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Okay, so I'm trying to implement frustum culling for my object quadtree, but keep running into a problem: How do I check if a quadtree is in the frustum if I don't know it's height? Since theres no limitation to it's height, I have to assume objects can be anywhere along the z-axis. So how do I perform this kind of test? Thanks in advance...

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 I don't know it's height? Since theres no limitation to it's height, I have to assume objects can be anywhere along the z-axis. So how do I perform this kind of test?

Why don't you know its height?

My quadtree/terrain implementation builds (at load time) the initial vertex buffer and then recursively subdivides this into the component index buffers and at the same time stores a bounding box for each node in the tree. Thus I know exactly the constraints for each node/leaf.

If you have a dynamic terrain then you can still process it at load time, just be sure to update your bounding box whenever you update the physical geometry.

Is that of any help? If not, provide some more details on why you cant get the bounding box and/or what information you have available...

hth
Jack

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Well, my terrain quadtree works like that too (only I use bounding spheres only).

However, my OBJECT quadtree doesn't keep track of it's height, since the objects are in linked lists and, well, frankly, doing that is a pain.

What I'm really looking for is a way to check the frustum against 4 rays (one in each corner), giving me a check against any height.

Each object has sphere based checking anyways, it's just that I need to quickly eliminate high-level nodes. I'm not too worried about the low-level incorrect trues it might return.

anyone?

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Hi,

Sorry for the slow reply... I was doing a recruitment event on Tuesday and been too busy since to look at GDNet [sad].

Anyway, for your ray-frustum check, I could see two possible ways of doing this:

Firstly; create a small sphere at each corner of your bounding area - you already have the code for sphere/frustum checking you say. Although this doesn't strike me as an elegant solution it should work.

Secondly; ray/plane collision tests are fairly simple mathematically (been a while since I messed with this though). You could generate a ray from your camera position to each corner; if the distance to the point is between the distances of the near plane and the far plane, and the intersection of the ray and the near plane is within the defined viewport, the corner must be visible. By implication, if all 4 corners are visible the whole area is, and if some of the corners are visible it's partially visible and you should recurse further...

I'm not too sure how you could compensate for the lack of a know height. Depending on whether your camera is fixed to a given DoF you might be able to improvise by only doing a ray-plane intersection on the XY axis' and ignoring the Z intersection (where Z is height). Or you could try and be clever and use plane-plane intersections; where you express your "ray" to each corner as a plane expanding infinitely up/down the height axis.

hth
Jack

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Thanks for responding, and I'm sorry for being pushy, I was too away from home and was expecting replies when I got back -- and got worried when there were none ;).

Anywho, back to my question, after really thinking it over, I've come to the conclusion that when I really need is some way to measure height, since an indefinate ray would alway intersect a plane if it wasn't exactly parallel, which would really never happen in my case.

So, I guess I'll have no choice but to set a maximum height for my engine, which is something I didn't want to do. Then, I'll have to write a box-frustum intersection function and use that to check if the whole quad is in view.

Kinda lousy, but I can't really think of doing anything else :).

Heres a link to my terrain engine (I'm adding the object tree to it now), if you're interested:
http://www.devimg.net/?View=484

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First off, maybe I dont understand ya. But why are you trying to check hieght at all? Unless you make an Oct-Tree which has the heights diveded into bboxs as well... I do 3 different tests on the bounding box's planes including sphere-box and box-box. I have never needed a reason to include height, I simply set hte heights to 1.0f :). If you want to look at some example code then I can post some if you tell me exactly what you are looking to test against and why height is a factor. This is a quad-tree right?

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That's perfect!

My problem was, that since I'm writting a 3D game -- I was thinking 3D. How over-complicated that is :).

I was thinking that since my frustum is 3D, and my QuadTree is 2D, I need to make my QuadTree 3D, but in reality, making the frustum 2D would be so much simpler.

So, now I need to find a way to get a 2D frustum, that is squash it down to z=0.0f; but how do I do that? I'd like to get all 8 corners, then move each of them to z=0.0f, and perform some kind of check against them. But how would I perform that? any suggestions?

Thanks a lot -- that was tremendously helpful ;).

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here is an example of how I built my frustum, I have a class called camera which holds this frustum in it....

class Frustum{	public:		Plane		plPlane[6];		Sphere		sphBoundingSphere;		Cone		cneBoundingCone;		float		fNearPlane;		float		fFarPlane;		float		fFieldOfView;		float		fHalfSHeight;		float		fHalfSWidth;		Vector		vecCamPos;		Vector		vecLook;		int		ContainsSphere(Sphere& pSphereInc);		int		ContainsAABB(AABB& pAABBInc);		void	Update(D3DXMATRIX& pmatView, D3DXMATRIX& pmatProj);		void	ConstructBoundingSphere(void);		void	ConstructBoundingCone(void);				Frustum();		virtual	~Frustum();};

this is where I update it, by ripping the planes from the matrix. Its pretty standard code for doing it in directx.

void Frustum::Update(D3DXMATRIX& pmatView, D3DXMATRIX& pmatProj){	D3DXVECTOR3 vecLocal(1.0f, 1.0f, 1.0f);	D3DXMATRIX matLocal = *D3DXMatrixIdentity(&matLocal);	D3DXMatrixMultiply(&matLocal, &pmatView, &pmatProj);	vecCamPos		= Vector( -matLocal._41, -matLocal._42, -matLocal._43 );	vecLook			= Vector( matLocal._13, matLocal._23, matLocal._33 );	// LEFT SIDE	plPlane = Plane( Vector( matLocal._14 + matLocal._11,									matLocal._24 + matLocal._21,									matLocal._34 + matLocal._31 ),									matLocal._44 + matLocal._41);	plPlane.Normalize();	// TOP SIDE	plPlane[TOP] = Plane( Vector(	matLocal._14 - matLocal._12,									matLocal._24 - matLocal._22,									matLocal._34 - matLocal._32 ),									matLocal._44 - matLocal._42);	plPlane[TOP].Normalize();	// RIGHT SIDE	plPlane = Plane( Vector( matLocal._14 - matLocal._11,									  matLocal._24 - matLocal._21,									  matLocal._34 - matLocal._31 ),									  matLocal._44 - matLocal._41);	plPlane.Normalize();	// BOTTOM SIDE	plPlane[BOTTOM] = Plane( Vector( matLocal._14 + matLocal._12,									   matLocal._24 + matLocal._22,									   matLocal._34 + matLocal._32 ),									   matLocal._44 + matLocal._42);	plPlane[BOTTOM].Normalize();	// NEAR SIDE	plPlane[CLOSE] = Plane( Vector( matLocal._13,								      matLocal._23,									  matLocal._33),									  matLocal._43);	plPlane[CLOSE].Normalize();	// FAR SIDE	plPlane[AWAY] = Plane( Vector( matLocal._14 - matLocal._13,									 matLocal._24 - matLocal._23,									 matLocal._34 - matLocal._33 ),									 matLocal._44 - matLocal._43);	plPlane[AWAY].Normalize();	ConstructBoundingSphere();	ConstructBoundingCone();}

1. if the box contains the camera pos, if it does then stop testing you know u need to render the quad, or keep traversing tree.

2. do a sphere-sphere test, if they do intersect continue

3. do a cone-sphere test, if that intersects continue

4. check if the frustum contains the sphere, if so then check
a. out frustum
b. in frustum
c. intersect frustum - then do a frustum box test
in? out? intersect?

By Z you mean height? different axis than me :). You still need to test the height against the infinite planes of the quadtree's quads because what if your frustum has turned? the Roll, in yaw-pitch-roll? IE. flying a plane

PS. You dont hafta squish yer frustum planes down. They form a 6 sided box already which means something hasta either be IN it or OUT of it, or INTERSECTING it. Which you would hafta render it if its IN or INTERSECTING.

In all reality all you have to do is a box-box test, but when traversing a quadtree, you dont want to test EVERY quad by a box-box because that would be VERY cpu intensive and waste alot of precious time. So test by easier methods first. Most of the quadtree will be outside and by using multiple easy tests then you save alot of time for when something is inside of it.

[Edited by - xsirxx on March 10, 2005 7:29:55 PM]

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Okay, a bit confusion there :)

The reason I can't do plane sphere tests, is that I don't want to limit myself to a certain height (although doing so might be wise, at this point).

However, since I think I may have found a simple solution, I'm gonna give it a try. What I intend to do (gonna write it today, when I can clear some free time), is to extract the 8 corners of my frustum (dunno how to do that yet).

Then , I'll move them to z = 0.0f, to flaten the frustum to 2D. Then, I'll construct a bounding square around the 8 points (that's simple) and finally, test all my quads against this specific square.

Sounds quite simple, all things considered. I just need to find out how to extract the 8 points.

What do you guys think, is this viable?

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I wouldnt do that, box-box testing is a HUGE waste of cpu time. Try it that way, then try it the other way if ya like. Its worth it now to write sphere-sphre, cone-sphere, box-box testing... You WILL use all this code over and over and over. I can give you examples of code for these tests, but I would study them, go over and really understand them. This is some easy 3d code comparable to what else is out there :). Also this is not the fastest, but its easy to understand for me and runs :).

