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Cryptopals Set 1 Challenge 4

Github - Lucasg/cryptopals: Matasano Crypto Challenges (implemented Mostly In C

Github - Lucasg/cryptopals: Matasano Crypto Challenges (implemented Mostly In C

Matasano crypto challenges ( ) implemented mostly in C. This is a serie of 56 technical challenges around software programming and cryptography made by some people at Matasano. Every solution can be built either on Windows or on Linux (Mint and Arch tested). The tools folder contains implementations of several standard cryptographic protocols and utilities : do not use them in production. You need to have access to the gcc toochain and basic shell commands (make, cd, sed, awk, etc.). On Windows, every challenge has been tested against the mingw32 compiler. Some challenges use pythons scripts : everything has been written for Python 3xx. Older versions may or may not work. On Linux, install pip3 : sudo (apt-get install | yum install | pacman -S) python3-pip to prevent name clashing with the system-wide pip binary. bottle web framework for challenges 31 & 32 libcurl : challenge 31 & 32 use libcurl to make requests to a remote webpage. Libcurl isn't installed by default on Windows, so you will need to download the static library compatible with your compiler (or build it yourself) and placing it in the corresponding's lib folder. Download page : On Unix systems, you can either use libcurl statically of dynamically (modify the Makefile to your needs). Don't forget to add the installed lib folder (something like /usr/lib/x86_64-linux-gnu ) in the LDFLAGS. The Makefile located in the root folder contains every useful commands, while global.mk has every project-wide variables. make 01 all will build all the targets in first challenge. make tools will build all the necessary libraries and bins in the tools folder. make exos clean will run make clean on every challenges. make exos build will run make build on every challenges, building the binary. (make whill aslo do the same) Continue reading >>

Cryptopals Set 1

Cryptopals Set 1

This is the start of a mini-series where I walk through the Cryptopals Challenges . There are 8 sets of exercises and Ill be tackling 1 per blog post. If youve already completed the challenges, this series might be useful for comparison (let me know if you have any suggestions). Ive chosen to do them in Python 2, but many of the solutions should also work for Python 3. Im doing this to better grasp practical cryptography attacks. Warning: There are spoilers (solutions) below! The very first exercise is to convert a hexadecimal string to base64. This is a trivial task using Python. hex_str = "49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d"decoded = hex_str.decode("hex")# I'm killing your brain like a poisonous mushroombase64_str = decoded.encode("base64")# SSdtIGtpbGxpbmcgeW91ciBicmFpbiBsaWtlIGEgcG9pc29ub3VzIG11c2hyb29t\n The second exercise is to write a function that takes two equal-length buffers and produces their XOR combination. The exercise provides hexadecimal strings as input so we must convert them to the raw byte representation in order to XOR them. I converted them to bytearrays but there may be a better Python type. Then, we just XOR each byte using Pythons in build XOR operator ^. def xor(b1, b2): b = bytearray(len(b1)) for i in range(len(b1)): b[i] = b1[i] ^ b2[i] return bb1 = bytearray.fromhex("1c0111001f010100061a024b53535009181c")b2 = bytearray.fromhex("686974207468652062756c6c277320657965")b = bytes(xor(b1, b2))# the kid dont playb.encode("hex")# 746865206b696420646f6e277420706c6179 This is when the Cryptopals Challenge starts to get interesting! In this exercise, the plaintext has been encrypted with one character (known as a Caesar cipher ). The goal is to find this character (the key), given a ciphe Continue reading >>

Github - Jesseemond/matasano-cryptopals: My Solutions And Explanations Of The Challenges.

Github - Jesseemond/matasano-cryptopals: My Solutions And Explanations Of The Challenges.

