Thanks for the great explanation! This also provides sufficient knowledge to learn ECC as it covers all possibilities of point addition and its properties. Understanding the math with clarity was very helpful.
By defining a binary operator, point addition, we can see that all the points on an elliptic curve form a group, which has four basic properties: the operation is closed and associative, there is an identity and each point has its inverse. Great presentation! Thanks!
Thanks. In particular I appreciate that you walked through the equations and solved for (x3,y3). I've seen a number of videos which show the graphical solution but hadn't seen the basic equation solving. I was frustrated by this and, as you showed, it is not that complicated to solve the equation.
@muffemod Good question! It's so that the points form a group under point addition. Hopefully with this example it will make more sense. A group needs a single, unique element e, the identity, so that a * e = a for all a in group. Define ' * ' so that a * b = -(a + b) over the integers. Now a * 0 = -a Is it possible to define an identity? a * e = a => -a + -e = a =>e = -2a Operation * with integers has no id. Taking out the reflection means no identity, and no group. Does that make sense?
Not to me. I understand that reflecting it is just a part of the procedure. But what i concern about is why this point addition is associative. In your video, you just say that it is associative without any proof. I know that if we use algebra, we can definitely prove it. But that would be a very ugly way to prove it. Is there any better way?
that's just a property that needs to be fulfiled ,we need this elements to form a group in order to do the cryptography stuff with it ,in order for it to be a group the elements(in this case the curve points) needs to be associative ,the inverse must exist ... and it needs to have an identity element mainly an element "a "that has the property P*a =P for all elements. and for P*infinity=P you need to reflect .its just about the group operation ,since you need reflecting to get the identity..
Great video!!! I love the Wiles based modular semistable elliptic curved shapes to describe quantum fundamental - indivisible particles as the potential range of spacetime gushing to existence.
Just a question: what is P_3 if you add two points P_1=(x_1,y_1), P_2=(x_2,y_2) and x_1>1, x_2>1, x_1 not equal to x_2? May be for some choice of P_1 and P_2 the third point does not always exist? May be I am missing something. Thank you in advance.
Its a theorem. In cubic polynomials, sum of roots equals -ba. More precisely, a polynomial ax3 + bx2 + cx + d with roots r1, r2, r3; will satisfy the following: r1 + r2 + r3 = -ba
Ok, so stupid me has to ruin the party by saying that this doesn't make sense. How does a line that passes through a and a' going to intersect the structure on the positive side of the graph? I can agree that the slope of the structure on the right side will eventually approach infinity because the function's derivative results in the numerator having a higher polynomial degree than the denominator but this will result in a vertical line at infinity, but I don't think that it will ever pass the y axis ever again. Because the round structure is to the left of the y axis, it doesn't seem like the line that passes through those two aforementioned points is ever going to touch the structure on the right.
You said that for every two points, there's always a third point. I don't really get it. In your examples you chose the two points on the small circle, and you got a point on the shape on the right of the curve. But, what if you take two points on the right shape, that miss the circle? For instance take a line that is almost vertical. Where is the 3rd point? Is it at infinity?
Very good question. I was confused too when I read it, so I went over to desmos. The answer is, although it may not look like it, the almost vertical line will intersect the curve again at a third point, it might be very high up, but it will happen. This is due to the way the curve bends
@@willnewman9783 No, you can go to Wolfram Alpha and compute the intersection of y=10-9x and the elliptic curve, and you will see that one of the solutions is complex number.
@@borisreitman this is false. The line y=10-9x and the elliptic curve y^2=x^3-x intersect each other at 3 points, all of which are real. Wolfram alpha told me that all of them are complex, but the imaginary parts were on the order of 10^-13, which is just a round-off type error
@@willnewman9783 You are right. I went to the Desmos calculator, and saw it. I now understand the principle. Instead of thinking y as a function of x, things become clear if we think of x as a function of y, and look at the graph sideways. The curve grows slower than a line. Therefore, any line x=ay+b that grows will always catch up with it. On the other hand, a line with a fixed x value, such as x=2, will stay behind the curve, because the curve is growing, but the line is not.
Well, some of it you can learn in Abstract Algebra, which will probably be your junior year. That's where you learn about groups. What is interesting about these curves is that they form a group. (One thing interesting about them.) But you can learn a lot about algebra as an undergrad--groups, rings, fields, and Galois theory. Then, from one point of view, these curves described here are just one example of a group. That's what she's showing.
greg55666 RE: " What is interesting about these curves is that they form a group." To be specific, the curves themselves don't form a group. Rather, the points of the curve form an (abelian) group. And moreover, if you make restrictions on the field in which the coordinates of the points lie, then you get interesting subgroups.
