DEFCON OpenCTF is an amateur CTF that is open to anyone who signs up and is “at” DEFCON during the event. I say “at” because you just need to be there to sign up and set up your equipment. Your team doesn’t need to be in the room at the time, since they have no way of enforcing that.

The CTF is set up in a Jeopardy style question mode where there are a few problems open at the beginning and when enough problems get solved the people who solve the last problems get to open new questions of their choice. The reason for this I’m told is to give an advantage to the smaller teams, so large teams can’t just put one person on each problem and work it out.

The team I am a part of along with co-workers and friends is xip. This year we placed 9th overall. We scored much more points this year than last year, so I feel we did an excellent job.

I worked on a number of problems. The first one is usually a gimme and worth practically nothing but makes sure you can access the scoreboard. We logged into the scoreboard this year by an ssh private key generated for us by the OpenCTF admins. This was encoded in a QR code onto a receipt printed on receipt paper along with other information which I will get to later.

I used my iPhone’s camera to read the QR code and copy it over to . One of the downsides of going to DEFCON is if you really want to stay secure you don’t bring a laptop with anything important on it and you don’t log into any services using WIFI or other wireless protocols. Although I hear LTE is generally okay. After copying down the private key to my laptop I put it on our CTF box for everyone to copy themselves and use. I will do a write-up on the CTF box some other time.

On the receipt we were given two IP addresses. One for the scoreboard which shows how many points each team currently has, and the questions. And one for a box the admins set up if we needed one. Using the ssh key we then log into the scoreboard box and select the question. Questions range from an IP address and port, just a hint at what you need to find or where to look, or a binary and any combination of the above.

Sanity Check

The first question this year as it was last year I believe was shown like so. ` Hack the __! ` If anyone has seen the movie Hackers you should know the answer is ‘Planet’. The OpenCTF group had also posted a hint on twitter earlier in the day about something needing to be capitalized.


The other problem I successfully solved was a file that contained a hex value and directional IDs (top, bottom, left, right). This file translated to a bitmap image, the IDs were unique to two lines. For example on one line there would be a top ID of 1234, and the only other place 1234 was found would be another line with 1234 as the bottom ID. Essentially a scrambled matrix.

Given file.

Value,Right Edge Id,Bottom Edge Id,Left Edge Id,Top Edge Id

To solve this I created a class block. A forewarning, CTF code is hastily written and thrown together. This is in no way the best way to solve the problem. The full files will be available on my github


class block:
    color = 0
    bottom = 0
    top = 0
    left = 0
    right = 0

    def __init__(self, c,r,b,l,t):
        self.color = c
        self.right = r
        self.bottom = b
        self.left = l = t

    def __str__(self):
        return "{0}:{1},{2},{3},{4}".format(self.color, self.right, self.bottom, self.left,

I then opened the file and iterated over the contents

for l in content:
    splitted = l.split(',')
    color = splitted[0]
    right = int(splitted[1])
    bottom = int(splitted[2])
    left = int(splitted[3])
    top = int(splitted[4])
    blocks.append(block(color, right, bottom, left, top))

One can easily find the top left block by finding the line with top and left IDs which are zero.

top_left = [x for x in blocks if ( == 0 and x.left == 0)][0]

Then just find the next one to the right and shift down when you get to the end.

while current.bottom is not 0 and current.right is not 0:
    row = []
    output.write(current.color + ' ')
    while current.right is not 0:
        n = [x for x in blocks if (x.left == current.right)][0]
        current = n
        output.write(n.color + ' ')
    current = row[0]
    n = [x for x in blocks if ( == current.bottom)][0]
    current = n
    #print('added row')

This will write the hex output in ascii (probably making the next part harder than it needs to be but what I did at the time). The lines are like so

0x799394 0x678182 0x5b7478 0x587175 0x48636a 0x2e4950 0x1d3843 0x1c3742 ...
0x6e8688 0x607a7b 0x566f73 0x516a6e 0x486166 0x365158 0x2c444e 0x28434e ...

Now we have all the color values in the right places and know how big the image is by how many hex values we have in a row and column.

This then needs to be written as actual hex and a header and footer added so an image view knows what it is. This unfortunately was the most frustrating part as getting the hex just right is important. I ended up just finding a BMP image on the internet that was the correct size and copy-pasting the header and footing into the hex file.

So first we open the file and write the hex for the header.

output = open('raw_hex.bmp', 'w')

I did it in these chucks because that is how many hex digits my hex editor was showing per line. The next step is just to iterate over the hex ascii code and print it as real hex. Then print the footer.

for l in content:
    hexes = l.strip('\n').split(' ')[:-1]
    for x in hexes:
        trimmed = binascii.unhexlify(x[2:])


This gives us a bmp picture. Of a kid (baby goat not human child) with the flag overlaid as.


The Price Is right

We did not get this question but it is infuriating enough that it is worth a write-up. It’s infuriating because once you know the solution its so simple but hard to get that first step.

On our receipt we were given a list of debits and credits for things such as custom team name, etc. These numbers were wildly different from each other and very large, they did not seem to have any pattern. The total also did not match, which was printed as ~3.50.

We first tried adding them up to find the actual total and tried multiple different rotations and bases to see if there was some code hidden in the total. We then tried concatenating them and seeing if they represented hex values or octal or something. We tried all the other things we tried on the first attempt on the concatination. We tried many different solution and a couple of us spend a few hours trying to find it. I eventually found that the first 4 numbers on the list 5117 Can sort of be changed to F 1 a g if you take the 5th letter, use 1 as an l because you are leet and many other flags are that way, and use the 7th letter g. This ended up being an unintentional red herring. Or an intentional one if the admins really hated us.

The end solution, which we asked the admins after the CTF was to, concatenate (not add) all the numbers together to get one large number. You then change this number to a hex value and each one of the hex values is a letter that became the flag. Seems simple once you know the answer but choosing from so many different obfuscation techniques leads to a multitude of wrong turns. The price is wrong Bob.

Overall OpenCTF was an awesome experience and gives a great introduction into reverse engineering problems and problem solving difficult questions. A big thank you (and screw you) to the problem creators: Kajer, soen, Yen, yotta, arronp, and Drifter.