Interesting! Breaking it amounts to breaking a single-byte XOR cipher for the1st character of every repeated key, then for the 2nd, the 3rd, etc. Guessing the keysize is done by picking the keysize that minimizes thenormalized hamming distance (distance / keysize) between"blocks" of the repeated key. This is because we can expect 2 different blocksencrypted with the right keysize to have similar bit patterns (e.g. matchingcharacters), so minimizing the normalized hamming distance gives us thekeysize that produces the most similar blocks. We need to normalize becausee.g. a keysize of 4 would have 2 times as many bits as a keysize of 2. When using ECB, two identical plaintext blocks will encrypt to the sameciphertext block. Therefore, a block that contains duplicate plaintext blocks will containduplicate ciphertext blocks once encrypted. Our oracle checks if the ciphertext contains duplicate blocks. If it does, weconsider the ciphertext to be encrypted using ECB. Otherwise, we consider thatit used CBC. 12. Byte-at-a-time ECB decryption (Simple) Detecting the blocksize is relatively easy. We start by notinglen(ciphertext). We continue adding prefix bytes until we notice a change inlen(ciphertext). When that happens, it means that we have crossed a blockboundary and we know the blocksize to be len(after) - len(start). Using the ECB property that we exploited earlier (identical plaintext blocksencrypt to identical ciphertext blocks), we can bruteforce individual bytesat the end of blocks. For example, assume that we want to bruteforce the (unknown) plaintext:hello this is a test. At first, we want to isolate the first character (h, but we don't know thatyet) at the end of a block, through a prefix of As. Like this: We get the encrypted result of this (call this C0). We can Continue reading >>

Cryptopals Set 1 Writeup

Cryptopals Set 1 Writeup

assert(hex2b64(given_input) == expected_output) Challenge 2 asks you to implement proper xor of two equal-length buffers. A simple check for equivalent lengths is all I felt necessary here: Already Pythons expressiveness gets a chance to shine. zip() creates a pairwise iterator from two seperate iterators, making this list comprehension seem effortless. Other than that, all thats going on here is a xor operation ^ on two bytes and a bytearray conversion. As Challenge 1 stated, Always operate on raw bytes, so Ive chosen to have my solutions only accept bytes as input. This paradigm allows my solutions to be useful outside of the challenge (and follows the SRP ). assert(xor(given_input1, given_input2) == expected_output) Challenge 3 , finally we get to break real crypto. The challenge asks for us to find the correct single char XOR encryption key from a given hex encoded string. Since the key / search space is small here, were free to brute force every possible key. But how do we programmatically know when weve found the correct key? Character frequency analysis , a subfield of cryptanalysis allows us to analyze the resulting output from our brute-force attempts in the way a human would. By assigning a score to all of the outputs, we can find the best. Using this frequency chart our scoring method should be something like this: given_input = "Burning 'em, if you ain't quick and nimble\n" \ "I go crazy when I hear a cymbal".encode("ascii") expected_output = unhexlify("0b3637272a2b2e63622c2e69692a2369...") assert repeating_key_xor(given_input, "ICE") == expected_output The string multiplication / list slicing & concatenation is doing is getting the longest possible string of full keys, plus the last partial substring of the key. Which is equal in length to the original buf Continue reading >>

[cryptopals, 1.6] Cracking Vigenre Cipher

[cryptopals, 1.6] Cracking Vigenre Cipher

[Cryptopals, 1.6] Cracking Vigenre cipher This challenge asks you to decrypt a blob encrypted XORing each byte with the corresponding byte of the key. That is, to encrypt the word foo using the key bar, the procedure to obtain the ciphertext is As we can see, in general the key must be extended such to be as long as the plaintext. If we were to encrypt an HTTP response with the key bar again, we have to concatenate the key to itself until the length of the plain HTTP response is reached; thus the key must be barbarbarbarbarbar The vulnerability of this scheme is hinted by Matasano itself. If we know the size of the key |k|, we can compose |k| chunks, each encrypted using a simple substitution cipher which we can break via frequency analysis. If we were to encrypt always the same byte, say AAAAAAAA, if we pick a keysize of 5 and compute the average Hamming distance among chunks of the plaintext each keysize long, and by picking a keysize of 4 or 6 we get a higher Hamming distance, it can be that the right keysize is 5 the Hamming distance of two bit patterns is the number of bits that are different. Now we wont deal with a plaintext of AAAAAA but intuitively the same principle applies. I didnt tried to prove it but I believe one could forge a specific (plaintext,key) pair such that this analysis is hijacked to wrong conclusions. So, the idea for solving this challenge is to guess the most likely keysize, collect a keysize number of chunks of characters taken from the ciphertexts then try to solve each of them by bruteforcing the (small) keyspace and perform frequency analysis to distinguish valid (i.e. likely to be in English) plaintexts. Yay!, how do you distinguish valid plaintexts from non valid ones? The good news is that we know which are the frequencies of a rando Continue reading >>

