Having looked through the paper cited in the wikipedia article on Bellman’s lost in a forest problem, it appears that the case of the triangle is actually solved. Like, wikipedia says the equilateral triangle case is still open, but the paper they cite presents a solution. (Edit: The wikipedia article has been fixed)
@@GarryDumblowski Simply because a better solution is known (and was proven to be optimal). It’s a zig-zag with three line segments at very specific lengths and angles that ends up being a bit shorter than a side length of the triangle, and hence better than a straight line.
@@ItsMeLunaAgain Solving the problem means to prove that certain solution is right and I don't think that has been done. What has been done is finding a solution that mathematicians think is right. That doesn't count as a mathematical proof.
What am I misunderstanding about Lesbegue’s problem? A circle with diameter 1 is a valid cover for all shapes with diameter 1, and for a cover to be a valid cover for all shapes of diameter 1, it has to be able to cover a circle with diameter 1, therefore the smallest possible cover is a circle with diameter 1…
In orienteering, lost in a forest is a common thing. That is why you are given a safety bearing which is a direction you are to walk to reach a known feature such as a road where rescuers will eventually find you. The only variation is when you are too injured to walk and you sit still and blow your whistle
7:28 This section confused me because you said the upper bound was "improved" (implying "reduced") to a bigger number. It turns out you had the initial number correct: 0.845 The other two however, you have erred by replacing the 4 with a second 8. "0.88414" should be 0.84414, and "0.88409" should be 0.84409. We also have a lower bound on this problem of 0.832, so we have quite a good answer even if it's not perfect/proven.
For unpacking squares... do you not add 1/2 of the area squared overall each time? So you add 1/2 of a square. Or a triangle. Meaning, tou can rotate in 15° segments as you did to get 45° to asd 1/root[2]. So... 11²=121 121-5² solved=96 96/3=32 (since when adding 1/4 of a 2x2 3 are empty, and the middle 5 is rotated to take root[2] more in its corners) 32/2=16 (l+w of unit square 1) Root[16]=4 3 from above and 1/root[2] omitted for simplicity are 3*(1/root[2]) A=4+(3/root[2]) 2/root[2] more than 10 squares. Then just use Penrose tiling?
Visual from 10 squares at 1:08, if moves 2/root[2] more in white spaces it has enough that we can move the top row all the way over, turn the (2,3) on an angle, and fit 1 more ar ~(3,3)
60 degree rotation, first row 3 last spot requires 2/root(3) as 90-60=30 for length, and 1/root(3) for height as opposite tan(30°|60°) 3+(2/root[3]), 3+(1/root[3]) Error in original comment as 4, at 15° it requires 5.121, at 45° 4.1414, at 60° it is 3.434 or 3+0.434=3+(root[2]/3) Inverted sign in my head. 90°*3=270°,90°*3.5=315°; -60°=210°,255°;360°-=120°,105°; 90°-=30°,15° 3 squares and adding half a square twice, 60° difference is 30°,15° as 3/4 of 4*90° missing and accounting for last 1/4 of 4*90°. First square moves 15°, second an additional 15° for 30°, adds 2/root(3) of the 3×3 for length and 1/root(3) of a 3×3 for height following Penrose tiling. I think, not validated in my head and on phone calculator seems to work.
@@jimmypickinsA lot of what you’ve written is not really interpretable as math, but it’s worth noting that just adding 1/sqrt(2) to the side length of the big square every time is far from optimal. Also, I’m sure you could try tiling smaller patterns to get a bound on larger ones, but it’s likely that the larger ones could optimized further than the small patterns that you’re tiling. Also, why specify that the tiling be a Penrose tiling? I don’t see how it being aperiodic makes the problem any simpler.
I"m confused by the cover thing. Isn't the smallest possible cover just the largest possible diameter 1 shape, which would be the circle, as any growth to the circle would make it larger, and any cover that could not cover the circle would not be a cover?
No, the cover has to have every possible shape of diameter 1 fit inside, not just take up at least as much space as every shape of diameter 1. An average broom takes up less volume than there is inside of an average backpack, but you can't fit a broom in a backpack. A disk of diameter 1 is not a valid cover; for example, an equilateral triangle with side lengths of 1 has a diameter of 1, and it cannot fit inside.
