The international law firm Holland & Knight used an NFT built and airdropped by its asset recovery team to serve a defendant in a hacking case.

The law firms Holland & Knight and Bluestone used a nonfungible token to serve a defendant in a hacking case with a temporary restraining order, marking the first documented legal process assisted by an NFT.

The so-called "service token" or "service NFT" was served to an unknown defendant in a hacking case involving LCX, a cryptocurrency exchange based in Liechtenstein that was hacked for over $8 million in January. The attack compromised the platform's hot wallets, resulting in the loss of Ether (ETH), USD Coin (USDC), and other cryptocurrencies, according to Cointelegraph at the time.

On June 7, LCX claimed that around 60% of the stolen cash had been frozen, with investigations ongoing in Liechtenstein, Ireland, Spain, and the United States. Based on a court judgment from the New York Supreme Court, Centre Consortium, a company created by USDC issuer Circle and crypto exchange Coinbase, has frozen around $1.3 million in USDC.

The monies were laundered through Tornado Cash, according to LCX, but were later tracked using "algorithmic forensic analysis." The organization was also able to identify wallets linked to the hacker as a result of the investigation.

In light of these findings, the law firms representing LCX, Holland & Knight and Bluestone, served the unnamed defendant with a temporary restraining order issued on-chain using an NFT. According to LCX, this system "was allowed by the New York Supreme Court and is an example of how innovation can bring legitimacy and transparency to a market that some say is ungovernable."

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.

Pixelynx, a music metaverse gaming platform, has teamed up with Beatport, an online music retailer focusing in electronic music, to establish a Synth Heads non-fungible token (NFT) Collection.

Richie Hawtin, aka Deadmau5, and Joel Zimmerman, nicknamed Pixelynx, have invented a new music metaverse game platform called Pixelynx. In January 2022, they released their first Beatport NFT drop, which saw 3,030 generative NFTs sell out in seconds.

The limited edition Synth Heads NFTs will be released in collaboration with Junction 2, the largest UK techno festival, and having one will grant fans special access tickets and experiences at the London-based festival.

Membership in the Synth Head community, day passes to the Junction 2 Festival 2022, Junction 2 and Beatport apparel, special vinyl releases, and continued access to future ticket drops are just a few of the experiences available.

Five lucky NFT holders will also receive a Golden Ticket, which includes access to a backstage artist bar and tickets to Junction 2's next large-scale London event this summer, in addition to full festival entrance for both days.

The Junction 2 festival will take place at Trent Park in London on June 18th and 19th, and will feature performances from Four Tet, Dixon, Amelie Lens, Robert Hood, and a slew of other artists. Holders of the original Synth Head NFT will be granted admission to the festival's guestlist as well as line-jumping privileges.

The new Synth Heads NFTs collection  contain 300 NFTs.

NFTs that provide IRL utility are in high demand.

The benefits of NFT drops related to In Real Life (IRL) utility aren't limited to Beatport and Pixelynx.

Coachella, a well-known music event, recently partnered with cryptocurrency exchange FTX to offer free NFTs to 2022 pass holders. Access to a dedicated entry lane, a meal and beverage pass, and limited-edition merchandise were all included with the NFTs.

Coachella also has its own NFT store on the Solana blockchain, where fans can buy Coachella NFTs and digital treasures that unlock exclusive on-site experiences, physical objects, lifetime festival passes, and "future adventures."

Individual artists and performers have begun taking advantage of NFT technology outside of large music festivals like Coachella.

DJ Tisto has revealed that he would release a VIP NFT for his upcoming "Eagle" collection during the EDC festival in Las Vegas in 2022. This NFT, dubbed "All Access Eagle," gives collectors the best chance to get NFTs from his first drop, as well as unique access to the music "Repeat It."