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Minswap Stableswap Formula

References: https://miguelmota.com/blog/understanding-stableswap-curve/

Base on the Article above, we finalize Stableswap formula is: $$A.n^{n}.\sum_{0}^{i}x + D = A.D.n^{n} + \frac{D^{n+1}}{n^{n}.\prod_{0}^{i}x} $$

Before we go to details, we need define some variables:

  • n: Number of Tokens
  • A: Amplification coefficient factor
  • D: Sum variant
  • $x$: Pool balance multiples: the multiple of stable assets, using for calculation between stable assets with different decimals

Core formula

Almost all Stableswap calculations are using 2 core formulas.

First is calculating Constant Sum Variant (get D) and calculating balance after exchaging (get Y)

get D

From the Stableswap formula, we change it to $$\frac{D^{n+1}}{n^{n}.\prod_{0}^{i}x} + D.(A.n^{n} - 1) - A.n^{n}.\sum_{0}^{i}x = 0$$

$$f(D) = \frac{D^{n+1}}{n^{n}.\prod_{0}^{i}x} + D.(A.n^{n} - 1) - A.n^{n}.\sum_{0}^{i}x$$ $$f'(D) = (n+1).\frac{D^{n}}{n^{n}.\prod_{0}^{i}x} + (A.n^{n} - 1)$$

Base on Newton method, we find the approximation root of the formula above:

$$x^{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}$$

$$D_{n+1} = D_{n} - \frac{f(D_{n})}{f'(D_{n})}$$ $$D_{n+1} = D_{n} - \frac{\frac{D_{n}^{n+1}}{n^{n}.\prod_{0}^{i}x} + D_{n}.(A.n^{n} - 1) - A.n^{n}.\sum_{0}^{i}x}{(n+1).\frac{D^{n}}{n^{n}.\prod_{0}^{i}x} + (A.n^{n} - 1)}$$

$$D_{n+1} = \frac{D_{n} * ((n+1).\frac{D^{n}}{n^{n}.\prod_{0}^{i}x} + (A.n^{n} - 1)) - \frac{D_{n}^{n+1}}{n^{n}.\prod_{0}^{i}x} + D_{n}.(A.n^{n} - 1) - A.n^{n}.\sum_{0}^{i}x}{(n+1).\frac{D^{n}}{n^{n}.\prod_{0}^{i}x} + (A.n^{n} - 1)}$$

Let $DP = \frac{D_{n}^{n+1}}{n^{n} . \prod_{0}^{i}x}$ and we multiple $D_{n}$ on both numerator and denominator of right side, we have:

$$D_{n+1} = \frac{(n . DP + A.n^{n}.\sum_{0}^{i}x).D_{n}}{(n + 1).DP + (A.n^{n} - 1).D_{n}}$$

get Y

Let say, after exchanging, the new balance sum and new balance prod is: $\sum_{0}^{i}x'$ and $\prod_{0}^{i}x'$

$$A.n^{n}.\sum_{0}^{i}x' + D = A.D.n^{n} + \frac{D^{n+1}}{n^{n}.\prod_{0}^{i}x'}$$ Equal to $$A.n^{n}.(\sum_{0}^{i}x' - y + y) + D = A.D.n^{n} + \frac{D^{n+1}}{n^{n}.\frac{\prod_{0}^{i}x'}{y}.y}$$ Equal to $$A.n^{n}.y^{2} + y . (A.n^{n}.(\sum_{0}^{i}x' - y) + D - A.n^{n}.D) - \frac{D^{n+1}}{n^{n}.\frac{\prod_{0}^{i}x'}{y}} = 0$$ Equal to $$y^{2} + y . ((\sum_{0}^{i}x' - y) + \frac{D}{A.n^{n}} - D) - \frac{D^{n+1}}{A.n^{n}.n^{n}.\frac{\prod_{0}^{i}x'}{y}} = 0$$

Use Newton method, we have $$f(y) = y^{2} + y . ((\sum_{0}^{i}x' - y) + \frac{D}{A.n^{n}} - D) - \frac{D^{n+1}}{A.n^{n}.n^{n}.\frac{\prod_{0}^{i}x'}{y}}$$ $$f'(y) = 2.y + ((\sum_{0}^{i}x' - y) + \frac{D}{A.n^{n}} - D)$$ Then $$y_{i} = \frac{y_{i-1}^{2} + \frac{D^{n+1}}{A.n^{n}.n^{n}.\frac{\prod_{0}^{i}x'}{y_{i-1}}}}{2.y_{i-1} + ((\sum_{0}^{i}x' - y_{i-1}) + \frac{D}{A.n^{n}} - D)}$$

Use case formula

On our Stableswap, we do not use Token Balance for calculation because Stableswap Tokens can have different decimals. So:

  • Calculation Balance: Token Balance * Multiple
  • $A.n^{n}$: A variable on Pool Datum

You can checkout the code to see more about each use case formula:

  • Deposit: calculate_deposit
  • Exchange: calculate_exchange
  • Withdraw: calculate_withdraw
  • Withdraw Imbalance: validate_withdraw_imbalance
  • Withdraw One Coin: calculate_withdraw_one_coin