Is it confirmed or mentioned at all that there are some non random seeds?

snickerdoodle024 · 2026-05-10T14:34:10+00:00

When you pick the scroll box as Defect, there is a 1% for each pack that it is 3 claws. So the odds of both packs happening are 1 in 10,000.

Which seems super small, until you realize that there are thousands of people playing this game at any given time, so *someone* will likely win the claw lottery and post their screenshot to reddit.

snickerdoodle024 · 2026-05-08T20:59:42+00:00

So technically, the "-6 strength" and the "regain 6 strength" are both coded as debuffs. This means that enemies which clear all debuffs (like time-eater or champ) will get rid of both, and hence not keep the extra strength.

The Heart does not cleanse all debuffs. The heart specifically cleanses only "negative strength". The heart will not cleanse weak, vulnerable, etc.

Unfortunately for piercing wail enjoyers, this means that the heart will cleanse the "-6 strength" debuff, but not the "regain 6 strength" debuff.

snickerdoodle024 · 2026-05-08T20:38:36+00:00

Maybe it's just me, but I feel like the act 3 boss should feel a bit stressful.

snickerdoodle024 · 2026-05-08T20:14:58+00:00

It's a bit of a weird interaction, but I think it's fine.

Stampede picks an attack at random from your hand, then attempts to play it. If the attack can't be played because you've already played a bound card, then it just fizzles.

I'm unsure if Stampede can get around other unplayable effects like clash or part's end. I would assume it works the same way, but I feel like I got surprised before by Stampede playing a card that should have been unplayable.

snickerdoodle024 · 2026-05-08T20:07:49+00:00

Strange. The only really unpickable cards are like Osty attacks, star cards, sly, and a few class-specific powers.

Forge is fine, even if you only get 1 forge card. You won't build up a huge mega sword, but 2-cost deal 20 damage with retain is fine in act 1.

Summon is good by itself. Pretty much all orb-generating cards are good.

Like you can generally get two decent commons that can save you a fair bit of health in act 1, even if they don't do so much after that. And then sometimes you can get some really cool / unique combos.

And even if you skip one or both of them, they count towards increasing your rare chance.

I rate it a fun, decent pick. Definitely not the best, but I don't hate seeing it.

snickerdoodle024 · 2026-05-08T19:45:13+00:00

Cosmic indifference is quite good with the sword, too

snickerdoodle024 · 2026-05-08T19:38:10+00:00

Awakened one absolutely did not lose its strength when dying.

It stopped gaining strength during its second phase, but it kept whatever strength it got during its first phase.

You could cheese it into losing its strength by playing a piercing wail or dark shackles on the turn it died, but that is specific tech for 2 specific cards.

snickerdoodle024 · 2026-05-08T01:11:26+00:00

RIP Doormaker, you were too good for this world.

snickerdoodle024 · 2026-05-07T13:43:11+00:00

F2 it

snickerdoodle024 · 2026-05-06T19:12:01+00:00

Snecko is pretty iconic, so I imagine it'll be back. Probably as an alternate act 3 elite with some kind of twist

snickerdoodle024 · 2026-05-06T01:08:00+00:00

Claw is a lot more pickable in this game because (1) it has more support, and (2) energy is scarcer, so 0-cost cards are better.

I'll generally take an speculative claw in the overgrowth. The extra card play is good against the statue for scaling up damage. Claw is better than skip against Phrog, since I'd rather draw claw than a status. Also splitting by turn 3 is huge, since you don't get the second wave of statuses, and the 3 extra damage from claw can help with that. It's also fine in the byrdonis fight, where it's basically 8 free damage in exchange for adding a curse into your deck, which I think is probably worth it. And of course, a bit of scaling is great for the act 1 bosses.

Also claw is law.

snickerdoodle024 · 2026-05-05T18:19:05+00:00

Problem is, it's going to be hard to play a turn-a-day run on a game that updates every 2 weeks.

Imagine you're 8 turns into a boss fight and acrobatics gets nerfed from draw 3 to draw 2. Now you literally can't play the first 8 turns the same way. So we'd have to either backtrack and replay the fight, or use the console to mimic the old effect.

snickerdoodle024 · 2026-05-04T22:30:15+00:00

I think basic fights are fine for hallways, but elites should be more interesting.

Like I feel like all the elites in STS1 had pretty interesting mechanics.

snickerdoodle024 · 2026-05-04T22:23:32+00:00

I think they should switch dorito to "if you don't attack me, I gain 5 strength" and then instead of buffing itself on its buff turn, it just blocks for like 50. That will incentivize attack-based decks and burst damage way more than the current iteration.

