# Cheatsheet - Crypto 101

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## Crypto 101

**This is a beginner guide, not an Academic paper
Also this is very light an CTF-specific. If you are interested in Crypto check out crypto101.io**

**This Cheatsheet will be updated regularly**

When you are trying to solve a Crypto Challenge for a CTF, first of all, you need to detect which Cipher is used.

For example if it’s a Symmetric or Asymmetric cipher, a Classic cipher or if it’s just an Hash.

*Keep in mind that CTFs are meant to be broken so think before implementing a bruteforce over an AES-128 key.
Bruteforce may be present but only over a limited key-space or time-span*

**NOTE: Most of the time the encrypted message is the flag, and most of the time the flag is in a known format like ctfname{data} so this will leak part of the plaintext. Be smart**

### Misc

If your ciphertext has lots of numbers and `@\>%`

, don’t worry it’s **rot47**

If your ciphertext has only uppercase letters and numbers test it as **base32**
If your ciphertext has alphanumeric chars and ends with `=`

or `==`

it’s **base64**
It might be that they used their own alphabet for the base64. Check this page to better understand this.
Always test for **rot13** if the challenge gives only few points.

### Hash

There are some tool to detect the hash type (if it’s not known), but a search on DuckDuckGo with the prefix `hash`

will also do the trick.

If organizers are using a custom hashing algo you won’t find it, but I hope you will find out it’s custom (and broken).

Now that you know what hashing algo it is, most of the time you will need to find some sort of collision.

#### Collision

If it’s a (not-full) prefix/postfix/pattern collision, you can solve it by **bruteforce**.

Sometimes this falls under the term PoW (Proof-of-Work).

e.g. `Calculate a string str such that sha1('SaltyToast'+str) start with '00000'`

See this python script that solve the example above

#### Merkle-Damgård based hash functions

Merkle-Damgård based hash functions (MD5, SHA1, SHA2) uses this construction:

##### Reusable hash collision

Let `a`

and `a’`

be strings with `length(a) = length(a’)`

= multiple of input block length,

Let `H0`

be the hash function without length padding

Let `||`

be the concatenation function

`H0(a) = H0(a’)`

implies `H(a || b) = H(a’ || b)`

for an arbitrary string `b`

Then the pair `a`

/`a’`

is a reusable hash collision

In this case if there is a public known existing collision, you can **append** the same message to get another collision.

e.g. With this Python script

See this SharifCTF 7 Challenge

##### Length Extension

Given the hash `H(X)`

of an unknown input `X`

, it is easy to find the value of `H(pad(X) || Y)`

, where `pad`

is the padding function of the hash. That is, it is possible to find hashes of inputs related to `X`

even though `X`

remains unknown.

A great tool for performing automatic Length Extension attack is HashPump.

Alternatively if the challenge is using a custom hash or if you want to implement your own length ext. attack you can follow this writeup from team **spritzer**.

I will report and decribe here their code for the extend function:

*NOTE: SLHA1 is a custom hashing algorithm derived from SHA1*

The extend code takes as input the old *digest*, the plaintext *length*, and the data to extend as *ext*.

In the first 3 lines we are reconstructing the padding that was applied to the text before the old digest.

In the fourth line we create a new SLHA1 object.

In the sixth-seventh line we use library internals to recreate the internal state from the old digest and apply the length

Then we perform the update with new data and we return the newly appended data (`pad + ext`

) and the new digest.

#### MD5

Some tools for generating MD5 collision (hashclash, fastcoll, SAPHIR-Livrables) or SingleBlock collision (very HOT).

### Classical

#### Caesar cipher

##### Caesar KCA

On Caesar you can try a *CiphertextOnly Attack (COA) aka Known Ciphertext Attack (KCA)* by analizing the statistics of each character
I’ve made a website that does this automatically by comparing the IC of each letter with ones from
Italian and English dictionary.

**This will only works if your alphabet is limited to**: `A-Z`

or `a-z`

An alternative can be xortool.

##### Caesar KPA

Also Caesar can be broken with a *Known-plaintext attack (KPA)*.

If some messages are encrypted with the same key, you can recover the key from a Plaintext-Ciphertext pair and decrypt the other messages.
(or part of messages).

e.g. have the following ciphertext: `dahhksknhzwoaynapodkqhznaiwejoaynap`

, we know that the crib (plaintext) starts with `helloworld`

.

We can see that shifting by `4`

return `dahhksknhz`

, the starting ciphertext. So now we know that the key is shift by `4`

and we can decrypt all the message. (Try it yourself ;)

Another demo is available here in this SECCON2016 challenge

##### Caesar CPA

In a *Chosen-plaintext Attack (CPA)* scenario, where you can input a plaintext in a Caesar encryption oracle, remember that shifting `A`

by `C`

will result in `C`

, so a plaintext made of `A`

’s will expose the Key as ciphertext.

This also works as a *Chosen-ciphertext Attack (CCA)*
Like in this HackThatKiwi2015 CTF challenge

#### Vigenere cipher

Vigenere is done by repeating Caesar for each letter in the plaintext with a different shift according to the key.
You can try a *COA* once you have guessed the keylength (with Kasisky).

Or You can try a *KPA* or *CPA/CCA* like in Caesar.

Also sometimes you will need to bruteforce a partial key like in this SECCON2016 challenge

### Symmetric

#### XOR

See Caesar/Vigenere section

The XOR (symbol `⊕`

) is based on the Exclusive disjunction. (returns `1`

only if inputs are different)

So note this rules:

`A ⊕ 0 = A`

`A ⊕ A = 0`

`(A ⊕ B) ⊕ A = B`

XORing a Plaintext with a Key will output a Ciphertext that XORed again with the Plaintext will return the Key.

