MicroCorruption is a "game" made by **Matasano** in which you will have to debug some programs in **assembly**. There is a total of 19 levels and it gets harder and harder by the number. The first levels are made for beginners though! So it seems like a great tool to learn, and that's what our prof Emmanuel Fleury must have thought when he gave us this as homework. One rule: every level is worth one point.

I started writing the solutions here but as people told me "it was unethical to post solutions of a challenge online", I promptly removed them. If someday the challenge will shut down I will post the write ups online so that people can still learn from it :)

I feel like I don't write much about my formation, and that it could be useful to people who are wondering about **studying Cryptography at Bordeaux University**.

There is a good article from a M1 student here: http://journaldumaster.stats.yt/master-csi-presentation/

And as it says there, the master 1 is do-able both for maths and CS people as long as you're willing to catch up in the other subject. There's a lot of theory that will allow you to study more interesting subjects in the second year of Master.

I've talked about some of the subjects but one subject I forgot to talk about was a M1 class: Elliptic Curves, taught by Fabien Pazuki and if you have the chance of taking a class from the guy just do it. He's one of the best math teacher I have had in my life, along with Vincent Borrelli (Surfaces & Curves at Lyon 1) and **some dude I can't remember the name of**. Each one of them were both really passionate and making true efforts to be pedagogical.

One of my professor organized a Hacking Week this semester but I didn't have time to do it. Since I'm in holidays I thought I would take a look at it and write a bit about how I solved them.

Here's the **Crypto Challenge number 2** (out of 5) from this **CTF** (Capture The Flag):

user0:$apr1$oTsx8NNn$bAjDZHpM7tCvHermlXKfZ0
user1:$apr1$UxOdpNtW$funTxZxL/8y3m8STvonWj0

user2:$apr1$w7YNTrjQ$0/71H7ze5o9/jCnKLt0mj0
user3:$apr1$AIw2h09/$Ti0TRlU9mDpCGm5zg.ZDP.
user4:$apr1$048HynE6$io7TkN7FwrBk6PmMzMuyC.
user5:$apr1$T2QG6cUw$eIPlGIXG6KZsn4ht/Kpff0
user6:$apr1$2aLkQ0oD$YRb6aFYMkzPoUCj70lsdX0

You have 7 different users with their respective password hashed and you have to find them. It's just the 2nd out of 5 crypto problems, it's pretty basic, but I never brute forced passwords for real before (I remember using **John The Ripper** when I was in middle school but that's for script kiddies).

What's Apr1 ? It's a hash function that uses md5. And md5 is pretty weak, lots of **rainbow tables** on google.

This is how Apr1 looks in PHP according to Wikipedia, also the passwords are supposed to be alpha (a to z) in lowercase.

```
function apr1($mdp, $salt) {
$max = strlen($mdp);
$context = $mdp.'$apr1$'.$salt;
$binary = pack('H32', md5($mdp.$salt.$mdp));
for($i=$max; $i>0; $i-=16)
$context .= substr($binary, 0, min(16, $i));
for($i=$max; $i>0; $i>>=1)
$context .= ($i & 1) ? chr(0) : $mdp{0};
$binary = pack('H32', md5($context));
for($i=0; $i<1000; $i++) {
$new = ($i & 1) ? $mdp : $binary;
if($i % 3) $new .= $salt;
if($i % 7) $new .= $mdp;
$new .= ($i & 1) ? $binary : $mdp;
$binary = pack('H32', md5($new));
}
$hash = '';
for ($i = 0; $i < 5; $i++) {
$k = $i+6;
$j = $i+12;
if($j == 16) $j = 5;
$hash = $binary{$i}.$binary{$k}.$binary{$j}.$hash;
}
$hash = chr(0).chr(0).$binary{11}.$hash;
$hash = strtr(
strrev(substr(base64_encode($hash), 2)),
'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/',
'./0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'
);
return '$apr1$'.$salt.'$'.$hash;
}
```

It seems pretty difficult to reverse. Let's not forget that hashes are one-way functions and that they also lose information. I don't know if they do lose information on a 7-letters-password though, but it seemed quite stupid to go down this road when I could just brute force it.

What language offers a good library to hash with Apr1? Well I didn't know, and I felt like maybe Unix could do it well for me.

Turns out that **OpenSSL** has a command line for it:

`openssl passwd -apr1 -salt SALT PASSWD`

A quick bash script later:

```
#!/bin/bash
test[1]='$apr1$oTsx8NNn$bAjDZHpM7tCvHermlXKfZ0'
salt[1]='oTsx8NNn'
test[2]='$apr1$UxOdpNtW$funTxZxL/8y3m8STvonWj0'
salt[2]='UxOdpNtW'
test[3]='$apr1$w7YNTrjQ$0/71H7ze5o9/jCnKLt0mj0'
salt[3]='w7YNTrjQ'
test[4]='$apr1$AIw2h09/$Ti0TRlU9mDpCGm5zg.ZDP.'
salt[4]='AIw2h09/'
test[5]='$apr1$048HynE6$io7TkN7FwrBk6PmMzMuyC.'
salt[5]='048HynE6'
test[6]='$apr1$T2QG6cUw$eIPlGIXG6KZsn4ht/Kpff0'
salt[6]='T2QG6cUw'
test[7]='$apr1$2aLkQ0oD$YRb6aFYMkzPoUCj70lsdX0'
salt[7]='2aLkQ0oD'
while read line
do
if [ "${#line}" == 7 ]
then
for num in {1..7}
do
noob=$(openssl passwd -apr1 -salt $salt[$num] $line)
if [ "$noob" == "$test[$num]" ];
then
echo $line;
fi
done
fi
done < /usr/share/dict/words
```

I read the `/user/share/dict/words`

that contains a simple dictionary of words on Unix, I try only the 7-letters-words.

