Of course it is false! If you plug it in for a 1-gon, it'll say you need 1.5 points. Clearly, since you can't have half of a point, then it simply must be wrong. ...I'll show myself out.
+RandomPanda0 but a "1-gon" is not a plane/ 2-dimensional. it's only apoint without even 1 dimension. that is kinda like 0! is not 0 because you didn't start out any other faculty calculation with 0 ("0*1*2..."), because they would all stay at 0.
this is such a cool theorem/hypothesis, I love when there is something profound, you can say about groups of objects regardless of the configuration of those objects. Math is awesome!
In a recursive manner, tackling the problem backwards. Building a convex shape with n sides (Dn lines) with Sn vertices, Dn not parallel to Dn-1 will yield a couple of things: A set number of intersections A set number of zones If you "pull" a point on a segment of the n gone, into the zone, outside the n-gone, directly adjacent, only then can you get a n+1 convex n+1 gone. This point, now Sn+1 is the intersection of Dn and Dn+1. Since Dn+1 is not parallel to Dn, it will intersect all the other lines (new intersections, new zones). This is the only way to build a convex shape. Any other point will yield something concave. It does work with parallel lines, not at the start of course, lines must intersect. But Dn-3 can be parallel to Dn. Not with the optimum solution. So the constellation of candidate points for the new gone is limited, and the minimum number of points is: The sum of all the zones made up by Dn lines minus all the intersections of Dn lines minus 1 (this is the zone inside the n gone) I'm french, I can clarify if needed.
Very interesting! There was some discussion about practical applications of researching mathematical stuff like this. At least one part of it involves motivation to develop computationally more efficient computers, since maths stuff like checking how many decimals of pi or how far in this problem can a computer compute gives some estimate about it. Also many of these kinda math theorems are in a way tied together, in other words even if this problem wouldn't have any easy-to-see practical applications, it's theoretically very possible that these theorems are analogous to some other math theorems, which could potentially have many many applications.. and so on. Cheers!
That's interresting. I like this problem. This kind of problem is really in the trail of Erdös's usual problems or the problems he was keen on: a simple/straightforward combinatoric question told and often involving an asymptotic bound.
is there some part of the conjecture that states that three points or more points in plane can't form a line because otherwise you could just have a line with 33 points and it would be impossible to create a 7-gone with that?
If you take any three points you can have a wave structure. Two for half wave. A straight line not touching more than two is a collapsed half wave. Seven sides have 14 collapses. So 28 plus 5 for structure.
I'm having a go at trying to solve the problem... I have tried my best to try and focus the problem down by working with how the various positions of the points on the plane, which I worked on by scribbling out the equations of lines, and the two coordinates the points would take up, and the ordering of the end points of which the next line would be drawn... I'm going to see if this line of thinking is going to work... maybe I am trying to see where I can go with looking at the various possible combinations that could exist between the points.
Hey Brady. I don't think he meant that the Erdos conjecture formula gives you the minimum number required. I think it's just a number that is enough. Since he said sometimes it may not have the +1.
a practical application of convex polygons off the top of my head would be in solving a linear program. Convexity is important when looking at a feasible region of solutions to find the optimal solution since that region will have a finite amount of sides and points.
Ramsey theory, which tells us about how order arises in chaotic systems. If we understand simple problems like these, we can better understand the weather, the stock market, and many other chaotic systems.
I'm pretty sure the solution to this is to determine all of the different ways that you can form a concave n-gon from convex (n - k)-gons, where n - k > 2 through indenting vertices. And by pretty sure, I mean I'll have to play around a bit more.
