I have been tinkering with the tesseract a bit: Take a 4x4 chess board and stretch and join it to form a torus. Then the squares of the chess board correspond to the vertices of a tesseract, with the edges corresponding to knight moves. Then allocate 4 digit binary strings to each vertex according to the 4 coordinates of each vertex of the hypercube. The flattened out chess board then becomes a magic square, with each row, column, major diagonal and 2x2 square having sum 30. Regular 4 digit Gray code gives you a nice symmetrical vertex traversal of the tesseract, whereas for an edge traversal you can use balanced Gray code.
@@blablablablablablablablablbla On a flat 4x4 chess board, knight moves are quite limited. If you roll the chess board into a cylinder, then knight moves can go beyond the side of the original board, giving a few more options. If you stretch the cylinder, bend it around and join it together to form a torus, then you have even more possible moves. On a toroidal 4x4 board, there are 4 possible squares to which a knight can move from each square. On a tesseract, each of the 16 vertices is joined to 4 other vertices by an edge. Topologically, the graph of knight moves on a 4x4 board on a torus is identical to the vertex and edge graph of a tesseract.
@@blablablablablablablablablbla Next suppose your tesseract has vertices at (0,0,0,0), (0,0,0,1), (0,0,1,0),..., (1, 1, 1, 1). Treat those as 4 digit binary strings 0000, 0001, 0010,..., 1111. The knight moves on the toroidal 4x4 board correspond to edges, which correspond to changing one of the 4 bits. Label the squares of the board with the 4 digit binary strings and flatten it out. The resulting square is a magic square whose rows, columns, major diagonals and 2x2 subsquares all sum to 30.
@@blablablablablablablablablbla Gray code involves changing one bit at a time to enumerate all values in a range from 0 to 2^n-1 inclusive. Since changing one bit corresponds to a knight move or edge, Gray code can be used to provide a traversal of all the vertices of the tesseract. If you want to traverse all the edges precisely once, then regular Gray code will not suffice - you need to use something called balanced Gray code.
Underrated. I did actually discovered this and something similar. You basically just covered over the concept of what I just discovered and extra concepts, and in another video from someone, he talks about how there’s more freedom or mobility for higher dimensions, including the increases of vectors and it’s own unique matrices. The pattern that’s related to the Pascal’s triangle, would be expressed in a summative equation of k=0 (on the bottom of the summation), d on top (represents the dimensions) then right after is c(k pick d)x^d-k as x is expressed as y-r (r is the range or +/- distance from x to r or y=x+r), and the (k pick d) represents the coefficients which discreetly contributes to that pattern of Pascal’s triangle, and d-k is the dimension subtracted by the numerical value of k since you add the previous dimensions (Ex: 1x^2+2x^1+1x^0 which simplifies to x^2+2x+1 to represents 2d as a polynomial equation for n-dimensional structures such as hypercubes. The d-k for the power of x represents the order of degree of the layer out equation and as d can be as high as it is like using tree(3) but it’s power will always approach zero, this can be expressed as a limit of the powers approaching 0 and it’s true in many cases, such as whole numbers, negatives, decimals, and etc… If expressed as a a new summation equation: P(x) = summation of k=0 (on the bottom) and d (on the top) c(k pick d)x^lim n to infinity (1+ xln(a)/n)^n. I think that’s right? Correct me if I’m wrong (doing this with the lack of fundamentals of limits). If it’s wrong then you can use d-k as the power as it works universally. I will try my best using quantifiers: Universal quantifier (upside down capital A) x element of (€; I am using this because I don’t have access to the real symbol),, P(x) is true, and Universal quantifier (upside down A)x, y € R (Real numbers),, existence quantifier r (Backwards E) € R such that y=x+r and x=y-r are true. If you’re lost then this will make sense! Consider x^2+2x+1 and y=x+r and x=y-r, then r is fixed to equal to 1 as for bases (x as in being 4 and y= sqrt of 25 which is the total of x^2+2x+1 then you finding the square root of it or for the base) as portrayed by the highest power - 1 = sqrt of the total of the dimensional equation subjected to x or d-1= sqrt of the total of the dimensional equation subjected to x (r=d-1; the same in value but in different contexts so that you’re not confused) which is a difference of one from those bases but is a range of 2 when x=4 and y=6 to make our statement true, this can be applied to other polynomials respective to their dimensional polynomic equation even 10d. And let’s say that x in this example is 4 and by plugging it in we get 25 since the original equation in vertex form is (x+1)^2 + 0, after x or y has a value, then the other variable has a fixed value or is dependent on the other values like r and including x or y (r is mandatory! And x or y can work as it’s interchangeable but follows its own separate rules as y has to be greater than x, and x has to be less than y, that’s a fixed given!) I hope that covers it! I think I deserve a pin 📌.