Writeup For Cryptopals, Set 1

Writeup For Cryptopals, Set 1

Writeup of the first (warmup) set in the Matasano Crypto Challenges . This set is not supposed to take long, but it took me 5 sittings of 45 minutes apiece (mornings before work may not be the most efficient time to work on these) The task calls for fromHex and toBase64, but we end up also needing toHex and fromBase64 within the same set. Recall that hex and base64 are just different views on binary data; since were writing crypto in Haskell we use ByteString. Text is usually a more sensible interchange format for strings, but isnt very efficient to iterate over char-by-char; might as well just use String. So, the types look like fromHex :: String -> Maybe ByteStringtoBase64 :: ByteString -> String and the implementations are straightforward from Wikipedia. Check the length of each argument. If they match, run B.zipWith xor a b. We need to bruteforce some ciphertext where every byte has been XORd against a constant value. For every possible byte value [0..255], perform the decryption, and rank the result according to ASCII character frequency. I just added 1 to the score each time a human-readable character was found. Make sure you include space (0x20), known to be the most common character in most English plaintext corpus. If your code above wasnt returning the score along with the key, update it to do so. Bruteforce every line in the collection and the one with the best score is your answer. If your scoring function doesnt consider spaces youll get a wrong answer. Repeat the key until its as long as the input string, then run fixed XOR. Uses a little more space than it needs to, but quick. The hardest exercise in the set by far, despite the problem description giving you a clear set of steps. Use the length-normalised Hamming distance to guess the keysize. Contrary t Continue reading >>

Cryptopals Challenge 4 Concern

Cryptopals Challenge 4 Concern

i am not getting the desired results for Cryptopals challenge 4 set 1. The concept of the program to check to see if any of these 300ish strings have been XORd by a single character. So with a brute force, my solution is take every string, XOR it with every character on the keyboard, and check to see if any of these results produce an english sentence. if not, then check the next string. Here is my code: MY_DICT = {}index = 0my_plaintext = "Now that the party is jumping"#fills the dictionary with hex strings from the txt filewith open("hexstrings.txt") as f: my_list = f.readlines() for x in my_list: MY_DICT[index] = x.rstrip('\n') index = index + 1i=0input() #this is just here to help me keep track of where i am when running it#this loop fills possible_plaintext with all the possible 255 XORs of the i'th string#of the dictionary that was previously filler from the txt filefor i in range(326): possible_plaintexts = brute_force_singlechar_xor(MY_DICT[i]) print(possible_plaintexts) if possible_plaintexts == my_plaintext: #line of concern print("ya found it yay :) ") Im sure that myBruteForce function works because it worked properly on the last problem where i XORd every possible char against a string. and i also know that the plaintext is the one provided bc i saw the solution. im just not sure why my program isnt recognizing that the plaintext is not in the dictionary. (i am aware that using a scoring system to score every string to see if its close to english would be easier, but this is the way i chose to do it for now until i figure out how to get my scoring function to work /: ) Continue reading >>

