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143 : true
234,518 Views ā¢ Jan 11, 2024 ā¢ Click to toggle off description
Or, for reference: Ā Ā Ā ā¢Ā TheĀ hardestĀ problemĀ onĀ theĀ hardestĀ testĀ Ā
Editing from the original video into this short by Dawid KoÅodziej
Views : 234,518
Genre: Education
Uploaded At Jan 11, 2024 ^^
warning: returnyoutubedislikes may not be accurate, this is just an estiment ehe :3
Rating : 4.871 (420/12,555 LTDR)
96.76% of the users lieked the video!!
3.24% of the users dislieked the video!!
User score: 95.14- Overwhelmingly Positive
RYD date created : 2024-09-02T00:19:32.736818Z
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YouTube Comments - 129 Comments
Top Comments of this video!! :3
As a computer engineer into 3D gaming, we generally use binary space partition method recursively to find collision detection of two arbitrary objects.
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I know this question. It's 12.5%. I didn't work that out but I remember hearing it
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11-dimensional alien beings:
"It's usually easier to think of it in 10 dimensions"
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This is one of my favorite videos of yours
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My guess would be that all points need to be in the "same half", first point has a probability of 100% to be with itself, second 50%, 3rd 50% and 4th 50%. And by having all of those together youd get a probability of 1/8?
Maybe it's harder than this though not sure
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those flashy subtitles are killing me
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First idea:
Create a function F take 4 pointsā coordinate as the input variable and output binary data if the tetrahedral contain the center
Use multi variable calculus to find the volume under F and divide by its domain
35 |
it's 1/2^n in the general n-dimensional case
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for the center to be in the triangle there must be 2 points in one half and the third one in the other half and also it must be orthogonal to the line between the other 2 points (sorry for my awful English)
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We calculate the maximum area of triangle which is not touching the center and we calculate the maximum area of triangle without any restrictions and then we use thise to vslues to make a percentage
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I love that you put these shorts out nowāand I wanna highlight a benefit that I donāt know if youāve considered quite yet:
Iāve been a consistent viewer of the long-form videos for a while, but I would sort of āpick and chooseā which videos to watch based on what topics seems interesting to me.
With shorts though, I often get a taste of a video thatās been out for a long time but I previously thought I wouldnāt be interested it, and I click through to the full vid.
Sometimes, even if Iām not in the mood for a whole analytical video right away (sometimes I just donāt have the mental energy to watch something that high level) Iāll save it to my watch later playlist.
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Wish this video was longer so he could explain! Jk, the link is at the bottom of the video and he made another short explaining exactly this.
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š An excellent condition and video. We highly appreciate it. Thank you
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They actually re-used this type of question in my A-Grade in Germany, but instead of a circle they used a Pentagon shape, where the 5 Points were the Corners.
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For the 2D example, fix the first point at the top since it can always be rotated so point 1 doesn't matter.
The second point can be anywhere in a 180Ā° arc and will determine the possible range of the third with a low of a 0Ā° arc at the bottom to a max of a 180Ā° arc. Assuming the average possible range for point 3 equals 90Ā° we can determine a 25% chance that a random set of three points will encompass the center.
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I think it's 1/8? My thought is each point has to be on the other side of the sphere, so point 1 > point 2 has 50% chance. Point 2 > point 3 50% chance. Point 3 > point 4 50% chance. 1/2 ^3 is 1/8. Maybe?
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@3blue1brown the answer is There are two points because of magnetic superposition the first point is located at the center of the object which is PI and the second point is located in your imagination.
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Sweet Jeez. I just made a guess of 12% before watching the whole video. I figured there's so much room you could fit a tetrahedron without touching the center. When I saw the answer for a circle and triangle was 25%, I said, 'If the answer is actually 12.5%, I'm gonna shit myself.'
I know no one would believe me, but at least I know it's true. I however, in fact, did not shit my pants.
I didn't guess the right answer, but it's interesting to see the power of an educated guess.
Also, I probably got lucky. I wouldn't be bragging either if I hadn't got close. Just ignore all my previous bad guesses and I look like a genius!
At least I feel like a genius today! š
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Considering the easier problem with circle , the probability would be 1 - P( three points are in a semi circle ).I think the later probability is quite easy to figure out as a triangle inside a semicircle will never include the center.
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@bfoo
7 months ago
āIf you canāt solve a difficult problem, there is an easier problem you canāt solve.ā
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