Also your in NO WAY limiting height at all, nor do ya need to know a height. thats the beauty, its height free :).

bool Cone::Intersects(const Sphere& sphereInc) const{	Vector vecU, vecD;	float fSin = sinf(this->fAngle);	float fCos = cosf(this->fAngle);	vecU = this->vecEye - (sphereInc.r / fSin) * this->vecAxis;	vecD = sphereInc.c - vecU;	if( vecAxis.Dot(vecD) >= (vecD.Magnitude() * fCos) )	{		// center is inside K''		vecD = sphereInc.c - this->vecEye;		if( -this->vecAxis.Dot(vecD) >= (vecD.Magnitude() * fSin) )		{			//center is inside K'' and inside K'			return (vecD.Magnitude() <= sphereInc.r);		}		else		{			//center is inside K'' and outside K'			return true;		}	}	else	{		//center is outside K''		return false;	}}

bool Sphere::Intersects(const Sphere& sphereInc) const{	Vector vSepAxis	= (this->c - sphereInc.c);	float fRadiiSum	= (this->r + sphereInc.r);	if( vSepAxis.MagnitudeSQ() < (fRadiiSum * fRadiiSum) )return (true);		return(false);}

eCD AABB::ContainsPoint2d(Vector &pvecInc){	Plane planeL;	Plane planeR;	Plane planeT;	Plane planeB;	planeL = Plane( vPoint[LBL], Vector( vPoint[LBL].x, (vPoint[LBL].y - 1), ((vPoint[LBL].z + vPoint[LTL].z) / 2)), vPoint[LTL]);	planeT = Plane( vPoint[LTL], Vector( ((vPoint[LTL].x + vPoint[LTR].x) / 2), (vPoint[LTL].y - 1), vPoint[LTL].z), vPoint[LTR] );	planeR = Plane( vPoint[LTR], Vector( vPoint[LTR].x, (vPoint[LTR].y - 1),((vPoint[LTR].z + vPoint[LBR].z) / 2)), vPoint[LBR] );	planeB = Plane( vPoint[LBR], Vector( ((vPoint[LBR].x + vPoint[LBL].x) / 2), (vPoint[LBR].y - 1), vPoint[LBR].z), vPoint[LBL] );	if( planeL.ClassifyPoint(pvecInc) == Plane::eCP::BACK )return OUTSIDE;	else if( planeT.ClassifyPoint(pvecInc) == Plane::eCP::BACK )return OUTSIDE;	else if( planeR.ClassifyPoint(pvecInc) == Plane::eCP::BACK )return OUTSIDE;	else if( planeB.ClassifyPoint(pvecInc) == Plane::eCP::BACK )return OUTSIDE;	else	{		return INSIDE;	}}

I REALLY REALLY REALLY cant stress how important it is that you go the distance on this quadtree. If the quadtree is halfassed or the right managing doesnt go into it, it can kill your game. Your going to run this code THOUSANDS and THOUSANDS of times per render, more than any other code(most likely) :).

PS. I used a VERY nice tutorial at one point for alot of these ideas, so if any1 can recognize what tutorial it was and has it, please post a link! :)

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Okay, my problem isn't the part of checking against the frustum. Really, I have all the code I need to check both sphere, a box and whenever I need against the frustum.

What I'm missing is a way to convert my quadtree node to either a sphere, a box or any other shape I know how to check against the frustum.

In fact, speed is not (yet) an issue.

I need to find a way to check a quadtree node against the frustum, and naturally, since it's a quadtree, and not an octree, it isn't limited in height (I don't plan to limit it).

I hope this clears up my issue some more.

I'll attemp to implement my method tomorrow, and post back some results.

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Yea, but you really dont have a problem at all. You already have all your box-sphere-cone testing code, alls ya hafta do is take the box against the frustum by the most expensive test there is:

int Frustum::ContainsAABB(AABB& pAABBInc){	Plane::eCP localeeCP;	int iPointOut	= 0;	int iPlaneIn	= 0;	for(int i = 0; i < MAX_PLANES; i++)	{		iPointOut = 0;		for(int ii = 0; ii < MAX_POINTS; ii++)		{			localeeCP = plPlane.ClassifyPoint(pAABBInc.vPoint[ii]);			if(localeeCP == Plane::eCP::BACK)iPointOut++;		}				if(iPointOut == MAX_POINTS) return(OUTFRUSTUM);		else if(iPointOut == 0) iPlaneIn++;	}	if(iPlaneIn == MAX_PLANES) return(INFRUSTUM);		return(INTERSECTFRUSTUM);}

and test like so

					switch(m_Camera->frFrustum.ContainsSphere(pTL->sphere))					{						case OUTFRUSTUM:							bSkip = true;							break;						case INFRUSTUM:							bLocalTest = false;							break;													case INTERSECTFRUSTUM:													switch(m_Camera->frFrustum.ContainsAABB(pTL->aabb))							{								case INFRUSTUM:									bLocalTest = false;									break;								case OUTFRUSTUM:									bSkip = true;									break;								case INTERSECTFRUSTUM:									break;							}												break;					}

but in all reality testing against sphere-cone is much nicer and doesnt have the height problem either. This tests against points of the box against the planes of the frustum.

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But where do I get the AABB box from? If this were an octree, I'd have the AABB, but in a quadtree, I don't have an AABB box, all I have is a square with unlimited height.

Heres a diagram I made to show the difference.

Gonna try to implement my method today and see how it goes.

Blue is an AABB box,

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Success! or at least, sort of :).

I finally got about an hour of free time, and managed to write all the code I needed. Though it's not working with my object tree, I was so pleased with how fast this check was, that I decided to add it to my terrain tree too, as the first check. It works very well in my terrain tree, so I suspect an error in my object tree is what's causing problems, and not the actual testing code.

Actually, the hardest part turned out to be finding the frustum's AABB. Once I had that, the checks were very simple.

Creating the frustum's AABB:
	D3DXMATRIX tmpMat;	float flt;	D3DXMatrixInverse(&tmpMat, &flt, &mat); // mat is the proj * view * world matrix.		// Calculate AABB;	D3DXVECTOR3 vec = D3DXVECTOR3(1.0f, 1.0f, 1.0f);	D3DXVec3TransformCoord(&vec, &vec, &tmpMat );	m_AABB[0] = vec.x;	m_AABB[1] = vec.x;	m_AABB[2] = vec.y;	m_AABB[3] = vec.y;	vec = D3DXVECTOR3(-1.0f, 1.0f, 1.0f);	D3DXVec3TransformCoord(&vec, &vec, &tmpMat );	m_AABB[0] = max(vec.x, m_AABB[0]);	m_AABB[1] = min(vec.x, m_AABB[1]);	m_AABB[2] = max(vec.y, m_AABB[2]);	m_AABB[3] = min(vec.y, m_AABB[3]);	vec = D3DXVECTOR3(1.0f, -1.0f, 1.0f);	D3DXVec3TransformCoord(&vec, &vec, &tmpMat );	m_AABB[0] = max(vec.x, m_AABB[0]);	m_AABB[1] = min(vec.x, m_AABB[1]);	m_AABB[2] = max(vec.y, m_AABB[2]);	m_AABB[3] = min(vec.y, m_AABB[3]);	vec = D3DXVECTOR3(-1.0f, -1.0f, 1.0f);	D3DXVec3TransformCoord(&vec, &vec, &tmpMat );	m_AABB[0] = max(vec.x, m_AABB[0]);	m_AABB[1] = min(vec.x, m_AABB[1]);	m_AABB[2] = max(vec.y, m_AABB[2]);	m_AABB[3] = min(vec.y, m_AABB[3]);	vec = D3DXVECTOR3(1.0f, 1.0f, -1.0f);	D3DXVec3TransformCoord(&vec, &vec, &tmpMat );	m_AABB[0] = max(vec.x, m_AABB[0]);	m_AABB[1] = min(vec.x, m_AABB[1]);	m_AABB[2] = max(vec.y, m_AABB[2]);	m_AABB[3] = min(vec.y, m_AABB[3]);	vec = D3DXVECTOR3(-1.0f, 1.0f, -1.0f);	D3DXVec3TransformCoord(&vec, &vec, &tmpMat );	m_AABB[0] = max(vec.x, m_AABB[0]);	m_AABB[1] = min(vec.x, m_AABB[1]);	m_AABB[2] = max(vec.y, m_AABB[2]);	m_AABB[3] = min(vec.y, m_AABB[3]);	vec = D3DXVECTOR3(1.0f, -1.0f, -1.0f);	D3DXVec3TransformCoord(&vec, &vec, &tmpMat );	m_AABB[0] = max(vec.x, m_AABB[0]);	m_AABB[1] = min(vec.x, m_AABB[1]);	m_AABB[2] = max(vec.y, m_AABB[2]);	m_AABB[3] = min(vec.y, m_AABB[3]);	vec = D3DXVECTOR3(-1.0f, -1.0f, -1.0f);	D3DXVec3TransformCoord(&vec, &vec, &tmpMat );	m_AABB[0] = max(vec.x, m_AABB[0]);	m_AABB[1] = min(vec.x, m_AABB[1]);	m_AABB[2] = max(vec.y, m_AABB[2]);	m_AABB[3] = min(vec.y, m_AABB[3]);

the checkAABB function:
bool cFrustum::CheckAABB(D3DXVECTOR3 Center, D3DXVECTOR3 halfSize){	if (m_AABB[0] > Center.x - halfSize.x)		if (m_AABB[1] < Center.x + halfSize.x)			if (m_AABB[2] > Center.y - halfSize.y)				if (m_AABB[3] < Center.y + halfSize.y)					return true;	return false;}

And heres how I call a check:
// Render this quadtree.int cQuadTree::Render(LPDIRECT3DDEVICE9 d3dDevice, cCamera *Camera, cFrameManager *FrameManager){	if (!m_HasChildren)	{		cTerrainCullData CullData = m_Terrain->GetCullData();		if (Camera->GetFrustum()->CheckSphere(CullData.Center.x, CullData.Center.y, CullData.Center.z, CullData.Radius))		{			return m_Terrain->Render(d3dDevice, Camera, FrameManager);		}		return 0;	} 	else	{		D3DXVECTOR3 vec = D3DXVECTOR3(m_TSets.Width / 10.0f, m_TSets.Height / 10.0f, 0.0f);		if (Camera->GetFrustum()->CheckAABB(m_CullData.Center, vec))		{			if (Camera->GetFrustum()->CheckSphere(m_CullData.Center.x, m_CullData.Center.y, m_CullData.Center.z, m_CullData.Radius))			{				int sum = 0;				for(short i=0; i<2; i++)					for(short j=0; j<2; j++)					{						sum += m_Children[i + j * 2].Render(d3dDevice, Camera, FrameManager);					};				return sum;			}			return 0;		}		return 0;	};}

BTW, I got my last post wrong, the red is the AABB, and the blue is the quadtree node.


text:
# Revision notes.

Revision History
Revision 0.11611 April 2005
• New material has been added to Chapter 2 and Chapter 4, including Figure 2.1. Current development work is mainly focusing on Section 4.4.4.

• The third basis set choice, "3) To use 6-31G** type the number 3 (and see Revision Note 0.112)," has been commented out of the example program freeMQM-1.cpp, since a bug seems to have been introduced into the code involving the 6-31G** basis set. The option can still be used; the program just will not tell you that the option is available.