The test ran in a few minutes and gave me nothing.

Well, I guess with a 7 letters password they must have used gibberish words. Let's try a real bruteforce:

```
for a in {a..z}
do
for b in {a..z}
do
for c in {a..z}
do
for d in {a..z}
do
for e in {a..z}
do
for f in {a..z}
do
for g in {a..z}
do
truc=$a$b$c$d$e$f$g;
for num in {1..7}
do
noob=$(openssl passwd -apr1 -salt $salt[$num] $truc)
if [ "$noob" == "$test[$num]" ];
then
echo $truc;
fi
done
done
done
done
done
done
done
done
```

It ran and ran and... nothing.

Well. Let's not spend too much on this. There is John The Ripper that does this well and even oclHashcat that does this with the GPU.

Let's create a `john.conf`

with the following to limit the password to 7 letters:

```
[Incremental:Alpha7]
File = $JOHN/alpha.chr
MinLen = 7
MaxLen = 7
CharCount = 26
```

Let's launch John:

`john -i=Alpha7 hackingweek.txt`

(don't forget to put the hashed password in hackingweek.txt).

Wait and wait and wait.. and get the passwords =)

I got asked this question in an interview. And I knew this question beforehands, and that it had to deal with hashtables, but never got to dig into it since I thought nobody would asked me that for a simple internship.

I didn't know how to answer, in my mind I just had a simple php script that would have looked like this:

```
$arr = array(-5, 5, 3, 1, 7, 8);
$target = 8;
for($i = 0; $i < sizeof($arr) - 1; $i++)
{
for($j = $i + 1; $j < sizeof($arr); $j++)
{
if($arr[$i] + $arr[$j] == $target)
echo "pair found: ${arr[i]}, ${arr[j]}";
}
}
```

But it's pretty slow, it's mathematically correct, but it's more of a CS-oriented question. How to implement that quickly for machines? The answer is **hash tables**. Which are implemented as **arrays** in PHP (well, arrays are like super hash tables) and as **dictionaries** in Python.

I came up with this simple example in python:

```
arr = (-5, 5, 3, 1, 7, 8)
target = 8
dic = {}
for i, item in enumerate(arr):
dic[item] = i
if dic.has_key(target - item) and dic[target - item] != i:
print item, (target - item)
```

- iterate the list
- assign the hash of the value to the index of the value in the array
- to avoid finding a pair twice, we do this in the same for loop:

we do the difference of the target sum and the number we're on, we hash it, if we find that in the hash table that's good!
- but it could also be the number itself, so we check for its index, and it has to be different than its own index.

VoilĂ !
We avoid the n-1! additions and comparisons of the first idea with hash tables (I actually have no idea how fast they are but since most things use hash tables in IT, I guess that it is pretty fast).

One last exam, ECC, and then I'm free to do whatever I want (no I still haven't found an internship, but I talked with **TrueVault**, **Cloudflare**, **MatterMark**, **Spotify** and maybe **Matasano** so this has been a good experience nonetheless).

I stumbled upon the notes of Ben Lynn an ex Stanford's student that took an ECC class there. They're pretty awesome and I kinda want to do something like that on this blog. Maybe next year it's a bit late for that :)

The notes are here

We're learning a lot of algorithm in my **algebre et calcul formel** class. One of them is the **Toom-Cook algorithm** used for multiplication of large integers.

I found a super simple explanation of it on a forum, it helps:

Say, we want to multiply 23 times 35.

We write,

p(x) = 2x + 3,

q(x) = 3x + 5.

We are using our realization that any integer can be written as a polynomial.

Here, p(x), represents 23, and q(x), represents 35, when x equals 10.

We write,

p(x)q(x) = r(x).

That is, p(x) times q(x), equals r(x).

So,

(2x + 3)(3x + 5) = ax^2 + bx + c = r(x).

Now,

p(0)q(0) = r(0).

So,

(2*0 + 3)(3*0 + 5) = a*0 + b*0 + c.

Therefore,

c = 15.

Now,

p(1)q(1) = r(1).

Therefore, when we do the substitutions (for x and c),

a + b = 25.

Now,

p(-1)q(-1) = r(-1).

Therefore, when we do the substitutions (for x and c),

a - b = -13.

Now, we already know c, and we just need to find a and b.

We have two linear equations and two unknowns,

a + b = *25,

a - b = -13.

We just add the two equations and we get,

2a = 12.

Therefore,

a = 6.

Now, we can substitute 6 for a in,

a + b = 25,

and we get,

b = 19.

So,

r(x) = 6x^2 + 19x + 15.

Now, we substitute 10 for x in r(x), and we are done,

r(10) = 600 + 190 + 15 = 805.

Believe it or not!