11 ms·

Presumably generated lots of variations of that sentence. It's 28 bits, so you only need to have around 14 places where 4 variations are possible. For instance

by devit 3y ago

Presumably generated lots of variations of that sentence. It's 28 bits, so you only need to have around 14 places where 4 variations are possible. For instance, it could start with "Was", "I was", "verifying" could be "checking" or "computing" or "testing", etc. A bit tight, but it seems feasible with some work.

fbdab103 3y ago
Throwing a loop of numbers on the end seems far more feasible. I put together a hasty Python implementation which can immediately find three hits at 5 characters. TQDM is reporting ~450k tries per second, so depending on how lucky you are, would probably want to redo in a faster language to solve for 7+ import hashlib import itertools import tqdm BLOCKS_SIZE = 5 sentence_prefix = "The SHA256 for this sentence begins with:" itos = {0x00:"zero", 0x01:"one", 0x02:"two", 0x03:"three", 0x04:"four", 0x05:"five", 0x06:"six", 0x07:"seven", 0x08:"eight", 0x09:"nine", 0x0a:"a", 0x0b:"b", 0x0c:"c", 0x0d:"d", 0x0e:"e", 0x0f:"f"} for nums in tqdm.tqdm(itertools.product(itos.keys(), repeat=BLOCKS_SIZE)): sentence = f"{sentence_prefix} {', '.join(itos[num] for num in nums[:-1])}, and {itos[nums[-1]]}." hash_true = hashlib.sha256(bytes(sentence, "utf8")).hexdigest() guessed_prefix = "".join(f"{n:x}" for n in nums) true_prefix = hash_true[:BLOCKS_SIZE] if guessed_prefix == true_prefix: print("collision") print(sentence) print(hash_true)
aib 3y ago
Indeed, this is how I did it. 2^28 is around 270 million. With little more than a handful options, it should be possible. Although I have to say, it turned out to be more difficult than I'd initially thought. Maybe it's better to think of it as 28 different boolean choices. echo -n "Indeed. This is how I managed to do it. 2^28 is around 300 million. With only a handful options, it's possible. Although, I must say that it turned out to be more difficult than I'd initially thought. Perhaps it's better to think of it as 28 different alternatives." | sha256sum
- Alextigtig 3y ago
  This is crazy! The hash of this comment once again begins 182a7c9!!! Well done indeed!

by devit 3y ago

fbdab103 3y ago

Throwing a loop of numbers on the end seems far more feasible. I put together a hasty Python implementation which can immediately find three hits at 5 characters. TQDM is reporting ~450k tries per second, so depending on how lucky you are, would probably want to redo in a faster language to solve for 7+ import hashlib import itertools import tqdm BLOCKS_SIZE = 5 sentence_prefix = "The SHA256 for this sentence begins with:" itos = {0x00:"zero", 0x01:"one", 0x02:"two", 0x03:"three", 0x04:"four", 0x05:"five", 0x06:"six", 0x07:"seven", 0x08:"eight", 0x09:"nine", 0x0a:"a", 0x0b:"b", 0x0c:"c", 0x0d:"d", 0x0e:"e", 0x0f:"f"} for nums in tqdm.tqdm(itertools.product(itos.keys(), repeat=BLOCKS_SIZE)): sentence = f"{sentence_prefix} {', '.join(itos[num] for num in nums[:-1])}, and {itos[nums[-1]]}." hash_true = hashlib.sha256(bytes(sentence, "utf8")).hexdigest() guessed_prefix = "".join(f"{n:x}" for n in nums) true_prefix = hash_true[:BLOCKS_SIZE] if guessed_prefix == true_prefix: print("collision") print(sentence) print(hash_true)

aib 3y ago

Indeed, this is how I did it. 2^28 is around 270 million. With little more than a handful options, it should be possible. Although I have to say, it turned out to be more difficult than I'd initially thought. Maybe it's better to think of it as 28 different boolean choices. echo -n "Indeed. This is how I managed to do it. 2^28 is around 300 million. With only a handful options, it's possible. Although, I must say that it turned out to be more difficult than I'd initially thought. Perhaps it's better to think of it as 28 different alternatives." | sha256sum

Alextigtig 3y ago
This is crazy! The hash of this comment once again begins 182a7c9!!! Well done indeed!