The previous posts covered mathematical curiosities. Now we can leave that realm, find an application and turn them into tools to work with elliptic curve cryptography. That means finding and choosing which particular values of the parameters bring us into a curve we can use for cryptography.

The Bitcoin curve

The Bitcoin curve uses the secp256k1 curve parameters.

\(G\) and \(n\) expressed here as hexadecimal numbers are despite the small font rather large numbers. The prime number \(p\) is almost \(2^{256}\), which means most numbers under \(2^{256}\) are also in the prime field. What is also impressive is that \(n\) is also really close to \(2^{256}\). This means we can apply scalar multiplication to \(G\) as much as \(n\) times before we end at the point at infinity.

It escapes my current knowledge: how are these particular parameters found? They are publicly documented everywhere, yet I could not easily find which is the method to find them, that being the principal question of its security. Cryptography bases its security on the infeasibility of practicing some mathematical operations in a short time.

Testing that \(p\) is prime is already a challenge, naively you have to test the divisors up to \(\sqrt{p}\), and that is already a lot to verify and unfeasible under that approach. Yet, once validated that \(p\) is prime, you can work on that large finite field. Where the difficult problem arises is then choosing parameters \(a\) and \(b\). I have found the comment: for Bitcoin they are so simple, there shouldn’t be any backdoor in them, but there is no proof of that. With that choice of parameters, you keep compounding the problem. Now, you have the task to find the generator or basis point \(G\). In essence, it could be any point on the curve. It doesn’t seem like an easy task to find it, \(x^3\) grows much faster than \(y^2\), thus you need to start with a large number, before it shrinks through the modulo operation to and verifies the equality of our elliptic curve. Once you have that point, under scalar multiplication at a certain multiple of this point you reach the point at infinity, forming a finite cyclic group, that is the set:

\[ \{ G, 2G, 3G, 4G, …, nG \} \text{ where only } nG=0\]

\(n\) for Bitcoin is also a large number, and it is prime too. I keep wondering how this came out to be, testing that no other \(n\) takes you to the point at infinity is equivalent to having tested all Bitcoin private keys, which we assume to be computationally unfeasible. Also, how is it that they exactly found \(G\), the first point, and not stumbled into \(42G\) or any other point first. Even starting with that point, you can’t find \(n\) and give an order until you scan the entire space, which is unfeasible. Under the assumption that the parameters are secure, with the hint of evidence that the Bitcoin network has survived years of attacks. I, for the moment, move forward.

The numbers below \(n\) will become your private keys. Since \(n\) is large, there are a lot of numbers to pick your private keys from, and if you randomly pick a number(with an appropriate source of randomness), you’ll end up with a key, that nobody else has come up to choose by coincidence. That’s why you instantiate your wallet with that randomly chosen key, it is empty.

There have been some stories of hacked Bitcoin wallets, mainly hacks to your computing device, then guessing brain wallets which by construction have a low entropy, then hacks because some devices had poor randomness. After all those hacks, the hint of the safety of Bitcoin is that year after year, there has not been a large moment of funds from old wallets. Those old large funded wallets remain the bounty for which Bitcoin is constantly attacked and its security constantly audited. As the beginner I am, I still need to catch up on experience to convince my healthy skepticism that it is safe.

Specializing the code

The code presented in the previous posts is generic, and now we need to provide some auxiliary functions and some specialization to use the Bitcoin curve easily while programming. I introduce the new type aliases, function constructors and constants:

 1type S256Field = FieldElement (2 ^ 256- 2^ 32 - 977)
 2type NField
 3  = FieldElement
 4      0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
 5type S256Point = ECPoint S256Field
 7s256point :: S256Field -> S256Field -> S256Point
 8s256point x y =
 9  let p = ECPoint x y 0 7
10  in  if validECPoint p then p else error "Invalid point"
12ncons = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
13gcons = s256point
14  0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
15  0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8

For some reason, the compiler does not allow me to define type Nfield = FieldElement ncons, thus I have to explicitly declare that constant repeatedly.

Now we can also test that the order of \(G\) is \(n\), that is \(nG=0\):

1scalarProduct ncons gcons -- ECPoint(Infinity)

Digital signature and verification

The key relationship of public-key cryptography is \(P=eG\). It is easy to calculate \(P\) from \(e\) and \(G\), but we cannot easily find \(e\), given \(P\) and \(G\). \(P\) will be your public key, and \(e\) is your private key. \(e\) is an integer bellow \(n\) and \(P\) is a coordinate \((x,y)\) on the elliptic curve. \(P=eG\) is rather boring, you have only gone from private to public key, and know that such calculation is asymmetric. Easy from the scalar multiplication side, infeasible to divide.