Given the backstory with the two mathematicians suddenly getting married and sticking together for 90 years, plus the name of the problem, I was half-expecting this to be some kind of secret mathematical proof about happiness :P
Seems like it began as a juggling problem, it was mentioned in the other video that Graham was into circus performance. Just so we can abort the whole "what would I use this for???" debate
O.K., I know where the +1 bit comes from, I think. If you have n points and they set up so that all the points are being used up in an n-gon and you move one of the points away from the center of the 'gon then it shouldn't effect the concavity of the 'gon uless it crosses a line formed by two other points in the 'gon. The reason would be this, if you take any four points in your 'gon, they form at least two triangles. If three points are in a line--this could happen if you move one of the two points shared by both triangles towards the other--then there are two right triangles. If you continue to bring those two triangles together, they both become obtuse and then you have to take that point out of your 'gon to maintain concavity and the 'gon becomes an n-1-gon. You can keep doing this and at some point, doing this brings enough points into the inside of the 'gon then you have the situation where you can take one point on the outside of the 'gon out of the connection and instead connect two of the points in the interior to the convex 'gon. because you have a potential acute triangle that does not cut though a potential line in the 'gon. But where does the 2^(n-2) come from? I don't know. If you start with an 8-gon and move one point into the middle, you have a 7-gon and an extra point. Say point one (they're numbered sequentially around the rim). if you move #3 into the middle, you have an accute triangle with the to moved points and #2, so you can take #2 out of your 'gon and restore #1 and #3 to your 'gon. But only if they have not crossed another potential line between two points still in 'gon. But this is intersting. If you move #1 in and it crosses the 8-2 line, then we go from 8 to 7 sides. if we bring #2 in it may become a 6-gon, but only if we also moved #1 past the 8-2 line. If we move #3 in past the 8-4 line, it becomes a 5-gon but only if both #1 and #2 have also been moved past that line as well. If we move the #'s 1-4 past the 8-5, then the the formula suggests that there must now be two points in the middle of the 4-gon that form an accute triangle with #5, #6 or #7 and does not cut through the 8-6 or the 7-5 potential lines. It has to be acute be cause an obtuse potential line in one of these potential triangles would violate concavity. And I'm liable either wrong or just figured out something already know. I bet wrong.
+Frank Harr Gotit! At least for 9 points. 5 6 7 8 from above forms a 4-gon. And that has 4 internal potential triangles. You can put an extra point in each of those and very carefully arrange it so that none of those give you an opportunity to make a 5-gon. You cannot put a point outside of the 4-gon because no matter where you put it, you could form a 90 degree angle with the perimiter and that allows you to incorporate it into the perimeter. And no matter where you put it withing the 4-gon, it shares a potential triangle with another point and THAT forms a triangle with a vertex on the perimiter, and another point in the same potential triangle and that creates an acute potential triangle that is within another and thus you cant take that vertex off of your perimiter and incorporate those two points as two new vertices within the perimiter. Now if you have a 5-gon, don't bother putting any points outside of the perimiter because it can only form a right angle with the perimiter and end this quickly or you can just move your vertex and thus make your old vertex a point within the 5-gon. You can carefully fill each potential triangle with one point each and then. . . I am working on the next part. This is fun. I thought I had a handle on something, but it broke off. But I honestly think that what we're dealing with here is overlapping triangles and quadralaterals. You have those on the perimiter, those inside the n-gon and every time you add a point, you get more triangles. And eventually you get so many that you you can replace one vertex with two from the points inside the n-gon.
+Frank Harr There's another way to think about it. If you inscribe a pentagram in a pentagon (let's call it 1-5 after the points) you have two regions the center pentagon and everything else. If this was a hexa- septa or any n>5-gon, you'd have more regions, but I don't know anything about those yet. Now, if you a random point a in pentagon 1-5, there are two regions where you cannot put another point without automatically allowing the pentagon to become a hexagon. And I'm still working on defining these regions. If your a is in section 1-3 of 1-5 then you draw a line from 1 through a and out of the pentagon and another line from 3 through a and out of the pentagon. The two triangular-shaped regions between those two lines and the perimiter are part of your "danger zone" where, if you place point b, you can automatically eliminate vertex 2 and create a hexagon. Point a could also be part of section 5-2 or 2-4 (but not both), in which case you have two more danger zones. This is, of course, because of the definition of convex here. That is, as you go around the perimiter, you only turn in one direction. Which, incidently, get you your shortest perimiter per area. But I doubt that's helpful. Inside the center, you need at least four points to get rid of three vertices. I'm still working on figuring those danger zones out.