• Olinking between documents is in the process of being implemented. This is anticipation of the need to link several tutorials together. Now, just the main tutorial, Topics in Molecular Quantum Mechanics, the revision notes, and the development notes are cross referenced using olinks. Thus, chapter numbers, figure numbers, equations numbers, etc., that appear as links, will kept current as the documents expand. On the other hand, references to chapter numbers, equations numbers, etc., that don't appear as links should be assumed to be outdated as the document expands [e.g., see the revision notes 0.100 - 0.115].

• There have been several minor corrections.

Revision 0.11525 March 2005
• Chapter 3 has been split into two chapters and a fair amount of material on the angular momentum properties of Slater determinants has been added to the new Chapter 4.

• The discussion of the spin angular momentum properties of closed shell Slater determinants in Appendix B has been extended to include orbital angular momentum. Several typos have also been corrected in Appendix B.

Revision 0.11401 March 2005
• The Chapter 1 from the last revision is in the process of being split into two chapters. After the discussion of hydrogen-like atoms has been completed, the development will focus on the basics of atomic and molecular spectroscopy.

• Due to the large number of equations, the automatic numbering system of DocBook has been implemented. See the development notes for details.

Revision 0.11319 February 2005
• A new appendix has been created in which example calculations and details on equations in the main text will be presented. Some material from this appendix may be moved to other chapters in the future.

• Chapter 1 has undergone a fair amount of revising, and the equations involving the calculation of the matrix elements of one-electron operators has been moved to Chapter 2.

• The search for typos has been postponed until the current development work on Chapter 1 and Appendix B has been completed.

Revision 0.11215 January 2005
• In the example program freeMQM-1.cpp, the names of the basis set files H-6-31G*.dat, H-6-31G**.dat, He-6-31G*.dat, and He-6-31G**.dat have been changed to H-6-31Gx.txt, H-6-31Gxx.txt, He-6-31Gx.txt, and He-6-31Gxx.txt to allow them to be used on Windows systems. In addition, a platform dependent instability has been observed in the calculation of the molecular hydrogen ground state energy when one chooses to use the 6-31G** basis set (i.e., H-6-31Gxx.txt). The p-type polarization functions (L = 1) in the 6-31G** basis set appear to prevent the program from finding a solution on some platforms. The program, freeMQM-1, ran flawlessly when compiled with gcc on a Pentium II machine running Redhat Linux 8.0 for both the Hartree-Fock calculation and the density functional theory calculation. The program fails to find a solution for some values of the interatomic bond distance, such as the default value of 1.40 a.u., when compiled with gcc on a Mobile AMD64 machine running Fedora Core 3 x86_64. It also fails to find a solution for all values of the interatomic bond distance on a Mobile AMD64 machine using the cygwin gcc compiler on Windows XP. The program converges to an incorrect solution on some systems when the 6-31G** basis set is used in a Hartree-Fock calculation. Although the ground state of the hydrogen molecule is a  $\Sigma$ state, one would hope that the program would to be able to use a complete set of basis functions in the search for a solution, and let the program find that the expansion coefficients are zero for some functions as part of the solution, i.e., C[I]=0. This appeared to be the case on the Pentium II system running Redhat 8.0, since, in spite of intensive testing over a very broad range of input values, reasonable solutions were always produced and no anomalies observed. This system was used to produce the numbers discussed in the tutorial. Those numbers, however, were generated during the time between September 2003 and December 2003, and it is possible that a bug has been introduced since then. Although the problem only appears when the polarization functions in 6-31G** are used, it is, of course, unsatisfactory that the same program can appear to be completely robust on one system, and manifestly pathological on another. Since the original system is not currently available, and because demonstrating the use of polarization functions in the calculation of the ground state energy of two-electron systems, a conceptually trivial extension of the problem, is not a priority of this tutorial, the tracking of the bug in the program will not be pursued at this time.

Revision 0.11115 January 2005
• Many typographical errors have been corrected in Chapter 2, along with other minor corrections and additions. During the next couple of weeks, more material will be added to the sections on angular momentum in many-electron atoms and molecules, in addition to correcting typographical errors in other chapters.

Revision 0.11013 January 2005
• Many typographical errors have been corrected in Chapter 1, along with other minor corrections and additions.

Revision 0.10930 December 2004

Revision 0.10819 August 2004
• The development has been temporarily diverted to the basics of molecular spectroscopy with the intention of returning focus to exact-exchange methods and TDDFT immediately afterward. The Jmol applet has been updated and probably will be updated regularly. In addition, a substantial number of corrections and additions have been made.

Revision 0.10714 March 2004
• Chapter 2.1, on the formalism of density-functional theory, has been expanded and split into two new Chapters 2.1 and 2.2; plus substantial corrections and additions. The development will now focus on exact-exchange methods.

Revision 0.10608 March 2004
• Additional material regarding the Hartree-Fock theory has been added to Sections 1.1 and 1.2. Introduction to Kohn-Sham theory is in development; plus minor corrections and additions

Revision 0.10524 February 2004
• Introduction to density-functional theory has begun development; plus minor corrections and additions.

Revision 0.10412 February 2004

Revision 0.10319 January 2004
• A few small changes to the example program freeMQM-1.cpp, including more useful output information.

• Sections on exact-exchange methods and TDDFT have begun development; plus minor changes and corrections.

Revision 0.10203 October 2003
• Perdew, Zunger correlation functional PZ81 and the Perdew, Burke, Ernzerhof exchange functional PBE have been added to the functional routine functionals.cpp to allow comparisons with various ab initio packages.

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Date of registration
: Apr 26th 2013

Platform: PS4

Location: Arizona, USA

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Sunday, April 16th 2017, 2:54pm

### Quoted from "CaptaPraelium"

So, the problem is well defined: Y axis paranoia, plus, tragedy of the commons. These are both psychological issues with players and I've no clue how to snap them out of it. But at least we've defined it.

We're definitely onto something here Capta, but now we must identify why BF1 players exhibit such a strong case of those issues. I think Noctyrne may have provided the answer we' ve been looking for, albeit unintentionally.

### Quoted from "NoctyrneSAGA"

I don't think a force multiplier would necessarily count as a "force multiplier" if it is not able to stand up to the pressure of more than one player.

-----------------------------

BF4's air and armor could be soloed in pubs; they weren't actually force multipliers unless in control by very experienced pilots and tankers, which wasn't necessarily a regular occurrence by any means I would imagine. A single Engineer with a dumb fire launcher could kill an out of position tank or helicopter with relative ease if the operators were bad enough (which in many cases they were), because dumb fires could one hit kill helis and seriously fuck up or outright destroy a tank in a couple well placed hits.

BF1's vehicles, in general, are true force multipliers now; no one infantryman can easily solo any vehicle in BF1 especially planes in pubs (except maybe the light tank and artillery truck). I think this has led to Y axis paranoia and tragedy of the commons. With tanks now requiring a full squad's amount of effort to neutralize and infantry being devoid of a gadget that can one hit kill pesky aircraft we have players who exhibit severe cases of these two symptoms because they are now just learning about how force multipliers work and what they are. Prior titles didn't really teach them because they didn't truly exist in many cases.

I have never felt so helpless against vehicles as I have with BF1. In BF3/4 I felt I had the necessary tools to combat a vehicle on my own power which led me to attack vehicles as often as I could because my tank/ heli competition wasn't all that good. I have to imagine that if I feel this way, then other players must be as well. The thing is countering vehicles in coordination in Battlefield hasn't ever been an absolute necessity, but now it is and I think many are still adapting to this change. I know for certain that I won't engage a tank unless I see fellow infantry with AT equipment nearby in hopes they'll join in.

But it's different with tanks; they're right there at eye level, stuck to the ground, and they're slow so accumulating the effort of randoms with AT isn't as difficult because once a tank is spotted it generally gets players' attention. Planes on the other hand are above eye level, located in the air (duh), and are just fast. Even if you can gather the attention of randoms away from the ground it's probably already too late because you've either been bombed or they flew out of range. So far, the only success stories I've seen regarding a combined effort to shoot down planes has involved communication. This is a problem. DICE said themselves that the game should work as intended without communication and as we can see it basically doesn't, otherwise we would see players regularly working in conjunction to counter force multipliers. Right now planes are perceived as strong and untouchable because they necessitate that more than one player shoots at it at the same time which is rare unless communication is involved.
To Aim Assist or not to Aim Assist, that is the question.

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### My "Contributions"

Can't get a title

Posts: 1,524

Date of registration
: Dec 23rd 2013

Platform: Xbox One

Location: The Land of Multitudinous Kangaroos

Reputation modifier: 12

Sunday, April 16th 2017, 5:20pm

The problem with vehicles in general is that they require a certain level of teamwork to destroy, a level which is just not seen on pubs.

I really don't envy DICE's balancing job - how to retain a vehicle's role as a force multiplier, but with lessening the amount of teamwork required to destroy one (or encouraging more players to demonstrate the level of teamwork required to destroy one).
something something Model 8 bestgun

### Quoted from "Pastafarianism"

Next, wanna try adding a guy that you KNOW is bad, and just testing to see that? Example: PP-2000 (god I so wanna love this gun, and yet...)

### Quoted from "Pastafarianism"

Example: PP-2000 (god I so wanna love this gun, and yet...)

Yes, it comes in last so far, but that is mostly because I'm making it shoot at 100m ADS - Not Moving as one of the criteria. Even then, between 50-100m Not Moving, when you include Useability, it is only 1.37% worse than the MTAR-21. Within 50m then it even beats the A-91.

Have a look, vs. the A-91 Carbine:

Using it with Muzzle Brake and Compensator is a wash in terms of overall performance. Comp is SLIGHTLY more accurate, while MB is SLIGHTLY more easy to use. Their overall scores are basically tied, with MB just ahead. I guess either can be recommended.

### Quoted from "Pastafarianism"

But... You can't be counting for the fact that it takes 9 bullets to kill at "long" range... Don't you dare tell me my A-91 is worse than a 9 BTK 650 RPM mediocre PDW.

Also. Just go heavy barrel. The recoil is low enough.

### Quoted from "Zer0Cod3x"

Well, technically...