A good idea for a video is to give problems in math and then reveal whether its unsolved or not. "10 Simple to State Problems in Geometry, guess if its solved or not"
why cant you prove ulam's packing conjecture with something relating to the fact the spheres can only touch each other at points as opposed to other shapes which can have areas of their faces touching?
@@ojfehevery convex shape with nonzero sized flat faces can always at least pair off on single nonzero area faces of contact, yeah? When packing, anyway. There's at least an intuitive sense that at least pairing off some portion of the faces like this when possible would be important in most packing strategies, but i dont have the rigor to prove it personally
I suspect that the square packing problem might never be solved in general. I could see us solving individual cases of increasing size, but I feel it’s likely that the optimality proofs will differ in nature between each case. Not a professional mathematician though, so I could be very wrong.
If you're not satisfied by or interested in answers to math questions, I don't know what to tell you. You clicked on a math video. Or alternatively, you only feel that way about exactly these six math questions, for whatever reason you might have. Weirdly enough, you didn't provide any elaboration about this. But that's okay, you can have any opinion you want.
Looks like it’s unsolved as well. Erich Friedman has a website detailing many best known optima for packing problems of this sort, and mentions that many of the triangle-in-triangle examples are unproven.
Maybe I've misunderstood Lesbegue's universal covering problem, but why a circle with diameter of 1 (and area of pi/4 which is about 0.78) can't be a cover? I can't think of any shape with diameter of 1, that this circle wouldn't cover
Many maths problem look easy yet when attempted it doesn't seem so and the problem is after being solved many people see the solution a go like" iT wAs EAsY"
@unavail_edits I hope all your socks remain permanently wet to a degree where its chronically pestering you making sure you can never focus on objectives with the comfort of socks
I don't know where you got the terms at 5:22, but the left side is NOT the same as the right side. Tau is 2 * Pi, so the Denominator should be 2 * ³√2 and NOT ⁶√2. If you want to simplify it, it would be ³√16
You are using superscripts, which indicates nth root. The animation makes it look like nth root superscripts, but he says square root a.k.a. ²√ But nobody writes square root like that. Cube root, yes. ³√2 = 2^(⅓) Sixth root, yes ⁶√2 =2^(⅙)
I’m lost in the woods and I’m very glad I found this survival video.
Stay put! 😮
If mathmaticians dont solve them in the next year Ill have to get involved 🐺
As if you could prove any of them.
Also, mathematical conjectures can take eternities for proofs to be published. Just look at how long it took for someone to solve the FLT.
the chosen one 🙏
@@Gordy-io8sb Anyone can do math, it's only a matter of what tools they have available to them
@@Gordy-io8sb clearly you haven't met ben
Having looked through the paper cited in the wikipedia article on Bellman’s lost in a forest problem, it appears that the case of the triangle is actually solved. Like, wikipedia says the equilateral triangle case is still open, but the paper they cite presents a solution. (Edit: The wikipedia article has been fixed)
Huh, that's odd. I was actually just about to ask in the comments why moving in a straight line isn't the optimal solution for a triangle.
@@GarryDumblowski Simply because a better solution is known (and was proven to be optimal). It’s a zig-zag with three line segments at very specific lengths and angles that ends up being a bit shorter than a side length of the triangle, and hence better than a straight line.
2:51
Correction: the equilateral triangle is the only regular polygon that isn’t fat. its solution is known, but it isn’t a straight line path.
Moving Sofa problem is also an unsolved geometry problem
Well, only if you don't know how to PIVOT!!!!!!!!
I think the moving sofa problem was solved already
@@ItsMeLunaAgainWe have a solution that mathematicians *think* is optimal, but proving that it is is a considerably harder task.
@@ItsMeLunaAgain Solving the problem means to prove that certain solution is right and I don't think that has been done. What has been done is finding a solution that mathematicians think is right. That doesn't count as a mathematical proof.
What am I misunderstanding about Lesbegue’s problem?
A circle with diameter 1 is a valid cover for all shapes with diameter 1, and for a cover to be a valid cover for all shapes of diameter 1, it has to be able to cover a circle with diameter 1, therefore the smallest possible cover is a circle with diameter 1…
Okay, I think I found the flaw in my thinking. An equilateral triangle with diameter 1, for example, does not fit in the circle with diameter 1.
