# ImaginaryCTF 2022 Crypto Writeups

## Changelog (last updated 2022-07-22)

*Update 2022-07-22: Added Poker writeup, some touch-ups here and there.*

## Poker

By StealthyDev

In the early days of online poker rooms, the game would get cracked and players would play with perfect knowledge of the upcoming handsโฆ Can you deal out the next two games correctly?

Attachments: poker.py cards.22.07.16.txt

8 solves, 495 points

## Challenge poker.py

`from random import getrandbits from math import prod HEARTS = "๐ฑ๐ฒ๐ณ๐ด๐ต๐ถ๐ท๐ธ๐น๐บ๐ป๐ฝ๐พ" SPADES = "๐ก๐ข๐ฃ๐ค๐ฅ๐ฆ๐ง๐จ๐ฉ๐ช๐ซ๐ญ๐ฎ" DIAMONDS = "๐๐๐๐๐ ๐๐๐๐๐๐๐๐" CLUBS = "๐๐๐๐๐๐๐๐๐๐๐๐๐" DECK = SPADES+HEARTS+DIAMONDS+CLUBS # Bridge Ordering of a Deck ALPHABET52 = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnop_rstuvw{y}" CARDS_PER_DEAL = 25 assert CARDS_PER_DEAL % 2 == 1 MAX_DEAL = prod(x for x in range(len(DECK) - CARDS_PER_DEAL + 1, len(DECK) + 1)) DEAL_BITS = MAX_DEAL.bit_length() def text_from_cards(string): return string.translate(string.maketrans(DECK, ALPHABET52)) def deal_game(): shuffle = getrandbits(DEAL_BITS) % MAX_DEAL deck = list(DECK) deal = "" while shuffle > 0: deal += deck.pop(shuffle % len(deck)) shuffle //= len(deck) + 1 while len(deal) < CARDS_PER_DEAL: deal += deck.pop(0) return deal def print_puzzle(): with open("cards.txt", "w") as cards_file: for i in range(750): table = deal_game() cards_file.write(f"Game {i+1}:\n") for i in range((CARDS_PER_DEAL - 5) // 2): cards_file.write(f"{i + 1}: {table[i * 2]}{table[i * 2 + 1]} ") cards_file.write("\nTable: {}{}{} {} {}\n\n".format(*table[CARDS_PER_DEAL - 5:]))`

We get a poker simulator which deals cards based on the python `random()`

PRNG and are told to predict two deals after the end of the file.

By re-running some of `poker.py`

locally, we find:

`DEAL_BITS = 133 2 ** DEAL_BITS = 10889035741470030830827987437816582766592 MAX_DEAL = 7407396657496428903767538970656768000000`

With some regex and careful implementation, we can convert the dealt cards back to a number `x`

for each deal such that `x = getrandbits(DEAL_BITS) % MAX_DEAL`

.

For each deal, one of two cases have occurred:

`getrandbits(DEAL_BITS) < MAX_DEAL`

and`x = getrandbits(DEAL_BITS)`

`getrandbits(DEAL_BITS) >= MAX_DEAL`

and`x = getrandbits(DEAL_BITS) - MAX_DEAL`

We can split the first case depending on the value of `x`

and rearrange to get three cases:

`x < 2 ** DEAL_BITS - MAX_DEAL`

and`getrandbits(DEAL_BITS) = x`

`x >= 2 ** DEAL_BITS - MAX_DEAL`

and`getrandbits(DEAL_BITS) = x`

`x < 2 ** DEAL_BITS - MAX_DEAL`

and`getrandbits(DEAL_BITS) = x + MAX_DEAL`

Note that from our perspective with only knowledge of $x$, we are unable to differentiate between case 1 or 3. If our $x$ has a value that may have come from case 1 or 3, there are two possible values of `getrandbits(DEAL_BITS)`

. In case 2, this is only one possible value of `getrandbits(DEAL_BITS)`

. We can put this into a symbolic python PRNG cracker that supports unknown bits to recover the PRNG state. For case 1 and 3, we will **only report bits which are the same** in both `x`

and `x + MAX_DEAL`

, and any disagreeing bits will be labelled as unknown. In case 2, we know that `getrandbits(DEAL_BITS) = x`

for sure and can report all bits of `x`

to the PRNG solver.

To simplify implementation, only the low 128 bits were considered when recovering PRNG output and the upper 5 bits were discarded. Some quick estimations suggest that case 2 occurs around 36% of the time, giving an expected leak of 87 bits of PRNG state per deal. After testing, around 400 deal outputs are needed as input for the PRNG cracker to successfully recover the PRNG state, which can be confirmed by predicting the remaining 350 deals correctly. Finally, the flag is obtained by converting the 751st and 752nd deals to text.