Comparing a PP2K with HB and an A-91 with comp and stubby (as you suggested in an earlier post), at 50m not moving, the A-91 is only better by 4 damage per hitrate. While at 75m and 100m, surprisingly the PP2K does better than the A-91 (I'm pretty damn surprised as well).

And 10m and 50m moving the PP2K also does more damage per hitrate than the A-91. At 25m the A-91 is only better by about half a bullet's damage as well.

In addition, the PP2K has a much larger mag size and substantially less recoil. And it looks hella awesome. So comparing the A-91 to a PDW is of some worth after all, as the PP2K is better (technically, not practically) than the A-91.

Mind blown.

### Quoted from "Pastafarianism"

I... I...

*cries in a corner*

### Quoted from "Veritable"

Zer0Cod3x explained it very well. If you look at the raw numbers right here on Symthic Comparison, you can see how that happened:

A-91 vs PP-2000 | BF4 Weapon Comparison | Symthic

A-91's "23%" RPM advantage only afforded it 1 extra round.

Velocities are wash.

V-Recoil are wash (and this is HBar on PP2k vs. A-91 without).

Hipfire and ADS - Moving are better on the PP2k, but it's a PDW and not the surprising part.

The surprising part is that, as equipped (and we see above that PP2k HBar has almost same V-Recoil as A-91 without HBar so why not?), the PDW performs better at 50 - 100m than a bloody Carbine. Why?

SIPS, 42% better on the PP2k.

And here is the most important part. ADS - Not Moving Spread, 0.35 vs. 0.2, 43% improvement.

Without HBar then of course the PP2k loses, which is why when I add all the attachments together for an Overall Ranking, it would slot below the A-91. Run HBar on it, though, then... I'm sorry

### Quoted from "Pastafarianism"

@Veritable
@Zer0Cod3x
I... I...
But...
Wha...
I AM HAVING AN EXISTENTIAL CRISIS IN SCHOOL BECAUSE OF YOU TWO.

FUCK YOU NERDS AND YOUR FANCY NUMBERS

SEXY RUSSIAN BULLPUPS FTW.

In all seriousness, thank you both so much for giving me the numbers. I still don't want to accept them. You have led the horse to water. I still need to drink.

Up and down. Bounce all around

Posts: 3,099

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: Apr 15th 2013

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Sunday, April 16th 2017, 5:51pm

### Quoted from "Ritobasu"

What separates BF1 from the other games is how much more difficult it is for infantry to apply their own anti-vehicle tools effectively, and the larger gap between BF1's anti-vehicle gadgets vs solo driving BF1's vehicles.
BF3 says hi. Remember how stiglas were just a waste of the gadget slot that were unable to do anything against any competent pilot? Or how the javelin also was a waste of the gadget slot that couldn't get a hit on any player who knew how to hit "x" outside ~60 meters because IR-flare cooldown vs traveltime & reload? Or how RPG/SMAW could be dodged by armor outside throwing distance and even if you scored a hit you could have to throw 3-4 more downrange before you got through ERA?

### Quoted from "NoctyrneSAGA"

It really is quite frustrating when Helen Keller sets up her LMG in the only doorway in/out of an area.

### Quoted from "Watcher-45"

What kind of question is that? Since when is cheese ever a bad idea?

### Quoted from "LeGarcon"

Hardline is a fun and sometimes silly Cops and Robbers sorta thing and I think that's great. Or it would be if it didn't suck.

Posts: 1,730

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: Jan 12th 2014

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Reputation modifier: 13

Sunday, April 16th 2017, 7:38pm

### Quoted from "Ritobasu"

What separates BF1 from the other games is how much more difficult it is for infantry to apply their own anti-vehicle tools effectively, and the larger gap between BF1's anti-vehicle gadgets vs solo driving BF1's vehicles.
BF3 says hi. Remember how stiglas were just a waste of the gadget slot that were unable to do anything against any competent pilot? Or how the javelin also was a waste of the gadget slot that couldn't get a hit on any player who knew how to hit "x" outside ~60 meters because IR-flare cooldown vs traveltime & reload? Or how RPG/SMAW could be dodged by armor outside throwing distance and even if you scored a hit you could have to throw 3-4 more downrange before you got through ERA?
BF2 still had the most blatantly imbalanced combined arms meta because of the air balance. BF3 managed to come very close with the air balance and also had absurdly powerful armor because of ERA.

I don't think they ever got ground fired stingers or iglas to be really viable air killers, they just worked to deny some airspace between flared runs.

Javelin was added as a choice in BF3/BF4 as a team weapon, not a solo weapon. You are supposed to fire on designated targets painted by SOFLAM third seaters, armor gunners, or recons, or attack helos. It is exceptionally powerful in that capacity.

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: Mar 25th 2014

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Sunday, April 16th 2017, 7:58pm

### Quoted from "tankmayvin"

I don't think they ever got ground fired stingers or iglas to be really viable air killers, they just worked to deny some airspace between flared runs.

Javelin was added as a choice in BF3/BF4 as a team weapon, not a solo weapon. You are supposed to fire on designated targets painted by SOFLAM third seaters, armor gunners, or recons, or attack helos. It is exceptionally powerful in that capacity.
Ya.. no.

The AA missiles never did deny airspace in BF3. Given if a pilot ever entered the stigla range. At least not if the pilot was running ECM and was slightly good. The countermeasure reload time of the airvehicles was so fast any follow up stinger was senseless. The pilot would run off and wait until CM's reloaded again and press X. Then to turn around and kill every spotted icon out of the stigla range. It just would not hit good pilots. The same can be seen in BF4. Good pilots run ECM, stigla locks on, ECM pops up, airvehicle finishes its run and is out of range/behind cover/in high alttidue before 2nd stigla missile could reacquire target. They had no impact on good pilots.

Same with javelins against tanks. Soflam+Jav combo against tanks didnt work. The IR Smoke reloaded too fast, and the javelin magically exploded.

This included Guided bombs by airvehicles:

There were real chimpansees at work at DICE working on BF3 balance.

I'm afraid they hired the same chimps to work on BF1's vehicle balance.
RIP Sraw

Salt Miner

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Sunday, April 16th 2017, 8:59pm

A significant problem with everything in BF1 is a lack of tools that deliver high burst-damage. In BF4 you could OHK all aircraft with dumbfire rockets (even if hard or luck-based), you could get behind an MBT and ram two RPGs up its ass before it could react, and so many other similar things.

In BF1 we seem to have a huge shift to slowing down all TTKs, from things like the "but they're really accurate" Lewis and Huot with their atrocious damage and velocity, to the Rocket Gun's pathetic 15-ish damage no matter what angle you hit an A7V, to getting at best 15-ish damage on a Fighter with the best AA tools infantry can use (guns), to the Tank Hunter Attack Plane with its average at-best AT damage output, and so on.

Even Ammo 2.0 is based around this philosophy, having constant ammo slowly instead of a lot at once. People don't want a slow, constant supply of Smoke Grenades, they want two Smoke Grenades right now. High burst usage. Why do you think the Automatico is such a popular weapon? It kills very fast, faster than most people can do anything about in most situations, is perfectly usable at nearly any relevant range, and is also very easy to use. The Rem 8 .35 is also another fantastic high burst damage weapon, one that is definitely effective at all relevant ranges, but is just a bit harder to use.

There are very few things that come close to allowing this kind of high burst damage output in BF1, one of them is the Tank Hunter Landship, and another is getting a Tank within AT Grenade range, but even that is pretty lacking. I was trying to take A on Grappa and there was a pesky Arty Truck hanging around. After a few deaths that saw him just repair all his damage before I could get back on him again, I managed to spawn outside his LoS and sneak up about 15m away, beside him. AT Grenade, Light AT Grenade, AT Grenade, drop prone, Rocket Gun shot, he died.

It felt fast, but really only by BF1 standards. I expended all three of my grenades and a Rocket Gun shot (which I have to be prone and stationary for) just to take down a silly little Arty Truck, when in BF4 I could have dropped an MBT in a mere two RPG shots. And the first RPG inflicts a Disable, meaning he's neither leaving nor turning to minimize damage, while I'm pretty sure none of my hits on the Arty Truck were Disables.

And that's best case scenario against the weakest Tank in BF1. Everything in BF1 is slow, and individual actions/shots rarely have significant impact/damage on their own. That's why people so often feel like they're not accomplishing anything.
Who has fun, wins.

This post has been edited 1 times, last edit by "BleedingUranium" (Apr 16th 2017, 9:05pm)

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Sunday, April 16th 2017, 9:06pm

### Quoted from "Zer0Cod3x"

The problem with vehicles in general is that they require a certain level of teamwork to destroy, a level which is just not seen on pubs.

I really don't envy DICE's balancing job - how to retain a vehicle's role as a force multiplier, but with lessening the amount of teamwork required to destroy one (or encouraging more players to demonstrate the level of teamwork required to destroy one).
People just need to grow the fuck up and stop whining about vehicles being fundamentally asymmetric. You get infinite lives in BF, dying doesn't matter unless you lose position. You're not sitting out until round end, etc.

The biggest blow to BF balance IMO was the implementation of detailed stats tracking where the CS KDR status-whoring could be transplanted to a game like BF where it doesn't and shouldn't matter.

BF is a teamwork centric game, but where the baseline selling point encourages you to play it like you're the lone wolf hero of a single player game that has a sort of dull NPC squad attached. Supporting roles are neither glamourized nor rewarded points wise.

The balancing job is one thing, but BF needs to sit down and decide what sort of game it wants to be again. It's designed around coordinated squads and complimentary plays between squads. Instead we generally get packs of lone wolves zerg rushing single flags. etc

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Sunday, April 16th 2017, 9:42pm

Mother of god tank, NO ONE LIKES TO RESPAWN. Stats don't have a goddamn thing to do with it. I don't care if I have practically limitless lives, I don't want to die at all; it is not fun to spawn in over and over again, but it's a necessary evil with how BF1's vehicles work. Infantry against vehicles is like the equivalent of facing an end of level boss in an RPG; you gotta throw everything you have at them in a very specific manner otherwise you'll just die and have to start all over again.