I commented on the Polytopes among other interesting Mathematical shapes, figures & curiousities at length in the other Geometry video, so I very much appreciate the analysis here. Keep up the good work indeed as u have some very, very important topics covered in your Channel here! 🤔
"We have discovered an new form, what should we call it ?" " an igga !" "Wait what ? Why ?" "You didn't learn about names for higher dimensionnals beings didn't you ?"
I never knew shapes in higher dimensions could be so interesting! the tesseract and hypercubes are really cool, but I can’t even imagine what a 5D shape would look like. It makes me wonder how people use these ideas in today’s world, like in science or even movies! Do you think we’ll ever be able to really see these shapes, or are they just too hard to understand? I want to learn more about this!
I think that "octochoron" sounds cooler then "teseeract". Altough, i think that polychoron means only the 3d surface of the hypersolid, for example octochoron means only the 8 cubes, not the inside. Its like a cube is the entire shape, but hexahedron is only the surface
@@tr2k500no, yes and maybe? Until we can find proof, it’s subjective rhetorical question like asking if god exists or alien life exists somewhere in the universe. Though what happens after death is also another but more spread out there through the help of religion. I am atheist, there’s a very low probability that there’s any action after, as no one could see anything from birth as our eyes were in the midst of being developed, you wouldn’t see black or white, you’d see nothing and a metaphor of what it looks like is the color of oxygen without any high tech or methods, just looking in a random direction, there’s no color, that’s how I think of death. If you’re religious then I won’t criticize you, as it’s your life, live the way you want and feel.
So, what would a 4D hemisphere (hollow) be called? 3-hemisphere, hyper hemisphere, hemihypershere, half hypersphere, half 3-sphere, 3-half sphere or something else? I got the distance of the center of mass of it from the centre as 2R/π. Can anyone confirm this?
i feel this explanation is a bit confusing a tesseract (4d cube) is just a cube, but all sides of the shape are 3d-cubes, just like how all sides of 3d cubes are 2d squares. looking at an image you can count the cubic shapes formed [in the extrusion that connects both cubes]. this applies to all 3d translations, eg. a 4d octahedron will have 8 octahedral sides this idea applies only to 4d volume and not spacetime dimensions, because of spacetime being constant, but you can create a 4d shape based on the path of travel the object takes in time
I watch a lot of your videos and still don't understand any of them.
I discovered some of these on concepts when expressing the pattern of each specific hyperplane. Like. x^2 + 2x + 1.
I have been tinkering with the tesseract a bit: Take a 4x4 chess board and stretch and join it to form a torus. Then the squares of the chess board correspond to the vertices of a tesseract, with the edges corresponding to knight moves. Then allocate 4 digit binary strings to each vertex according to the 4 coordinates of each vertex of the hypercube. The flattened out chess board then becomes a magic square, with each row, column, major diagonal and 2x2 square having sum 30. Regular 4 digit Gray code gives you a nice symmetrical vertex traversal of the tesseract, whereas for an edge traversal you can use balanced Gray code.
I have no idea what abut of that means but it sure sounds cool
Could you elaborate?
@@blablablablablablablablablbla On a flat 4x4 chess board, knight moves are quite limited. If you roll the chess board into a cylinder, then knight moves can go beyond the side of the original board, giving a few more options. If you stretch the cylinder, bend it around and join it together to form a torus, then you have even more possible moves. On a toroidal 4x4 board, there are 4 possible squares to which a knight can move from each square. On a tesseract, each of the 16 vertices is joined to 4 other vertices by an edge. Topologically, the graph of knight moves on a 4x4 board on a torus is identical to the vertex and edge graph of a tesseract.
@@blablablablablablablablablbla Next suppose your tesseract has vertices at (0,0,0,0), (0,0,0,1), (0,0,1,0),..., (1, 1, 1, 1). Treat those as 4 digit binary strings 0000, 0001, 0010,..., 1111. The knight moves on the toroidal 4x4 board correspond to edges, which correspond to changing one of the 4 bits. Label the squares of the board with the 4 digit binary strings and flatten it out. The resulting square is a magic square whose rows, columns, major diagonals and 2x2 subsquares all sum to 30.