Matasano Crypto Challenges, Set 1

Matasano Crypto Challenges, Set 1

The first challenge is pretty straightforward, using pythons built-in functions or pwntools , as I use heavily in these challenges. The functions a2b_hex, unhexlify, and ''.decode("hex") all do the same thing. They take in an ascii string of the form [0-9a-f]* and decode it. The string decodes to I'm killing your brain like a poisonous mushroom. We can then use binascii.b2a_base64() to convert the byte string to a base64 representation. This challenge uses iterators to do a fixed-length xor. The chr() function takes an ASCII integer code in the range 0-255 and returns a character string. The ord() function does the inverse, taking a character and returning its integer. We want to xor these integers. Thus, ''.join([chr(ord(i)^ord(j)) for i, j in zip(s1, s2)]) will first xor the integers of each character of the input strings with ord(). Then it will convert back into a character string with chr(). Extending the last challenge, we can create a variable-length input xor function. Or, we can use pwntools xor(). I iterate through all 255 possible single character xor keys, scoring each decrypted word based on character frequency. cand_words.append((result, word_score, char)) Challenge 1.4 Detect single-character XOR This challenge asks us to find which of the 60-character strings in the file has been encrypted by single-character XOR. Using the find_singlechar_key_xor utility function created in the last challenge, I can iterate through all the file strings and find the possible single-character xor key and word score of each line. The line turns out to decode to Now that the party is jumping\n, xored against the byte 5. Challenge 1.5 Implement repeating-key XOR This is a cop-out. Pwntools implements repeating-key xor in its utils.fiddling library. We can use python itertoo Continue reading >>

Cryptopals Walk-through Series Set 1 Challenge 6

Cryptopals Walk-through Series Set 1 Challenge 6

0000 0000 0000 1010 < We end up with this 16 bit number ( 10d, since 5d + 5d = 10d) Since our intention was to end up with a 16 bit sequence like this: 0000 0000 0101 0101, adding them directly does not help us. Instead we must shift the bits first, in order to arrive at an addition operation such: 0000 0000 0101 0000 < (Shifted to the left by 4) 0000 0000 0101 0101 < Our desired result ( 85d) In our case this is exactly what we are doing in lines 344 to 351. Next we put into motion a reversed process of challenge 1 , and AND our newly created 64 bit string (48 bits total, comprised of eight 6 bit sequences, resulting from 8 base64 characters read), 6 times against the respective position of the sequence we want to extract. Those values are then stored into the b64decryptStr array and are now hexadecimal representatives of the base64 string. (f was originally being returned to utilize the length of the block, which I later found was unnecessary for the later steps of the exercise, the value is discarded upon returning to main() ) Now that we have accomplished the base64 decoding, the challenge page goes on to detail the method we must use to find the key and decrypt the message. First we must devise a function to calculate the hamming distance of two strings. I would imagine there are probably multiple ways to do this, but a simple implementation of this function can be done as follows: Take two strings to compare as parameters, find the length of the block using strlen(), and XOR each character of the two blocks against each other and store the result in a new array result. Next we take the result of each XORd character (iterating through the length of the block, line 290) and count the 1 bits in the byte (iterate through the bits in the byte, line 292). We do this by Continue reading >>

Cryptopals Challenge 4: Detect Single-character Xor Encryption

Cryptopals Challenge 4: Detect Single-character Xor Encryption

Cryptopals challenge 4: Detect single-character XOR encryption This is my write up of the fourth Cryptopals challenge , using Python3 as my language of choice. The challenge: One of the 60-character strings in this file has been encrypted by single-character XOR. I believe there are two ways to solve this challenge: 1) Read the file line be line and detect the string that is single-byte XOR encrypted, then pass that single string to a function that does a single-byte XOR brute force; or 2) Read the file and perform a single-byte XOR brute force on each string, selecting the string with the highest English frequency score. I solved the challenge both ways: Detect the string in the file that is single-byte XOR encrypted I had previous written about a tool I made to help detect encryption by analyzing byte positions, so I attempted to use that for this challenge. Unfortunately, by looking at each line in the file and looking at the number of missing bytes, the overall entropy of each line looked pretty much the same: I decided to put all of these values in a list and sort the list to check for outliers. Here is how this looks in Python: # Create a list to store the line number and entropy results for each linedistribution_list = []# Iterate and analyze each line in the filefor item in data: # Store the results in a dictionary results = {"line": line_number, "missing bytes": missing_byte_count} # Append the dictionary to the list created outside the for loop distribution_list.append(results) # Sort the distribution_variable by the key missing bytes in the results dictionary, # only providing the top result most_missing = sorted(distribution_list, key=lambda x: x['missing bytes'], reverse=True)[0] least_missing = sorted(distribution_list, key=lambda x: x['missing bytes'])[0 Continue reading >>

