where in the Bitcoin Protocol is SHA256(SHA256(x)) performed and why?

  • i just found one area; the Merkle Tree – joe Mar 17 '13 at 17:23

Bitcoin uses double hashing almost everywhere it hashes in one of two variants:

  • RIPEMD160(SHA256(x)) called Hash160 which produces a 160 bit output

    • hashing the public key to generate part of a Bitcoin addresses
  • SHA256(SHA256(x)) called Hash256 which produces a 256 bit output

    • generating the checksum in a Bitcoin address
    • hashing the block in a merkle tree
    • linking transaction outputs and inputs
    • hash of the block header (and thus the proof of work and the link to the previous block)

It seems like Satoshi chose Hash256 whenever collisions are a problem, and Hash160 when only (multi target) second pre-images matter. This is consistent with a goal of achieving 128 bits of security.

You need a 2*n bit hash to achieve n bit collision resistance, and you need a t*n bit hash to achieve n bit second pre-image resistance. If we assume a conservative 4 billion targets, and a 128 bit security level, this leads to 256 bit hashes for collision resistance and 160 bit hashes for multi-target second-preimages.

So why does he hash twice? I suspect it's in order to prevent length-extension attacks.

SHA-2, like all Merkle-Damgard hashes suffers from a property called "length-extension". This allows an attacker who knows H(x) to calculate H(x||y) without knowing x. This is usually not a problem, but there are some uses where it totally breaks the security. The most relevant example is using H(k||m) as MAC, where an attacker can easily calculate a MAC for m||m'. I don't think Bitcoin ever uses hashes in a way that would suffer from length extensions, but I guess Satoshi went with the safe choice of preventing it everywhere.

To avoid this property, Ferguson and Schneier suggested using SHA256d = SHA256(SHA256(x)) which avoids length-extension attacks. This construction has some minor weaknesses (not relevant to bitcoin), so I wouldn't recommend it for new protocols, and would use HMAC with constant key, or truncated SHA512 instead.

Some related reading:

Here's the main hashing function:

template<typename T1>
inline uint256 Hash(const T1 pbegin, const T1 pend)
{
    static unsigned char pblank[1];
    uint256 hash1;
    SHA256((pbegin == pend ? pblank : (unsigned char*)&pbegin[0]), (pend - pbegin) * sizeof(pbegin[0]), (unsigned char*)&hash1);
    uint256 hash2;
    SHA256((unsigned char*)&hash1, sizeof(hash1), (unsigned char*)&hash2);
    return hash2;
}

I'd say anyplace that calls that uses SHA256 twice.

As for why, see this.

Hashing the block in a Merkle tree and with two rounds of SHA256 increases in security, so both function would have to be broken to undo the hash.

Senders specify a locking script and recipients specify an unlocking script. These stacked hashes confirm proof of work by the sending UTXO and legitimatecy of the recipient to receive the transaction.

  • This doesn't make sense and doesn't answer the question – MeshCollider Aug 13 at 0:30

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