Problem

C? Who uses that anymore. If we really want to be secure, we should all start learning lambda calculus. http://2018shell2.picoctf.com:41367

Solution

An extremely large payload to the interpreter results in an node.js error message

PayloadTooLargeError: request entity too large
    at readStream (/problems/lambdash-3_0_867a993b23b277b2e144cc3e2d73f6e4/node_modules/raw-body/index.js:155:17)
    at getRawBody (/problems/lambdash-3_0_867a993b23b277b2e144cc3e2d73f6e4/node_modules/raw-body/index.js:108:12)
    at read (/problems/lambdash-3_0_867a993b23b277b2e144cc3e2d73f6e4/node_modules/body-parser/lib/read.js:77:3)
    at urlencodedParser (/problems/lambdash-3_0_867a993b23b277b2e144cc3e2d73f6e4/node_modules/body-parser/lib/types/urlencoded.js:116:5)
    at Layer.handle [as handle_request] (/problems/lambdash-3_0_867a993b23b277b2e144cc3e2d73f6e4/node_modules/express/lib/router/layer.js:95:5)
    at trim_prefix (/problems/lambdash-3_0_867a993b23b277b2e144cc3e2d73f6e4/node_modules/express/lib/router/index.js:317:13)
    at /problems/lambdash-3_0_867a993b23b277b2e144cc3e2d73f6e4/node_modules/express/lib/router/index.js:284:7
    at Function.process_params (/problems/lambdash-3_0_867a993b23b277b2e144cc3e2d73f6e4/node_modules/express/lib/router/index.js:335:12)
    at next (/problems/lambdash-3_0_867a993b23b277b2e144cc3e2d73f6e4/node_modules/express/lib/router/index.js:275:10)
    at expressInit (/problems/lambdash-3_0_867a993b23b277b2e144cc3e2d73f6e4/node_modules/express/lib/middleware/init.js:40:5)

Problem

Can you crack the key to decrypt map2 for us? The key to map1 is 11443513758266689915.

Hint

z3

Solution

Given the problem and the hint, it is clear that we can use z3 to solve this problem. We can create a z3 BitVec and pass it into the verify function to avoid writing a custom decrypter. Because we don't know the length of the bit vector, I used a conservative estimate of 128.

Problem

As the song draws closer to the end, another executable be-quick-or-be-dead-3 suddenly pops up. This one requires even faster machines. Can you run it fast enough too?

Solution

After decompiling the program with Snowman, we can see pseudocode for the calc function:

uint32_t calc(uint32_t edi) {
    uint32_t eax2;
    uint32_t eax3;
    uint32_t eax4;
    uint32_t eax5;
    uint32_t eax6;
    uint32_t v7;

    if (edi > 4) {
        eax2 = calc(edi - 1);
        eax3 = calc(edi - 2);
        eax4 = calc(edi - 3);
        eax5 = calc(edi - 4);
        eax6 = calc(edi - 5);
        v7 = eax6 * 0x1234 + (eax2 - eax3 + (eax4 - eax5));
    } else {
        v7 = edi * edi + 0x2345;
    }
    return v7;
}

Problem

Dressing up dogs are kinda the new thing, see if you can get this lovely girl ready for her costume party. Dog Or Frog

Solution

When we visit the page, we are greeted with a submission form. The instructions tell us that this is a machine learning problem and to not approach it as a web-exploitation problem. The goal is to submit an image that is similar (on a byte-by-byte basis) that is similar to the dog, but is recognized as a tree frog.

Problem

"You came across a custom server that Dr Xernon's company eleCTRic Ltd uses. It seems to be storing some encrypted files. Connect with nc 2018shell2.picoctf.com 15037. Can you get us the flag?" Source

Solution

The title makes a clear reference to AES-CTR. We can see that this mode of AES requires a unique nonce for each encrypted value to remain cryptographically secure. If we look at the problem's source code, we can see that the counter remains constant for all values. This breaks the encryption used and we are able to reveal the keystream and encrypt arbritary data.

class AESCipher(object):
    def __init__(self):
        self.bs = 32
        random = Random.new()
        self.key = random.read(AES.block_size)
        self.ctr = random.read(AES.block_size)

    def encrypt(self, raw):
        cipher = AES.new(self.key, AES.MODE_CTR, counter=lambda: self.ctr)
        return cipher.encrypt(raw).encode('base64').replace('\n', '')

Problem

"There seem to be a few more files stored on the flash card server but we can't login. Can you?"

Solution

The site appears to be the same as the previous flaskcard challenges Flaskcards Skeleton Key and Flaskcards.

When we create an account, we are presented with the following screen:

We can try server side injection. If we type in {{1+1}}, the webpage displays 2. Any values in double brackets is being executed on the server. After trying to find hidden variables, such as config, flag, etc; I decided to look for a remote code execution vulnerability.