Cryptopals Walk-through Series Set 1 Challenge4

Cryptopals Walk-through Series Set 1 Challenge4

Cryptopals Walk-through Series Set 1 Challenge4 Welcome back to the fourth write up of the ongoing series through the matasano cryptopals challenges! Our next task isnt too much different from the one before it, as we are tackling another single character XOR cipher. Set 1 Challenge 4 Detect single character XOR The source code for this challenge can be found here In this challenge, we are given a file of about a few hundred hexadecimal strings. One of them has been encrypted with a single character XOR. In order to find this string, we are going to implement a near copy of our code from the previous challenge, and modify to handle multiple strings. Here we declare several familiar buffers: buf1 to hold the contents of the input string decbuf to hold the byte conversion of the input string tmessage to hold the decrypted plain-text for a particular key hexDecode to store the byte converted input string Essentially the same as the last challenge. Here we read from the file, using a filename supplied by a command line argument. We also declare an array to hold all of the input strings that are extracted from the file. Stepping into our read function, we get each line until the file reaches EOF, storing them into our filecontents array. We now get each string into its hexadecimal equivalent, using the same buffering + strtol() process as in previous exercises. We now return to our scoring method like before, adding points to our string score based on the position found (or not found) in the *freq literal, containing the list of the most common English letters. If the result of the XOR with the key is not found in the list, the score is hit hard with 255 points, surely making it impossible to match or go below the score of the actual plain text string. This process continue Continue reading >>

Cryptopals Set 1 Challenge 3: Single-byte Xor Cipher : Crypto

Cryptopals Set 1 Challenge 3: Single-byte Xor Cipher : Crypto

Don't forget to read our RULES PAGE ! The rules listed there are also used as this sub's report reasons. The quick version; Assume good faith and be kind. This is a friendly subreddit. Codes, simple ciphers, ARGs, and other such "weak crypto" don't belong here. (Rule of thumb: If a desktop computer can break a code in less than an hour, it's not strong crypto.) You're probably looking for /r/codes . Do not ask people to break your cryptosystem without first sharing the algorithm. Sharing just the output is like... "Crack this cipher" challenges also belong in /r/codes unless they're based on interesting crypto implementation bugs, e.g. weak RSA keys. Familiarize yourself with the following before posting a question about a novel cryptosystem, or else the risk is nobody will take their time to answer: Don't use this sub to cheat on competitions or challenges! You're expected to solve such problems yourself. You may ask for help to understand it, but you should disclose the source. Systems that use crypto are not necessarily relevant here, e.g. Bitcoin. Maybe try /r/cryptocurrency ? Political news also very rarely belong here. See the list of related subs below for alternatives. Remember that this sub is focused on the algorithms, and isn't political. Continue reading >>

Working Cryptopals Set 1 Challenges In F#

Working Cryptopals Set 1 Challenges In F#

Working CryptoPals Set 1 challenges in F# The CryptoPals challenges are a great hands-on introduction to cryptography and cryptanalysis . This post will outline some of my approaches to the challenges, written in F#. My solutions for Set 2 challenges are in this post . This one is easy if you know how hexadecimal representation works . Well go from hex to bytes, then from bytes to base 64. This function converts one hex char to a byte. There are BCL methods to do that for you, but redundancy can be educational. let hexToByte = function | '0' -> 0uy | '1' -> 1uy | '2' -> 2uy | '3' -> 3uy | '4' -> 4uy | '5' -> 5uy | '6' -> 6uy | '7' -> 7uy | '8' -> 8uy | '9' -> 9uy | 'a' -> 10uy | 'b' -> 11uy | 'c' -> 12uy | 'd' -> 13uy | 'e' -> 14uy | 'f' -> 15uy | _ -> failwith "Invalid hexadecimal character" Astute observers might say thats only using half the byte, stop wasting my electricity but thats why we have this function: let hexToBytes (hexChars: string) : byte array = [| for i in 0 .. 2 .. hexChars.Length - 1 do let hi = hexToByte hexChars.[i] let lo = hexToByte hexChars.[i+1] yield (hi <<< 4) ||| lo |] A byte is 8 bits. A hex digit represents 4 bits, affectionately referred to as a nibble. You can fit two hex digits in a byte by storing one hex digit on the left 4 bits, and another on the right. Shifting hi <<< 4 scoots the bits over four places, e.g. 0000 1010 becomes 1010 0000. ORing with the other 4 bits simply sets the low bits, e.g. 1010 0000 ||| 0000 1100 becomes 1010 1100. Can you smell the silicon yet? I didnt bother implementing base 64 conversion. let bytes = hexToBytes "49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d"Convert.ToBase64String bytes // expected: SSdtIGtpbGxpbmcgeW91ciBicmFpbiBsaWtlIGEgcG9pc29ub3VzIG11c Continue reading >>