And it really isn't all of DICE's fault that Battlefield doesn't work how they want it to in pubs. The community shares the blame as well because a lot of them don't know the best way to play. They probably have no idea what coordination looks like because pub Battlefield is very casual in nature with a small portion of players who actually understand the gist of the game mechanics and how they actually function.
To Aim Assist or not to Aim Assist, that is the question.

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text:
# euclidean geometry! help

• June 8th 2008, 11:26 AM
skystar
euclidean geometry! help {solved}
Suppose that in a triangle ABC the angle ABC=angle ACB show that side AB=AC. In triangle ABC let X lie on BC in such a way that angle AXB is a right angle. Show X is the midpoint of BC.

ok its clearly isosceles so i thought mentioning Angle-side-angle then to show right angle i got confused where to go! (Headbang)
• June 8th 2008, 11:50 AM
skystar
• June 8th 2008, 11:57 AM
masters
Quote:

Originally Posted by skystar
Suppose that in a triangle ABC the angle ABC=angle ACB show that side AB=AC. In triangle ABC let X lie on BC in such a way that angle AXB is a right angle. Show X is the midpoint of BC.

ok its clearly isosceles so i thought mentioning Angle-side-angle then to show right angle i got confused where to go! (Headbang)

Since triangle ABC is isosceles, and $\overline{AB}\cong{\overline{AC}} \; \; and \; \; \angle {B} \cong{\angle{C}} \; \; then \; \; \triangle{ABX}\cong{\triangle{ACX}}$ by AAS (angle-angle-side) or HL (hypotenuse-leg) theorems.

$\overline{CX} \cong{\overline{BX}}$ by CPCTC (Corresponding Parts of Congruent Triangles are Congruent.

X is the midpoint by definition of midpoint.
• June 8th 2008, 12:08 PM
masters
If two angles of a triangle are congruent, then the sides opposite those angles are congruent.

Since we knew that $\angle{B}\cong{\angle{C}} \; \; then \; \; we \; \;conclude \; \;that \; \; \overline{AB}\cong{\overline{AC}}$


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# Question aaac7

Jul 21, 2015

Velocity: -4.0 m/s

#### Explanation:

The first thing you need to do to solve this problem is figure out whether or not the stone will be moving upwards or downwards after 0.97 seconds.

Since upward is set as positive and downward as negative, the sign of the gravitational acceleration, $g$, will have to match this convention as well.

At the top of its motion, i.e. at maximum height, the velocity of the stone will be zero. This means that you can determine the time it takes the stone to reach that height by

underbrace(v_"top")_(color(blue)("=0")) = v_i - g * t

t = v_i/g = (5.5cancel("m")/cancel("s"))/(9.8cancel("m")/"s"^cancel(2)) = "0.561 s"

After 0.561 s the stone begins to fall towards the ground. This means that the sign of the velocity after 0.97 s will be negative.

To determine its velocity, calculate the time it spends in free fall

${t}_{\text{fall" = 0.97 - 0.561 = "0.409 s}}$

The stone is in free fall for 0.409 seconds, which means that its velocity at that point in time will be

v = underbrace(v_"top")_(color(blue)("=0")) - g * t_"fall"

v = -9.8"m"/"s"^cancel(2) * 0.409cancel("s") = color(green)(-"4.0 m/s")#


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## Elementary Algebra

$3\sqrt[3] {3}$
Recall, $\sqrt[3] a \times \sqrt[3] b = \sqrt[3] {a \times b}$. Thus, we can multiply 9 and 9 to obtain the simplified expression: $\sqrt[3] {9 \times 9} = \sqrt[3] {81}$ In order to simplify a cube root, we consider the factors of the number inside of the cube root. If any of these factors are perfect cubes, meaning that their cube root is an integer, then we can simplify the expression. We know that 27 and 3 are factors of 81. We know that 27 is a perfect cube, so we simplify: $\sqrt[3] {27} \sqrt[3] {3}=3\sqrt[3] {3}$


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# Tag Info

9

You can linearize the objective as follows. Let binary decision variable $x_i$ indicate whether item $i$ is chosen, and let binary decision variable $y_c$ indicate whether the count of chosen items is $c$. Let $w_i$ be the weight of item $i$, and let $b_c$ be the bonus/penalty for choosing $c$ items. The problem is to maximize $$\sum_i w_i x_i + \sum_c b_c ... 7 If I understand correctly, binary variable g_j indicates whether group j is special, binary variable a_{i,j} indicates whether exactly i items are taken from group j, and binary variable b_i indicates whether exactly i items are taken from the special group. You want to enforce the logical implication$$b_i \implies \bigvee_j (g_j \land a_{i,j}...

7

So it costs \$13 for the two of them on a weekday and \$24 for the two of them on a weekend. If you want to maximize the number of movies, skip the expensive ones and the \$100 budget yields$\lfloor 100/13 \rfloor = 7$movies. If you prefer writing it explicitly as an optimization (yes, knapsack) problem, let integer decision variables$x$and$y$be the ... 7 Even though knapsack problems are relatively easy to solve in practice, there does not exist a polynomial-time algorithm to solve even the standard knapsack problem, unless$\mathcal{P}=\mathcal{NP}$. The knapsack problem can be solved in pseudo-polynomial time$\mathcal{O}(nC)$. Even when the input length (=number of digits describing the problem) is small, ... 6 BPPLIB – A Bin Packing Problem Library: http://or.dei.unibo.it/library/bpplib 5 It seems what you are looking for, is the maximum dispersion problem. The following blog post discusses a MIQP formulation along with a number of different MILP formulations http://yetanothermathprogrammingconsultant.blogspot.com/2019/06/maximum-dispersion.html?m=1 4 Let$x_{i}$be a binary variable that takes value$1$if item$i \in I_k$is selected. You want to choose$n$items from each set$I_k$, so impose $$\sum_{i\in I_k} x_{i} = n \quad \forall k$$ You can optimize whatever you want with these variables. If you want to maximize the$\ell_1$distance between each pair of items, you are going to need a pairwise ... 4 You have an instance of the 0-1 knapsack problem where you want to determine which teams to select to maximize the number of wins, subject to a budget. The linked page provides a DP recurrence, which is obtained by conditioning on whether or not you select the next team. The optimal objective value for your instance is 28: \begin{matrix} \text{team} & ... 3 I think the problem is NP-hard since: it will reduce to the 0-1 Knapsack problem with an equality constraint; and, changing$\leq$to$=\$ in the 0-1 Knapsack constraint does not change its complexity (see explanation below). So, your problem is NP-hard as 0-1 Knapsack is. P.S. To see why (2) is correct, suppose all weights and values are equal. Then the ...

2

For the 2D and 3D variant you can find multiple instances as well as instance generators here: https://github.com/Oscar-Oliveira/OR-Datasets/tree/master/Cutting-and-Packing https://www.euro-online.org/websites/esicup/data-sets/

Only top voted, non community-wiki answers of a minimum length are eligible


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# Stock prices using a monte carlo simulation with a normal inverse gauss distribution

I am supposed to model daily stock prices with a normal inverse gauss distribution in excel. I feel like I am misssing some basics because I cant transform the information from the academic papers into an excel formula. Does anyone have any experience with this distribution? How do I go from the PDF to the St = So * exp (X) formula?

-

Note: In your text you refer to the stock prices and using a normal inverse gaussian. This would correspond to a normal model. However the formula you write suggests a log-normal (Black-Scholes) like model (not sure what X is), i.e. using a normal inverse gaussian for the stock returns.

For Excel: The spreadsheet at http://www.christian-fries.de/finmath/spreadsheets/ does a Monte-Carlo Simulation of a Black-Scholes model and the corresponding risk neutral valuation of a derivative. To convert the log-normal process to a normal one: convert the Euler scheme

=D18+$C$6*D18*$C$8+$C$7*D18*NORMINV(RANDOM();0;1)*SQRT($C$8)

you have

=D18+$C$6*$C$8+$C$7*NORMINV(RANDOM();0;1)*SQRT($C$8)

where this is cell E18 and

• D18 denotes the previous cell / realization of the stock
• \$C\$6 is the drift r
• \$C\$8 is the time step dt
• \$C\$7 is the volatility $\sigma$

and each row then corresponds to a Monte-Carlo path. (Remark: In the normal case case, the Euler scheme is the exact solution. In the log normal case it is much better to use the Euler scheme for log(S), i.e. $S(t+\Delta t) = S(t) * \exp(r \Delta t - 0.5 \sigma \Delta t + \Delta W(t))$ - I assume you can guess the corresponding Excel formula.

-


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Department of

# Mathematics

Seminar Calendar
for events the day of Wednesday, September 4, 2013.

.
events for the
events containing

More information on this calendar program is available.
Questions regarding events or the calendar should be directed to Tori Corkery.
     August 2013           September 2013          October 2013
Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa
1  2  3    1  2  3  4  5  6  7          1  2  3  4  5
4  5  6  7  8  9 10    8  9 10 11 12 13 14    6  7  8  9 10 11 12
11 12 13 14 15 16 17   15 16 17 18 19 20 21   13 14 15 16 17 18 19
18 19 20 21 22 23 24   22 23 24 25 26 27 28   20 21 22 23 24 25 26
25 26 27 28 29 30 31   29 30                  27 28 29 30 31

Wednesday, September 4, 2013

3:00 pm in 143 Altgeld Hall,Wednesday, September 4, 2013

#### Integrability of generalized pentagram maps and cluster algebra

###### Michael Shapiro (Michigan State University)

Abstract: The pentagram map that associates to a projective polygon a new one formed by intersections of short diagonals was introduced by R. Schwartz and was shown to be integrable by V. Ovsienko, R. Schwartz and S. Tabachnikov. M. Glick demonstrated that the pentagram map can be put into the framework of the theory of cluster algebras.

We extend and generalize Glick's work by including the pentagram map into a family of discrete completely integrable systems. Our main tool is Poisson geometry of weighted directed networks on surfaces.. The ingredients necessary for complete integrability -- invariant Poisson brackets, integrals of motion in involution, Lax representation -- are recovered from combinatorics of the networks. Our integrable systems depend on one discrete parameter $k>1$. The case $k=3$ corresponds to the pentagram map. For $k>3$, we give our integrable systems a geometric interpretation as pentagram-like maps involving deeper diagonals. If $k=2$ and the ground field is $\mathbb C$, we give a geometric interpretation in terms of circle patterns.