We can thus build upon that relationship and construct a linear combination and have the equation

\[ uG + vP = kG \]

Re arranging the terms \[ vP = (k-u)G \]

Now requiring that \(v\neq0\)

\[ P = eG = v^{-1}(k-u) G \]

That implies for any combination of \((k, u, v)\) that satisfies the equation, we are also in knowledge of the private key \(e\). What we need is to put constraints into \((k, u, v)\) in such a way that they are meaningful. Also, revealing \((k,u,v)\) as shown now, reveals your secret key, thus it is important to provide these constants in a way that you don’t reveal your key.

The goal of a digital signature is to certify that a message originates from the owner of a specific key and nobody else. Thus, we need to mix information of the public key, the message and some additional arbitrary target to make it infeasible to forge the signature.

We start selecting the arbitrary target: \(R=kG\). That is an arbitrary point on the curve, with key: \(k\). However, we are only interested in the \(x\) coordinate of this point and call that number \(r\). The goal of the digital signature is that having \(r\) on the start, then mixing it with the message and the signer’s public key, we can find again the point in the curve with \(x\) coordinate being \(r\).

The message(\(m\)) we want to sign, is first fingerprinted through a hash function(\(HASH\)), which is also an asymmetric function that converts any arbitrary data into a fingerprint(\(z\)) (the hash) of fixed length. That equation resumes to \(z = HASH(m)\)

With that, we define our constants to include those values

\[ u = z / s , v = r / s \]

Here \(s\) is the signature we need to calculate. It is also important to realize that the division takes place on the prime field defined by the group order \(n\). That confused me for a while, but since all these constants are the factors we use to multiply the basis point \(G\), then it makes sense to define them over the group order and not over the prime field of the elliptic curve.

Calculating \(s\) from our previous equations:

\begin{align} uG &+ vP & = & kG \\\ uG &+ veG & = & kG \\\ u &+ ve & = & k \\\ z/s &+ re/s & = & k\\\ (z &+re)/s & = & k\\\ \end{align}

Finally, we obtain: \[s = (z +re)/k\]

A signature is then described by the pair of numbers \((r,s)\). That is the random target and the signature itself. The person willing to validate our signature needs those plus our public key and the signed message to calculate its hash. The verification of the signature results from using those values to calculate \(R=uG+vP\) and verifying that the point obtained has its \(x\) coordinate equal to \(r\).

It is important that we only reveal \(r\) and not \(k\), because as stated earlier, knowing \((k,u,v)\) reveals our private key. It is also important to never reuse \(k\) for different messages, because that will lead the signatures to leak the private key.

Implementing the digital signature algorithm

After such a long introduction it’s time to go back to Haskell code. First, I define a new datatype for the signature to hold the pair of numbers \((r,s)\)

1data Signature = Signature
2  { r :: S256Field
3  , s :: NField
4  } deriving (Show)

I’m really happy how Haskell made me aware that \(r\) and \(s\) are elements of different fields. In all the previous explanations, I superficially treat them as integers, but it is important when calculating division to be sure in which finite field you are working, and realizing that all the relevant calculations for multipliers of the point \(G\) happen on the finite field of the group order \(n\).

To use the scalarProduct defined in the previous posts, you also needed to convert back from the Finite Field to an integer. I tried to first implement Haskell’s standard toInteger, but I couldn’t. So I defined my auxiliary function:

1asInt :: KnownNat n => FieldElement n -> Integer
2asInt (FieldElement n) = n

With that, I have enough to implement the signature verification algorithm.

1verifySignanture :: NField -> Signature -> S256Point -> Bool
2verifySignanture z (Signature r s) pub = x target == r
3 where
4  target =
5    scalarProduct (asInt $ z / s) gcons
6      <> scalarProduct (asInt $ fromInteger (asInt r) / s) pub

There is a some back and forth conversion of types with asInt and fromInteger depending on each use case. I like that in Haskell I need to be explicit about it. It is more work, and certainly some noise, but I became aware of what I was doing, by understanding why the compiler complained and fixing it.

Using Haskell’s packaging

Everything up to this point, you could do by firing up your text editor and ghci. It all worked. However, to continue implementing the digital signature algorithm, additional modules become necessary and managing those with a package manager of some sort. I selected cabal because it seemed simple enough to get started.

I start by creating a cabal file with the following contents:

 1cabal-version:      2.4
 2name:               programmingbitcoin
 4author:             Óscar Nájera
 5maintainer:         [email protected]
12    hs-source-dirs:   src
13    exposed-modules:  Ecc
14    build-depends:    base ^>=
15                    , bytestring >=
16                    , cryptohash-sha256
17    default-language: Haskell2010

This is only to include the dependencies cryptohash-sha256 and bytestring. I get the feeling base is what I was already using directly loaded in ghci. I must also move my Haskell source file into src/Ecc.hs for it to be correctly found and compiled. That file now needs some extra things on its header:

1module Ecc where
3import qualified Crypto.Hash.SHA256 as SHA256
4import qualified Data.ByteString    as BS

That defines the Ecc module and exports everything I have defined. Then it imports the 2 packages that are in my new dependencies.