In orienteering, lost in a forest is a common thing. That is why you are given a safety bearing which is a direction you are to walk to reach a known feature such as a road where rescuers will eventually find you. The only variation is when you are too injured to walk and you sit still and blow your whistle
Kepler's conjecture is proved.
there is a difference between a formal proof and an algorithmic justification.
My toxic trait is believing each question will take me 5 minutes to solve
I like tiling and packing problems, yet I like neither tiling nor packing.
It's kind of interesting that getting to the edge of an equalateral triangle fastest from a random point does not encourage going in a straight line.
7:28
This section confused me because you said the upper bound was "improved" (implying "reduced") to a bigger number.
It turns out you had the initial number correct: 0.845
The other two however, you have erred by replacing the 4 with a second 8.
"0.88414" should be 0.84414, and "0.88409" should be 0.84409.
We also have a lower bound on this problem of 0.832, so we have quite a good answer even if it's not perfect/proven.
Lebesgue
Reuleaux
He also missed an 8 after the comma for the original upper bound on the Lebesgue problem
For unpacking squares... do you not add 1/2 of the area squared overall each time?
So you add 1/2 of a square. Or a triangle. Meaning, tou can rotate in 15° segments as you did to get 45° to asd 1/root[2].
So...
11²=121
121-5² solved=96
96/3=32 (since when adding 1/4 of a 2x2 3 are empty, and the middle 5 is rotated to take root[2] more in its corners)
32/2=16 (l+w of unit square 1)
Root[16]=4
3 from above and 1/root[2] omitted for simplicity are 3*(1/root[2])
A=4+(3/root[2])
2/root[2] more than 10 squares.
Then just use Penrose tiling?
Visual from 10 squares at 1:08, if moves 2/root[2] more in white spaces it has enough that we can move the top row all the way over, turn the (2,3) on an angle, and fit 1 more ar ~(3,3)
60 degree rotation, first row 3 last spot requires 2/root(3) as 90-60=30 for length, and 1/root(3) for height as opposite tan(30°|60°)
3+(2/root[3]), 3+(1/root[3])
Error in original comment as 4, at 15° it requires 5.121, at 45° 4.1414, at 60° it is 3.434 or 3+0.434=3+(root[2]/3)
Inverted sign in my head. 90°*3=270°,90°*3.5=315°; -60°=210°,255°;360°-=120°,105°; 90°-=30°,15°
3 squares and adding half a square twice, 60° difference is 30°,15° as 3/4 of 4*90° missing and accounting for last 1/4 of 4*90°.
First square moves 15°, second an additional 15° for 30°, adds 2/root(3) of the 3×3 for length and 1/root(3) of a 3×3 for height following Penrose tiling.
I think, not validated in my head and on phone calculator seems to work.
This is the first comment I didn't understand after reading three times
@@jimmypickinsA lot of what you’ve written is not really interpretable as math, but it’s worth noting that just adding 1/sqrt(2) to the side length of the big square every time is far from optimal.
Also, I’m sure you could try tiling smaller patterns to get a bound on larger ones, but it’s likely that the larger ones could optimized further than the small patterns that you’re tiling.
Also, why specify that the tiling be a Penrose tiling? I don’t see how it being aperiodic makes the problem any simpler.
i swear all of these are just “Lorleantz’s icosahedron-tangent line conjecture” or whatever
Thank god, i was wondering how I would find my way out the forest.
You know, you really could've bothered to look up how to spell Stanisław Ulam's name if you already decided to write it out in the video
Wow that last one is tantalizing
Lebesgue*
Reuleaux*
Please proofread French names, you cannot just put the letters in any order...
Sure you can. It's not exactly Welsh, but it's close.
Please write your "french" names with 4 letters at most, good practice for not misusing them while keeping all the sounds intact.
Isn't that how French names were created initially?
Also pronounce Hungarian names right. "Pál" (Hungarian for "Paul") is not /pe:l/.
Stanisław, not Stanislov
JUSTICE FOR THE OPAQUE SET PROBLEM
It's Stanisław Ulam, not Stanislov
7:26 the second figure is a worse upper bound, not an improvement
Could you make a video about problems which went unsolved for a very long time, similar to these, but then were solved?
I love to hear about unsolved problems :))
Very interesting... It's like a treasure hunt :))
Lebesgue
I"m confused by the cover thing. Isn't the smallest possible cover just the largest possible diameter 1 shape, which would be the circle, as any growth to the circle would make it larger, and any cover that could not cover the circle would not be a cover?