## Exploit code

`from math import prod from random import seed, getrandbits import re # !wget \ # https://raw.githubusercontent.com/icemonster/symbolic_mersenne_cracker/main/main.py \ # -O untwister.py 2>/dev/null from untwister import Untwister HEARTS = "๐ฑ๐ฒ๐ณ๐ด๐ต๐ถ๐ท๐ธ๐น๐บ๐ป๐ฝ๐พ" SPADES = "๐ก๐ข๐ฃ๐ค๐ฅ๐ฆ๐ง๐จ๐ฉ๐ช๐ซ๐ญ๐ฎ" DIAMONDS = "๐๐๐๐๐ ๐๐๐๐๐๐๐๐" CLUBS = "๐๐๐๐๐๐๐๐๐๐๐๐๐" DECK = SPADES+HEARTS+DIAMONDS+CLUBS CARDS_PER_DEAL = 25 MAX_DEAL = prod( x for x in range(len(DECK) - CARDS_PER_DEAL + 1, len(DECK) + 1) ) DEAL_BITS = MAX_DEAL.bit_length() print("deal bits", DEAL_BITS) def deal_game_leak(frombits=None): if frombits is None: getrand = getrandbits(DEAL_BITS) else: getrand = frombits shuffle = getrand % MAX_DEAL shuffle_orig = shuffle deck = list(DECK) deal = "" deal_idxs = [] while shuffle > 0: deal_idx = shuffle % len(deck) deal_idxs.append(deal_idx) deal += deck.pop(deal_idx) shuffle //= len(deck) + 1 while len(deal) < CARDS_PER_DEAL: deal += deck.pop(0) return deal, getrand, shuffle_orig, deal_idxs def cards_to_deal_idxs(deal): deck = list(DECK) deal_idxs = [] for c in deal: deal_idx = deck.index(c) deck.pop(deal_idx) deal_idxs.append(deal_idx) return deal_idxs def deal_idxs_to_getrand(deal_idxs): cur_deck_len = len(DECK) - len(deal_idxs) + 1 acc = 0 for deal_idx in deal_idxs[::-1]: acc = acc * cur_deck_len + deal_idx cur_deck_len += 1 return acc def getrand_to_randbits(getrand): cap = DEAL_BITS // 32 * 32 bits = bin(getrand + 2 ** cap)[-cap:] return bits def getrand_to_randbits_mod(getrand): if (getrand + MAX_DEAL) >= 2 ** DEAL_BITS: return getrand_to_randbits(getrand) a = getrand_to_randbits(getrand) b = getrand_to_randbits(getrand + MAX_DEAL) res = "".join([ i if i == j else "?" for i, j in zip(a, b) ]) return res # Sanity checks test_seed = 0 seed(test_seed) for _ in range(10): deal = deal_game_leak() deal_idxs = cards_to_deal_idxs(deal[0]) assert(deal_idxs == deal[3]) getrand = deal_idxs_to_getrand(deal_idxs) randbits = getrand_to_randbits_mod(getrand) assert(all([ i == j for i, j in zip(randbits, bin(deal[1])[-128:]) if i != "?" ])) def pipeline(cards): known = getrand_to_randbits_mod( deal_idxs_to_getrand( cards_to_deal_idxs( cards ) ) ) return [ known[i:i+32] for i in range(0, 128, 32) ][::-1] + ["?" * 32] with open("poker/cards.22.07.16.txt", "r") as f: raw = f.read() deals = raw.strip().split("\n\n") print("num deals", len(deals)) def parse_deal(deal, check_num): deal_num = re.search( r"Game (\d+):", deal, re.UNICODE | re.MULTILINE ).group(1) deal_num = int(deal_num) assert check_num == deal_num cards = re.sub( r"[ -~\n]*", "", deal, re.UNICODE | re.MULTILINE ) assert(len(cards) == CARDS_PER_DEAL) return cards all_deals = [ parse_deal(deal, deal_idx + 1) for deal_idx, deal in enumerate(deals) ] print() print("deals") for i in all_deals[:5]: print(i) all_deals_pred = [ i for deal in all_deals for i in pipeline(deal) ] print() print("pipeline output") for i in all_deals_pred[:5]: print(i) use = 400 print() print("submitting") ut = Untwister() for i in all_deals_pred[:use * 5]: ut.submit(i) print("solving") rng = ut.get_random() print("verify") for i in all_deals[use:]: test = deal_game_leak(rng.getrandbits(DEAL_BITS))[0] assert(test == i) print("flag") ALPHABET52 = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnop_rstuvw{y}" def text_from_cards(string): return string.translate(string.maketrans(DECK, ALPHABET52)) for i in range(3): deal = deal_game_leak(rng.getrandbits(DEAL_BITS))[0] print(text_from_cards(deal), end="") print()`

