If tasked with the problem of coming up with a zk-SNARK protocol, many people would make their way to this point and then get stuck and give up. How can a verifier possibly check every single piece of the computation, without looking at each piece of the computation individually? But it turns out that there is a clever solution.

Polynomials

Polynomials are a special class of algebraic expressions of the form:

• x+5
• x^4
• x^3+3x^2+3x+1
• 628x^{271}+318x^{270}+530x^{269}+…+69x+381

i.e. they are a sum of any (finite!) number of terms of the form cx^k

There are many things that are fascinating about polynomials. But here we are going to zoom in on a particular one: polynomials are a single mathematical object that can contain an unbounded amount of information (think of them as a list of integers and this is obvious). The fourth example above contained 816 digits of tau, and one can easily imagine a polynomial that contains far more.

Furthermore, a single equation between polynomials can represent an unbounded number of equations between numbers. For example, consider the equation A(x)+ B(x) = C(x). If this equation is true, then it's also true that:

• A(0)+B(0)=C(0)
• A(1)+B(1)=C(1)
• A(2)+B(2)=C(2)
• A(3)+B(3)=C(3)

And so on for every possible coordinate. You can even construct polynomials to deliberately represent sets of numbers so you can check many equations all at once. For example, suppose that you wanted to check:

• 12+1=13
• 10+8=18
• 15+8=23
• 15+13=28

You can use a procedure called Lagrange interpolation to construct polynomials A(x) that give (12,10,15,15) as outputs at some specific set of coordinates (eg. (0,1,2,3)), B(x) the outputs (1,8,8,13) on thos same coordinates, and so forth. In fact, here are the polynomials:

• A(x)=-2x^3+\frac{19}{2}x^2-\frac{19}{2}x+12
• B(x)=2x^3-\frac{19}{2}x^2+\frac{29}{2}x+1
• C(x)=5x+13

Checking the equation A(x)+B(x)=C(x) with these polynomials checks all four above equations at the same time.

Comparing a polynomial to itself

You can even check relationships between a large number of adjacent evaluations of the same polynomial using a simple polynomial equation. This is slightly more advanced. Suppose that you want to check that, for a given polynomial F, F(x+2)=F(x)+F(x+1) with the integer range {0,1…89} (so if you also check F(0)=F(1)=1, then F(100) would be the 100th Fibonacci number)

As polynomials, F(x+2)-F(x+1)-F(x) would not be exactly zero, as it could give arbitrary answers outside the range x={0,1…98}. But we can do something clever. In general, there is a rule that if a polynomial P is zero across some set S=\{x_1,x_2…x_n\} then it can be expressed as P(x)=Z(x)*H(x), where Z(x)=(x-x_1)*(x-x_2)*…*(x-x_n) and H(x) is also a polynomial. In other words, any polynomial that equals zero across some set is a (polynomial) multiple of the simplest (lowest-degree) polynomial that equals zero across that same set.

Why is this the case? It is a nice corollary of polynomial long division: the factor theorem. We know that, when dividing P(x) by Z(x), we will get a quotient Q(x) and a remainder R(x) is strictly less than that of Z(x). Since we know that P is zero on all of S, it means that R has to be zero on all of S as well. So we can simply compute R(x) via polynomial interpolation, since it's a polynomial of degree at most n-1 and we know n values (the zeros at S). Interpolating a polynomial with all zeroes gives the zero polynomial, thus R(x)=0 and H(x)=Q(x).

Going back to our example, if we have a polynomial F that encodes Fibonacci numbers (so F(x+2)=F(x)+F(x+1) across x=\{0,1…98\}), then I can convince you that F actually satisfies this condition by proving that the polynomial P(x)=F(x+2)-F(x+1)-F(x) is zero over that range, by giving you the quotient:
H(x)=\frac{F(x+2)-F(x+1)-F(x)}{Z(x)}
Where Z(x) = (x-0)*(x-1)*…*(x-98).
You can calculate Z(x) yourself (ideally you would have it precomputed), check the equation, and if the check passes then F(x) satisfies the condition!

Now, step back and notice what we did here. We converted a 100-step-long computation into a single equation with polynomials. Of course, proving the N'th Fibonacci number is not an especially useful task, especially since Fibonacci numbers have a closed form. But you can use exactly the same basic technique, just with some extra polynomials and some more complicated equations, to encode arbitrary computations with an arbitrarily large number of steps.

Everyone is talking about the $UST attack right now, including Janet Yellen. But no one is talking about how much money the attacker made (or how brilliant it was). Lets dig in.

Our story starts in late March, when the Luna Foundation Guard (or LFG) starts buying BTC to help back $UST. LFG started accumulating BTC on 3/22, and by March 26th had a $1bn+ BTC position. This is leg #1 that made this trade (or attack) brilliant.

The second leg comes in the form of the 4pool Frax announcement for $UST on April 1st. This added the second leg needed to help execute the strategy in a capital efficient way (liquidity will be lower and then the attack is on).

We don't know when the attacker borrowed 100k BTC to start the position, other than that it was sold into Kwon's buying (still speculation). LFG bought 15k BTC between March 27th and April 11th, so lets just take the average price between these dates ($42k).

So you have a ~$4.2bn short position built. Over the same time, the attacker builds a $1bn OTC position in $UST. The stage is now set to create a run on the bank and get paid on your BTC short. In anticipation of the 4pool, LFG initially removes $150mm from 3pool liquidity.

The liquidity was pulled on 5/8 and then the attacker uses $350mm of UST to drain curve liquidity (and LFG pulls another $100mm of liquidity).

But this only starts the de-pegging (down to 0.972 at the lows). LFG begins selling $BTC to defend the peg, causing downward pressure on BTC while the run on $UST was just getting started.

With the Curve liquidity drained, the attacker used the remainder of their $1b OTC $UST position ($650mm or so) to start offloading on Binance. As withdrawals from Anchor turned from concern into panic, this caused a real de-peg as people fled for the exits

So LFG is selling $BTC to restore the peg while the attacker is selling $UST on Binance. Eventually the chain gets congested and the CEXs suspend withdrawals of $UST, fueling the bank run panic. $UST de-pegs to 60c at the bottom, while $BTC bleeds out.

The crypto community panics as they wonder how much $BTC will be sold to keep the peg. There are liquidations across the board and LUNA pukes because of its redemption mechanism (the attacker very well could have shorted LUNA as well). BTC fell 25% from $42k on 4/11 to $31.3k

So how much did our attacker make? There aren't details on where they covered obviously, but if they are able to cover (or buy back) the entire position at ~$32k, that means they made $952mm on the short.

On the $350mm of $UST curve dumps I don't think they took much of a loss, lets assume 3% or just $11m. And lets assume that all the Binance dumps were done at 80c, thats another $125mm cost of doing business. For a grand total profit of $815mm (bf borrow cost).

BTC was the perfect playground for the trade, as the liquidity was there to pull it off. While having LFG involved in BTC, and foreseeing they would sell to keep the peg (and prevent LUNA from dying) was the kicker.

Lastly, the liquidity being low on 3pool in advance of 4pool allowed the attacker to drain it with only $350mm, causing the broader panic in both BTC and $UST. Any shorts on LUNA would've added a lot of P&L here as well, with it falling -65% since 5/7.

And for the reply guys, yes I know a lot of this involves some speculation & assumptions. But a lot of money was made here either way, and I thought it would be cool to dive into how they did it.

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