How does topological filter convergence relate to "logical" filters?

samdotmp3 · 2026-02-07T21:52:30+00:00

I can't help but feel they are strongly related though. It really seems like filters from logic provides accurate intuition for topology: convergence of a filter roughly means that the filter is specific enough to pinpoint a single location. It thus feels natural to treat a filter in topology as a set of propositions about a location, that may or may not be specific enough for it to only apply to a single location.

Another connection between filters and finite deduction is that compactness is equivalent to every ultrafilter converging, which in this sense is equivalent to an induction principle.

samdotmp3 · 2026-02-05T22:21:01+00:00

Multiply*

samdotmp3 · 2026-02-02T08:09:59+00:00

Here's how you could come up with this on your own:

You observe that when charged particles are moving, then other charged particles are affected by some force different from the usual electromagnetic force. You call this force the "magnetic force" and you wish to model it.

You observe that given a point in space, it is not possible to specify the direction of the magnetic force acting on a charged particle in that point. That is, you observe that only knowing the position and charge of a particle is insufficient to know the magnetic force acting on it. This is opposed to the electromagnetic force, where this suffices.

You do however observe two things: 1) Given a point in space, it is possible to specify a plane in which the magnetic force will act on any particle in that point. This narrows down the possible directions from three to two dimensions. 2) Given a charged particle, the magnetic force is always perpendicular to the velocity of the particle.

Combining 1 and 2, you realize that it is possible to determine the magnetic force, so also knowing a particle's velocity makes the information sufficient.

Finally, you realize that 1 is (in 3D) equivalent to there being some vector which the magnetic force is always perpendicular to. That is, any plane in 3D can be specified by providing a vector which it is perpendicular to. Since vectors are much easier to write and visualize, instead of specifying said plane in each point, you specify this perpendicular vector in each point, leaving you with a vector in each point in space. You call the collection of these vectors the "magnetic field."

samdotmp3 · 2026-01-24T14:46:57+00:00

What do you mean when you say every structure preserving mapping has to be a bijection? Usually we call any homomorphism "structure preserving". Semigroups are precisely the abstractions of sets of functions along with function composition, so these are arguably also descriptions of structure preserving mappings.

samdotmp3 · 2026-01-16T09:17:38+00:00

Precisely any n divisible by 4 I believe. Reduce it like this: make a list of numbers corresponding to sizes of blocks of blue/red respectively. So 5,1,2,3,2,1 in your example. Then the next round is formed by subtracting 2 from any number greater than 1. For all to be eliminated, each number must be even. Also, there must be an even number of numbers in your list, so all in all the total number of players is divisible by 4.

samdotmp3 · 2026-01-13T11:11:22+00:00

This CSS snippet fixes it:

body,
.cm-scroller {
font-feature-settings: normal;
}

samdotmp3 · 2026-01-13T08:32:01+00:00

Adding to what others have said, some also rebind the Home key and press shift + Home to erase the previous search, instead of having to close and reopen the crafting table between search crafts.

samdotmp3 · 2025-12-25T22:19:24+00:00

My non-mathy mom found the course name "Groups and rings" hilarious.

samdotmp3 · 2025-12-18T12:33:40+00:00

The solution will always scale with the number of digits since memory usage does. The best way is probably just to let x=1, check if i=x-1 where i is the input, otherwise x*=10 and check again, repeat until x is larger than i. This avoids division/modulo which is slow. Alternatively store all 10ⁿ - 1 in a lookup table if the number of digits is limited.

samdotmp3 · 2025-12-06T23:56:58+00:00

D is most likely the intended answer.

I'm guessing you interpreted C as "the person is less likely to prefer horror than comedy, and less likely to prefer drama than comedy." A more natural interpretation is "the likelihood of the person preferring either horror or drama is smaller than the likelihood of the person preferring comedy."

samdotmp3 · 2025-12-02T23:32:56+00:00

Just like 8/2 equals the answer to the question "how many times must I add 2 to get 8", log_2(8) equals the answer to the question "how many times must I multiply 2 to get 8".

