n00b19CTF(Easy-Flipp, 100 pt) This is simple crypto challenge created by me. This past weekend, Sarah and I took part in the first CTF competition hosted by the cybersecurity company Tenable. Participated in CTF events. And of course, always … The encryption is AES-ECB of a nonce, which is then XOR’d with the blocks of the message. Poor encryption. python3, from Crypto.Util.Padding import padimport stringfrom pwn import remote, log, ATTEMPTS = 10CHARS = string.printable[:-6]CHARSX2 = [c1+c2 for c1 in CHARS for c2 in CHARS]PRE_FLAG = "utflag{"POST_FLAG = guessed = "}"push = 2 + len(guessed), def attempt(payload): p.recvline() p.sendline(payload) p.recvline() res = p.recvline()[:-1] return res[:16*2] == res[48*2:64*2] # NOTE: response is in hexadecimal format, two digits for each byte, def prepare_payload(guess): first_block = pad(guess.encode(), 16) if len(guess)<16 else guess[:16].encode() return first_block + push*b'P', progress = log.progress('FLAG')for pos in [i for i in range(len(guessed)+1, 24) if i!=7]: chars, push = (CHARS, push+1) if pos!=6 else (CHARSX2, push+2) for c in chars: progress.status(c+guessed) payload = prepare_payload(c+guessed) res = False for j in range(ATTEMPTS): if attempt(payload): res = True break if res: guessed = c+guessed break else: progress.failure(guessed) break, p.sendline("q")progress.success(PRE_FLAG+guessed)```. That means that the encrypted data can be changed in a controlled way, and the decrypted version of the data will reflect that change. CTF events / UTCTF 2020 / Tasks / Random ECB / Writeup; Random ECB by FedericoCioschi / Salted Crhackers. m0leCon CTF 2020 Teaser — King Exchange. The service is an ECB encryption oracle that randomly prefixes the plaintext with an 'A' with a probability of ~1/2. 16 random messages are generated and encrypted and signed. decrypt (iv2) key2 = xor (to_blocks (ctxt)) return str (base64. we are given a source code , and a listen server: nc core.ccug.my.id 39002, Source : chall.py. Crypto-CSAW-CTF-2017-Baby-Crypt. The 128 in AES-128 indicates that the encryption is using 16 byte blocks (16 * 8 = 128). Hosting provided by Transdata. Let’s take a look at what’s going on. These two challenges were part of the side channels category of Ledger Donjon CTF, and involved exploiting fault attacks. aes ecb crypto csaw. ... │ │ 4 But that looks random to another person. When an input string like `s0m3_t3xt` is sent to the server, this is how plaintext looks like before server-side encryption: | B0 | B1 | B2 | B3 | B4 | B5 | B6 | B7 | B8 | B9 | B10 | B11 | B12 | B13 | B14 | B15 || :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: || **s** | **0** | **m** | **3** | **_** | **t** | **3** | **x** | **t** | u | t | f | l | a | g | { || F | L | A | A | A | A | A | A | A | A | A | A | A | A | A | A || A | A | A | A | A | G | } | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* |. khash was then used as a key to encrypt the msg containing the IV and key used in the file encryption. [Pwn] Pwn2Win 2019 CTF - Random Vault 2019-11-11 Pwn x64 Format String formatstring , pwn2win2019 Comments Word Count: 1,528 (words) Read Time: 10 (min) Random Vault To exploit it, we use a slightly modified version of the ECB byte at a time attack.. We see that for any input I we provide to the oracle, we will get the encryption of either I+flag or 'A'+I+flag. The text sent to the server is appended before/in front ofthe flag. Starting Variable (SV): Similar to the IV, except it is used during the first block to provide a random seed during encryption The first step in attacking a block-based cipher is to determine the size of the block. Finally, the khash and the encrypted msg are encoded to hex, concatenated, encoded to base64, … The Remote class handles setting up a listening server on a randomized port for you; all we must do is specify the program_name to execute when receiving a connection.. Country place: 22. 2019-10-21 At the time of CTF, this challenge had zero solved. ... For AES-OFB mode, the first step is the same as AES-CBC mode, we select a random IV of 16 bytes. This block is the encryption of 15*'A'+flag[0]. The ciphertext being 16 bytes, it means that in fact the IV … The plaintext will be randomly concatenated to the char ‘A’. formatstring stack_overflow nullcon2020. 2018-11-26 [Crypto] TUCTF 2018 - AESential Lesson. The guys over at Matasano have made a set of 48 challenges that show attacks on real-world crypto. Classical cryptography. First of all, it is necessary to discover the flag length.I have used this simple script.```python#! So in order to get a crib for each byte, we first begin by seeing what the encryption of a full block of 'A's looks like. So the idea is to exploit this weakness to brute force flag bytes one by one. Follow @CTFtime !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="//platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); All tasks and writeups are copyrighted by their respective authors. Replay Attack (Known Block Ciphertext) Crypto 11; Electronics 1; Forensics 9; Misc 6; Network 2; Pwn 34. So, I thought to write a writeup for beginners to understand how CBC bit flipping works. That is, sometimes the program will encrypt be ‘A’+plaintext+flag and other simply plaintext+flag. Proceeding with the following bytes is quite simple.Therefore, to guess the second byte, you have to make sure that server-side plaintext takes this form: | | B0 | B1 | B2 | B3 | B4 | B5 | B6 | B7 | B8 | B9 | B10 | B11 | B12 | B13 | B14 | B15 || :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: || **=>** | ***GUESS*** | **}** | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* || | PUSH | PUSH | PUSH | PUSH | u | t | f | l | a | g | { | 3 | c | b | _ | w || | 1 | 7 | h | _ | r | 4 | n | d | 0 | m | _ | p | r | 3 | f | 1 || **=>** | **x** | **}** | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* |, | | B0 | B1 | B2 | B3 | B4 | B5 | B6 | B7 | B8 | B9 | B10 | B11 | B12 | B13 | B14 | B15 || :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: || **=>** | ***GUESS*** | **f** | **1** | **x** | **}** | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* || | PUSH | PUSH | PUSH | PUSH | PUSH | PUSH | PUSH | u | t | f | l | a | g | { | 3 | c || | b | _ | w | 1 | 7 | h | _ | r | 4 | n | d | 0 | m | _ | p | r || **=>** | **3** | **f** | **1** | **x** | **}** | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* | *PAD* |, Here the code able to capture the whole flag:```python#! Specifically, we can move blocks of data (16 bytes at a time) around - rearrange, delete, etc.
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