Wombat—An Efficient Stableswap Algorithm

Abstract

Curve Finance invented the first stableswap-focused algorithm. However, its algorithm involves (1) solving complex polynomials and (2) requiring assets in the pool to have the same size of liquidity. This paper introduces a new stableswap algorithm–Wombat, to address these issues. Wombat uses a closed-form solution, so it is more gas efficient and adds the concept of asset-liability management to enable single-side liquidity provision, which increases capital efficiency. Furthermore, we derive efficient algorithms from calculating withdrawal or deposit fees as an arbitrage block. Wombat is named after the short-legged, muscular quadrupedal marsupials native to Australia. As Wombats are adaptable and habitat-tolerant animals, the invariant created is also adaptable and tolerant to liquidity changes.

Appendices

A Screenshots of Uniswap and Curve

See Figs. 2 and 3.

Fig. 2

Uniswap screenshot

Fig. 3

Curve screenshot

B Proof of Theorem 2

Proof

If $r^*=1$, then $D=\left( \sum _{k\in \mathcal {T}}L_k\right) (1-A)$ by Eq. (5), ${r^*}'=1$ by Eq. (6), and $D'=\left( \delta ^L_i+\sum _{k\in \mathcal {T}}L_k\right) (1-A)$ by Eq. (7). Hence, Eq. (8) becomes

$$\begin{aligned} (L_i+\delta ^L_i)\left( \frac{A_i+\delta ^A_i}{L_i+\delta ^L_i}-\frac{A(L_i+\delta ^L_i)}{A_i+\delta ^A_i}\right) +\sum _{k\in \mathcal {T}\setminus \{i\}}L_k\left( r_k-\frac{A}{r_k}\right)&=\left( \delta ^L_i+\sum _{k\in \mathcal {T}}L_k\right) (1-A)\nonumber \\ A_i+\delta ^A_i-\frac{A(L_i+\delta ^L_i)^2}{A_i+\delta ^A_i}+D-L_i\left( r_i-\frac{A}{r_i}\right)&=\delta ^L_i(1-A)+D\nonumber \\ \delta ^A_i-\frac{A(L_i+\delta ^L_i)^2}{A_i+\delta ^A_i}+\frac{AL_i^2}{A_i}&=\delta ^L_i(1-A). \end{aligned}$$

(13)

If $\delta ^L_i<0$ and $\delta ^A_i\ge 0$, then the left hand side (LHS) of Eq. (13) is positive while the right hand side (RHS) is negative, a contradiction. Similarly, if $\delta ^L_i>0$ and $\delta ^A_i\le 0$, then the LHS is negative while the RHS is positive, a contradiction again. Hence, $\delta ^L_i$ and $\delta ^A_i$ always share the same sign.

To compare $\delta ^L_i$ and $\delta ^A_i$, we first rewrite Eq. (9) as

$$\begin{aligned} b'&=\frac{1}{L_i+\delta ^L_i}\left( \left( \sum _{k\in \mathcal {T}}L_k\right) (1-A)-L_i\left( r_i-\frac{A}{r_i}\right) -\left( \delta ^L_i+\sum _{k\in \mathcal {T}}L_k\right) (1-A)\right) \\&=-\frac{1}{L_i+\delta ^L_i}\left( L_i\left( r_i-\frac{A}{r_i}\right) +\delta ^L_i(1-A)\right) . \end{aligned}$$

Therefore, Eq. (10) yields

$$\begin{aligned}&\delta ^A_i-\delta ^L_i\\&=\frac{-(L_i+\delta ^L_i)b'+\sqrt{\left( (L_i+\delta ^L_i)b'\right) ^2+4A(L_i+\delta ^L_i)^2}}{2}-L_ir_i-\delta ^L_i\\&=\frac{-L_i\left( r_i+\frac{A}{r_i}\right) -\delta ^L_i(1+A)+\sqrt{\left( L_i\left( r_i-\frac{A}{r_i}\right) +\delta ^L_i(1-A)\right) ^2+4A(L_i+\delta ^L_i)^2}}{2}. \end{aligned}$$

Note that

$$\begin{aligned} \left( r_i-\dfrac{A}{r_i}\right) ^2+4A=r_i^2-2A+\dfrac{A^2}{r_i^2}+4A=\left( r_i+\dfrac{A}{r_i}\right) ^2\end{aligned}$$

and

$$\begin{aligned} (1-A)^2+4A=1-2A+A^2+4A=(1+A)^2.\end{aligned}$$

Furthermore, by the AM-GM inequality, we have $r_i+\dfrac{1}{r_i}\ge 2$, so

$$\begin{aligned} \left( r_i-\frac{A}{r_i}\right) (1-A)+4A&=r_i+\frac{A^2}{r_i}-A\left( r_i+\frac{1}{r_i}\right) +4A\\&\le r_i+\frac{A^2}{r_i}-A\left( r_i+\frac{1}{r_i}\right) +2A\left( r_i+\frac{1}{r_i}\right) \\&=\left( r_i+\frac{A}{r_i}\right) (1+A), \end{aligned}$$

where the equality holds if and only if $r_i=1$. As a result,

$$\begin{aligned}&\left( L_i\left( r_i-\frac{A}{r_i}\right) +\delta ^L_i(1-A)\right) ^2+4A(L_i+\delta ^L_i)^2\\&=L_i^2\left( r_i-\frac{A}{r_i}\right) ^2+4AL_i^2+2L_i\left( r_i-\frac{A}{r_i}\right) \delta ^L_i(1-A)+8AL_i\delta ^L_i\\&\qquad +(\delta ^L_i)^2(1-A)^2+4A(\delta ^L_i)^2\\&=L_i^2\left( r_i+\frac{A}{r_i}\right) ^2+2L_i\delta ^L_i\left( \left( r_i-\frac{A}{r_i}\right) (1-A)+4A\right) +(\delta ^L_i)^2(1+A)^2. \end{aligned}$$

If $\delta ^L_i<0$, then

$$\begin{aligned}&\sqrt{\left( L_i\left( r_i-\frac{A}{r_i}\right) +\delta ^L_i(1-A)\right) ^2+4A(L_i+\delta ^L_i)^2}\\&\ge \sqrt{L_i^2\left( r_i+\frac{A}{r_i}\right) ^2+2L_i\delta ^L_i\left( r_i+\frac{A}{r_i}\right) (1+A)+(\delta ^L_i)^2(1+A)^2}\\&=\left| L_i\left( r_i+\frac{A}{r_i}\right) +\delta ^L_i(1+A)\right| , \end{aligned}$$

so $\delta ^A_i-\delta ^L_i\ge 0$, with the equality holds if and only if $r_i=1$; if $\delta ^L_i>0$, then

$$\begin{aligned}&\sqrt{\left( L_i\left( r_i-\frac{A}{r_i}\right) +\delta ^L_i(1-A)\right) ^2+4A(L_i+\delta ^L_i)^2}\\&\le \sqrt{L_i^2\left( r_i+\frac{A}{r_i}\right) ^2+2L_i\delta ^L_i\left( r_i+\frac{A}{r_i}\right) (1+A)+(\delta ^L_i)^2(1+A)^2}\\&=\left| L_i\left( r_i+\frac{A}{r_i}\right) +\delta ^L_i(1+A)\right| , \end{aligned}$$

so $\delta ^A_i-\delta ^L_i\le 0$, again with the equality holds if and only if $r_i=1$.