#### SBox-Based Block Cipher

Block cipher Analysis example SharifCTF 7 - Blobfish

#### Feistel

A reverse-crypto SECCCON2016 challenge for Feistel

#### ECB Mode

ECB is the basic of block cipher mode. Every block of plaintext is encrypted with the same key.

This cause that identical plaintext blocks are encrypted into identical ciphertext blocks; thus, it does not hide data patterns well.

##### Replay Attack (Known Block Ciphertext)

Also if you know what Plaintext resulted in a certain Ciphertext, you can replay that Ciphertext or when you see that Ciphertext you know what was the Plaintext. (Only when the Key is constant!)

e.g. In this ABCTF2016 Challenge, you are given a bunch of AES-128 ECB Plaintext-Ciphertext pairs, and the encrypted flag `e220eb994c8fc16388dbd60a969d4953f042fc0bce25dbef573cf522636a1ba3fafa1a7c21ff824a5824c5dc4a376e75`

You can divide the encrypted flag into 16-byte blocks:

```
e220eb994c8fc16388dbd60a969d4953
f042fc0bce25dbef573cf522636a1ba3
fafa1a7c21ff824a5824c5dc4a376e75
```

And search each block individually in the Plaintext-Ciphertext pairs. This will give you the 16-byte corresponding plaintext.

##### Encryption Oracle Attack

Alternatively, if you have an **Encryption oracle Attack** available (that always use the same key) and the service append to your input a secret (**secret suffix**), you can go for a *CPA*.

You start by sending a message that is `1 byte`

shorted then the `block length`

.

In this case the first character of the secret will fall in the first block,

Save the encrypted value of the first block, then send a new message like the first one but appending a random final byte.

You can then bruteforce the final byte.

When the encrypted value match the saved one, you have guessed the fisrt secret character.

e.g. Let’s use AES-128 ECB that has `16 byte`

blocks

We send the garbage message length `15`

-> `AAAAAAAAAAAAAAA`

The server will reply with `Enc('AAAAAAAAAAAAAAA'+secret)`

but the first block is equal to `Enc('AAAAAAAAAAAAAAA'+secret[0])`

.

So now we can bruteforce the last character until the ciphertext match

```
AAAAAAAAAAAAAAAa
AAAAAAAAAAAAAAAb
AAAAAAAAAAAAAAAc
...
```

This method will take max `256`

attempt for the secret’s length, **4096 tries for a 16 byte blocks**

There is a script that do this for the EncryptionService - ABCTF2016 challenge

#### CBC Mode

In CBC mode, each block of plaintext is XORed with the previous ciphertext block before being encrypted. This way, each ciphertext block depends on all plaintext blocks processed up to that point. To make each message unique, an initialization vector (IV) must be used in the first block.

##### Predictable IV Attack

In a CBC Encryption oracle, If you can predict the IV for the next message (or simply it’s a fixed IV) you can verify if a previous ciphertext comes from a guess plaintext.

You have a ciphertext block `C1`

and it’s IV `IV1`

, you want know if that is the encryption of the plaintext `P1`

.

If you can predict the next IV (`IV2`

), you can made up your plaintext like: `P2 = P1 ⊕ IV1 ⊕ IV2`

.

So you can send this new message to the server that encrypt it this way:

`C2 = Enc(k,IV2 ⊕ P2)`

but with the forged `P2`

this become

`C2 = Enc(k,IV2 ⊕ P1 ⊕ IV1 ⊕ IV2) = Enc(k,P1 ⊕ IV1)`

.

`C2 == C1`

only if you have guessed `P1`

.

##### key used as the IV

IV is not meant to be secret, because you can easily recover it.

If you have a ciphertext block, you can recover the key by letting someone decrypt `C = C1 || Z || C1`

(where `Z`

is a block filled with null/`0`

bytes)

The decrypted blocks are the following

```
P1 = D(k, C1) ⊕ IV = D(k, C1) ⊕ k = P1
P2 = D(k, Z) ⊕ C1 = R
P3 = D(k, C1) ⊕ Z = D(k, C1) = P1 ⊕ IV
```

`R`

is a random block that we can throw away.
Finally we can recover the IV with `P1 ⊕ P2 = P1 ⊕ P1 ⊕ IV = IV`

##### Bit Flipping attacks

Note that if we XOR a ciphertext block, the next plaintext block will be XORed as well (X propagates like in the image).

If you have control over the plaintext `P`

, you can fill a block `P[i]`

with filler data `Z`

that will be encrypted.

Once it’s encrypted you must find that ciphertext block `C[i]`

and replace it with a XORed version of `C[i]`

, `Z`

and your desired value `G`

, like `X = C[i] ⊕ Z ⊕ G`

.
When the server will decrypt the ciphertext, will get a different plaintext `P’`

where `P’[i]`

will be indecipherable nonsense, but `P’[i+1]`

will be your desided value `G`

.

```
P’[i+1] = P[i+1] ⊕ X
= P[i+1] ⊕ Z ⊕ G
= Z ⊕ Z ⊕ G
= G
```

e.g. This scenario is common in website where cookies that are stored encrypted in the client and the server will decrypt them for every request.

You can select a very long username or password that will act as `Z`

and set `G`

to something like `;admin=true;`

. :wink: