We participated in darkCTF. Here are the writeups for the blockchain challenges I solved.

Table of Contents§

[Cryptography] Duplicacy Within§

Looks like Mr. Jones has found a secret key. Can you retrieve it like him?

Format : darkCTF{hex value of key}

https://www.blockchain.com/btc/tx/83415dded4757181c6e1c55104e2742a6f8cff05a9a46fbf029ae47b0054d511

z1 = 0xc0e2d0a89a348de88fda08211c70d1d7e52ccef2eb9459911bf977d587784c6e

z2 = 0x17b0f41c8c337ac1e18c98759e83a8cccbc368dd9d89e5f03cb633c265fd0ddc

I googled the value of z1 and found the following tool.

daedalus/bitcoin-recover-privkey: Proof of concept of bitcoin private key recovery using weak ECDSA signatures

The ECDSA random number generation of Bitcoin was incomplete, and the vulnerability existed that allowed for the calculation of a private key. Bitcoin has already fixed this vulnerability by implementing "RFC 6979: Deterministic Usage of the Digital Signature Algorithm (DSA) and Elliptic Curve Digital Signature Algorithm (ECDSA)."

I run the following script.

#!/usr/bin/env python
#
# Proof of concept of bitcoin private key recovery using weak ECDSA signatures
#
# Based on http://www.nilsschneider.net/2013/01/28/recovering-bitcoin-private-keys.html
# Regarding Bitcoin Tx https://blockchain.info/tx/9ec4bc49e828d924af1d1029cacf709431abbde46d59554b62bc270e3b29c4b1.
# As it's said in the previous article you need to poke around into the OP_CHECKSIG function in order to get z1 and z2,
# in other hand for every other parameters you should be able to get them from the Tx itself.
#
# Author Dario Clavijo <dclavijo@protonmail.com> , Jan 2013
# Donations: 1LgWNdNTnzeNgNMzWHtPtXPjxcutJKu74r
#
# This code is licensed under the terms of the GPLv3 license http://gplv3.fsf.org/
#
# Disclaimer: Do not steal other peoples money, that's bad.

# The math
# Q=dP compute public key Q where d is a secret scalar and G the base point
# (x1,y1)=kP where k is random choosen an secret
# r= x1 mod n
# compute k**-1 or inv(k)
# compute z=hash(m)
# s= inv(k)(z + d) mod n
# sig=k(r,s) or (r,-s mod n)
# Key recovery
# d = (sk-z)/r where r is the same 

import hashlib

tx = "83415dded4757181c6e1c55104e2742a6f8cff05a9a46fbf029ae47b0054d511"
p  = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
#r  = 0xd47ce4c025c35ec440bc81d99834a624875161a26bf56ef7fdc0f5d52f843ad1
#s1 = 0x44e1ff2dfd8102cf7a47c21d5c9fd5701610d04953c6836596b4fe9dd2f53e3e
#s2 = 0x9a5f1c75e461d7ceb1cf3cab9013eb2dc85b6d0da8c3c6e27e3a5a5b3faa5bab
z1 = 0xc0e2d0a89a348de88fda08211c70d1d7e52ccef2eb9459911bf977d587784c6e
z2 = 0x17b0f41c8c337ac1e18c98759e83a8cccbc368dd9d89e5f03cb633c265fd0ddc

# r1 and s1 are contained in this ECDSA signature encoded in DER (openssl default).
der_sig1 = "30440220d47ce4c025c35ec440bc81d99834a624875161a26bf56ef7fdc0f5d52f843ad102202f88bf73d0f94a1e917d1a6e65ba15a9dbf52d0999c91f2c2c6bb710e018f7e001"

# the same thing with the above line.
der_sig2 = "30440220d47ce4c025c35ec440bc81d99834a624875161a26bf56ef7fdc0f5d52f843ad102203602aff824a32c19825425704546145d5fbc282ee912089923e824f46867647b01"

params = {'p':p,'sig1':der_sig1,'sig2':der_sig2,'z1':z1,'z2':z2}

def hexify (s, flip=False):
    if flip:
        return s[::-1].encode ('hex')
    else:
        return s.encode ('hex')

def unhexify (s, flip=False):
    if flip:
        return s.decode ('hex')[::-1]
    else:
        return s.decode ('hex')