+Frank Harr O.K., if you put points a and b in the center area of of 5-gon 1-5 draw a line through them to the perimiter you'll have 3 vertices on one side and two on the other. Let's say it's vertices 1-3 and 4-5. Now, if you draw a line from 5 to a and from 4 to b (and they don't cross) this creates two "danger zones", but not as dangerous as before. If you put one point in each of these two zones (c and d), you can draw your convex hexagon. However, you can put both points in the same zone if you put point d in the two triangles formed by a line drawn through 5 and a (extending to line ab) and a line drawn from a through c to the perimiter of the 5-gon. You can also put c and d in the small triangle formed by the line from 5 through a, the line from 5 through b and line ab if point d is on the same side of line ac that b is.
+Frank Harr You know, if you draw a line from 1 though a and from 3 through b, you'll get three triangles based on lines 1a, 3b, the perimiter and line ab that are also danger zones. These things just multiply. ---------2 -------/ \ -----1 3 -----| | -----| a b | -----| | ----5-------4
I'm pretty sure the definition of convex and concave polygons is convex: all exterior angles of a polygon are greater than 180. and concave: one or more exterior angles are less than 180
I believe that one of the professors at my university said that he worked with Graham on the mathematics of juggling. Would you be willing to ask him to do a talk about that problem?
correct me if im wrong but can't you just make a figure with more sides like go from pentagon to hexagon , but use like 39 points, or something ridiculous and still be with in the rules?
I think I may have managed to prove the conjecture, took some days of work, and 11 full A4 pages of endless maths, but I think I've got the proof :) I'll need to get someone who knows mathematics deeper than me to check it and make sure it is solid before I hand it out or message anyone about it though.
Interesting.. why does the audio for this video come out of my center speaker? Mono audio, or is everything mixed in 5.1 and sent to the center channel for some reason?
So a convex 2-gon (A line segment?) would have 2 points. That makes sense since line segments can't be concave, if you would even call a line segment a polygon.
So the conjecture is for a set of points, what is the maximum number of sides can you draw, in a concave way, when the points do not have straight lines between them? Is it 2^(n-2)+1=x where n is the number of points and x is the number of sides?
No, the conjecture is that in order to guarantee there's at least one convex N-gon (an N-sided polygon that has no "dents"), you need at least 2^(N-2)+1 points.
What if you had a three dimensional object, however only one vertex was present in the two dimensional plane you were working in, if the object is still considered a polygon (and it might not, I don't know if it would) then it you could use the theorem on that single vertex while it might appear to be a 1D object while in reality a 3D object, resulting in 1.5 dots.
Brady, ask them mathematicians about how each of those discussed crazy math problems are connected to the real world. If you do that, i'll subscribe for sure:)
Assuming you've done high school calculus, the derivative of e^x happens to be e^x, and that doesn't really occur with any other function making it useful for logarithms and such.
There are several things that are special about e. One example is that the derivative of e^x is e^x itself. This means that for the function f(x) = e^x, the slope of the function is the same as the value of the function. This makes e extremely important in an extremely useful branch of mathematics, differential equations, which describes, among many other things, heat transfer, population increase, compound interest, and many aspects of movement.
Why can't I draw a 7-gon with the 9-point layout? If I connect all the points except the two almost at the centre, I end up with a 7-gon, with only right turns?
The question is how many points GUARANTEE that you can draw the n-gon. Sure, in some configurations, you will be able to draw some huge n-gons, but that's not the point.
The happy ending problem based on 4 degree equation the dots is basically a root of the surface basically in the 4 points is the based on three degree equation is the imposible to made gon if we 2^n-2+1 is the four degree equation
+zoranhacker I guess he could have said irregular pentagon but wanted to keep his words consistent in the video so people understood what he said as clearly as possible.
