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chore: make calc_withdraw_one_coin simulation consistent with _calc_withdraw_one_coin #86

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chore: make calc_withdraw_one_coin simulation consistent with _calc_withdraw_one_coin #86

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c45b360
Make calc_withdraw_one_coin consistent with _calc_withdraw_one_coin
michaelhly Nov 21, 2020
d110681
Fix
michaelhly Nov 21, 2020
36191e2
Add test
michaelhly Nov 21, 2020
a5d0083
Fix
michaelhly Nov 21, 2020
a3d1d56
-1 to account for rounding errors
michaelhly Nov 21, 2020
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4 changes: 2 additions & 2 deletions tests/simulation.py
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Original file line number Diff line number Diff line change
Expand Up @@ -148,7 +148,7 @@ def remove_liquidity_imbalance(self, amounts):

def calc_withdraw_one_coin(self, token_amount, i):
xp = self.xp()
xp_reduced = xp
xp_reduced = list(xp)
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@halito22 halito22 Nov 17, 2021

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thank you


D0 = self.D()
D1 = D0 - token_amount * D0 // self.tokens
Expand All @@ -166,6 +166,6 @@ def calc_withdraw_one_coin(self, token_amount, i):
self.x = [x // (p // 10 ** 18) for x, p in zip(xp_reduced, self.p)]
dy = xp_reduced[i] - self.y_D(i, D1)
self.x = [x // (p // 10 ** 18) for x, p in zip(xp, self.p)]
dy_0 = (xp[i] - new_y)
dy_0 = xp[i] - new_y
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@halito22 halito22 Nov 17, 2021

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thank you


return dy, dy_0 - dy