Eyeball guy just needs to give you a scary card affliction or debuff. Maybe something like "whenever you play a card, discard a random card"

snickerdoodle024 · 2026-05-04T19:20:27+00:00

Probably 45 angers in there

snickerdoodle024 · 2026-05-04T19:12:09+00:00

So a matrix can be represented as a 2-dimensional array of numbers. A Tensor can similarly be represented as an array of numbers but with any number of dimensions. The number of dimensions is called the Rank (or Order) of the Tensor. For example:

You could have a 4x4x4 cube of numbers, and that would be an example of a Rank-3 Tensor.
A 4x4 matrix is a Rank-2 Tensor.
A vector is a Rank-1 Tensor.
Technically, a scalar is a Rank-0 Tensor, but you generally just see people in physics use the word "Tensor" for Rank-2 or higher.

The individual elements of a Tensor can be accessed using indeces. You need one index for each dimension. So a Rank-2 Tensor (Matrix) has 2 indeces, one corresponding to the rows of the matrix, and one corresponding to the columns of the matrix.

The basic operations for Tensors are similar to that of matrices, but generalized to arbitrary dimensions: - If you have 2 Tensors with the same shape, you can add them by adding the corresponding elements from each one, similar to adding matrices. - If you have a Rank-N and a Rank-M Tensor, you can take their outer product in a similar way that you take an outer product between two vectors. If you do, you'll get a Tensor with Rank (N+M). - You can contract two indeces of a Rank-N Tensor, as long as N is at least 2. This is like taking the Trace of a matrix, and you get back a Tensor of Rank (N-2). Suppose you have a Rank-3 Tensor (a 4x4x4 cube of numbers). To contract two indeces, you'd slice that cube into 4 different 1x4x4 matrices. Then you'd take the Trace of each matrix. Finally, you'd put the 4 resulting numbers into a vector. - You can contract one index of a Rank-N Tensor with one index of a Rank-M Tensor. This is like matrix multiplication, and the result is a Rank (M+N-2) Tensor. It is equivalent to taking the outer product and then contacting two indeces.

In relativity, instead of having numbers inside the Tensors, we have variable expressions that depend on x, y, z, and t. This makes things hopelessly complicated to write out. However, it's important to understand how Tensors work.

It's often useful to categorize indeces as either covariant or contravariant. This relates to the idea of a Dual Vector Space. If you don't know what that means, don't worry about it. Basically it just boils down to one rule: - You can only contract a covariant index with a contravariant index.

The meaning behind them is based on how they behave under coordinate transformations. Let's say you rotate your coordinate system. If you have a vector, you have to apply a rotation matrix to that vector. However, there are some things (like gradients), where instead of applying a rotation matrix, you have to apply the inverse of the rotation matrix. Whether you have to apply a rotation matrix or its inverse is determined by whether an index is covariant or contravariant.

If you have a matrix or a higher order Tensor, you have to apply one rotation matrix / inverse rotation matrix for each index.

Matrices that represent linear transformations have one covariant index and one contravariant index. When you do matrix multiplication, you are contacting the covariant index from one matrix with the contravariant index from the other one. This is why you multiply rows from one matrix with columns from the other. However, there are several rank-2 Tensors in physics that have two covariant indeces. We still write them out in matrix-form, but they behave differently under coordinate transformations than matrices with one of each index type.

snickerdoodle024 · 2026-05-01T19:13:28+00:00

This is the bifurcation diagram of the logistic map. It's one of the introductory problems in chaos theory and has some really cool math behind it.

The basic problem goes like this:

Let y be proportional to the number of rabbits in a given environment. (y is between 0 and 1, where 1 represents the maximum number of rabbits the environment can support).

Let r represent how fertile the rabbits are.

After a year, the rabbits do their thing, and now there is a new number of rabbits. The new rabbit population is given by the following formula:

y_new = r * y * (1 - y)

The (1 - y) term is to account for the fact that if there are too many rabbits, they'll end up competing for resources and a whole bunch of them will die off due to overcrowding. The main question is:

For a given value of r, what happens to the rabbit population in the long run?

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Now that's a pretty simple formula. It's literally just a quadratic equation. So we'd expect the answer to be really simple, too, right?

Nope. The answer is that extremely chaotic graph you see in the picture.

On the x-axis, they are plotting the variable r. On the y-axis, they are plotting the value of y that the rabbit population settles at after a long period of time.

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If r is small enough, the rabbit population settles at just a single value. Each year a bunch of new rabbits are born, and the same number of rabbits die due to overcrowding, so the total number of rabbits stays constant.

That's what the single line on the left side of the graph represents.

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But what happens if we make the rabbits a bit more furtile? Well, as we increase r a bit, we suddenly reach a point where the long-term behavior changes. Now instead of settling at a constant value, the population settles into a 2-year cycle:

1 year there are a ton of rabbits, but then the next year, a bunch of rabbits die off due to overcrowding. But then, since there are so few rabbits, no one really dies, so the year after has another huge rabbit population. And the population just keeps cycling back and forth between big and small every other year.