def inttohexstr(i):
	tmpstr = hex(i)
	hexstr = tmpstr.replace('0x','').replace('L','').zfill(64)
	return hexstr

b58_digits = '123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz'

def dhash(s):
    return hashlib.sha256(hashlib.sha256(s).digest()).digest()

def rhash(s):
    h1 = hashlib.new('ripemd160')
    h1.update(hashlib.sha256(s).digest())
    return h1.digest()

def base58_encode(n):
    l = []
    while n > 0:
        n, r = divmod(n, 58)
        l.insert(0,(b58_digits[r]))
    return ''.join(l)

def base58_encode_padded(s):
    res = base58_encode(int('0x' + s.encode('hex'), 16))
    pad = 0
    for c in s:
        if c == chr(0):
            pad += 1
        else:
            break
    return b58_digits[0] * pad + res

def base58_check_encode(s, version=0):
    vs = chr(version) + s
    check = dhash(vs)[:4]
    return base58_encode_padded(vs + check)

def get_der_field(i,binary):
        if (ord(binary[i]) == 02):
                length = binary[i+1]
                end = i + ord(length) + 2
                string = binary[i+2:end]
                return string
        else:
                return None

# Here we decode a DER encoded string separating r and s
def der_decode(hexstring):
        binary = unhexify(hexstring)
        full_length = ord(binary[1])
        if ((full_length + 3) == len(binary)):
                r = get_der_field(2,binary)
                s = get_der_field(len(r)+4,binary)
                return r,s
        else:
                return None

def show_results(privkeys):
		print "Posible Candidates..."
		for privkey in privkeys:
        		hexprivkey = inttohexstr(privkey)
			# print "intPrivkey = %d"  % privkey
			print "darkCTF{%s}" % hexprivkey
			# print "bitcoin Privkey (WIF) = %s" % base58_check_encode(hexprivkey.decode('hex'),version=128)
			# print "bitcoin Privkey (WIF compressed) = %s" % base58_check_encode((hexprivkey + "01").decode('hex'),version=128)


def show_params(params):
	for param in params:
		try:
			print "%s: %s" % (param,inttohexstr(params[param]))
		except:
			print "%s: %s" % (param,params[param])

# By the Fermat's little theorem we can say that:
# a * pow(b,p-2,p) is the same as (a/b mod p) 
# This is needed to avoid floating numbers since we are dealing with prime numbers 
# and beacuse this the python built in division isn't suitable for our needs,
# it returns floating point numbers rounded and we don't want them.
def inverse_mult(a,b,p):
	y =  (a * pow(b,p-2,p))  #(pow(a, b) modulo p) where p should be a prime number
	return y

# Here is the wrock!
def derivate_privkey(p,r,s1,s2,z1,z2):
        privkey = []

        s1ms2 = s1-s2
        s1ps2 = s1+s2
        ms1ms2 = -s1-s2
        ms1ps2 = -s1+s2
        z1ms2 = z1*s2
        z2ms1 = z2*s1
        z1s2mz2s1 = z1ms2-z2ms1
        z1s2pz2s1 = z1ms2+z2ms1
        rs1ms2 = r*s1ms2
        rs1ps2 = r*s1ps2
        rms1ms2 = r*ms1ms2
        rms1ps2 = r*ms1ps2

        privkey.append(inverse_mult(z1s2mz2s1,rs1ms2,p) % p)
        privkey.append(inverse_mult(z1s2mz2s1,rs1ps2,p) % p)
        privkey.append(inverse_mult(z1s2mz2s1,rms1ms2,p) % p)
        privkey.append(inverse_mult(z1s2mz2s1,rms1ps2,p) % p)
        privkey.append(inverse_mult(z1s2pz2s1,rs1ms2,p) % p)
        privkey.append(inverse_mult(z1s2pz2s1,rs1ps2,p) % p)
        privkey.append(inverse_mult(z1s2pz2s1,rms1ms2,p) % p)
        privkey.append(inverse_mult(z1s2pz2s1,rms1ps2,p) % p)

        return privkey

def process_signatures(params):

	p = params['p']
	sig1 = params['sig1']
	sig2 = params['sig2']
	z1 = params['z1']
	z2 = params['z2']

	tmp_r1,tmp_s1 = der_decode(sig1) # Here we extract r and s from the signature encoded in DER.
	tmp_r2,tmp_s2 = der_decode(sig2) # Idem.