For a second I was like "can't they at least check it for some big n with computers?", but then I hear that just for n=7 it's already getting too hard!... Just how big the computational complexity for that problem is? How fast does it rise or what kind of numbers we are talking about?
Well, for the n=7 case, if we wanted to check all possible 7-gons for 2^(7-2) + 1 = 33 points, there would be (33 choose 7) possible shapes to check, which is over 4 million! And that would just be for a single arrangement of points! In general, since we'd want to check configurations of 2^(n-2) + 1 points for n-gons, there would be (2^(n-2) + 1 choose n) shapes to choose for each possible arrangements of points, which is a function that grows faster than any exponential!
O.K., if you have a triangle with two points inside of it, you can make one convex 4-gon. but if you make a convex 4-gon with two points in it, you don't have to make a convex 5-gon. This is hard.
I like how he described the concave/convex polygon thing using the turns, I never thought about that but it's so simple and make sense.
Is anyone else happy that this channel exists?
I am!
I'm having a nerdgasm by hearing such problem exists and not proven yet.
It's on that little corner of youtube with all those things we missed when we were at school.
WleoW here u can be as nerdy as u want as no one cares
Me.
Rip Ron graham . One of the greatest modern mathematicians
Rest in peace.
Ron Graham is a boss. His talk at ICCM 2013 was one of the best talks I've been too.
Of course it is false! If you plug it in for a 1-gon, it'll say you need 1.5 points. Clearly, since you can't have half of a point, then it simply must be wrong.
...I'll show myself out.
+RandomPanda0 but a "1-gon" is not a plane/ 2-dimensional. it's only apoint without even 1 dimension. that is kinda like 0! is not 0 because you didn't start out any other faculty calculation with 0 ("0*1*2..."), because they would all stay at 0.
BloCKBu5teR That was the joke.
1 gon is not real
RIP Ron Graham. What a legend
"It's actually called the 'Happy Ending Theorem'"
Me: *pauses the video and laughs loudly*
RIP
Ron Graham
Ron Graham must have some kind of prestige to be able to hire Robert Duvall to portray him in TH-cam videos.
this is such a cool theorem/hypothesis, I love when there is something profound, you can say about groups of objects regardless of the configuration of those objects. Math is awesome!
Ron Graham's office is awesome. Brady, it would be great to see more of it.
easy... hold my beer
IT's Graham's NUMBER GUY!
Pretty amazing. His name slipped my mind just at the moment though.
A southern way of saying convex turn is the NASCAR turn. RIP Sir!
This is a very simple and clever explanation of convex and concave polygons, didn't know this before. Greets
(1:42) I thought the "PROVEN" green box would take me to the video that proves it :(
In a recursive manner, tackling the problem backwards.
Building a convex shape with n sides (Dn lines) with Sn vertices, Dn not parallel to Dn-1 will yield a couple of things:
A set number of intersections
A set number of zones
If you "pull" a point on a segment of the n gone, into the zone, outside the n-gone, directly adjacent, only then can you get a n+1 convex n+1 gone. This point, now Sn+1 is the intersection of Dn and Dn+1. Since Dn+1 is not parallel to Dn, it will intersect all the other lines (new intersections, new zones).
This is the only way to build a convex shape. Any other point will yield something concave.
It does work with parallel lines, not at the start of course, lines must intersect. But Dn-3 can be parallel to Dn. Not with the optimum solution.
So the constellation of candidate points for the new gone is limited, and the minimum number of points is:
The sum of all the zones made up by Dn lines minus all the intersections of Dn lines minus 1 (this is the zone inside the n gone)
I'm french, I can clarify if needed.
Can you clarify the part where you didn't prove the conjecture
He has a marvelous proof of this conjecture, but the margins of the youtube comments section is too small for him to write it out.
Graham himself... damn Brady when we will see Hawking on Sixty Simbols? :D
Dudok22 not anytime soon lol
Sadly never.
Rip
Jan 8 1942-mar 14 2018
:(
Born on the 300th anniversary of Galileo’s death and died on the 139th anniversary of Einstein’s birth.
