text:
# American Institue of Mathematical Sciences

ISSN:
1534-0392

eISSN:
1553-5258

All Issues

## Communications on Pure & Applied Analysis

2008 , Volume 7 , Issue 3

Select all articles

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J. Colliander , Justin Holmer , Monica Visan and  Xiaoyi Zhang
2008, 7(3): 467-489 doi: 10.3934/cpaa.2008.7.467 +[Abstract](33) +[PDF](272.2KB)
Abstract:
We consider the Cauchy problem for a family of semilinear defocusing Schrödinger equations with monomial nonlinearities in one space dimension. We establish global well-posedness and scattering. Our analysis is based on a four-particle interaction Morawetz estimate giving a priori $L_{t,x}^8$ spacetime control on solutions.
B. E. Ainseba and  Sebastian Aniţa
2008, 7(3): 491-512 doi: 10.3934/cpaa.2008.7.491 +[Abstract](26) +[PDF](243.0KB)
Abstract:
The internal zero-stabilization of the nonnegative solutions to some parabolic equations is investigated. We provide a necessary and a sufficient condition for nonnegative stabilizability in terms of the sign of the principal eigenvalue of a certain elliptic operator. This principal eigenvalue is related to the rate of the convergence of the solution. We give some evaluations of this principal eigenvalue with respect to the geometry of the domain and of the support of the control. A stabilization result for an age-dependent population dynamics with diffusion is also established.
2008, 7(3): 513-532 doi: 10.3934/cpaa.2008.7.513 +[Abstract](33) +[PDF](238.0KB)
Abstract:
We consider the generalized Benjamin-Ono equation, regularized in the same manner that the Benjamin-Bona-Mahony equation is found from the Korteweg-de Vries equation [3], namely the equation $u_t + u_x +u^\rho u_x + H(u_{x t})=0,$ where $H$ is the Hilbert transform. In a second time, we consider the generalized Kadomtsev-Petviashvili-II equation, also regularized, namely the equation $u_t + u_x +u^\rho u_x - u_{x x t} +\partial_x^{-1}u_{y y} =0$. We are interested in dispersive properties of these equations for small initial data. We will show that, if the power $\rho$ of the nonlinearity is higher than $3$, the respective solution of these equations tends to zero when time rises with a decay rate of order close to $\frac{1}{2}$.
Roberta Bosi , Jean Dolbeault and  Maria J. Esteban
2008, 7(3): 533-562 doi: 10.3934/cpaa.2008.7.533 +[Abstract](44) +[PDF](369.2KB)
Abstract:
By expanding squares, we prove several Hardy inequalities with two critical singularities and constants which explicitly depend upon the distance between the two singularities. These inequalities involve the $L^2$ norm. Such results are generalized to an arbitrary number of singularities and compared with standard results given by the IMS method. The generalized version of Hardy inequalities with several singularities is equivalent to some spectral information on a Schrödinger operator involving a potential with several inverse square singularities. We also give a generalized Hardy inequality for Dirac operators in the case of a potential having several singularities of Coulomb type, which are critical for Dirac operators.
M. Pellicer and  J. Solà-Morales
2008, 7(3): 563-577 doi: 10.3934/cpaa.2008.7.563 +[Abstract](42) +[PDF](219.8KB)
Abstract:
We consider a model for a damped spring-mass system that is a strongly damped wave equation with dynamic boundary conditions. In a previous paper we showed that for some values of the parameters of the model, the large time behaviour of the solutions is the same as for a classical spring-mass damper ODE. Here we use spectral analysis to show that for other values of the parameters, still of physical relevance and related to the effect of the spring inner viscosity, the limit behaviours are very different from that classical ODE.
Ling Hsiao , Fucai Li and  Shu Wang
2008, 7(3): 579-589 doi: 10.3934/cpaa.2008.7.579 +[Abstract](41) +[PDF](172.8KB)
Abstract:
The combined quasineutral and inviscid limit for the Vlasov-Poisson-Fokker-Planck (VPFP) system is rigorously derived in this paper. It is shown that the solution of VPFP system converges to the solution of incompressible Euler equations with damping. The proof of convergence result is based on compactness arguments and the so-called relative-entropy method.
Marianne Korten and  Charles N. Moore
2008, 7(3): 591-600 doi: 10.3934/cpaa.2008.7.591 +[Abstract](33) +[PDF](152.1KB)
Abstract:
We consider the two-phase Stefan problem $u_t=\Delta\alpha(u)$ where $\alpha(u) =u+1$ for $u<-1$, $\alpha(u) =0$ for $-1 \leq u \leq 1$, and $\alpha(u)=u-1$ for $u \geq 1$. We show that if $u$ is an $L_{l o c}^2$ distributional solution then $\alpha(u)$ has $L_{l o c}^2$ derivatives in time and space. We also show $|\alpha(u)|$ is subcaloric and conclude that $\alpha(u)$ is continuous.
Massimiliano Berti , M. Matzeu and  Enrico Valdinoci
2008, 7(3): 601-615 doi: 10.3934/cpaa.2008.7.601 +[Abstract](39) +[PDF](212.6KB)
Abstract:
We construct entire solutions of $\Delta u=f(x,u,\nabla u)$ which are superpositions of odd, periodic functions and linear ones, with prescribed integer or rational slope.
2008, 7(3): 617-630 doi: 10.3934/cpaa.2008.7.617 +[Abstract](39) +[PDF](174.6KB)
Abstract:
In this paper, we study a fourth order degenerate parabolic equation, which arises in epitaxial growth of nanoscale thin films. We establish the existence of weak solutions, based on the uniform estimates for the approximate solutions. The nonnegativity and the finite speed of propagation of perturbations of solutions are also discussed.
Li Ma and  Lin Zhao
2008, 7(3): 631-643 doi: 10.3934/cpaa.2008.7.631 +[Abstract](37) +[PDF](183.8KB)
Abstract:
In this paper, we first derive a new non-linear type inequality for Newtonian potential and then we study the regularity problem for positive weak solutions to the non-linear Laplace equation:

