The Pacific island state recognizes decentralized autonomous organizations.

The Republic of the Marshall Islands has recognized decentralized autonomous organizations (DAOs) as legal entities, giving collectively owned and managed blockchain projects global recognition.

The Marshall Islands' amended the Non-Profit Entities Act 2021 that now recognizes DAOs, which are blockchain-based entities governed by self-organizing communities. Incorporating Admiralty LLC, the island country's first DAO, was made possible thanks to the amendement. MIDAO Directory Services Inc., a domestic organization established to assist DAOs in the Marshall Islands, assisted in the incorporation.

The new law currently allows any DAO to register and operate in the Marshall Islands.

“This is a unique moment to lead,” said Bobby Muller, former Marshall Islands chief secretary and co-founder of MIDAO. He believes DAOs will help create “more efficient and less hierarchical” organizations.

A global hub for DAOs, the Marshall Islands hopes to become a global hub for DAO registration, domicile, use cases, and mass adoption. He added:

According to the World Bank, the Marshall Islands is an independent island state in the Pacific Ocean near the Equator. To create a blockchain-based cryptocurrency that would be legal tender alongside the US dollar, the island state has been actively exploring use cases for digital assets since at least 2018.

In February 2018, the Marshall Islands approved the creation of a new cryptocurrency, Sovereign (SOV). As expected, the IMF has criticized the plan, citing concerns that a digital sovereign currency would jeopardize the state's financial stability. They have also criticized El Salvador, the first country to recognize Bitcoin (BTC) as legal tender.

Marshall Islands senator David Paul said the DAO legislation does not pose the same issues as a government-backed cryptocurrency. “A sovereign digital currency is financial and raises concerns about money laundering,” . This is more about giving DAOs legal recognition to make their case to regulators, investors, and consumers.

After Terra's algorithmic stablecoin collapsed in May, TRON announced a plan to increase the capital backing its own stablecoin.

USDD, a near-carbon copy of Terra's UST, arrived on the TRON blockchain on May 5. TRON founder Justin Sun says USDD will be overcollateralized after initially being pegged algorithmically to the US dollar.

A reserve of cryptocurrencies and stablecoins will be kept at 130 percent of total USDD issuance, he said. TRON described the collateral ratio as "guaranteed" and said it would begin publishing real-time updates on June 5.

Currently, the reserve contains 14,040 bitcoin (around $418 million), 140 million USDT, 1.9 billion TRX, and 8.29 billion TRX in a burning contract.

algorithmic failure

USDD was designed to incentivize arbitrageurs to keep its price pegged to the US dollar by trading TRX, TRON's token, and USDD. Like Terra, TRON signaled its intent to establish a bitcoin and cryptocurrency reserve to support USDD in extreme market conditions.

Still, Terra's UST failed despite these safeguards. The stablecoin veered sharply away from its dollar peg in mid-May, bringing down Terra's LUNA and wiping out $40 billion in value in days. In a frantic attempt to restore the peg, billions of dollars in bitcoin were sold and unprecedented volumes of LUNA were issued.

Sun believes USDD, which has a total circulating supply of $667 million, can be backed up.

"Our reserve backing is diversified." Bitcoin and stablecoins are included. USDC will be a small part of Circle's reserve, he said.

TRON's news release lists the reserve's assets as bitcoin, TRX, USDC, USDT, TUSD, and USDJ.

All Bitcoin addresses will be signed so everyone knows they belong to us, Sun said.

Not giving in

Sun told that the crypto industry needs "decentralized" stablecoins that regulators can't touch.

Sun said the Luna Foundation Guard, a Singapore-based non-profit that raised billions in cryptocurrency to buttress UST, mismanaged the situation by trying to sell to panicked investors.

He said, "We must be ahead of the market." We want to stabilize the market and reduce volatility.

Currently, TRON finances most of its reserve directly, but Sun says the company hopes to add external capital soon.

Before its demise, UST holders could park the stablecoin in Terra's lending platform Anchor Protocol to earn 20% interest, which many deemed unsustainable. TRON's JustLend is similar. Sun hopes to raise annual interest rates from 17.67% to "around 30%."

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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.