How to get uncompressed public key from compressed one ?

[Farghaly]

Uncompressed public key is:
0x04 + x-coordinate + y-coordinate

Compressed public key is:
0x02 + x-coordinate if y is even
0x03 + x-coordinate if y is odd

How to use this equation to derive the uncompressed public key

y^2 mod p = (x^3 + 7) mod p

 

[TimS]

From https://en.wikipedia.org/wiki/Quadratic_residue and http://mersennewiki.org/index.php/Modular_Square_Root, if

r^2 = a mod m where m = 3 mod 4 (as secp256k1's p does)
then
r = +-a^((m+1)/4) mod m

So:

y^2 mod p = (x^3 + 7) mod p
y mod p = +-(x^3 + 7)^((p+1)/4) mod p

So calculate (x^3 + 7)^((p+1)/4) mod p, and if the parity of the first answer you get is wrong, then take the negative of that answer (since we're working modulo an odd number, taking the negative will flip the even/odd parity).

Just for fun, here's some Python code that'll do the calculation in the blink of an eye, preset to work on the public key of (the random) private key 55255657523dd1c65a77d3cb53fcd050bf7fc2c11bb0bb6edabdbd41ea51f641

Code:

def pow_mod(x, y, z):
    "Calculate (x ** y) % z efficiently."
    number = 1
    while y:
        if y & 1:
            number = number * x % z
        y >>= 1
        x = x * x % z
    return number

p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
compressed_key = '0314fc03b8df87cd7b872996810db8458d61da8448e531569c8517b469a119d267'
y_parity = int(compressed_key[:2]) - 2
x = int(compressed_key[2:], 16)
a = (pow_mod(x, 3, p) + 7) % p
y = pow_mod(a, (p+1)//4, p)
if y % 2 != y_parity:
    y = -y % p
uncompressed_key = '04{:x}{:x}'.format(x, y)
print(uncompressed_key)

[deepceleron]

Be mindful that there is no practical reason to "convert" between uncompressed and compressed public keys. Each has their own Bitcoin address - you cannot spend funds sent to the uncompressed public key's address with the compressed public key.

[ftc-c]

Who can give a C++ implementation ?

[fbueller]

@ftc-c: Use libsecp256k1, AFAIK it has preprocessor stuff for C++ support, and it's easy enough to work through.

[samson]

The following link shows how to do it in C using the polarssl / mbedTLS library, you should be able to port this across a different crypto library so long as it has a proper 'bignum'/mpi implementation.

https://gist.github.com/flying-fury/6bc42c8bb60e5ea26631

The above example contains test data as well, from the comments '// Set y2 = X^3 + B' and '// Compute square root of y2' you can see where the important part is done.

I used the above to make a dll/so/dylib which does this useful conversion based on the above code.

What deepceleron said also applies, there are two distinct coin addresses which can be created from a single private key depending on whether the public key is 'compressed' or not.

The above conversion is still useful if you use ECC for things like verifying signatures.

[CIYAM]

Welcome to use the CIYAM code which includes parts of the Bitcoin code as well (but might be easier to follow in terms of the C++ classes as it doesn't involve any Boost stuff).

https://github.com/ciyam/ciyam/blob/master/src/crypto_keys.cpp

[Blockchain Mechanic]

Be mindful that there is no practical reason to "convert" between uncompressed and compressed public keys. Each has their own Bitcoin address - you cannot spend funds sent to the uncompressed public key's address with the compressed public key.

Hi

So I could use both addresses to receive funds so long as i have the one private key?

[ranochigo]

Be mindful that there is no practical reason to "convert" between uncompressed and compressed public keys. Each has their own Bitcoin address - you cannot spend funds sent to the uncompressed public key's address with the compressed public key.

Hi

So I could use both addresses to receive funds so long as i have the one private key?

Yes. The compressed public key and the uncompressed ones have different public keys but they can be derived from the same private key. However, they may be different in the Wallet Import Format, private keys starting with 5 are compressed while the uncompressed ones starts from K or L.

[chutium]

I wrote a converter in javascript language, may will be useful for mobile apps or browser extensions.