`deal bits 133 num deals 750 deals ๐ฆ๐๐ญ๐๐๐ฝ๐ก๐๐๐พ๐๐ ๐๐ฒ๐ง๐๐๐๐๐๐ณ๐ฑ๐ธ๐๐ช ๐๐๐ก๐๐๐ฃ๐๐๐๐ฅ๐๐ง๐๐๐๐ฑ๐จ๐ฎ๐๐ข๐ณ๐ซ๐๐ฉ๐ด ๐บ๐ฑ๐๐๐๐ณ๐จ๐๐๐ช๐ข๐๐ฒ๐ด๐๐๐๐ฅ๐ฎ๐น๐ป๐ต๐๐ง๐ฉ ๐๐๐ข๐๐๐ต๐๐ฝ๐๐ ๐ฒ๐๐ฎ๐ท๐๐บ๐ช๐๐ณ๐ฆ๐ธ๐๐๐ค๐ด ๐ข๐๐๐ ๐ป๐๐ธ๐๐๐น๐๐๐ฑ๐๐ฅ๐ซ๐ง๐ต๐๐ญ๐๐ฉ๐ท๐๐ pipeline output 0??11?00010101101101110001001101 1?111??1?????0000??01?0?100?01?? 0010????1100???0????00???011??0? 0?010???1????1????0???010111???? ???????????????????????????????? submitting STT> Solving... solving STT> Solved! (in 8.13s) verify flag ictf{CheaTINg_PsuEdoRAnDOm_BITgenEratioN}bcdfhjklpGJTYgCOrVswXhAj{aQnSPckfW`

## Secure Encoding: Base64

By puzzler7

Base64 is my favorite encryption scheme.

Attachments: encode.py out.txt

10 solves, 492 points

## Challenge encode.py

`from base64 import b64encode from random import shuffle charset = b'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789+/=' shuffled = [i for i in charset] shuffle(shuffled) d = {charset[i]:v for(i,v)in enumerate(shuffled)} pt = open("flag.txt").read() while "\n\n\n" in pt: pt = pt.replace("\n\n\n", '\n\n') while ' ' in pt: pt = pt.replace(' ', ' ') assert all(ord(i)<128 for i in pt) ct = bytes(d[i] for i in b64encode(pt.encode())) f = open('out.txt', 'wb') f.write(ct)`

Most fun challenge of the CTF IMO.

We get a file `encode.py`

and a very large base64 encoded text `out.txt`

. By reading `encode.py`

we understand that it does the following:

- Reads text from a file
- Trims every instance of 3 or more newlines to 2 newlines
- Trims every instance of 2 or more spaces to 1 space
- Asserts that all text has
`ord(c) < 128`

- Encodes text as base64
- Substitutes each base64 character to another unique base64 character

Our task is to recover the original text despite the base64 character set being substituted.

The first observation is to ensure valid base64 padding. We see the `=`

padding character in the middle of the text which is invalid base64, and also that the `2`

character only appears at the end of `out.txt`

. We can make a naive swap of `=`

with `2`

to make the padding valid.

We need to solve for some permutation / mapping of 64 encrypted characters to 64 original characters. The next observation to make is that the more correct assignments of shuffled charset to original charset we have, the more the base64 decoded text will โlookโ like a valid input text. This suggests a simulated annealing approach to recover the shuffle permutation.

First we make a mutation function which swaps 2 (or rarely 3 or 4) characters with each other inside the ciphertext, allowing us to iterate through permutation search space.

We then make a fitness function which grades decryptions on how โvalidโ it is:

- We assign a heavy penalty to decryptions containing patterns that definitely cannot appear in the plaintext, such as
- Illegal characters
`ord(c) >= 128`

- Illegal substrings such as three newlines
`"\n\n\n"`

and two spaces`" "`

(but I skipped these in the final exploit)

- Illegal characters
- Assuming the text is printable ASCII, we assign a small penalty to non-printable characters that have
`ord(c) < 128`

- Assuming the text is english, we add a bonus to decryptions containing text patterns which frequently appear in english such as
- Whitespace (
`" "`

,`"\n"`

) - Alphabets
- Common english bigrams (
`"th"`

,`"he"`

) and trigrams (`"the"`

,`"and"`

) - Valid english punctuation (
`", "`

,`". "`

) - And several other patterns. See here for the stats used.