@@blablablablablablablablablbla Gray code involves changing one bit at a time to enumerate all values in a range from 0 to 2^n-1 inclusive. Since changing one bit corresponds to a knight move or edge, Gray code can be used to provide a traversal of all the vertices of the tesseract. If you want to traverse all the edges precisely once, then regular Gray code will not suffice - you need to use something called balanced Gray code.
Underrated. I did actually discovered this and something similar. You basically just covered over the concept of what I just discovered and extra concepts, and in another video from someone, he talks about how there’s more freedom or mobility for higher dimensions, including the increases of vectors and it’s own unique matrices. The pattern that’s related to the Pascal’s triangle, would be expressed in a summative equation of k=0 (on the bottom of the summation), d on top (represents the dimensions) then right after is c(k pick d)x^d-k as x is expressed as y-r (r is the range or +/- distance from x to r or y=x+r), and the (k pick d) represents the coefficients which discreetly contributes to that pattern of Pascal’s triangle, and d-k is the dimension subtracted by the numerical value of k since you add the previous dimensions (Ex: 1x^2+2x^1+1x^0 which simplifies to x^2+2x+1 to represents 2d as a polynomial equation for n-dimensional structures such as hypercubes. The d-k for the power of x represents the order of degree of the layer out equation and as d can be as high as it is like using tree(3) but it’s power will always approach zero, this can be expressed as a limit of the powers approaching 0 and it’s true in many cases, such as whole numbers, negatives, decimals, and etc…
If expressed as a a new summation equation: P(x) = summation of k=0 (on the bottom) and d (on the top) c(k pick d)x^lim n to infinity (1+ xln(a)/n)^n. I think that’s right? Correct me if I’m wrong (doing this with the lack of fundamentals of limits). If it’s wrong then you can use d-k as the power as it works universally.
I will try my best using quantifiers: Universal quantifier (upside down capital A) x element of (€; I am using this because I don’t have access to the real symbol),, P(x) is true, and Universal quantifier (upside down A)x, y € R (Real numbers),, existence quantifier r (Backwards E) € R such that y=x+r and x=y-r are true. If you’re lost then this will make sense! Consider x^2+2x+1 and y=x+r and x=y-r, then r is fixed to equal to 1 as for bases (x as in being 4 and y= sqrt of 25 which is the total of x^2+2x+1 then you finding the square root of it or for the base) as portrayed by the highest power - 1 = sqrt of the total of the dimensional equation subjected to x or d-1= sqrt of the total of the dimensional equation subjected to x (r=d-1; the same in value but in different contexts so that you’re not confused) which is a difference of one from those bases but is a range of 2 when x=4 and y=6 to make our statement true, this can be applied to other polynomials respective to their dimensional polynomic equation even 10d. And let’s say that x in this example is 4 and by plugging it in we get 25 since the original equation in vertex form is (x+1)^2 + 0, after x or y has a value, then the other variable has a fixed value or is dependent on the other values like r and including x or y (r is mandatory! And x or y can work as it’s interchangeable but follows its own separate rules as y has to be greater than x, and x has to be less than y, that’s a fixed given!)
I hope that covers it! I think I deserve a pin 📌.
I have more, but it’s related to phi and Pascal’s triangle. Might be too complicated to explain although simple looking like the collatz conjecture.
nice text wall.
Nice. Dimensional analogy is a very handy tool in this topic.
My favourite: 6:18 120-Cell
Dot -> line ‐> pentagon ‐> dodecahedron
The Hypersphere and Hyperball section made my brain explode with a lack of understanding.
hass your brain ball healed by now?
I coin the term ultracube for a 5-cube
what about megacube? or even rampagecube
And a supercube for 6-cubes.
I commented on the Polytopes among other interesting Mathematical shapes, figures & curiousities at length in the other Geometry video, so I very much appreciate the analysis here.
Keep up the good work indeed as u have some very, very important topics covered in your Channel here! 🤔
Since N-dimentional hypercube is called n-cube, that means none of them is called hypercube except the 4d one.
What?
2:55 Japan jumpscare
lmao
"We have discovered an new form, what should we call it ?"
" an igga !"