Hashing the message

The first step to sign a message is to calculate the hash of that message. In Bitcoin, the hash function is two rounds of the sha256 hash. I use the hash function from the package I imported cryptohash-sha256.

1hash256 :: BS.ByteString -> BS.ByteString
2hash256 = SHA256.hash . SHA256.hash

This function works on ByteString thus it is also a good idea, to put the language extension {-# LANGUAGE OverloadedStrings #-} in the header of the file, to be able to directly use Strings on the source code and have the compiler transform them to ByteString.

This hash function returns a ByteString, however we need \(z\) as a number, because we operate mathematically with it to calculate the signature. Thus the auxiliary function:

1fromBytes :: BS.ByteString -> Integer
2fromBytes = BS.foldl' f 0 where f a b = a `shiftL` 8 .|. fromIntegral b

as the function signature indicates, fromBytes takes me from ByteString to Integer. Then I can calculate the hash of the message directly to a number in the finite field defined by the group order with:

1sighash :: BS.ByteString -> NField
2sighash = fromInteger . fromBytes . hash256

Getting the random target deterministically

To create a signature we use a random number \(k\), which should be a cryptographically secure random number. Even if you had a good source of randomness, it seems to be a better practice to generate this number deterministically following the specification defined in RFC 6979 .

I first need the inverse function fromBytes to transform numbers into ByteString, which an extra constrain to always have them be 256 Bits (32 Bytes) long, for this I include the zeroPad function and use the greatness of function composition in Haskell.

 1integerToBytes :: Integer -> BS.ByteString
 2integerToBytes = BS.pack . go
 3 where
 4  go c = case c of
 5    0 -> []
 6    c -> go (c `div` 256) ++ [fromIntegral (c `mod` 256)]
 8zeroPad :: Integer -> BS.ByteString -> BS.ByteString
 9zeroPad n s = BS.append padding s
10 where
11  padding = BS.pack (replicate (fromIntegral n - fromIntegral (BS.length s)) 0)
13toBytes32 :: Integer -> BS.ByteString
14toBytes32 = zeroPad 32 . integerToBytes

The RFC describes the algorithm to generate \(k\) in an imperative way, and it reuses variables. Haskell forbids that, because everything is a constant. Thus each new transformation of a value gets a new name, until I reach some form of recursion. Have a look at my implementation.

 1deterministicK :: NField -> NField -> NField
 2deterministicK (FieldElement priv) (FieldElement z) = fromInteger
 3  $ candidate k2 v2
 4 where
 5  k    = BS.pack $ replicate 32 0
 6  v    = BS.pack $ replicate 32 1
 7  zbs  = toBytes32 z
 8  skbs = toBytes32 priv
 9  k1   = SHA256.hmac k $ v `BS.append` "\NUL" `BS.append` skbs `BS.append` zbs
10  v1   = SHA256.hmac k1 v
11  k2   = SHA256.hmac k1 $ v1 `BS.append` "\SOH" `BS.append` skbs `BS.append` zbs
12  v2   = SHA256.hmac k2 v1
14  candidate k v =
15    let vNew = SHA256.hmac k v
16        can  = fromBytes vNew
17    in  if can >= 1 && can < ncons
18          then can
19          else
20            let kp = SHA256.hmac k $ vNew `BS.append` "\NUL"
21                vp = SHA256.hmac kp vNew
22            in  candidate kp vp

Here I use also the SHA256.hmac from the library I imported and BS.pack transforms a list of 8bit integers into a ByteString. \NUL and \SOH are the zero and the one transformed into a ByteString, I just copied that directly from the ghci repl.

Signing the message

Signing the message follows the equations described in the previous section . The only thing extra is requiring as integers that \(s<n/2\).

1signMessage :: NField -> BS.ByteString -> Signature
2signMessage priv mesg =
3  let z               = sighash mesg
4      k               = deterministicK priv z
5      rm              = scalarProduct (asInt k) gcons
6      FieldElement sm = (z + fromInteger (asInt (x rm)) * priv) / k
7      ss              = if sm > div ncons 2 then ncons - sm else sm
8  in  Signature (x rm) (fromInteger ss)


This was another leap learning Haskell. Figuring out how to manage dependencies in a project. Learning how to convert between types, it is really good that I must do that explicitly in Haskell, and that it is really strict preventing you from mixing objects. I became aware of what I was trying to do by those simple type enforcements.

The next step is to worry about testing. At the moment all logic from these blog posts fits in 187 lines of code. I can still read it, and I have my manual test cases. However, as the project grows in complexity, keeping track of it becomes really hard and exhausting, test help to track features and document how to use the software.