- Whitespace (

Next we create a temperature evaluation function to probabilistically accept or reject new permutations based on a โtemperatureโ parameter and the improvement of fitness from last iteration. We should accept any permutations that improve the fitness, but at the same time we want to give permutations with poorer fitness a small chance to be accepted to avoid being stuck in local minima.

Finally we create a scheduler function which keeps track of overall best permutations and decryption so far while adjusting temperature over time, also known as a cooling schedule.

To keep runtime fast, we trim the ciphertext to the first 40 thousand base64 characters (i.e. first 30 thousand bytes of plaintext).

After around 5 minutes of compute over 4 threads, most of the text is readable and we can find most of the flag as well!

## Exploit code

`from collections import Counter from base64 import b64decode from random import choice, sample, random, seed from string import printable, ascii_letters from numpy import linspace from tqdm.auto import tqdm from multiprocessing import Pool with open("b64/out.txt", "r") as f: raw = f.read() # 2 is the original padding character print("Least common", Counter(raw).most_common()[-3:]) print("Tail", raw[-5:]) # swap 2 and = raw = raw.replace("2", "@").replace("=", "2").replace("@", "=") print("Total length", len(raw)) # https://www3.nd.edu/~busiforc/handouts/cryptography/cryptography%20hints.html non_printable_charset = set(range(128)) - set(printable.encode()) whitespace_charset = set(" \n".encode()) alphabet_charset = set(ascii_letters.encode()) one_score = { i: 0 for i in range(256) } for i in range(128, 256): one_score[i] = -1000 for i in non_printable_charset: one_score[i] = -10 for i in whitespace_charset: one_score[i] = 4 for i in alphabet_charset: one_score[i] = 2 two_patterns_more = [ ". ", ", ", ".\n", "\n" # whoops the last "\n" should have been "\n\n" ] two_patterns = [ "th", "er", "on", "an", "re", "he", "in", "ed", "nd", "ha", "at", "en", "es", "of", "or", "nt", "ea", "ti", "to", "it", "st", "io", "le", "is", "ou", "ar", "as", "de", "rt", "ve", ] two_patterns_less = [ " T", " O", " A", " W", " B", " C", " D", " S", "E ", "S ", "T ", "D ", "N ", "R ", "Y ", "F ", ] two_score = {} for i in two_patterns_more: two_score[i.encode()] = 10 for i in two_patterns: two_score[i.encode()] = 5 for i in two_patterns_less: two_score[i.encode()] = 1 three_patterns = [ "the", "and", "tha", "ent", "ion", "tio", "for", "nde", "has", "nce", "edt", "tis", "oft", "sth", "men", " a ", " i ", ] three_score = { i.encode(): 10 for i in three_patterns } four_patterns = [ " of ", " to ", " in ", " it ", " is ", " be ", " as ", " at ", " so ", " we ", " he ", " by ", " or ", " on ", " do ", " if ", " me ", " my ", " up ", " an ", " go ", " no ", " us ", " am ", ] four_score = { i.encode(): 10 for i in four_patterns } def swap(x, *swaps): assert(len(swaps) >= 2) sub = {i: j for i, j in zip(swaps, swaps[1:] + swaps[:1])} return "".join([sub.get(i, i) for i in x]) b64_charset = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789+/' def mutate(x): while True: swap_len = choice([2, 2, 2, 2, 3, 4]) swaps = sample(b64_charset, swap_len) x = swap(x, *swaps) if choice([True, True, True, False]): return x def fitness(x): x = b64decode(x) n = len(x) acc = 0 for i in range(n): c = x[i] cc = x[i:i+2] ccc = x[i:i+3] cccc = x[i:i+4] acc += ( one_score[c] + two_score.get(cc, 0) + three_score.get(ccc, 0) + four_score.get(cccc, 0) ) return acc / n def mutate_and_fitness(x): x = mutate(x) return x, fitness(x) def check_temp(fit_delta, temp): if fit_delta > 0: return True return - temp * random() < fit_delta take = 40000 print("Using first b64 characters", take) cur = raw[:take] cur_fit = fitness(cur) best = cur best_fit = cur_fit # Nothing up my sleeve, https://remywiki.com/1116 seed(1116) schedule = sum([ list(linspace(1, 0.0001, 200)) * 30, list(linspace(0.5, 0.0001, 200)) * 10, list(linspace(0.1, 0.0001, 200)) * 10, ], []) since_new_best = 0 stall_limit = 5 pbar = tqdm(schedule) with Pool(processes=4) as pool: for temp in pbar: muts = pool.map(mutate_and_fitness, [cur] * 4) mut, mut_fit = max(muts, key=lambda x: x[1]) pbar.set_description( f"Best: {best_fit:.03f} " f"Temp: {temp:.02f} " f"Cur: {cur_fit:.03f} " f"Stall: {since_new_best} " f"Mut: {mut_fit:.03f} " ) if mut_fit > best_fit: best, best_fit = mut, mut_fit since_new_best = 0 continue since_new_best += 1 if since_new_best == stall_limit: since_new_best = 0 cur, cur_fit = best, best_fit continue if check_temp(mut_fit - cur_fit, temp): cur, cur_fit = mut, mut_fit pln = b64decode(best) print("First 1000 characters") print("=" * 80) print(pln[:1000].decode()) print("=" * 80) print("Found ictf", b"ictf" in pln) if b"ictf" in pln: flag_pos = pln.index(b"ictf") print(pln[flag_pos-10:flag_pos+50])`