Watched it twice but still have trouble accepting why do you reflect? It seems like you're justifying reflecting by saying a reflection of a reflection is itself. Which is true but doesn't explain why you reflect. In the slide around 3:40 to 3:59 why can't just -P be the third point? Why isn't the 3rd point of intersection just simply -P ? P + infinity = -P
Very nice into video but @5:37 doesn't prove associativity. That's the hardest one to prove. Some folks like J.W.S. Cassels, J.S. Milne and L. Washington use the nine point theorem for cubic (pg 13) citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.437.4096&rep=rep1&type=pdf others use the Weierstrass p function.
ganondorfchampin, 5:38, you provide an illustration and postulate “we get exact same point thus associativity holds” - however, it does NOT follow from the drawing that you get exact same point. You don’t use the fact that the curve is elliptical ;) P.S. I’m doing well reading, thanks.
".. Competition. I won!"
I see why you won this contest! Great explanation, congratulations!
After 10 years , thank you very much
Thanks for the great explanation! This also provides sufficient knowledge to learn ECC as it covers all possibilities of point addition and its properties. Understanding the math with clarity was very helpful.
By defining a binary operator, point addition, we can see that all the points on an elliptic curve form a group, which has four basic properties: the operation is closed and associative, there is an identity and each point has its inverse. Great presentation! Thanks!
Thanks. In particular I appreciate that you walked through the equations and solved for (x3,y3). I've seen a number of videos which show the graphical solution but hadn't seen the basic equation solving. I was frustrated by this and, as you showed, it is not that complicated to solve the equation.
very clear and informative knowledge about the background to elliptic curve cryptography. you rock!
No, thank YOU!
Crystal clear explanation, found it very helpful. Thanks!
Excellent presentation!
Riverninj4: "So not only is this a group, but the curve E under point addition is an Abelian group..."
My Brain: 🎶🎵🎷🐴 🎶🎺🐒 ✨🪘🦫
@muffemod Good question! It's so that the points form a group under point addition. Hopefully with this example it will make more sense.
A group needs a single, unique element e, the identity, so that a * e = a for all a in group.
Define ' * ' so that a * b = -(a + b) over the integers.
Now a * 0 = -a
Is it possible to define an identity?
a * e = a
=> -a + -e = a
=>e = -2a
Operation * with integers has no id.
Taking out the reflection means no identity, and no group. Does that make sense?
Not to me. I understand that reflecting it is just a part of the procedure. But what i concern about is why this point addition is associative. In your video, you just say that it is associative without any proof. I know that if we use algebra, we can definitely prove it. But that would be a very ugly way to prove it. Is there any better way?
we can use an elliptic curve to define some range of points in an elliptic curve, not only two points of an Eucledean straight line!
Wow, really good explanation. Thank you!
that's just a property that needs to be fulfiled ,we need this elements to form a group in order to do the cryptography stuff with it ,in order for it to be a group the elements(in this case the curve points) needs to be associative ,the inverse must exist ... and it needs to have an identity element mainly an element "a "that has the property P*a =P for all elements.
and for P*infinity=P you need to reflect .its just about the group operation ,since you need reflecting to get the identity..
Can you PLEASE create more videos? You are GREAT and THE BEST!!!
thanks!!!!!!!
:) choose even other math subjects! ;)
Good presentation.
Thanks. Nicely done. It helped me understand this a little better.
You're just.. so good.
My neanderthal brain does not understand
Great video!!! I love the Wiles based modular semistable elliptic curved shapes to describe quantum fundamental - indivisible particles as the potential range of spacetime gushing to existence.
My Professor spent an entire two hr lecture, and I didn't learn anything. Well until now.
You are better than Wikipedia. Thanks.
So basically the reason for reflecting is to satisfy the conditions of a group.
Very good explaination. Thanks Very Much !!!
Thanks for presentation
Just a question: what is P_3 if you add two points P_1=(x_1,y_1), P_2=(x_2,y_2) and x_1>1, x_2>1, x_1 not equal to x_2? May be for some choice of P_1 and P_2 the third point does not always exist? May be I am missing something. Thank you in advance.
Can someone please explain what is meant by *_the coefficient of x2 is the opposite sum of the roots_* ?
Its a theorem. In cubic polynomials, sum of roots equals -ba. More precisely, a polynomial ax3 + bx2 + cx + d with roots r1, r2, r3; will satisfy the following: r1 + r2 + r3 = -ba
What is the value of lambda if P1=P2?