(joint with M.Gektman, S.Tabachnikov, and A.Vainshtein)

4:00 pm in 245 Altgeld Hall,Wednesday, September 4, 2013

#### Representations via geometry

###### Thomas Nevins (Department of Mathematics, University of Illinois at Urbana-Champaign)

Abstract: Representations of groups (and Lie algebras) naturally emerge from studying physical systems with symmetry. I will explain one path by which an algebraic geometer approaches the study of representations via the example of the group of 2x2 matrices of determinant one. I hope to briefly indicate connections not only to algebraic geometry and mathematical physics but also to symplectic geo


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5.1: What is a matrix?

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

To start with, a matrix is nothing more than a rectangular arrangement of a set of elements (actually, it's a bit more complicated than that, but we will return to matrices of more than two dimensions in a little bit). Rectangles have sizes that are described by the number of rows of elements and columns of elements that they contain. A "3 by 6" matrix has three rows and six columns; an "I by j" matrix has I rows and j columns.  A matrix that has only one row is called a "row vector."  A matrix that has only one column is called a "column vector.

Figure 5.1 shows a two-by-four matrix.  Figure 5.2 shows a four by two matrix.  Just for the moment, ignore the contents of the cells (e.g. 1,1).

Figure 5.1.  Example of a "two-by-four" matrix

 1,1 1,2 1,3 1,4 2,1 2,2 2,3 2,4

Figure 5.2.  Example of at "four-by-two" matrix

 1,1 1,2 2,1 2,2 3,1 3,2 4,1 4,2

The elements (cells) of a matrix are identified by their "addresses." Element 1,1 is the entry in the first row and first column; element 13,2 is in the 13th row and is the second element of that row. The cell addresses have been entered as matrix elements in the two examples above.

Matrices are often represented as arrays of elements surrounded by vertical lines at their left and right, or square brackets at the left and right. In web pages it's easier to use "tables" to represent matrices. Matrices can be given names; these names are usually presented as capital bold-faced letters. Social scientists using matrices to represent social networks often dispense with the mathematical conventions, and simply show their data as an array of labeled rows and columns. The labels are not really part of the matrix, but are simply for clarity of presentation. The matrix in figure 5.3 for example, is a 4 by 4 matrix, with additional labels.

Figure 5.3.  Four-by-four matrix with additional row and column labels

 A B C D A --- 1 0 0 B 1 --- 1 0 C 1 1 --- 1 D 0 0 1 ---

The matrices used in social network analysis are frequently "square."  That is, they contain the same number of rows and columns.  But "rectangular" matrices are also used, as are row and column vectors.  The same conventions apply to all these variations.

Occasionally, social network analysts will use a "3-dimensional" matrix.  A three dimensional matrix has rows, columns, and "levels" or "slices."  Each "slice" has the same rows and columns as each other slice.  UCINET thinks about these more complicated 3-dimensional arrays of data as a collection of two-dimensional matrices.


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Homework Help: Integration of a fraction

1. Nov 7, 2011

syj

1. The problem statement, all variables and given/known data

i need to integrate:
$\frac{1}{r(u-r)}dr$

2. Relevant equations

u is a constant

3. The attempt at a solution

im not sure if i should decompose the fraction. i tried that, but it didnt seem to be of any help.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 7, 2011

Dick

Yes, use partial fractions. Show us how it didn't help. It should have.

3. Nov 7, 2011

syj

ok
so i get
$\frac{1}{r(u-r)}=\frac{A}{r}+\frac{B}{u-r}$

so that
$B-A=0$
and
$Au=1$

so i got
$A=\frac{1}{u}=B$

this gives me
$\frac{1}{ur}+\frac{1}{u(u-r)}$

Last edited by a moderator: Nov 7, 2011
4. Nov 7, 2011

HallsofIvy

Good! Now, what do you get when you intgrate those?

5. Nov 8, 2011

syj

$\int(\frac{1}{ur}+\frac{1}{u(u-r)})$

$=\frac{1}{u}\int\frac{1}{r}+\int\frac{1}{u-r}$

$=\frac{1}{u}(ln(r)-ln(u-r))$

$=\frac{1}{u}(ln\frac{r}{u-r})$

my problem is that i dont know where to put $r_0$ here. The solution has $r_0$

6. Nov 8, 2011

HallsofIvy

Then perhaps you should have told us from the start what the problem really is. The problem you posted does not have any r0.

7. Nov 8, 2011

syj

oh, sorry what i mean is that when it integrates they have $r_0$, sorry, i meant the solution has $r_0$


text:
MathSciNet bibliographic data MR406947 (53 #10732) 10F40 Levin, M. B. The uniform distribution of the sequence $\{\alpha \lambda \sp{x}\}$$\{\alpha \lambda \sp{x}\}$. (Russian) Mat. Sb. (N.S.) 98(140) (1975), no. 2 (10), 207–222, 333. Links to the journal or article are not yet available

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


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# Wikipedia:Reference desk/Archives/Mathematics/2009 May 10

Mathematics desk
< May 9 << Apr | May | Jun >> May 11 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

# May 10

## Suggestion for a good book on axiomatic set theory

What is a good book on axiomatic set theory? I am not familiar iwth first-order predicate calculus so I want the book to have some discussion on that. I also want the book to be at graduate level. I know sets but don't know the axioms for them. I also don't know the continuum hypothesis well. Thanks for help. —Preceding unsigned comment added by 58.161.138.117 (talk) 02:03, 10 May 2009 (UTC)

Paul Halmos' "Naive Set Theory" gives quite a good treatment of the subject. There is very little first-order predicate calculus in it (in fact, I don't think there's any) - I found that to make it a good, intuitive introduction to the subject, as someone who also was not familiar with FOPC. (Don't be misled by the name - the book IS about axiomatic set theory; the 'naive' in the title is just referring to the intuitive treatment of the axioms.) Icthyos (talk) 07:54, 10 May 2009 (UTC)
I think the standard graduate texts on set theory are Kunen's and Jech's. I'm not at that level so I haven't done more than flip through them for a minute. "Naive set theory" is more of an undergraduate book. At the undergraduate level, Enderton's is a bit more modern. The class I took in college used it. If you're not familiar with first-order predicate calculus, you're likely to find any serious graduate set theory text to be absolutely impenetrable. I would start with Enderton's "Introduction to Mathematical Logic" or something comparable. Finally, Wikipedia has some very enthusiastic and knowledgeable editors writing about set theory and logic, so I've learned a lot from the articles here. 67.122.209.126 (talk) 11:49, 10 May 2009 (UTC)
I attempted to read Jech's book a few months ago, but found it far too terse when it came to explaining the FOPC - he just explained what the symbols he was using signified, then launched straight into long strings of them. I've yet to come across a book that sets it out at a nice, even pace - I might give the other ones mentioned above a look. Icthyos (talk) 16:15, 10 May 2009 (UTC)
The standard introductory graduate-level books for general set theory are indeed Jech's book and Kunen's book. Jech's is more thorough and much longer; Kunen's book spends more time being pedagogical but is not as good as a reference. In general I would recommend Kunen's book to start with for a graduate-level text. If you make it through that, you will have learned a significant chunk of basic set theory. You are unlikely to find a set theory book that spends a long time on first order logic; you will want a more general-purpose mathematical logic text for that. — Carl (CBM · talk) 04:35, 11 May 2009 (UTC)

## Numerical Methods for Solving PDEs

Consider the PDE ${\displaystyle u_{t}+au_{x}=f}$ and consider the scheme

${\displaystyle {\frac {11v_{m}^{n+1}-18v_{m}^{n}+9v_{m}^{n-1}-2v_{m}^{n-2}}{6k}}+a\left(1+{\frac {h^{2}}{6}}\delta ^{2}\right)\delta _{0}v_{m}^{n+1}=f_{m}^{n+1}.}$

We are supposed to show that this scheme is accurate of order (3,4) and it is UNSTABLE for all values of ${\displaystyle \lambda =k/h}$ where h is the space step size and k is the time step size. I have several questions about this scheme.

1.Is this scheme correct the way it is? I think that it is supposed to be

${\displaystyle {\frac {11v_{m}^{n+1}-18v_{m}^{n}+9v_{m}^{n-1}-2v_{m}^{n-2}}{6k}}+a\left(1+{\frac {h^{2}}{6}}\delta ^{2}\right)^{-1}\delta _{0}v_{m}^{n+1}=f_{m}^{n+1}.}$

The inverse is missing, right? Can anyone verify that?

2.Assuming the scheme is correct the way it is, for the stability, I perform von Neumann analysis and my amplification factor comes out to

${\displaystyle \phi _{3}(z)=(11+\alpha i)z^{3}-18z^{2}+9z-2,\,\,\alpha =6a\lambda \sin(\theta )\left(1-{\frac {2}{3}}\sin ^{2}\left({\frac {\theta }{2}}\right)\right)}$

and I want to see under what conditions is this polynomial a simple von Neumann polynomial. So I got all the way to ${\displaystyle \phi _{1}(z)}$ and I get that ${\displaystyle |\alpha |\geq 3{\sqrt {15}}}$ which then tells me that ${\displaystyle |a\lambda |\geq {\sqrt {30{\sqrt {3}}-45}}}$. My questions is what does this mean? Is this a condition on stability? Or does this somehow imply that the scheme is unconditionally UNSTABLE? How is this implied? This is kind of fishy because everything I have worked with so far, the stability condition has always been less than or equal to something. But this time I have greater then something so I am not sure. By the way, the same thing happens if I assume the missing inverse on the scheme so this scheme appears to be unconditionally unstable in either case.

