Now, Eddington's number purports (see Wiki for assumptions - hint - huge) to be the number of fundamental particles (protons + electrons, assuming that a neutron is made up of a proton and an electron)) in the universe. So, you'd basically only have 2 * 136 (= 272) particles per SHA-256 hash to be available to store your data for a collision (+ overhead)? I put in the 2 for a 50% change of a clash - is my maths OK? $\begingroup$ Consider that Eddington's number =136 * $2^ \approx 10^$.
new ( key , mode , *args , cryptocurrency **kwargs ) ¶ var MODE_ECB: Electronic Code Book (ECB) var MODE_CBC: Cipher-Block Chaining (CBC) var MODE_CFB: Cipher FeedBack (CFB) var MODE_OFB: Output FeedBack (OFB) var MODE_CTR: CounTer Mode (CTR) var MODE_OPENPGP: OpenPGP Mode var MODE_CCM: Counter with CBC-MAC (CCM) Mode var MODE_EAX: EAX Mode var MODE_GCM: Galois Counter Mode (GCM) var MODE_SIV: Syntethic Initialization Vector (SIV) var MODE_OCB: Offset Code Book (OCB) Crypto.Cipher.AES.
The counter number is encoded in big endian mode. initial_value : ( integer or bytes/bytearray/memoryview ) – (Only MODE_CTR ). If not present, the cipher will start counting from 0. The initial value for the counter. The value is incremented by one for each block.
The ciphertext file can be found here. Another Diffie-Hellman problem but not really because the numbers here are so small…. All possible pairs of generators and primes and random values of a and b can be bruteforced to obtain the key and if the chosen values successfully decrypt the ciphertext to output a message which starts with the flag format ‘uiuctf. A random pair of a generator and corresponding prime is selected. After that the standard implementation of Diffie-Hellman key exchange takes place.
will compel other countries as well to work on crypto regulations. The initiative from the US govt. Binance CEO CZ has hailed the White House crypto framework and called it a ‘responsible innovation.’ The report from the White House came recently as the result of the executive order that President Biden had issued back in March 2022.
The seven-day data shows a loss of 8.94% for this coin. Dash has also been in losses due to the continuing fluctuations in the market. The price value for DASH is currently in the $43.78 range. The latest data shows that it has shed 0.83% over the last 24 hours.
$\begingroup$ So, in 2600 universe-lifespans, cryptocurrency we would have a chance of finding a collision, but only if we saved/recorded/stored every single hash discarded by all the bitcoin
miners in the world, and, even then, it would just be two random block+nonce files with the same hash?
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With massive trading volumes and market capitalizations for bitcoin some of the larger crypto-currencies rivaling that of some of the worlds largest corporations. Today’s hype surrounding Bitcoin, Ethereum, cryptocurrency, and blockchain technologies rivals the dot-com bubble in the 90s.
$\begingroup$ galvatron: for your formula I get 3.59e13 (not e33). @James: your link divides by 1.37e9 and gives 26,254, but dividing by the correct 13.7e9 does give 2,625 like your text. Plus, as you say, storing and comparing. Also each bitcoin hash is two (albeit constrained) SHA256. Both: in Oct '17 bitcoin
is up to 10e18x2/s thus 5.4e11 years or 39 universes.
It’s your money, held in a digital form. It’s not a technology. The easiest way to define Bitcoin is to call it a "digital dollar." That’s really all it is — minus all the formal regulations that come with a bank (which is what makes it such a disruptive concept).
You would still have to find which two hashes these were. $\begingroup$ @JamesTheAwesomeDude During that time all the bitcoin miners would have a good chance that the two hashes that were calculated had the same hash. By "a given SHA-256 hash" I believe you mean Second Preimage Attack which cannot rely on birthday paradox.
The birthday paradox (as per the answer) states that you only need $2^$ hashes for a 50% chance of a collision. So you would have $136 * 2^$ particles to store each hash. Of course this is still wildly impractical! $\begingroup$ @Vérace your maths isn't quite right.
So, we can clearly see in the last line hexlify(hash[::-1]).decode("utf-8") that the hash is reversed. You can see it is hashed twice in the line hash = hashlib.sha256(hashlib.sha256(header_bin).digest()).digest() Well that plays a huge role in solving the challenge. It only has leading zeroes because it is reversed! Let me explain this more clearly by building on the example shown above : Remember that I said that it is important to remember that the header is hashed twice? What this means is that before reversing the final second hash, it had trailing zeroes, not leading zeroes which is exactly what we need. Since our input appended to the random nonce provided by the server is hashed, this means that we need to use the first hash of the header and provide that to the server.