 0×0í0Õ0¡0¤0ëu( RGB ‚r_icϏðPerfil RGB GenéricoAllgemeines RGB-Profilfn RGB cϏðe‡NöGenerel RGB-beskrivelseAlgemeen RGB-profielÇ|¼ RGB Õ¸\ÓÇ|Profilo RGB GenericoGeneric RGB Profile1I89 ?@>D8;L RGBUniwersalny profil RGBdescGeneric RGB ProfileGeneric RGB Profile
============ MATCH =============
€0cmap;¼cvt íþþÔ~fpgmÐØd2	TAglyfé¼]%˜ìheadïJT"„6hheaÂ:¼$hmtx¾àlocavvèmaxp	Ã>ð nameãiv prep’ì‡0è!
Uÿðàÿ ÿðàB}ccpVÂÎÉ»ƒ¼³¬¬µ…¼xYO„ OTxZIU^iq¤¥b`11>
=ÿó:ÿó‹–tŽ€t|žÞôþÂÿð˜K„ŽBZ=P`}·¤Árx³¦¬ž%š‡}¼2‘·Œ¦Ã¹¯¥UŽRs›¨BP(
¼·ªÆ¢›õáþ/¼)Ñ)ÆúºÁì׍Uº‹eêÿðÿð±þRÿð§œÂ}¬´»Á’¶»Œ´ …—­¦¦‘€¢Z3g3ÃÀ3ÿÑÿÍÿÍþŽÿÍÕiÛ>PI
============ MATCH =============
/CIDInit /ProcSet findresource begin
12 dict begin
begincmap
/CIDSystemInfo <<
  /Registry (Adobe)
  /Ordering (UCS)
  /Supplement 0
>> def
/CMapName /Adobe-Identity-UCS def
/CMapType 2 def
1 begincodespacerange
<00><FF>
endcodespacerange
1 beginbfrange
<21><21><00a0>
endbfrange
endcmap
CMapName currentdict /CMap defineresource pop
end
end

Okay, those are definitely not all going to work as coordinates.

Let's look closer at the second match:

In [10]:
m = matches[1]
decompressed = zlib.decompress(m.group(1))
print(''.join(chr(x) for x in decompressed[:20000]))
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That's actually very interesting. Most of the data consists of the same character over and over, but there are blocks of a few different characters repeated at different intervals. To me, that actually looks a lot like an image of some kind, but obviously not in the right format or shape.

Let's assume this is an image. Currently, the data just comes as one big line:

In [11]:
lines = ''.join(chr(x) for x in decompressed).split('\n')
print(len(lines), len(lines[0]))
1 55188

That's 1 line of 55188 characters. So it's definitely not nicely broken up by newlines for us. We're going to have to try to reshape it ourselves. First, let's convert it to a numpy array:

In [12]:
imdata = np.array(bytearray(decompressed), dtype=np.uint8)
print(len(imdata))
55188

Now we need to find the correct way to reshape the data. I'm assuming that the image being encoded is a rectangle of some kind, and the aspect ratio is probably not too weird (that is, it's probably not more than 5 times wider than it is high). So, it actually won't be that hard to just try every possible aspect ratio that evenly divides the number of elements in the image data. Let's do that:

In [13]:
for m in range(100,500):
    if mod(len(imdata), m) == 0:
        n = len(imdata) // m
        figure()
        title((m, n))
        imshow(imdata.reshape((m, len(imdata) // m)))

Bam! The second one is obviously correct, and it even looks like it might have readable text in it!

Let's try to clean it up a bit. The reshaped image is a 126x438 array of unsigned 8-bit integers, so it should probably be a grayscale image. But the imshow() command is interpreting it as some weird RGB image. To get around that, we'll just copy the data three times to get equal red, green, and blue channels and thus a nice grayscale image.

In [14]:
figure(figsize=(10,10))
imshow(np.repeat(imdata.reshape((126,438,1)),3, axis=2))
Out[14]:
<matplotlib.image.AxesImage at 0x114d8c7b8>

And there we have it: readable text extracted from the binary blob inside the PDF.

Cleaning it up

Looking back at the PDF source, the line just before the blob that created that image has the entries:

 /Subtype /Image /Width 146 /Height 126

Where that height of 126 is exactly what we found by searching. I can force the displayed image to use the width of 146 to fix the aspect ratio:

In [15]:
figure(figsize=(10,10))
imshow(np.repeat(imdata.reshape((126,438,1)),3, axis=2), extent=(0,146,0,126))
Out[15]:
<matplotlib.image.AxesImage at 0x115135f98>

This is exactly the image that can be found in a tiny dot in the lower half of the original PDF. Normally, you'd have to zoom in 6400% to find it, but we got it just by inspecting the binary data itself.

Solving the Puzzle

Now there's just one step left: actually solving the puzzle. The image we've found isn't an answer, it's just the next step. It's up to you, dear reader, to figure out what to do next. If you get stuck, the solution can always be found here.

Loose Ends

But what about the coordinates I found earlier? The ones that made such a nice letter L? In short, I have no idea what that meant. Perhaps it's used to position the image on the page? Perhaps it's just boilerplate of some kind. That's a puzzle for another day.