frajaq posted:

why would you play casual

why do you hate yourself

frajaq posted:

rydiafan posted:

Behind the mathematical power of nerds on the internet!

Iorveth: Meditation

Reddit nerd posted:

Hi guys, we just saw a new Iorveth card, and calculating the value you get is tricky. I did some quick maffs, and have discovered the ultimate ratio for two cards that battle each other. I assume that you choose two enemy units without armor or quen etc. Say unit 1 has power x and unit 2 has power y, with x < y. You would do maximum damage if y = phi x, with the golden ratio phi=1.6180... This is a limit situation, (only if the battle goes on for ever), but I just love it when the golden ratio pops up in a random place. Even so, it is a really good rule of thumb: Pick the highest enemy, divide its power by the golden ratio, and pick the other enemy closest to that value!

I could try to generalize this for quen and armor, but that is too much work for now, if someone is interested I'll give it a try!

Edit: Full explanation follows here. I tried to keep it simple, but it might be a little bit complicated (It is a proper mathematical proof now XD). We call the nth element of the fibonacci sequence an . Then an = an-1 + an-2 (definition of this sequence). The power of unit y after being attacked n times we call yn, and the same for x. As x attacks first, we have the following: (A) yn = yn-1 - xn-1, xn = xn-1 - yn with x0 = x and y0 = y

We now prove that (B) xn = a2n+1 x - a2n y AND yn = a2n-1 y - a2n x using induction: (B) Is our induction hypothesis and (A) is a useful identity (You can skip this part if you want, the result is that (B) is true.) Base case: n=1 y1 = 1 y - 1 x = a1 y - a2 x, and x1 =1 x - y1 = 2 x - y = a3 x - a2 y

We see that (B) holds for n=1.

We now assume that (B) holds for an integer k, we prove it holds for k+1 yk+1 = yk - xk = a2k-1 y - a2k x - ( a2k+1 x - a2k y) =a2k+1 y - a2k+2 x. And a similar calculation for x. We have proven (B) using induction

Now, what can we use this for? Well, we want the units to fight each other as long as possible, because then they take the most damage. In mathematics (and physics too btw.), as long as possible means infinitely long. The units must stay alive.

Therefore yn > 0 and xn > 0 for any n.

Using (B) 0 < yn = a2n-1 y - a2n x. We rewrite this to y > x (a2n / a2n-1). Taking the limit of n to infinity gives us the golden ratio This gives us one bound y => phi x

Using the other equation in (B) gives us 0 < xn = a2n+1 x - a2n y.

Rewriting this gives us y < x (a2n+1 / a2n), and this gives our other bound: y =< phi x. We conclude that in the limit situation, y = phi x

Edit 2: Some mistakes fixed, the last part should use boundaries.

frajaq posted:

finally a reason to ""watch"" gwent streamers

https://www.playgwent.com/en/twitch-drops

Megazver posted:

Out of curiosity, how do you actually get them twitch kegs? Do they show up when you launch the game?

frajaq posted:

drat they really want Gwent to be a thing huh

did they ever ask the salty writer of the Witcher book series how he felt about the card game?

Myrmidongs posted:

Yeah this is some next level nerd. First trackers, now spergulators.

Bacon Terrorist posted:

So is this actually a dead gay card game or what? I just downloaded it on PS4 and it’s fun enough I doing the solo challenges, is multiplayer full of whales? Or can fun be had with zero spend?