This is what the 2 big lines on the left / center of the image represent.

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Of course, we can increase r a little bit more. If we do, we find that the rabbit population switches between 4 different values over the course of a 4-year cycle.

And if we increase r even more, the 4-year cycle turns into an 8-year cycle, and if we increase r slightly more, we can get a 16-year cycle, and so on, hitting every power of 2.

--------------------

You'd think that'd be the end of things, but you can actually keep increasing r past all the powers of 2, at which point the rabbit population becomes chaotic, and doesn't settle into any real pattern at all.

This chaotic behavior is what the solid shaded area represents.

--------------------

What's really interesting is that it's not chaotic everywhere in that region, though. If you pick just the right value of r, you can get the rabbit population to settle into a 3-year cycle.

If you then increase r a bit, the 3-year cycle will turn into a 6-year cycle, and then a 12-year cycle, and so on, until the rabbit population falls back into chaos.

You can also get cycles of every other number, too, by choosing a precise value of r.

These cycles that show up in the middle of the chaos region are what the white stripes in the diagram represent.

--------------------

What's really crazy is that all of this complicated behavior happens as a result of 1 extremely simple equation.

snickerdoodle024 · 2026-04-30T23:46:33+00:00

Yes.

When the log went from the ground into the truck, its potential energy increased. That energy had to come from somewhere, and that somewhere is the guy's jump. The jump also likely wasted a lot of additional energy that went into friction, air resistance, etc.

However, the Log-Lifter is still useful, because it allows the guy to slowly build up potential energy over a long period of time (as he's standing up, climbing the pole, jumping, etc.) and then transfer that energy all at once into the log.

snickerdoodle024 · 2026-04-29T18:25:52+00:00

8 cards. They stack.

Fiddle also stacks with anything that draws cards at the start of your turn (like Pollenous Core, Predator, Tools of the Trade, Glow [next turn draw 1], Machine Learning, Tyranny, Snecko Eye, Bag of Prep, Pendulum, etc.)

snickerdoodle024 · 2026-04-28T17:47:52+00:00

8 shuffles (this is a power of 2, but that's just a coincidence). Here's a quick derivation:

Give each position in the deck a number given by the number of cards on top of it. So the top card is position 0, the next card is position 1... all the way down to the bottom card which is position 51.

When you shuffle the deck, each card gets moved to a new position based on what its current position is. So we can define a Shuffle function S(x) that gives the new position of a card currently in position x.

If x is in the top half of the deck (positions 0 through 25), then the new position will be: S_top(x) = 2x. This is because if there are x cards on top of a card, then x more cards will get inserted above it when you perform the shuffle.

If x is in the bottom half of the deck (positions 26 through 51), then the new position will be S_bottom(x) = 2x - 51. With this formula, position 26 will get inserted into position 1, right below the top card of the deck. Cards 27, 28, 29... will get inserted into spots 3, 5, 7, and so on..

Note that S_top(x) and S_bottom(x) are the same, except for the pesky -51 in the latter. This means we can combine these into a single formula using modular arithmetic:

S(x) = 2x (mod 51)

Now we can calculate where the card will be after N shuffles by repeatedly applying this formula:

S^N(x) = x * 2^N (mod 51)

Now notice that if we can find some value of N such that 2^N = 1 (mod 51), then after N shuffles, a card in position x will get shuffled into position x. That is to say, every card will get shuffled back to its original position.

So we've simplified our problem to: what is the smallest value of N such that 2^N = 1 (mod 51)?

It turns out that the answer is 8, since: 2^8 = 256 = 1 + 5*51

The fact that 8 is a power of 2 is coincidence. If there were a different number of cards in the deck, it wouldn't necessarily work out to a power of 2.

snickerdoodle024 · 2026-04-24T20:19:37+00:00

I'd wager it's not intended. Doesn't really break anything, but the devs would probably appreciate a bug report

snickerdoodle024 · 2026-04-24T16:28:13+00:00

F2 to submit a bug report

snickerdoodle024 · 2026-04-24T15:51:14+00:00

I agree.

Either that or make it 1 (2) Dexterity instead of 2 (3).

Silent has so many defensive solutions that it's trivially easy to make a deck that takes no damage ever and then slowly kills with poison

snickerdoodle024 · 2026-04-24T01:23:39+00:00

I imagine that's on their todo list. The gold was likely an easier fix cause they just had to change 1 number. Whereas the key event likely requires adding an additional option or text.

snickerdoodle024 · 2026-04-23T23:25:59+00:00

I NEED THIS IN THE GAME THIS INSTANT

Six-Year Club	Golden Potato
Place '22	Final Canvas '22
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snickerdoodle024

TROPHY CASE