Teacher: "big numbers aren't scary"
Big numbers:
Where is the ron graham talking about grahams number video??
I'm working on it - it's a BIG number to make a video about!
Numberphile looking forward to it!!
Numberphile: Take your time, Mr Brady! :)
Not only the problem is interesting, but also a thousand dollars?!
Now I'm so motivated, I've gotta figure this out somehow
Very interesting! There was some discussion about practical applications of researching mathematical stuff like this. At least one part of it involves motivation to develop computationally more efficient computers, since maths stuff like checking how many decimals of pi or how far in this problem can a computer compute gives some estimate about it. Also many of these kinda math theorems are in a way tied together, in other words even if this problem wouldn't have any easy-to-see practical applications, it's theoretically very possible that these theorems are analogous to some other math theorems, which could potentially have many many applications.. and so on. Cheers!
Oh, that kind of happy ending!
Rest in peace, Ronald. And thank you.
Wow. He's the guy who came up with Graham's number.
Oh man, I really hope you filmed more than two videos with Ronald Graham.
Grahams Number by Graham himself? AWESOME!!
That's interresting. I like this problem.
This kind of problem is really in the trail of Erdös's usual problems or the problems he was keen on: a simple/straightforward combinatoric question told and often involving an asymptotic bound.
Your video with Matt and Tony Padilla(?) on Graham's number is what drew me to Numberphile and (eventually) your other channels. Great job, Brady!
4:00, "In a complex situation, there is a certain order."
Modern developments in mathematics show the truth of that.
I was actually going to request a video on Graham's number!!! Thanks!
2:15 'If you put enough points down, must you get a *five-sided YES* ' XD
The title needs to be "WIN £1,000!!!"
Prof. Graham's handwriting is beautiful!
Hungarians again...
is there some part of the conjecture that states that three points or more points in plane can't form a line because otherwise you could just have a line with 33 points and it would be impossible to create a 7-gone with that?
Wow. The one and only professor Graham! Nice video!
If you take any three points you can have a wave structure. Two for half wave. A straight line not touching more than two is a collapsed half wave. Seven sides have 14 collapses. So 28 plus 5 for structure.
I hate the sound of felt pen on butcher paper. It is the same as fingernails on a chalkboard.
I like that sound
@@kourii same
@kourii I do too!
As a amateur Mathematician (Graph Theory), this is a problem which I would like to contribute to.
is there a way to generate sets of 2^(n-2) points which don't contain a convex n-gon
Heh! "Happy Endings"
Whoa...THE Ron Graham!
Conway, Graham, Grime...when will you get Perelman in a video to discuss something?
Omg !Upload the freaking graham's number video!I'm dying here Brady!Thanks for the video,nice job ;)
I'm having a go at trying to solve the problem...
I have tried my best to try and focus the problem down by working with how the various positions of the points on the plane, which I worked on by scribbling out the equations of lines, and the two coordinates the points would take up, and the ordering of the end points of which the next line would be drawn...
I'm going to see if this line of thinking is going to work... maybe
I am trying to see where I can go with looking at the various possible combinations that could exist between the points.
Just curious to know how it would be if all of the points were stacked upon one another?
Can't wait for the Graham's Number video by its "inventor"!!
Hey Brady. I don't think he meant that the Erdos conjecture formula gives you the minimum number required. I think it's just a number that is enough. Since he said sometimes it may not have the +1.
a practical application of convex polygons off the top of my head would be in solving a linear program. Convexity is important when looking at a feasible region of solutions to find the optimal solution since that region will have a finite amount of sides and points.
Happy Ending Problem - Numberphile
Ok, knowing this, what is the practical application for this aside from drawing convex polygons?
Ramsey theory, which tells us about how order arises in chaotic systems. If we understand simple problems like these, we can better understand the weather, the stock market, and many other chaotic systems.
using it to solve more complex problems
This problem is the coolest name I've heard.
How long will it take to reach the Happy Ending?
I'm confused, the new video features the Knuth arrow but Graham gets credit instead? Am I missing something?
