Netflix will stream Spiritfarer, Raji: An Ancient Epic, and Lucky Luna.

Netflix's Geeked Week brought a slew of announcements. The flurry of reveals for The Sandman, The Umbrella Academy season 3, One Piece, and more also included game and game-adjacent announcements.

Netflix released a teaser for Cuphead season 2 ahead of its August premiere, featuring more of Grey DeLisle's Ms. Chalice. DOTA: Dragon's Blood season 3 hits Netflix in August. Tekken, the fighting game that throws kids off cliffs, gets an anime, Tekken: Bloodline.

Netflix debuted a clip of Sonic Prime before Sonic Origins in June and Sonic Frontiers in 2022.

Castlevania: Nocturne will follow Richter Belmont.

Netflix is reviving licensed games with titles based on its shows. There's a Queen's Gambit chess game, a Shadow and Bone RPG, a La Casa de Papel heist adventure, and a Too Hot to Handle game where a pregnant woman must choose between stabbing her cheating ex or forgiving him.

Riot's rhythm platformer Hextech Mayhem debuted on Netflix last year, and now Netflix is adding games from Devolver Digital. Reigns: Three Kingdoms is a card game that lets players choose the fate of Three Kingdoms-era China by swiping left or right on cards. Spiritfarer, the "cozy game about death" from 2020, and Raji: An Ancient Epic are coming to Netflix. Poinpy, a vertical climber from the creator of Downwell, is now on Netflix.

Desta: The Memories Between is a turn-based strategy game set in dreams and memories.

Snowman's Lucky Luna will also be added soon.

With these games, Netflix is expanding beyond dinky mobile games — it plans to have 50 by the end of the year — and could be a serious platform for indies that want to expand into mobile. It takes gaming seriously.

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.

With the help of Virgil Abloh and Union LA‘s Chris Gibbs, it's now clear that Jordan Brand intended to bring the Air Jordan 2 back in 2022.
The “Future Is Now” collection includes two colorways of MJ's second signature as well as an extensive range of apparel and accessories.

The classic co-branding appears on short-sleeve tees, hoodies, and sweat shorts/sweat pants, all lightly distressed at the hems and seams.
Also, a filtered black-and-white photo of MJ graces the adjacent long sleeves, labels stitch into the socks, and the Jumpman logo adorns the four caps.
Liner jackets and flight pants will also be available, adding reimagined militaria to a civilian ensemble.
The Union LA x Air Jordan 2 (Grey Fog and Rattan) shares many of the same beats. Vintage suedes show age, while perforations and detailing reimagine Bruce Kilgore's design for the future.
The “UN/LA” tag across the modified eye stays, the leather patch across the tongue, and the label that wraps over the lateral side of the collar complete the look.
The footwear will also include a Crater Slide in the “Grey Fog” color scheme.

BUYING

On 4/9 and 4/10 from 9am-3pm, Union LA will be giving away a pair of Air Jordan 2s at their La Brea storefront (110 S. LA BREA AVE. LA, CA 90036). The raffle is only open to LA County residents with a valid CA ID. You must enter by 11:59pm on 4/10 to win. Winners will be notified via email.