The Bored Ape Yacht Club is one of the most prestigious NFT collections in the world. A collection of 10,000 NFTs, each depicting an ape with different traits and visual attributes, Jimmy Fallon, Steph Curry and Post Malone are among their star-studded owners. Right now the price of entry is 52 ether, or $210,000.

Which is why it's so painful to see that someone accidentally sold their Bored Ape NFT for $3,066.

Unusual trades are often a sign of funny business, as in the case of the person who spent $530 million to buy an NFT from themselves. In Saturday's case, the cause was a simple, devastating "fat-finger error." That's when people make a trade online for the wrong thing, or for the wrong amount. Here the owner, real name Max or username maxnaut, meant to list his Bored Ape for 75 ether, or around $300,000. Instead he accidentally listed it for 0.75. One hundredth the intended price.

It was bought instantaneously. The buyer paid an extra $34,000 to speed up the transaction, ensuring no one could snap it up before them. The Bored Ape was then promptly listed for $248,000. The transaction appears to have been done by a bot, which can be coded to immediately buy NFTs listed below a certain price on behalf of their owners in order to take advantage of these exact situations.

"How'd it happen? A lapse of concentration I guess," Max told me. "I list a lot of items every day and just wasn't paying attention properly. I instantly saw the error as my finger clicked the mouse but a bot sent a transaction with over 8 eth [$34,000] of gas fees so it was instantly sniped before I could click cancel, and just like that, $250k was gone."

"And here within the beauty of the Blockchain you can see that it is both honest and unforgiving," he added.

Fat finger trades happen sporadically in traditional finance -- like the Japanese trader who almost bought 57% of Toyota's stock in 2014 -- but most financial institutions will stop those transactions if alerted quickly enough. Since cryptocurrency and NFTs are designed to be decentralized, you essentially have to rely on the goodwill of the buyer to reverse the transaction.

Fat finger errors in cryptocurrency trades have made many a headline over the past few years. Back in 2019, the company behind Tether, a cryptocurrency pegged to the US dollar, nearly doubled its own coin supply when it accidentally created $5 billion-worth of new coins. In March, BlockFi meant to send 700 Gemini Dollars to a set of customers, worth roughly $1 each, but mistakenly sent out millions of dollars worth of bitcoin instead. Last month a company erroneously paid a $24 million fee on a $100,000 transaction.

Similar incidents are increasingly being seen in NFTs, now that many collections have accumulated in market value over the past year. Last month someone tried selling a CryptoPunk NFT for $19 million, but accidentally listed it for $19,000 instead. Back in August, someone fat finger listed their Bored Ape for $26,000, an error that someone else immediately capitalized on. The original owner offered $50,000 to the buyer to return the Bored Ape -- but instead the opportunistic buyer sold it for the then-market price of $150,000.

"The industry is so new, bad things are going to happen whether it's your fault or the tech," Max said. "Once you no longer have control of the outcome, forget and move on."

The Bored Ape Yacht Club launched back in April 2021, with 10,000 NFTs being sold for 0.08 ether each -- about $190 at the time. While NFTs are often associated with individual digital art pieces, collections like the Bored Ape Yacht Club, which allow owners to flaunt their NFTs by using them as profile pictures on social media, are becoming increasingly prevalent. The Bored Ape Yacht Club has since become the second biggest NFT collection in the world, second only to CryptoPunks, which launched in 2017 and is considered the "original" NFT collection.

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.

Apple is said to have opted for a standalone AR/VR headset over a more powerful tethered model.
It has had a tumultuous history.

Apple's alleged mixed reality headset appears to be the worst-kept secret in tech, and a fresh story from The Information is jam-packed with details regarding the device's rocky development.

Apple's decision to use a separate headgear is one of the most notable aspects of the story. Apple had yet to determine whether to pursue a more powerful VR headset that would be linked with a base station or a standalone headset. According to The Information, Apple officials chose the standalone product over the version with the base station, which had a processor that later arrived as the M1 Ultra. In 2020, Bloomberg published similar information.

That decision appears to have had a long-term impact on the headset's development. "The device's many processors had already been in development for several years by the time the choice was taken, making it impossible to go back to the drawing board and construct, say, a single chip to handle all the headset's responsibilities," The Information stated. "Other difficulties, such as putting 14 cameras on the headset, have given hardware and algorithm engineers stress."

Jony Ive remained to consult on the project's design even after his official departure from Apple, according to the story. Ive "prefers" a wearable battery, such as that offered by Magic Leap. Other prototypes, according to The Information, placed the battery in the headset's headband, and it's unknown which will be used in the final design.

The headset was purportedly shown to Apple's board of directors last week, indicating that a public unveiling is imminent. However, it is possible that it will not be introduced until later this year, and it may not hit shop shelves until 2023, so we may have to wait a bit to try it.
For further down the line, Apple is working on a pair of AR spectacles that appear like Ray-Ban wayfarer sunglasses, but according to The Information, they're "still several years away from release." (I'm interested to see how they compare to Meta and Ray-Bans' true wayfarer-style glasses.)