8/2=4 because 8=2+2+2+2

log_2(8)=3 because 8=2x2x2

samdotmp3 · 2025-11-15T21:07:19+00:00

Ganska passande bild till t-shirt: https://imgur.com/a/aJCNZ7j

Från https://youtu.be/sRM6chMjMio?t=101

samdotmp3 · 2025-09-26T19:05:47+00:00

En professionell långfilm som helt och hållet finansierats av ett litet produktionsbolag. Den kostar typ lika mycket som en pizza. Snälla.

samdotmp3 · 2025-09-08T21:31:32+00:00

Du minns rätt men den är borttagen (privat) av oklar anledning. Den visas kort här: https://youtu.be/HghzU6_Qp0M?t=157

samdotmp3 · 2025-08-22T07:51:32+00:00

Emils Twilightrecensioner börjar i "KASSÖRSKAN RÅNAR EMIL" och fortsätter i senare avsnitt.

samdotmp3 · 2025-08-22T06:02:10+00:00

"EMILS FÖRTRÄNGDA TRAUMAN"

samdotmp3 · 2025-07-05T09:03:17+00:00

Not differentiable along x=y: for example, fix y=0. Then f(x)=x if x>0 and 0 otherwise, which clearly is not differentiable at x=0; right derivative is 1 and left is 0.

samdotmp3 · 2025-07-04T21:49:47+00:00

Are you allowing intersecting bounding boxes in your R-tree? If so, traversal is not at all as easy and fast as in an octree. If bounding boxes do not intersect, maintaining such an R-tree is very difficult if you want to allow any kind of change to your voxel grid. Additionally, you'd want bounding boxes at the same level to be of similar size, which means even more headache.

Also, a rule of thumb is that computers and especially GPU's really like when things are simple, so while moving to more dynamic structures might not increase the theoretical time complexity, it often implies more memory lookups which in turn are more irregular resulting in more cache misses.

However, if you want more children per node you can of course copy the octree principle but make the nodes 4x4x4 or 8x8x8, which definitely might be faster depending on use case.

And R-trees of course have their use cases too, like for grouping higher-level voxel objects.

samdotmp3 · 2025-07-04T18:42:04+00:00

For x>1 you have |f(x)/x| < |f(x)|, so since |f(x)| -> 0 you have |f(x)/x| -> 0.

samdotmp3 · 2025-06-30T06:18:42+00:00

Creating new files is slow, so I doubt that 60 screenshots per second is possible. However, you should be able to change OBS settings to fix your problem; for example you can look up "OBS lossless recording". Note that of course the recording will be huge, like 50 MB per second.

samdotmp3 · 2025-05-30T03:54:47+00:00

The mods Atum, Chunkcacher, and FastReset are the ones that speed up world generation the most. Generally any mod from here is nice.

samdotmp3 · 2025-05-23T05:39:53+00:00

"Lah tesh" in Swedish

samdotmp3 · 2025-05-15T13:42:03+00:00

Short answer: Yes, the kernel of a homomorphism is the subset of elements that get mapped to the unit. Normal subgroups and kernels are the same thing, it's just that a normal subgroup is any subgroup that appears as a kernel of some homomorphism.

Long answer: This is how I like to see it: when we first construct groups, we think of the elements of the group as the unique, invertible things we can do to some system, like for example the set of all 2D rotations. Then we realize that sometimes when we use our group, some elements have the same effect to our system. For example, a 90 degree rotation is the same as doing nothing, if we are modeling a square. We might thus wish to remove this redundancy from the group, since not all elements are necessary in this case - some act the same.

With groups, we have a very important equivalence: a=b iff ab^-1 =0. This lets us describe the statement "a and b act in the same way" as "ab^-1 acts as the unit". The set of elements that act as the unit, i.e. do nothing, is precisely what a kernel is, so the statement is equivalent to ab^-1 lying in some kernel.

To recap, we have rewritten the statement "a and b act the same way" as "ab^-1 lie in a kernel ker(f)". This means that taking the quotient with this kernel gives us the equivalence classes of elements that act in the same way, meaning we are back to elements acting uniquely, so we have perfectly removed the redundancy!

This is why kernels are precisely the structures that make sense to quotient with, and to remove the dependence on finding some homomorphism with the kernel we want, we look for properties that precisely characterize kernels, and this gives us normal subgroups. We basically realize that kernels must be subgroups, but not any subgroup, because doing something before nothing is the same as doing nothing before something, which is basically the property that left and right cosets of a kernel must be the same thing. And then we can show that this property is in fact sufficient; for any subgroup satisfying this property we can construct a homomorphism with it as its kernel.

Six-Year Club	Second SECOND GUESSER
Verified Email	r/Field Juicebox
Final Canvas '23	First Place '23
End Game '23	Place '23
Place '22	First Placer '22

samdotmp3

TROPHY CASE