3.Furthermore, when I try to determine the order of accuracy, I need to take a smooth function, call in ${\displaystyle \phi }$ and then do its taylor expansion. The problem is that some terms (when I expand the delta expressions) I need to do taylor expansion in two variables simultaneously and I think that I need to do it up to ${\displaystyle O(h^{5})}$ and ${\displaystyle O(k^{5})}$ but it gets very cumbersome and complicated. First I didn't even know how to write the expansion but then the page here talks about multi-index notation but then I can't simplify using addition, subtraction, and division. Am I doing something wrong? Is it suppose to be much easier than this? This is actually preparation for the final in this class and I have spent the entire day today on this with no progress. Any help would be appreciated. Thanks! -Looking for Wisdom and Insight! (talk) 06:29, 10 May 2009 (UTC)

## Means

I am collecting ice data and I got my hands on sea-ice area data for each 10 degrees of longitude for Antarctica by month. I would like to remove the longitude dimension in order to have only one average value per month. What would be the natural method to do this? Arithmetic mean, geometric, harmonic, any other? Thank you. 213.204.122.116 (talk) 08:05, 10 May 2009 (UTC)

It depends on the distribution of the (random) variable, x. If x has a normal distribution, the arithmetic mean is natural. If x has a natural zero point and 0<x, transform into y=log(x) before taking the arithmetic mean and transforming back. (This is using the geometric mean). If x has both a natural zero and a natural unit and 0<x<1, then use the transformation y=log(x/(1-x)). Bo Jacoby (talk) 10:09, 10 May 2009 (UTC).
Thank you for your answer, but what do you mean by "natural"? I meant it as the "best" or "mostly used", but you seem to use it as a technical statistical term which I'm unfamiliar with. 213.204.119.133 (talk) 13:37, 10 May 2009 (UTC)
"Natural" means "occurring in nature". In this context, it means the value automatically comes up when you consider the source/meaning of the data. For example, if you were looking at the thickness of the ice, there would be natural zero point (no ice at all). The idea of negative thickness is meaningless, so you don't want the fact that there are no negative numbers in your data to affect your results. If you were looking at the ratio of two different isotopes of oxygen dissolved in the ice, there would be natural zero (all one isotope) and a natural one (all the other isotope). The idea of a value less than 0 or more than 1 would be meaningless in this case, so you don't want the lack of such values to affect your results. --Tango (talk) 13:59, 10 May 2009 (UTC)
Seems to me the obvious thing to do, if you want the total amount of ice, is to add up the areas in the wedges. Or are you given the proportion of ice to open water in each wedge? In that case you need to weight them by the amount of water (since Antarctica is not a circle). —Tamfang (talk) 05:28, 11 May 2009 (UTC)

Maybe you could answer my question on the science desk about the total volume of ice in the world please. 84.13.171.69 (talk) 15:00, 10 May 2009 (UTC)

## Top mathematical journals

Is there any impartial neutral agency that ranks the mathematical journals? Is there any pointer for this list? Thank you in advance. twma 09:23, 10 May 2009 (UTC)

as for your second question, no there is no point to having such a list. 94.27.208.52 (talk) 09:56, 10 May 2009 (UTC)
The ISI Science Citation Index (accessible through the Web of Science) is most commonly used in practice, even though the impartiality and neutrality of this organization as well as the rationale behind the algorithm it uses to compute its citation index are rather doubtful. — Emil J. 14:35, 10 May 2009 (UTC)

There is no "master list" that ranks all journals in order. However, because some sort of ranking is needed when universities evaluate faculty performance, there are many systems that are used by individual schools. Some use a measure of the number of citations to the journal that appear in print; the usefulness of this measure for mathematics is not as clear as its usefulness for lab science, but at least it's quantitative and objective. Other research universities simply present a list of journals broken down into groups, and expect faculty to publish in the higher groups and not the lower ones.

Typically, mathematicians who work in a particular subfield will have a sense of which journals are more prestigious and which are less so. If you have a particular field in mind, someone here might be able to give you a rough sense of the order. — Carl (CBM · talk) 14:58, 10 May 2009 (UTC)

Thanks for the general mathematical education. Based on Science Citation Index, mathematical Charlie Brown may be considered as good mathematics. Next question: How to determine the impact factor? Also based on Science Citation Index? Thank you in advance. twma 23:12, 10 May 2009 (UTC)
See Impact factor. — Carl (CBM · talk) 04:29, 11 May 2009 (UTC)

Thank you. In a hypothetical situation, a group of idiots making a lot of noise about a topic get their professorship while a clever person puts it into final definitive form but could never get a job because there is no more need to say anything about the topic. In a hypothetical situation, it could happen in scientific world. Thank you for the entertaining information about the reality. twma 16:39, 12 May 2009 (UTC)

## Simple vector equation: most general solution

Hi there guys, was just wondering what the most general solution of x+y=2a, x.y=c is in vector form? It's not half as tough as my usual questions but I'm feeling quite ill & my brain has switched off today! Thanks a lot, Spamalert101 (talk) 15:47, 10 May 2009 (UTC)

Well, if you wanted to solve only for x, or only for y, then you could express the equation as
x·(2ax) = c.
Then:
x·x −2a·x = −c.
Then complete the square:
x·x −2a·x + a·a = a·ac.
|xa|2 = |a|2c,
and you've got a sphere centered at a.
To solve for both variables, looking at (xy) space rather than just at x space, might involve more than just that terse conclusion; I'm not sure. Michael Hardy (talk) 17:03, 10 May 2009 (UTC)
Excuse me, what do you mean? Haven't you solved completely the problem in both variables? You said x is a point in the sphere, and the corresponding y is 2a - x ; what should one do better than that? --pma (talk) 18:59, 10 May 2009 (UTC)
I agree, he's solved the problem completely. There are two variables and two equations (which are neither contradictory or equivalent), so there is a unique solution. --Tango (talk) 19:53, 10 May 2009 (UTC)
¿unique solution? indeed he proves that the set (x,y) of solutions is affinely isomorphic to an Euclidean sphere of the x space, and it is also implicit in his solution that there is a unique solution iff |a|2 = c , and no solutions if |a|2 < c. --pma (talk) 20:36, 10 May 2009 (UTC)
Oops! The second equation is a scalar equation, not a vector one, so what I said was nonsense (even without that, it was still not quite right - the solution set would be discrete, it wouldn't necessarily be a single point). In terms of scalars, you have 2n variables and n+1 equations, so the solution set will be (n-1)-dimensional (which the sphere obviously is). I apologise for my idiocy! --Tango (talk) 20:49, 10 May 2009 (UTC)

The set of values of x satisfying this equation is a sphere. The set of y values is likewise a sphere. The equation gives a one-to-one correspondence between the two spheres. So how does one geometrically describe the set of pairs (xy)? Topologically it's just another sphere of the same dimension as either of the two aforementioned spheres. Michael Hardy (talk) 20:59, 10 May 2009 (UTC)

Wonderful, I'm appalled at quite how slow I was being but things are back on track now, thanks! Spamalert101 (talk) 22:01, 10 May 2009 (UTC)
We may just rephrase your solution formula saying that a solution (xy) is exactly an ordered pair of antipodal points on the sphere of center a and radius (|a|2 - c)1/2 --always assuming |a|2 > c --pma (talk) 06:47, 11 May 2009 (UTC)

## Areas of circles

Can someone help me with this problem? [1]I need to find the shaded region's area. Yes, it is a homework problem if you're wondering, I'm completely stumped by it. All the the middle circle is tangent to the rest of them, and the diameter of the large circle is split into 4 congruent parts by the centers of the circles. 24.6.46.177 (talk) 20:47, 10 May 2009 (UTC)

You are probably already aware, that the grey area can be expressed by sums and differences of the individual circle areas in the figure. (If you want to know more about that, see Constructive solid geometry). The core of the problem is to find the radius for every circle. One must be given in any case, two are easy. For the radius of the small center circle, you need to establish a relation (=equation) to other sizes in the figure which you already know and solve it.
There are probably many ways; my first idea was this: connect the center of one medium sized circle to the center of the small circle and also draw a vertical line through the center of the small circle across the whole figure. If you cannot see it immediately, try to label all resulting line segments and start collecting every bit of information you have about their relative lengths. You might need to use some ancient Greek mathematics in doing so. —Preceding unsigned comment added by 84.187.75.89 (talk) 01:24, 11 May 2009 (UTC)
I'd solve it as above since it is homework, but when you're finished you might be interested that it is an instance of Soddy circles, he produced an execrable poem with the solution which is in Problem of Apollonius Dmcq (talk) 11:46, 11 May 2009 (UTC)


text:
# Equations of Tangent and Normal to the Parabola

The equations of tangent and normal to the parabola ${y^2} = 4ax$ at the point $\left( {{x_1},{y_1}} \right)$ are ${y_1}y = 2a\left( {x + {x_1}} \right)$ and ${y_1}x + 2ay – 2a{y_1} – {x_1}{y_1} = 0$ respectively.

Consider that the standard equation of a parabola with vertex at origin $\left( {0,0} \right)$ can be written as
${y^2} = 4ax\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Since the point $\left( {{x_1},{y_1}} \right)$ lies on the given parabola, it must satisfy equation (i). So we have
${y_1}^2 = 4a{x_1}\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

Now differentiating equation (i) on both sides with respect to $x$, we have
$2y\frac{{dy}}{{dx}} = 4a \Rightarrow \frac{{dy}}{{dx}} = \frac{{2a}}{y}$

If $m$ represents the slope of the tangent at the given point $\left( {{x_1},{y_1}} \right)$, then
$m = {\frac{{dy}}{{dx}}_{\left( {{x_1},{y_1}} \right)}} = \frac{{2a}}{{{y_1}}}$

The equation of the tangent at the given point $\left( {{x_1},{y_1}} \right)$ is
$\begin{gathered} y – {y_1} = \frac{{2a}}{{{y_1}}}\left( {x – {x_1}} \right) \\ \Rightarrow {y_1}y – {y_1}^2 = 2ax – 2a{x_1} \\ \Rightarrow {y_1}y – 4a{x_1} = 2ax – 2a{x_1} \\ \Rightarrow {y_1}y = 2ax + 2a{x_1} \\ \Rightarrow \boxed{{y_1}y = 2a\left( {x + {x_1}} \right)} \\ \end{gathered}$

This is the equation of the tangent to the given parabola at $\left( {{x_1},{y_1}} \right)$.