`Least common [('F', 57), ('T', 46), ('2', 2)] Tail hoP22 Total length 597924 Using first b64 characters 40000 0%| | 0/10000 [00:00<?, ?it/s] First 1000 characters ================================================================================ The Project Gutenberg eBooj of The Picture of Dorian Gray, by Nscar Wilde This eBook is for the use of anyone anywhere in the United States and most other parts of the world at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Prozect Gutenberg License included with this eBook or online at www.gutenberg.org. If you are not located in the United States, you will have to check the laws of the country where you are located before using this eBooj. Title+ The Picture of Dorian Gray Author+ Oscar Wilde Release Date+ October, 19=4 [eBooj #174] [Most recently updated+ February 3, 2022] Language: English Produced by+ Judith Boss. HTML version by Al Haines. **: START OF THE PROJECT GUTEOBERG EBONK THE PICTURE NF DORIAN GRAY :** The Picture of Dorian Gray by Nscar Wilde Contents THE PREFACE CHAPTER I. CHAPTER II. CHAPTER III. CHAPTER IV. CHAPTER V. CHAPTER VI. CHAPTER VII. CHAPTER VIII. CHAPTER IX. CHAPTER ================================================================================ Found ictf True b'useless.\n\nictf~all_encodings_are_nothing_but_art}\n\nOSCAR WIL'`

## Living Without Expectations

By Robin_Jadoul

Sometimes you just gotta have some fun implementing bare hardness assumptions.

Attachments: lwe.py output.txt

10 solves, 492 points

`lwe.py`

has `rand()`

that provides a deterministic stream of integers from 0 to 7. The script generates a 69-dimension $s$ vector with elements 0 or 1. For each bit of the flag, we are given:

- 69x69 matrix $A$ with random elements $< q$
- 69-dimension vector $b$ equal to $A \times s + \text{rand()} \mod q$ if the flag bit is 0; or
- 69-dimension vector $b$ with random elements $< q$ if the flag bit is 1

This looks like learning with errors, where $s$ is the secret key.

Assuming that the flag begins with `ictf{`

, the first bit of `i`

should be 0 and we should get the $b = A \times s + \text{rand()} \mod q$ case. We can solve for $s$ using Babaiโs closest vector algorithm. Notably, both the noise as well as the secret key $s$ is small in comparison to the magnitude of $q$ and the elements in $A$ and $b$. The closest lattice point is therefore likely to reveal $s$. After this, we can check for each flag bit using $A \times s - b \mod q$ and inspecting if recovered vector is a possible error from $\text{rand()}$ by looking at its magnitude.