No, the cover has to have every possible shape of diameter 1 fit inside, not just take up at least as much space as every shape of diameter 1. An average broom takes up less volume than there is inside of an average backpack, but you can't fit a broom in a backpack. A disk of diameter 1 is not a valid cover; for example, an equilateral triangle with side lengths of 1 has a diameter of 1, and it cannot fit inside.
A good idea for a video is to give problems in math and then reveal whether its unsolved or not. "10 Simple to State Problems in Geometry, guess if its solved or not"
why cant you prove ulam's packing conjecture with something relating to the fact the spheres can only touch each other at points as opposed to other shapes which can have areas of their faces touching?
I mean… why would you be able to prove it using that? Also, there are many other shapes for which this happens, not just the sphere.
@@ojfehevery convex shape with nonzero sized flat faces can always at least pair off on single nonzero area faces of contact, yeah? When packing, anyway. There's at least an intuitive sense that at least pairing off some portion of the faces like this when possible would be important in most packing strategies, but i dont have the rigor to prove it personally
Prediction: none of these 'problems' if they are ever 'solved' will have satisfying solutions that are important or interesting.
I suspect that the square packing problem might never be solved in general. I could see us solving individual cases of increasing size, but I feel it’s likely that the optimality proofs will differ in nature between each case. Not a professional mathematician though, so I could be very wrong.
If you're not satisfied by or interested in answers to math questions, I don't know what to tell you. You clicked on a math video. Or alternatively, you only feel that way about exactly these six math questions, for whatever reason you might have. Weirdly enough, you didn't provide any elaboration about this. But that's okay, you can have any opinion you want.
ok?
26 tringles kek
Love your vids
If we consider triangles contained within other triangles is that a solved problem?
Looks like it’s unsolved as well. Erich Friedman has a website detailing many best known optima for packing problems of this sort, and mentions that many of the triangle-in-triangle examples are unproven.
Maybe I've misunderstood Lesbegue's universal covering problem, but why a circle with diameter of 1 (and area of pi/4 which is about 0.78) can't be a cover? I can't think of any shape with diameter of 1, that this circle wouldn't cover
Oh nevermind, it literally talks about shapes of constant width afterwards! Makes sense now
Yeah a triangle of edge length 1 is an example
For a simple counterexample, you can't cover an equilateral triangle of diameter 1 with a circle of diameter 1
Many maths problem look easy yet when attempted it doesn't seem so and the problem is after being solved many people see the solution a go like" iT wAs EAsY"
wouldn't the solution to the forest problem just be a fibonacci spiral?
Clearly not, given that the optimal solution for a “fat” forest (as he mentioned in the video) is a straight line.
The squares aren’t square ;-;
Upvoted for using tau.
Oh wow, never heard someone pronounce the d in width before.
6 views in 12 mins bro fell
no, the timing is different
@@ThoughtThrill365😂 can't tell if your familiar with that trend of "fell off" or if you're being sarcastic.
@unavail_edits I hope all your socks remain permanently wet to a degree where its chronically pestering you making sure you can never focus on objectives with the comfort of socks
Cringe
don’t call people “bro”
wow you must be into tau you used it without focusing on it as if it was well know
don't het me wrong tau isn't bad just uncommon
You speak too fast !!!
They absolutely don’t, but you do realize there’s settings to slow videos down, right?
I don't know where you got the terms at 5:22, but the left side is NOT the same as the right side. Tau is 2 * Pi, so the Denominator should be 2 * ³√2 and NOT ⁶√2. If you want to simplify it, it would be ³√16
It is not supposed to be the cubic root, but 3 times the square root, as he reads off.
It's 3*√2, not the third root, he says it in the video too
You are using superscripts, which indicates nth root. The animation makes it look like nth root superscripts, but he says square root a.k.a.
²√
But nobody writes square root like that.
Cube root, yes.
³√2 = 2^(⅓)
Sixth root, yes
⁶√2 =2^(⅙)
Your simplification is incorrect, also.
Since tau = 2pi
Simply substitute tau for 2pi
Tau/(6√2)
=>
2pi/(6√2)
=
pi/(3√2)
■
They're equal.
The terms he got have been conjectured since Kepler, without the tau conversion obv
5:42 japanese flag jumpscare