The slope of normal at $\left( {{x_1},{y_1}} \right)$ is $– \frac{1}{m} = – \frac{{{y_1}}}{{2a}}$

The equation of normal at the point $\left( {{x_1},{y_1}} \right)$ is $y – {y_1} = – \frac{{{y_1}}}{{2a}}\left( {x – {x_1}} \right)$
$\begin{gathered} \Rightarrow 2ay – 2a{y_1} = – {y_1}x + {x_1}{y_1} \\ \Rightarrow \boxed{{y_1}x + 2ay – 2a{y_1} – {x_1}{y_1} = 0} \\ \end{gathered}$

This is the equation of normal to the given parabola at $\left( {{x_1},{y_1}} \right)$.


text:
Groupe d’études et de recherche en analyse des décisions

# Decomposition methods for solving large-scale dynamic network flow problems

## Robert Curry – Mathematics Department, United States Naval Academy, États-Unis

Various network applications require dynamic flows to be transmitted according to a non-simultaneous schedule. In this talk, we consider a dynamic network flow problem that considers the presence of arc set up costs that may exist whenever an arc begins transmitting flow. This problem can be modeled by the minimum cost flow problem having arc-activation costs (MCF-A), in which arc $$(i,j)$$ is said to be activated on a path $$p$$ when $$(i,j)$$ is included on $$p$$ but not on $$p-1$$. As an alternative to this mixed-integer programming approach, we employe a relaxation-based decomposition algorithm for obtaining upper and lower bounds that increases the number of paths in a schedule, as needed. Computational results show that our approach determines an optimal solution in significantly less time than a mixed-integer programming approach. Finally, we extend our algorithm to consider other network flow problems.

Bio: Rob Curry is an Assistant Professor in the Mathematics Department at the United States Naval Academy. His research concentrates on methodology and algorithmic approaches in network optimization, combinatorial optimization, and integer programming. In particular, he studies exact algorithms for large-scale network flow optimization problems having applications in energy systems, wireless sensor networks, and humanitarian logistics. His most recent work focuses on decomposition methods for variations of maximum flow and minimum cost flow problems in wireless sensor network settings. He received his B.S. in Industrial Engineering from the University of Arkansas (2013), his M.S. in Industrial & Systems Engineering from the University of Florida (2014), and his Ph.D. in Industrial Engineering from Clemson University (2018).

Entrée gratuite.
Bienvenue à tous!


text:
# Please answer correctly and read it carefully (1096) Problem 3: An infinitely long rod lies along...

###### Question:

(1096) Problem 3: An infinitely long rod lies along the x-axis and carries a uniform linear charge density λ SpCm. A hollow cone segment of height H= 29 cm lies concentric with the x-axis. The end around the origin has a radius R- 10 cm and the far end has a radius R2- 19 cm. Refer to the figure. V-axis X-axis Otheexpertta.com 1496 Part (a) Consider the conic surface to be sliced vertically into an infinite number of rings, each of radius r and infinitesimal thickness d. Enter an expression for the electric flux differential trough one ofthese infinitesimal rings in terms of λ x, and the Coulomb constant dbs= Grade Summary Deductions 20% 0.955) 809% 7 8 9 HONE Attempts remaining: (4% per attempt) detailed view dx BACKSPACE L CEAR I give up! Hints: for a 12% deduction. Hints remaining: 0 Feedback: 56 deduction per What is the magnitude ofthe electrie field at a distance r from an infinite line of charge? -What is the expression for an element of surface area for a surface of Notice that the electric field and the normal to the surface area are no A surface of revolution is given by rotating the are length of the surface about an axis of revolution.

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UKS attack on KEA protocol

I'm trying to understand the UKS attack on KEA protocol.

This is described in this paper: Security Analysis of KEA Authenticated Key Exchange Protocol

So to summarize:

• KEA protocol:

• UKS attack on KEA:

"$\mathbb{M}$ registers a public key $g^a$ of some honest party $\mathbb{A}$ as $\mathbb{M}$'s own public key. Then $\mathbb{M}$ intercepts a key-exchange session between $\mathbb{A}$ and some other honest party $\mathbb{B}$ and at the same time starts a session between $\mathbb{M}$ and $\mathbb{B}$. Now $\mathbb{M}$ forwards ephemeral public key $g^x$ from $\mathbb{A}$ to $\mathbb{B}$ and ephemeral public key $g^y$ from $\mathbb{B}$ to $\mathbb{A}$. Since $\mathbb{M}$ has the same public key as $\mathbb{A}$, both $\mathbb{A}$ and $\mathbb{B}$ will compute identical session keys, however they participate in two different key-exchange sessions. $\mathbb{B}$ participates in a session with $\mathbb{M}$ while $\mathbb{A}$ participates in a session with $\mathbb{B}$. Finally, $\mathbb{M}$ reveals a session key of one of the sessions and announces the other session as a test session. Given a challenge key, $\mathbb{M}$ compares it to the revealed key. If they are the same, $\mathbb{M}$ decides that the challenge is a correct key for the test session and if different, $\mathbb{M}$ decides that the challenge key was chosen at random. The demonstrated attack breaks AKE security against a weak adversary (who can only reveal session keys)."

The thing that I don't understand in the attack description above, is the notion of "challenge key". It is said that the attacker $\mathbb{M}$ reveals a session key of one of the sessions, and then pick a challenge key to guess the key of the test session. But, I thought the key is the same in the two sessions so I don't understand what $\mathbb{M}$ want to guess? Secrets $a$ and $b$?

I think I am missing something, if someone could describe me in detail the attack it would be very nice. Thank you in advance.

Some background on formal key-exchange models

The goal of a key-exchange (KE) is to establish a session key between two parties. Naively, we could say that a KE is secure if no adversary will be able to figure out the session key (in full) established between two honest parties. However, in formal security models we take this a bit further and insist that the adversary should learn nothing about the session key, i.e., the adversary should not be able to glean any partial information about the session key from observing (and possibly interfering with) the KE.

This is formally captured by challenging the adversary on a target session of its choice. The challenge consists of first flipping a secret bit $b$ . If $b$ comes up as 0, then we return the actual session key to the adversary, while if it comes up as 1, then we return a randomly drawn key of the same length (completely independent of the KE). Now the adversary is supposed to guess which key he was given: real or random? If he cannot guess which key he was given with much better probability than $1/2$, then we say that the KE is secure. This "game" is completely artificial of course, but it nicely covers the idea that the adversary should learn nothing about the key, and is an example of the kind of "over-provisioning" present in formal security models.

In addition, we also allow the adversary to reveal the session keys of sessions unrelated to the target session (if we allowed the session key of the target session to be revealed, then the "game" above would ofc be completely trivial to win). What unrelated means can be quite technical to formulate, but intuitively it means that if $A$ and $B$ runs a KE between them, then this is unrelated to the KE session $A$ established with $C$ for instance, (or $B$ with $C$, or $D$ with $E$, etc...). Thus, if the attacker picks the $A \longleftrightarrow B$ exchange as its target, then it should still be allowed to reveal the session key from the $A \longleftrightarrow C$ exchange. If the KE is secure, then learning the session key from the $A \longleftrightarrow C$ exchange should not help the attacker learn anything about the session key in the $A \longleftrightarrow B$ exchange.

The UKS attack

In the UKS attack, there are two formally (in the sense of the security model) unrelated KE sessions: $A \longleftrightarrow B$ and $M \longleftrightarrow B$. This allows the adversary to reveal the session key from one, and still pick the other as its target session.

But, I thought the key is the same in the two sessions...

And this is exactly where the attack comes from! In practice these two exchanges are not unrelated at all, so when the adversary reveals the session key from, say, the $M \longleftrightarrow B$ session, then it also knows the session key of the $A \longleftrightarrow B$ session. From the point of view of the formal security model, the $A \longleftrightarrow B$ session is still a considered a valid target, so $M$ can trivially win the game (i.e., guess the value of the secret bit $b$).


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# What is the net area between f(x) = cosxsinx  and the x-axis over x in [0, 3pi ]?

Apr 7, 2017

We start by checking for x-intercepts within the given interval.

$f \left(x\right) = \cos x \sin x \to f \left(x\right) = \frac{1}{2} \sin 2 x$

If we set $f \left(x\right)$ to $0$ and solve for $x$.

$0 = \frac{1}{2} \sin 2 x$

$0 = \sin 2 x$

$2 x = 2 \pi \mathmr{and} 2 x = \pi$

$x = \pi \mathmr{and} x = \frac{\pi}{2}$

Since trig functions repeat, and the period of this function is $\pi$, we will have more than two crossings. If we analyze the graph, we realize that in $\left[0 , 3 \pi\right]$ there are $6$ intercepts.

graph{sinxcosx [-4.5, 15.5, -5.16, 4.84]}

This means that there are $3$ regions of equal area above the x-axis, and three areas of equal area below the x-axis. Since we're talking about net area, we should subtract the negative area (below the x-axis) from the positive area (above the x-axis).

Our expression for area will be

$A = 3 \left({\int}_{0}^{\frac{\pi}{2}} \frac{1}{2} \sin 2 x \mathrm{dx} - {\int}_{\frac{\pi}{2}}^{\pi} \frac{1}{2} \sin 2 x \mathrm{dx}\right)$

$A = 3 \left({\left[- \cos 2 x\right]}_{0}^{\frac{\pi}{2}} - {\left[- \cos 2 x\right]}_{\frac{\pi}{2}}^{\pi}\right)$

$A = 3 \left(- \cos \pi - \cos 0 - \left(- \cos 2 \pi - \cos \pi\right)\right)$

$A = 3 \left(1 - 1 + 1 - 1\right)$

$A = 0$

A geometric view of this gives sense to this result: whenever a positive area occurs, it is cancelled out by the negative area, giving a final result of $0$ (because there are an equal number of positive and negative areas).

Hopefully this helps!


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# Find the Middle Terms(S) in the Expansion Of: (I) ( X − 1 X ) 10 - Mathematics

Find the middle terms(s) in the expansion of:

(i) $\left( x - \frac{1}{x} \right)^{10}$

#### Solution

(i) $\left( x - \frac{1}{x} \right)^{10}$
$\text{ Here, n is an even number } .$
$\therefore \text{ Middle term } = \left( \frac{10}{2} + 1 \right) \text{ th = 6th term }$
$\text{ Now, we have }$
$T_6 = T_{5 + 1}$
$=^{10}{}{C}_5 x^{10 - 5} \left( \frac{- 1}{x} \right)^5$
$= - \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2}$
$= - 252$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 18 Binomial Theorem
Exercise 18.2 | Q 15.01 | Page 38