## Exploit code

`import numpy as np # https://jgeralnik.github.io/writeups/2021/08/12/Lattices/ def babai_closest_vector(B, target): # Babai's Nearest Plane algorithm M = B.LLL() G = M.gram_schmidt()[0] small = target for _ in range(1): for i in reversed(range(M.nrows())): c = ((small * G[i]) / (G[i] * G[i])).round() small -= M[i] * c return target - small q = 2**142 + 217 n = 69 nseeds = 142 with open("lwe/output.txt", "r") as f: raw = f.read() lines = raw.strip().split("\n") print("data size", len(raw), len(lines)) def parse_arr(arr): arr = arr.strip(" ][").split(", ") return np.array([ int(i.strip("'")[2:], 16) for i in arr ]) def parse_line(line): a, _, b = line.strip(" []").partition("] [") return parse_arr(a), parse_arr(b) # take first bit, should be 0 which = 0 a, b = parse_line(lines[which]) a = a.reshape((n, n)) # use only first 10 elements of b samples = 10 lock = 1 lockval = 2 ** 200 mat = [ [0 for x in range(n + samples + lock)] for y in range(n + samples + lock) ] for i in range(n): mat[i][i] = 1 for j in range(samples): mat[i][n+j] = a[j][i] for i in range(samples): mat[n+i][n+i] = q for i in range(samples): mat[n+samples][n+i] = -b[i] mat[n+samples][n+samples] = lockval vec = [0] * n + [0] * samples + [lockval] mat = Matrix(mat) vec = vector(vec) cv = babai_closest_vector(mat, vec) s = list(cv)[0:n] s = np.array(s) print("s", s) res = "" for which in range(len(lines)): a, b = parse_line(lines[which]) a = a.reshape((n, n)) if -7 * n < ((a @ s % q) - b).sum() <= 0: res += "0" else: res += "1" print("flag bin", res[:50], "...") pln = bytes.fromhex(hex(int(res, 2))[2:]) print(pln)`

`data size 67729514 336 s [1 1 1 0 0 0 0 0 0 0 1 1 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 0 0 0 0 1 1 0 0 1 1 1 1 1 0 1 0 0 1 0 0 0 0 1 0 0 1 0 1 0 1 1 0 0 0 0 1 1 1 1 1 0 1] flag bin 01101001011000110111010001100110011110110110110000 ... b'ictf{l4tt1c3_crypt0_t4k1ng_0v3r_th3_w0rld}'`

## Lorge

By Robin_Jadoul

I guess smoll needed a revenge after all ๐ญ

Attachments: lorge.py output.txt

29 solves, 433 points

From `lorge.py`

, we see that $p-1$ and $q-1$ are $2^{25}$ smooth. We first try to factor $n$ using Pollard p-1 attack and generator $g=2$, but at some point $g$ goes from a coprime number to $1$ straight away. This is because some largest prime factor $x$ of $p-1$ and $q-1$ are the same. Since the initial run stopped at the largest prime factor $x$, we can instead start with generator $g = 2^x \mod n$ and re-try the attack which succeeds.

## Exploit code

`from tqdm.auto import tqdm from gmpy2 import mpz, next_prime, log, floor, powmod, gcd with open("lorge/output.txt", "r") as f: raw = f.read() n, e, ct = [int(i.partition(" = ")[2]) for i in raw.strip().split("\n")] n = mpz(n) cap = mpz(2 ** 25) def pollard_rho_step(g, q): r = mpz(floor(log(cap) / log(q))) z = q ** r return powmod(g, z, n) leak = [] while True: g = mpz(2) cur = mpz(2) for i in leak: g = pollard_rho_step(g, i) pbar = tqdm(total=int(cap)) pbar.update(int(cur)) while cur < cap: g = pollard_rho_step(g, cur) check = gcd(g - 1, n) if check != 1: break nx = next_prime(cur) delta = nx - cur pbar.update(int(delta)) cur = nx if g == 1: print("leak common prime factor", cur) leak.append(cur) else: break p = gcd(g-1, n) q = n // p assert(p * q == n) phi = (p-1) * (q-1) d = powmod(e, -1, phi) pln = powmod(ct, d, n) bytes.fromhex(hex(int(pln))[2:])`

`0%| | 0/33554432 [00:00<?, ?it/s] leak common prime factor 19071329 0%| | 0/33554432 [00:00<?, ?it/s] b'ictf{why_d1d_sm0ll_3v3n_sh0w_up_on_f4ct0rdb???_That_m4d3_m3_sad!}'`

## stream

By Eth007

HELP! I encrypted my files with this program I downloaded from the internet. Can you recover my 42-byte flag?

37 solves, 390 points

Open `stream`

binary with ghidra and grok.

## Decompile of stream binary

The binary encrypts text with this procedure:

- Read an input file specified in
`argv[1]`

- Process 8 bytes of input at a time as a 64 bit unsigned integer
- Xor the input integer with a 64 bit unsigned integer key specified in
`argv[2]`

- Square the key
- Repeat until all of the input file is processed
- Write the output to a file specified in
`argv[3]`

What happens to input files that need padding to the nearest 8 bytes? We can run the script locally with a dummy key to test.