	# the key of ECDSA are the integer numbers thats why we convert hexa from to them.
	r1 = int(tmp_r1.encode('hex'),16)
	r2 = int(tmp_r2.encode('hex'),16)
	s1 = int(tmp_s1.encode('hex'),16)
	s2 = int(tmp_s2.encode('hex'),16)

	if (r1 == r2): # If r1 and r2 are equal the two signatures are weak and we can recover the private key.
 		if (s1 != s2): # This: (s1-s2)>0 should be complied in order be able to compute the private key.
			privkey = derivate_privkey(p,r1,s1,s2,z1,z2)
			return privkey
		else:
			raise Exception("Privkey not computable: s1 and s2 are equal.")
	else:
		raise Exception("Privkey not computable: r1 and r2 are not equal.")

def main():
	show_params(params)
	privkey = process_signatures(params)
	if len(privkey)>0:
		show_results(privkey)

if __name__ == "__main__":
    main()

The result is as follows.

$ pyenv local 2.7.18
$ python ProofOfConcept.py 
sig1: 30440220d47ce4c025c35ec440bc81d99834a624875161a26bf56ef7fdc0f5d52f843ad102202f88bf73d0f94a1e917d1a6e65ba15a9dbf52d0999c91f2c2c6bb710e018f7e001
p: fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
sig2: 30440220d47ce4c025c35ec440bc81d99834a624875161a26bf56ef7fdc0f5d52f843ad102203602aff824a32c19825425704546145d5fbc282ee912089923e824f46867647b01
z1: c0e2d0a89a348de88fda08211c70d1d7e52ccef2eb9459911bf977d587784c6e
z2: 17b0f41c8c337ac1e18c98759e83a8cccbc368dd9d89e5f03cb633c265fd0ddc
Posible Candidates...
darkCTF{791198f7b09c5e63fc5798df41c4090d2265d8066e4d4a917a9d604f17ccf856}
darkCTF{12cba205306996b4fc6d9f6a4b920cebecf0c7b88b2b773af0c3b6a551b16339}
darkCTF{ed345dfacf96694b03926095b46df312cdbe152e241d2900cf0ea7e77e84de08}
darkCTF{86ee67084f63a19c03a86720be3bf6f1984904e040fb55aa4534fe3db86948eb}
darkCTF{4d35700e35810e99564f9aeade7c0687298d0b401f2096845eb6f24285f71115}
darkCTF{a02e8abb6d006105a2325510e70f723f4f0ad8dbba57605a39cfec10d47e695c}
darkCTF{5fd1754492ff9efa5dcdaaef18f08dbf6ba4040af4f13fe18602727bfbb7d7e5}
darkCTF{b2ca8ff1ca7ef166a9b065152183f9779121d1a6902809b7611b6c4a4a3f302c}

The flag is darkCTF{791198f7b09c5e63fc5798df41c4090d2265d8066e4d4a917a9d604f17ccf856}

[Misc] Secret Of The Contract§

Ropsten network contains my dark secret. Help us find it. Name of the contract was 0x6e5EA18371748Db7F12A70037d647cDFCf458e45

I took a look at the transactions in the Ropsten network explorer (Etherscan).

https://ropsten.etherscan.io/tx/0x55cb76f021a69c1c5bae2830af946d4b11022c94dd91eeb65d520c11a3b5fca3

There are two successful transactions associated with this contract.

The first is the contract creation transaction. I looked at the change in state due to this transaction.

There is a hexadecimal string of alphabets. The decoding result:

$ echo 3337373233343665333533343633333733313330366537640000000000000030 | xxd -r -p
3772346e3534633731306e7d0
$ echo 3772346e3534633731306e7d0 | xxd -r -p
7r4n54c710n}

This string appears to be the end of the flag.

I looked at the change in the state due to the second transaction.

The decoding result:

$ echo 486d6d2d36343631373236423433353434363742333337343638333337323333 | xxd -r -p
Hmm-6461726B4354467B337468337233⏎
$ echo 6461726B4354467B337468337233 | xxd -r -p
darkCTF{3th3r3

$ echo 3735364435463335333733303732333436373333354600000000000000000000 | xxd -r -p
756D5F353730723467335F
$ echo 756D5F353730723467335F | xxd -r -p
um_570r4g3_

I joined together all the pieces and got the flag: darkCTF{3th3r3um_570r4g3_7r4n54c710n}