I'm pretty sure the solution to this is to determine all of the different ways that you can form a concave n-gon from convex (n - k)-gons, where n - k > 2 through indenting vertices. And by pretty sure, I mean I'll have to play around a bit more.
Given the backstory with the two mathematicians suddenly getting married and sticking together for 90 years, plus the name of the problem, I was half-expecting this to be some kind of secret mathematical proof about happiness :P
Seems like it began as a juggling problem, it was mentioned in the other video that Graham was into circus performance. Just so we can abort the whole "what would I use this for???" debate
Hold my beer, I can prove this.
I have to admit, I thought "happy ending" was about something else.
O.K., I know where the +1 bit comes from, I think.
If you have n points and they set up so that all the points are being used up in an n-gon and you move one of the points away from the center of the 'gon then it shouldn't effect the concavity of the 'gon uless it crosses a line formed by two other points in the 'gon.
The reason would be this, if you take any four points in your 'gon, they form at least two triangles. If three points are in a line--this could happen if you move one of the two points shared by both triangles towards the other--then there are two right triangles. If you continue to bring those two triangles together, they both become obtuse and then you have to take that point out of your 'gon to maintain concavity and the 'gon becomes an n-1-gon.
You can keep doing this and at some point, doing this brings enough points into the inside of the 'gon then you have the situation where you can take one point on the outside of the 'gon out of the connection and instead connect two of the points in the interior to the convex 'gon. because you have a potential acute triangle that does not cut though a potential line in the 'gon.
But where does the 2^(n-2) come from? I don't know. If you start with an 8-gon and move one point into the middle, you have a 7-gon and an extra point. Say point one (they're numbered sequentially around the rim). if you move #3 into the middle, you have an accute triangle with the to moved points and #2, so you can take #2 out of your 'gon and restore #1 and #3 to your 'gon. But only if they have not crossed another potential line between two points still in 'gon.
But this is intersting. If you move #1 in and it crosses the 8-2 line, then we go from 8 to 7 sides. if we bring #2 in it may become a 6-gon, but only if we also moved #1 past the 8-2 line. If we move #3 in past the 8-4 line, it becomes a 5-gon but only if both #1 and #2 have also been moved past that line as well. If we move the #'s 1-4 past the 8-5, then the the formula suggests that there must now be two points in the middle of the 4-gon that form an accute triangle with #5, #6 or #7 and does not cut through the 8-6 or the 7-5 potential lines. It has to be acute be cause an obtuse potential line in one of these potential triangles would violate concavity.
And I'm liable either wrong or just figured out something already know. I bet wrong.
+Frank Harr
Gotit! At least for 9 points.
5 6 7 8 from above forms a 4-gon. And that has 4 internal potential triangles. You can put an extra point in each of those and very carefully arrange it so that none of those give you an opportunity to make a 5-gon. You cannot put a point outside of the 4-gon because no matter where you put it, you could form a 90 degree angle with the perimiter and that allows you to incorporate it into the perimeter. And no matter where you put it withing the 4-gon, it shares a potential triangle with another point and THAT forms a triangle with a vertex on the perimiter, and another point in the same potential triangle and that creates an acute potential triangle that is within another and thus you cant take that vertex off of your perimiter and incorporate those two points as two new vertices within the perimiter.
Now if you have a 5-gon, don't bother putting any points outside of the perimiter because it can only form a right angle with the perimiter and end this quickly or you can just move your vertex and thus make your old vertex a point within the 5-gon. You can carefully fill each potential triangle with one point each and then. . .
I am working on the next part. This is fun.
I thought I had a handle on something, but it broke off. But I honestly think that what we're dealing with here is overlapping triangles and quadralaterals. You have those on the perimiter, those inside the n-gon and every time you add a point, you get more triangles. And eventually you get so many that you you can replace one vertex with two from the points inside the n-gon.
+Frank Harr
There's another way to think about it.