`$ python -c "print('\x00' * 42, end='')" > zeros.txt $ ./stream zeros.txt 7 zeros.enc.txt $ hexdump -C -v zeros.txt 00000000 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 00000010 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 00000020 00 00 00 00 00 00 00 00 00 00 |..........| 0000002a $ hexdump -C -v zeros.enc.txt 00000000 07 00 00 00 00 00 00 00 31 00 00 00 00 00 00 00 |........1.......| 00000010 61 09 00 00 00 00 00 00 c1 f6 57 00 00 00 00 00 |a.........W.....| 00000020 81 7d 05 a5 39 1e 00 00 01 3b 9b 6e 58 33 3c 30 |.}..9....;.nX3<0| 00000030 $ python -c "print(hex(pow(7, 2 ** 5, 2 ** 64)))" 0x303c33586e913b01`

It seems like the trailing bits are padded with zeros before xor. Since the flag is 42 bytes there will be 6 zero bytes of padding. These tail 6 bytes of `enc`

then leaks the high 6 bytes of `key`

during the final iteration.

Assuming the flag starts with `ictf{`

, we can use this to leak the low 5 bytes of `key`

during the first iteration. The `key * key`

operation is done on a `unsigned long long`

, which acts like multiplication modulo `2**64`

. As a result of this nice modulo, the $k$ lower bits of `key`

in later iterations only depends on the same $k$ lower bits of `key`

in previous iterations. We can manually square `key`

from first iteration to get the low 5 bytes of `key`

in the final iteration. This way, we have leaked `key`

.

We can recover `key`

of earlier iteration using iterative square root modulo prime power of `key`

from later iterations. The exploit uses `z3`

to calculate this square root. We can speed this up by early pruning if some `key`

produces a non-printable flag.

## Exploit code

`import struct from z3 import * from tqdm.auto import tqdm from string import printable from itertools import product # for some reason sage is slow, so z3 to the rescue def sqrt_mod_2_64(x): x = int(x) res = [] a = BitVec("a", 64) while True: s = Solver() s.add(a * a == x) for soln in res: s.add(a != soln) if str(s.check()) == "unsat": break soln = s.model()[a].as_long() res.append(soln) return res with open("stream/out.txt", "rb") as f: enc_raw = f.read() enc_raw, len(enc_raw) enc = struct.unpack("<" + "Q" * 6, enc_raw) key_low_first = (struct.unpack("<Q", b"ictf{\x00\x00\x00")[0] ^ enc[0]) & 0xffffffffff key_low_final = pow(key_low_first, 2 ** 5, 2 ** 64) & 0xffffffffff print("final key low 5 bits".ljust(22), hex(key_low_final).rjust(18)) key_high_final = enc[5] & 0xffffffffffff0000 print("final key high 6 bits".ljust(22), hex(key_high_final)) key_final = key_low_final | key_high_final print("final key".ljust(22), hex(key_final)) charset = set(printable.encode()) def search(idx, key, data=""): pln_chunk = bytes.fromhex(hex((enc[idx] ^ key) + 2 ** 64)[3:]) if idx == 5: pln_chunk = pln_chunk[-2:] if not all([i in charset for i in pln_chunk]): return None data = pln_chunk[::-1].decode() + data print(idx, data.strip()) if idx == 0: return prev_keys = sqrt_mod_2_64(key) for prev_key in prev_keys: search(idx-1, prev_key, data) search(5, key_final)`

`final key low 5 bits 0x241d06d81 final key high 6 bits 0x2297b40241d00000 final key 0x2297b40241d06d81 5 } 4 901bf2e4} 3 ystream_901bf2e4} 2 ed_my_keystream_901bf2e4} 1 _rec0vered_my_keystream_901bf2e4} 0 ictf{y0u_rec0vered_my_keystream_901bf2e4}`

## hash

By Eth007

My passwords are safe and secure with the use of sha42!

`nc hash.chal.imaginaryctf.org 1337`

39 solves, 378 points

`sha42`

has poor confusion. Each byte of the hash output is effectively the xor of some subset of the initial input bytes, depending on the length of the input byte array. We can do a (dumb) symbolic run of the hash to identify which input bytes affect each output byte, then use these correlations with `z3`

to solve for a possible input which produces some hash.

Since the challenge service does not compare input lengths, we can fix a length of 30 and will always get (some) solution since we have 30 unknowns and 21 equations.