Amazing vid. Thanks
Ok, so stupid me has to ruin the party by saying that this doesn't make sense. How does a line that passes through a and a' going to intersect the structure on the positive side of the graph? I can agree that the slope of the structure on the right side will eventually approach infinity because the function's derivative results in the numerator having a higher polynomial degree than the denominator but this will result in a vertical line at infinity, but I don't think that it will ever pass the y axis ever again. Because the round structure is to the left of the y axis, it doesn't seem like the line that passes through those two aforementioned points is ever going to touch the structure on the right.
Am I the only one who thinks that in 3:31 the equation for Y3 is wrong? Shouldn't it be y3=m(x1-x3)+y1?
It is already reflected.
Mam plz can you explain it more?
..then you need to include reflecting in the group operation(point addition) which i wrote as (*)
You said that for every two points, there's always a third point. I don't really get it. In your examples you chose the two points on the small circle, and you got a point on the shape on the right of the curve. But, what if you take two points on the right shape, that miss the circle? For instance take a line that is almost vertical. Where is the 3rd point? Is it at infinity?
Very good question. I was confused too when I read it, so I went over to desmos.
The answer is, although it may not look like it, the almost vertical line will intersect the curve again at a third point, it might be very high up, but it will happen.
This is due to the way the curve bends
@@willnewman9783 No, you can go to Wolfram Alpha and compute the intersection of y=10-9x and the elliptic curve, and you will see that one of the solutions is complex number.
@@borisreitman this is false. The line y=10-9x and the elliptic curve y^2=x^3-x intersect each other at 3 points, all of which are real.
Wolfram alpha told me that all of them are complex, but the imaginary parts were on the order of 10^-13, which is just a round-off type error
@@willnewman9783 You are right. I went to the Desmos calculator, and saw it. I now understand the principle. Instead of thinking y as a function of x, things become clear if we think of x as a function of y, and look at the graph sideways. The curve grows slower than a line. Therefore, any line x=ay+b that grows will always catch up with it. On the other hand, a line with a fixed x value, such as x=2, will stay behind the curve, because the curve is growing, but the line is not.
When will I learn about this? My high school's math isn't interesting like this :/
Maybe in college, but only if you specifically ask about Elliptic curves. Otherwise, you'll need to go to grad school to REALLY learn about these.
Well, some of it you can learn in Abstract Algebra, which will probably be your junior year. That's where you learn about groups. What is interesting about these curves is that they form a group. (One thing interesting about them.) But you can learn a lot about algebra as an undergrad--groups, rings, fields, and Galois theory. Then, from one point of view, these curves described here are just one example of a group. That's what she's showing.
greg55666 RE: " What is interesting about these curves is that they form a group."
To be specific, the curves themselves don't form a group. Rather, the points of the curve form an (abelian) group. And moreover, if you make restrictions on the field in which the coordinates of the points lie, then you get interesting subgroups.
Aaron Wolbach I'm sure that clarification was very helpful to the original poster.
greg55666 Fair point.
Please Can you give me the power point presentation ?
can anybody please explain how we derive the formulas for coordinates x3 & y3 when we are adding the SAME point to itself?? please and thank you!!
just consider the formula of a tangent line through the derivative of the curve
thx for this video i have a project programming Elliptic curve ECC cryptography on Stm32
Watched it twice but still have trouble accepting why do you reflect? It seems like you're justifying reflecting by saying a reflection of a reflection is itself. Which is true but doesn't explain why you reflect. In the slide around 3:40 to 3:59 why can't just -P be the third point?
Why isn't the 3rd point of intersection just simply -P ?
P + infinity = -P
Very nice into video but @5:37 doesn't prove associativity. That's the hardest one to prove.
Some folks like J.W.S. Cassels, J.S. Milne and L. Washington use the nine point theorem for cubic (pg 13) citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.437.4096&rep=rep1&type=pdf others use the Weierstrass p function.
hey keep posting thanks for the video more emphasis on security or cryptography
Hi
You can’t call this addition BEFORE you prove associative rule (A+B)+C=A+(B+C). P.S. and you didn’t prove it either.
I'm sure it wouldn't be too hard to prove algebraically if you just write it all out and simplify.
ganondorfchampin, it wouldn’t, yet your graphical “proof” doesn’t work. Just wanted to point it out.
Dmitrii Zholud You need to learn to read, as I provided no graphical proof.
ganondorfchampin, 5:38, you provide an illustration and postulate “we get exact same point thus associativity holds” - however, it does NOT follow from the drawing that you get exact same point. You don’t use the fact that the curve is elliptical ;)
P.S. I’m doing well reading, thanks.
Dmitrii Zholud No, I did not. Again, you need to learn to read.
Who else is here because of Bitcoin?