"Wait what ? Why ?"
"You didn't learn about names for higher dimensionnals beings didn't you ?"
ha ha
if you can portray 3d objects using a 2d screen then can you portray 4d objects using a 3d screen?
Yes, though a 3d being could only view the 3d screen on a 2d screen.
line ,"volume"=s
square "volume"=s^2=s squared
cube volume=s^3=s cubed
logic:
tesseract HYPERVOLUME=s^4=s tesseracted
1:13 Do I spot a 120 cell hyperdodecahedron? Is that a thing or am I conflating terms?
Yes it's that one on the bottom right
clickbait title, not every shape is covered, as that would take an uncountably infinite number of hours
Watched the entire video and understood non of it after the tesseract, great video
The higher dimensional cubes have names too: penteract, hexeract, (probably) s/hepteract, octaract, etc
I never knew shapes in higher dimensions could be so interesting! the tesseract and hypercubes are really cool, but I can’t even imagine what a 5D shape would look like. It makes me wonder how people use these ideas in today’s world, like in science or even movies! Do you think we’ll ever be able to really see these shapes, or are they just too hard to understand? I want to learn more about this!
I think that "octochoron" sounds cooler then "teseeract". Altough, i think that polychoron means only the 3d surface of the hypersolid, for example octochoron means only the 8 cubes, not the inside. Its like a cube is the entire shape, but hexahedron is only the surface
Linear Algebra makes higher dimensions looks too trivial
This video served as a humility test
5:03 "they are just points" nope, they are point. All of 0-d is a single point.
4:16 "and a square of radius r..." Ooook
You're the best, thanks!
I learned something
that higher dimensions and shapes there exist
@@tr2k500no, yes and maybe? Until we can find proof, it’s subjective rhetorical question like asking if god exists or alien life exists somewhere in the universe. Though what happens after death is also another but more spread out there through the help of religion. I am atheist, there’s a very low probability that there’s any action after, as no one could see anything from birth as our eyes were in the midst of being developed, you wouldn’t see black or white, you’d see nothing and a metaphor of what it looks like is the color of oxygen without any high tech or methods, just looking in a random direction, there’s no color, that’s how I think of death. If you’re religious then I won’t criticize you, as it’s your life, live the way you want and feel.
tRiNglE!!!
5:14
XD
Rod
so wouldnt s^4 be like... s hypercubed? or s tesseracted? lmai
A 0d particle would be a monad
anyone else think this guy sounds like Huggbees
"The line segment of lenght of S has a lenght of S" hmm, this floor is made out of floor
Why is it suddenly so interesting?
Other n-cubes have names: penteract, hexeract, hepteract...
lol, couldn't give two s**t's about maths but these are still pretty interesting
ÆÆÆÆÆÆÆ
@@lukatolstov5598 relatable
when the dimension add, would the math in our 3D world can not apply to higher dimension?
Hi! Your video is very good! Keep it up!
So, what would a 4D hemisphere (hollow) be called? 3-hemisphere, hyper hemisphere, hemihypershere, half hypersphere, half 3-sphere, 3-half sphere or something else?
I got the distance of the center of mass of it from the centre as 2R/π. Can anyone confirm this?
i feel this explanation is a bit confusing
a tesseract (4d cube) is just a cube, but all sides of the shape are 3d-cubes, just like how all sides of 3d cubes are 2d squares.
looking at an image you can count the cubic shapes formed [in the extrusion that connects both cubes]. this applies to all 3d translations, eg. a 4d octahedron will have 8 octahedral sides
this idea applies only to 4d volume and not spacetime dimensions, because of spacetime being constant, but you can create a 4d shape based on the path of travel the object takes in time
a line segment the length of s it's length is s never knew that interesting
Do you use ai for your voiceovers?
No
@@ThoughtThrill365you have a great voice!
@@ThoughtThrill365you did a good job covering the video! Check my comment for further explanation! And my hypothesis before this video.
more shapes
Do you want to have a channel with skydiving content? Or are you not motivated anymore?
so you are trying to explain higher dimensional geometry
with 4D looking formulas
in a 3D world
on a 2D screen
and with my 1D understanding of math?
😂😂😂 nice one
@@ThoughtThrill365that’s inception. Would you mind looking at my comment for further explanation from the video and my hypothesis?
belcyphre
1d 1dot
Every?
:D
This does not explain every higher dimensional shape.
bruh