## Exploit code

`from collections import defaultdict from tqdm.auto import tqdm from z3 import * from pwn import * # original implementation config = [ [int(a) for a in n.strip()] for n in open("hash/jbox.txt").readlines() ] def sha42(s: bytes, rounds=42): out = [0]*21 for round in range(rounds): for c in range(len(s)): if config[((c//21)+round)%len(config)][c%21] == 1: out[(c+round)%21] ^= s[c] return bytes(out).hex() # generate confusion relations confusion = defaultdict(list) # confusion[hash_index] = [input_char_indexes] length = 30 for target in range(length): probe = [0 for _ in range(length)] probe[target] = 1 probe = bytes(probe) res = sha42(probe) res = bytes.fromhex(res) for idx, f in enumerate(res): if f == 1: confusion[idx].append(target) # unhash solver def unhash(chal): solver = Solver() cs = [ BitVec(f"c{i}", 8) for i in range(length) ] for sink in range(21): rhs = 0 for s in confusion[sink]: rhs = rhs ^ cs[s] solver.add(rhs == chal[sink]) solver.check() model = solver.model() res = bytes([ model[s].as_long() for s in cs ]) return res # sanity check chal = sha42(b"some_16char_pass") chal = bytes.fromhex(chal) assert sha42(unhash(chal)) == chal.hex() # run conn = remote("hash.chal.imaginaryctf.org", 1337) for _ in tqdm(range(50)): conn.recvuntil(b"sha42(password) = ") chal = conn.recvline() chal = bytes.fromhex(chal.strip().decode()) soln = unhash(chal) conn.sendline(soln.hex().encode()) print(conn.recvall())`

`[x] Opening connection to hash.chal.imaginaryctf.org on port 1337 [x] Opening connection to hash.chal.imaginaryctf.org on port 1337: Trying 34.90.15.213 [+] Opening connection to hash.chal.imaginaryctf.org on port 1337: Done 0%| | 0/50 [00:00<?, ?it/s] [x] Receiving all data [x] Receiving all data: 16B [x] Receiving all data: 95B [+] Receiving all data: Done (95B) [*] Closed connection to hash.chal.imaginaryctf.org port 1337 b'hex(password) = Correct!\nCongrats! Your flag is: ictf{pls_d0nt_r0ll_y0ur_0wn_hashes_109b14d1}\n\n'`

## otp

By Eth007

Encrypt your messages with our new OTP service. Your messages will never again be readable to anyone.

`nc otp.chal.imaginaryctf.org 1337`

Attachments: otp.py

48 solves, 316 points

`secureRand`

returns more `1`

bits than `0`

bits, specifically giving `1`

bits around 70% of the time. Connect multiple times to get many `flag ^ secureRand()`

and the true flag bits will be the less common bit at each position.

## Exploit code

`from pwn import * from tqdm.auto import tqdm from collections import Counter def char_to_bin(x): return bin(0x100 + x)[3:] # simulate secureRand bias jumbler = [] jumbler.extend([2**n for n in range(300)]) jumbler.extend([3**n for n in range(300)]) jumbler.extend([4**n for n in range(300)]) jumbler.extend([5**n for n in range(300)]) jumbler.extend([6**n for n in range(300)]) jumbler.extend([7**n for n in range(300)]) jumbler.extend([8**n for n in range(300)]) jumbler.extend([9**n for n in range(300)]) print( "Probability of 1", sum([1 if int(str(i)[0]) < 5 else 0 for i in jumbler]) / len(jumbler) ) conn = remote("otp.chal.imaginaryctf.org", 1337) # leak many xor-ed flag n_leaks = 100 data_raw = [] for _ in tqdm(range(n_leaks)): conn.sendline(b"FLAG") conn.recvuntil(b"Encrypted flag: ") res = conn.recvline() data_raw.append( bytes.fromhex(res.decode().strip()) ) # recover flag data = [ "".join([char_to_bin(j)for j in i]) for i in data_raw ] pln_bin = "".join([ "1" if Counter(i).most_common(1)[0][0] == "0" else "0" for i in zip(*data) ]) pln = bytes.fromhex(hex(int(pln_bin, 2))[2:]) print(pln)`

`Probability of 1 0.69875 [x] Opening connection to otp.chal.imaginaryctf.org on port 1337 [x] Opening connection to otp.chal.imaginaryctf.org on port 1337: Trying 35.204.97.207 [+] Opening connection to otp.chal.imaginaryctf.org on port 1337: Done 0%| | 0/100 [00:00<?, ?it/s] b'ictf{benfords_law_catching_tax_fraud_since_1938}\n'`