If you inscribe a pentagram in a pentagon (let's call it 1-5 after the points) you have two regions the center pentagon and everything else. If this was a hexa- septa or any n>5-gon, you'd have more regions, but I don't know anything about those yet.
Now, if you a random point a in pentagon 1-5, there are two regions where you cannot put another point without automatically allowing the pentagon to become a hexagon. And I'm still working on defining these regions. If your a is in section 1-3 of 1-5 then you draw a line from 1 through a and out of the pentagon and another line from 3 through a and out of the pentagon. The two triangular-shaped regions between those two lines and the perimiter are part of your "danger zone" where, if you place point b, you can automatically eliminate vertex 2 and create a hexagon. Point a could also be part of section 5-2 or 2-4 (but not both), in which case you have two more danger zones. This is, of course, because of the definition of convex here. That is, as you go around the perimiter, you only turn in one direction. Which, incidently, get you your shortest perimiter per area. But I doubt that's helpful.
Inside the center, you need at least four points to get rid of three vertices. I'm still working on figuring those danger zones out.
+Frank Harr
O.K., if you put points a and b in the center area of of 5-gon 1-5 draw a line through them to the perimiter you'll have 3 vertices on one side and two on the other. Let's say it's vertices 1-3 and 4-5.
Now, if you draw a line from 5 to a and from 4 to b (and they don't cross) this creates two "danger zones", but not as dangerous as before. If you put one point in each of these two zones (c and d), you can draw your convex hexagon.
However, you can put both points in the same zone if you put point d in the two triangles formed by a line drawn through 5 and a (extending to line ab) and a line drawn from a through c to the perimiter of the 5-gon.
You can also put c and d in the small triangle formed by the line from 5 through a, the line from 5 through b and line ab if point d is on the same side of line ac that b is.
+Frank Harr
You know, if you draw a line from 1 though a and from 3 through b, you'll get three triangles based on lines 1a, 3b, the perimiter and line ab that are also danger zones. These things just multiply.
---------2
-------/ \
-----1 3
-----| |
-----| a b |
-----| |
----5-------4
I'm pretty sure the definition of convex and concave polygons is convex: all exterior angles of a polygon are greater than 180. and concave: one or more exterior angles are less than 180
I believe that one of the professors at my university said that he worked with Graham on the mathematics of juggling. Would you be willing to ask him to do a talk about that problem?
correct me if im wrong but can't you just make a figure with more sides like go from pentagon to hexagon , but use like 39 points, or something ridiculous and still be with in the rules?
I think I may have managed to prove the conjecture, took some days of work, and 11 full A4 pages of endless maths, but I think I've got the proof :)
I'll need to get someone who knows mathematics deeper than me to check it and make sure it is solid before I hand it out or message anyone about it though.
Interesting.. why does the audio for this video come out of my center speaker? Mono audio, or is everything mixed in 5.1 and sent to the center channel for some reason?
Ron Graham! Oh geez
Awesome problem! Thank You! :)
So a convex 2-gon (A line segment?) would have 2 points. That makes sense since line segments can't be concave, if you would even call a line segment a polygon.
So the conjecture is for a set of points, what is the maximum number of sides can you draw, in a concave way, when the points do not have straight lines between them? Is it 2^(n-2)+1=x where n is the number of points and x is the number of sides?
No, the conjecture is that in order to guarantee there's at least one convex N-gon (an N-sided polygon that has no "dents"), you need at least 2^(N-2)+1 points.
IceMetalPunk Right, got it.
Not quite sure mathematicians know what a "Happy Ending" is.
LOL
What if you had a three dimensional object, however only one vertex was present in the two dimensional plane you were working in, if the object is still considered a polygon (and it might not, I don't know if it would) then it you could use the theorem on that single vertex while it might appear to be a 1D object while in reality a 3D object, resulting in 1.5 dots.
Hypothetically, if you found a solution where would you submit it?
Collatz conjecture!! 3x+1 problem!!!
ben1996123 hi lolben
ben1996123
OMG its lolben!