It is using javascript big integer library latest version 1.6.36, and work with hex string directly:

Code:

const bigInt = require("big-integer");

function pad_with_zeroes(number, length) {
    var retval = '' + number;
    while (retval.length < length) {
        retval = '0' + retval;
    }
    return retval;
}

// Consts for secp256k1 curve. Adjust accordingly
// https://en.bitcoin.it/wiki/Secp256k1
const prime = new bigInt('fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f', 16),
pIdent = prime.add(1).divide(4);

/**
 * Point decompress secp256k1 curve
 * @param {string} Compressed representation in hex string
 * @return {string} Uncompressed representation in hex string
 */
function ECPointDecompress( comp ) {
    var signY = new Number(comp[1]) - 2;
    var x = new bigInt(comp.substring(2), 16);
    // y mod p = +-(x^3 + 7)^((p+1)/4) mod p
    var y = x.modPow(3, prime).add(7).mod(prime).modPow( pIdent, prime );
    // If the parity doesn't match it's the *other* root
    if( y.mod(2).toJSNumber() !== signY ) {
        // y = prime - y
        y = prime.subtract( y );
    }
    return '04' + pad_with_zeroes(x.toString(16), 64) + pad_with_zeroes(y.toString(16), 64);
}

Test with private key from Tim's example 55255657523dd1c65a77d3cb53fcd050bf7fc2c11bb0bb6edabdbd41ea51f641:

Code:

ECPointDecompress('0314fc03b8df87cd7b872996810db8458d61da8448e531569c8517b469a119d267')

returns:

Code:

'0414fc03b8df87cd7b872996810db8458d61da8448e531569c8517b469a119d267be5645686309c6e6736dbd93940707cc9143d3cf29f1b877ff340e2cb2d259cf'

refer to https://stackoverflow.com/a/53480175/5630352

From https://en.wikipedia.org/wiki/Quadratic_residue and http://mersennewiki.org/index.php/Modular_Square_Root, if

r^2 = a mod m where m = 3 mod 4 (as secp256k1's p does)
then
r = +-a^((m+1)/4) mod m

So:

y^2 mod p = (x^3 + 7) mod p
y mod p = +-(x^3 + 7)^((p+1)/4) mod p

So calculate (x^3 + 7)^((p+1)/4) mod p, and if the parity of the first answer you get is wrong, then take the negative of that answer (since we're working modulo an odd number, taking the negative will flip the even/odd parity).

Just for fun, here's some Python code that'll do the calculation in the blink of an eye, preset to work on the public key of (the random) private key 55255657523dd1c65a77d3cb53fcd050bf7fc2c11bb0bb6edabdbd41ea51f641

Code:

def pow_mod(x, y, z):
    "Calculate (x ** y) % z efficiently."
    number = 1
    while y:
        if y & 1:
            number = number * x % z
        y >>= 1
        x = x * x % z
    return number

p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
compressed_key = '0314fc03b8df87cd7b872996810db8458d61da8448e531569c8517b469a119d267'
y_parity = int(compressed_key[:2]) - 2
x = int(compressed_key[2:], 16)
a = (pow_mod(x, 3, p) + 7) % p
y = pow_mod(a, (p+1)//4, p)
if y % 2 != y_parity:
    y = -y % p
uncompressed_key = '04{:x}{:x}'.format(x, y)
print(uncompressed_key)

[Frodek]

We have:
const prime = new bigInt('fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f', 16),
    if( y.mod(2).toJSNumber() !== signY ) {
        // y = prime - y
        y = prime.subtract( y );

I wonder, how to compute y = prime - y when y will larger than prime?
(may be problem with signed/big unsinged)
Very large case, I don't know how to reproduce,
this cases will throw away when generated private keys?

[pooya87]

I wonder, how to compute y = prime - y when y will larger than prime?

in modular arithmetic, the numbers in the group defined by p aren't bigger than p, if they were their remainder (mod p) is calculated hence the finite group. if prime was 5 the group is like this:
{0, 1, 2, 3, 4} when we have 5 we calculate its mod prime and put 0 in its place. for 6 we put 1,...

as for y, it is the vertical coordinate of the points on the curve. and -y is not like regular arithmetic where you flip the sign, -y is defined as (prime - y) so that we mirror the point over x axis and find the other point on curve.

[brainless]

https://bitcointalk.org/index.php?topic=5244940.msg57700007#msg57700007

easy script mention here, where you can load pubkeys from file and result print back into new file

[brainless]

compress to uncompress

Code:

import binascii

p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F

def decompress_pubkey(pk):
    x = int.from_bytes(pk[1:33], byteorder='big')
    y_sq = (pow(x, 3, p) + 7) % p
    y = pow(y_sq, (p + 1) // 4, p)
    if y % 2 != pk[0] % 2:
        y = p - y
    y = y.to_bytes(32, byteorder='big')
    return b'\x04' + pk[1:33] + y

with open('add.txt') as f:
  for line in f:
    line=line.strip()
    print(binascii.hexlify(decompress_pubkey(binascii.unhexlify(line))).decode(),file=open("uncomp.txt", "a"))

uncompress to compress

Code:

def cpub(x,y):
 prefix = '02' if y % 2 == 0 else '03'
 c = prefix+ hex(x)[2:].zfill(64)
 return c
with open('add.txt') as f:
  for line in f:
    line=line.strip()
    x = int(line[2:66], 16)
    y = int(line[66:], 16)
    pub04=cpub(x,y)

    print(pub04,file=open("comp.txt", "a"))