Cryptopals Challenges

Cryptopals Challenges

A collection of 48 exercises that demonstrate attacks on real-world crypto. This is a different way to learn about crypto than taking a class or reading abook. We give you problems to solve. They're derived from weaknesses inreal-world systems and modern cryptographic constructions. We give you enoughinfo to learn about the underlying crypto concepts yourself. When you'refinished, you'll not only have learned a good deal about how cryptosystems arebuilt, but you'll also understand how they're attacked. This is the qualifying set. We picked the exercises in it to rampdevelopers up gradually into coding cryptography, but also to verify that wewere working with people who were ready to write code. This set is relatively easy. With one exception, most of these exercisesshould take only a couple minutes. But don't beat yourself up if it takeslonger than that. It took Alex two weeks to get through the set! If you've written any crypto code in the past, you're going to feel likeskipping a lot of this. Don't skip them. At least two of them (we won'tsay which) are important stepping stones to later attacks. SSdtIGtpbGxpbmcgeW91ciBicmFpbiBsaWtlIGEgcG9pc29ub3VzIG11c2hyb29t So go ahead and make that happen. You'll need to use this code for the restof the exercises. Cryptopals Rule: Always operate on raw bytes, never on encoded strings. Only use hex and base64 for pretty-printing. (ns hale.cryptopals.set1.challenge1 (:require [clojure.set :refer [map-invert]] [clojure.string :as str] [hale.cryptopals.utils :as utils] [clojure.test :as t])) Axiomatic Base64 map of six-bit bytes (indices) to ASCII characters The following table visually explains the process of Base64 encoding: Get the underlying bytestream of the input data. (In the table they areencoding the ASCII word "Man" whereas Continue reading >>

Matasano Solutions

Matasano Solutions

Original Challenge Description from matasano In this file are a bunch of hex-encoded ciphertexts. Remember that the problem with ECB is that it is stateless and deterministic; the same 16 byte plaintext block will always produce the same 16 byte ciphertext. Since one byte represents two nibbles, in our give file we need to find duplicate patterns in the the block of 32. "fold" command is our friend Now what we need to do is get these blocks for every line of the input and see if there are any duplicates. "sort" and "wc" can be used in a loop. Original Challenge Description from matasano Here is the opening stanza of an important work of the English language: Burning 'em, if you ain't quick and nimbleI go crazy when I hear a cymbal Encrypt it, under the key "ICE", using repeating-key XOR. In repeating-key XOR, you'll sequentially apply each byte of the key; the first byte of plaintext will be XOR'd against I, the next C, the next E, then I again for the 4th byte, and so on. 0b3637272a2b2e63622c2e69692a23693a2a3c6324202d623d63343c2a26226324272765272a282b2f20430a652e2c652a3124333a653e2b2027630c692b20283165286326302e27282f Encrypt a bunch of stuff using your repeating-key XOR function. Encrypt your mail. Encrypt your password file. Your .sig file. Get a feel for it. I promise, we aren't wasting your time with this. Now lets create the file with given stuff Based on the first and last few bytes it looks good but lets make sure it really matches with given output. We will copy & paste the given output in a file and then compare with what was generated by "my-encrypt" Original Challenge Description from matasano One of the 60-character strings in this file has been encrypted by single-character XOR. As the hint suggested previous code would help. I realized that my English sc Continue reading >>

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