Who is lolben??
lol, you are like a super-master of Rubik cubes. Pleasure to meet you, ben1996123
Where is the new video about odd equations?
There've been some seriously big names on Numberphile lately.
how's this a problem?
Unsolved for a hundred years but doesn't qualify as a problem in your book...?
Brady, ask them mathematicians about how each of those discussed crazy math problems are connected to the real world. If you do that, i'll subscribe for sure:)
please explain e, the natural number. what's so special about e?
When you differentiate e^x it remaps itself to get e^x.
Assuming you've done high school calculus, the derivative of e^x happens to be e^x, and that doesn't really occur with any other function making it useful for logarithms and such.
There are several things that are special about e. One example is that the derivative of e^x is e^x itself. This means that for the function f(x) = e^x, the slope of the function is the same as the value of the function. This makes e extremely important in an extremely useful branch of mathematics, differential equations, which describes, among many other things, heat transfer, population increase, compound interest, and many aspects of movement.
The thing that's special about it is that the exponential function e^x is equal to its own derivative.
Zura ja nai e is not natural
Thanks !
I'd like to donate to proving/disproving this ----- link please
I'm very interested to know the formula for 3d shapes
I'm glad I found this comment ! Agreed much
If you give me 4 non-colinear points I can build you a convex tetrahedron! 3^(n-3)+1
Any 3-d version of this?
Why can't I draw a 7-gon with the 9-point layout? If I connect all the points except the two almost at the centre, I end up with a 7-gon, with only right turns?
The question is how many points GUARANTEE that you can draw the n-gon. Sure, in some configurations, you will be able to draw some huge n-gons, but that's not the point.
Only writers and mathematicians can find a problem with a happy ending
So is the plane infinitely large? And therefore can points be placed infinitely far apart from each other?
Yes, No. Since it's infinitely vast the points can be as far away from each other as you like, but still (in this case) finitely far apart.
Interestingly, convex polygons also consist only of right turns.
what's the configuration of 16 that doesn't give you a hexagon
hey what if the 5 point are colinear. join 4 point will just form a line so what say .
The happy ending problem based on 4 degree equation the dots is basically a root of the surface basically in the 4 points is the based on three degree equation is the imposible to made gon if we 2^n-2+1 is the four degree equation
Was there any advancements with this in the last 7 years?
Will you post video if somebody figures this out? Also if i want to get Ron Grahams 1000 dollar(or euro) where do i post the proof?
Can´t wait for Graham´s number by himself...
2:13 He can't say "pentagon"? :(
+RonJohn63 pentagons have equal side lengths
+Kaustubh Dongre doesn't only the regular pentagon have them?
+zoranhacker I guess he could have said irregular pentagon but wanted to keep his words consistent in the video so people understood what he said as clearly as possible.
***** But the way to learn new, "ooh scary complicated" stuff is to see and hear it used. Otherwise, you keep people permanently infantilized.
One point can be more than one since it is 0 dimensions the dots don't take up any space
Wow! It hasn't even been proven for the must be a convex 7-gon inside 2^(7-2) + 1, ie 33, points. It's beyond computation!
The god of numbers!
Has this been solved yet?
For a second I was like "can't they at least check it for some big n with computers?", but then I hear that just for n=7 it's already getting too hard!... Just how big the computational complexity for that problem is?
How fast does it rise or what kind of numbers we are talking about?
Well, for the n=7 case, if we wanted to check all possible 7-gons for 2^(7-2) + 1 = 33 points, there would be (33 choose 7) possible shapes to check, which is over 4 million! And that would just be for a single arrangement of points! In general, since we'd want to check configurations of 2^(n-2) + 1 points for n-gons, there would be (2^(n-2) + 1 choose n) shapes to choose for each possible arrangements of points, which is a function that grows faster than any exponential!
I'd like to see the 16 points exemple where you can't get a 6points convex in it
O.K., if you have a triangle with two points inside of it, you can make one convex 4-gon.
but if you make a convex 4-gon with two points in it, you don't have to make a convex 5-gon.
This is hard.