$-\Delta u= f(u)\quad$ in $\Omega,$

with $f(u)\in L^1(\Omega)$. Here $\Omega$ is a bounded domain in $R^n$, and $f(u)$ is a regular function with respect to $u$. We give an apriori estimate for positive weak solutions. We show that under some appropriate assumptions on the non-linear term $f$, the positive weak solutions are in fact in some local Sobolev space $W_{l o c}^{1,\tau}(\Omega)$. We also derive a very general local monotonicity formula for variational solutions to the equation above with special nonlinear term $f$.

Mircea Sofonea , Cezar Avramescu and  Andaluzia Matei
2008, 7(3): 645-658 doi: 10.3934/cpaa.2008.7.645 +[Abstract](41) +[PDF](219.3KB)
Abstract:
Let $C(\mathbb R_+;X)$ denote the Fréchet space of continuous functions defined on $\mathbb R_+=[0,\infty)$ with values on a real Banach space $(X,$ ||$\cdot$||$_X)$. We prove a fixed point theorem for operators $\Lambda:C(\mathbb R_+;X)\to C(\mathbb R_+;X)$ which satisfy a sequence of inequalities involving an integral term. Then we consider a mathematical model which describes the frictionless contact between a viscoplastic body and a deformable foundation. The process is quasistatic and is studied on the unbounded interval of time $[0,\infty)$. We provide the variational formulation of the problem, then we use the abstract fixed point theorem to prove the existence of a unique weak solution to the model. We complete our study with a regularity result.
Najat Ziyadi , Said Boulite , M. Lhassan Hbid and  Suzanne Touzeau
2008, 7(3): 659-675 doi: 10.3934/cpaa.2008.7.659 +[Abstract](142) +[PDF](216.8KB)
Abstract:
The aim of this paper is to analyse a dynamic model which describes the spread of scrapie in a sheep flock. Scrapie is a transmissible spongiform encephalopathy, endemic in a few European regions and subject to strict control measures. The model takes into account various factors and processes, including seasonal breeding, horizontal and vertical transmission, genetic susceptibility of sheep to the disease, and a long and variable incubation period. Therefore the model, derived from a classical SI (susceptible-infected) model, also incorporates a discrete genetic structure for the flock, as well as a continuous infection load structure which represents the disease incubation. The resulting model consists of a set of partial differential equations which describe the evolution of the flock with respect to time and infection load. To analyse this model, we use the semigroup and evolution family theory, which provides a flexible mathematical framework to determine the existence and uniqueness of a solution to the problem. We show that the corresponding linear model has a unique classical solution and that the complete nonlinear model has a global solution.
Walter Allegretto , Yanping Lin and  Zhiyong Zhang
2008, 7(3): 677-698 doi: 10.3934/cpaa.2008.7.677 +[Abstract](22) +[PDF](287.3KB)
Abstract:
In this paper we consider a biosensor model in $R^3$, consisting of a coupled parabolic differential equation with Robin boundary condition and an ordinary differential equation. Theoretical analysis is done to show the existence and uniqueness of a Holder continuous solution based on a maximum principle, weak solution arguments. The long-time convergence to a steady state is also discussed as well as the system situation. Next, a finite volume method is applied to the model to obtain an approximate solution. Drawing in part on the analytical results given earlier, we establish the existence, stability and error estimates for the approximate solution, and derive $L^2$ spatial norm convergence properties. Finally, some illustrative numerical simulation results are presented.
Alexander Arbieto and  Carlos Matheus
2008, 7(3): 699-713 doi: 10.3934/cpaa.2008.7.699 +[Abstract](39) +[PDF](207.4KB)
Abstract:
We study local and global well-posedness of the initial value problem for the Schrödinger-Debye equation in the periodic case. More precisely, we prove local well-posedness for the periodic Schrödinger-Debye equation with subcritical nonlinearity in arbitrary dimensions. Moreover, we derive a new a priori estimate for the $H^1$ norm of solutions of the periodic Schrödinger-Debye equation. A novel phenomenon obtained as a by-product of this a priori estimate is the global well-posedness of the periodic Schrödinger-Debye equation in dimensions $1$ and $2$ without any smallness hypothesis of the $H^1$ norm of the initial data in the "focusing" case.
Emmanuel Hebey and  Jérôme Vétois
2008, 7(3): 715-741 doi: 10.3934/cpaa.2008.7.715 +[Abstract](40) +[PDF](393.9KB)
Abstract:
We discuss and prove existence of multiple solutions for critical elliptic systems in potential form on compact Riemannian manifolds.
Daniele Bartolucci and  Luigi Orsina
2008, 7(3): 743-744 doi: 10.3934/cpaa.2008.7.743 +[Abstract](21) +[PDF](83.7KB)
Abstract:
We correct the statements of Proposition 4.1 and Theorem 4.1 in [1]: "Uniformly elliptic Liouville type equations: Concentration compactness and a priori estimates," Comm. Pure and Appl. Analysis, 4 (2005), 499--522.

2016  Impact Factor: 0.801


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Homotopy class of a homeomorphism

I was reading a book and find this statement:

''It is well kown that each homeomorphism $f:S\rightarrow R$ between compact surfaces is homotopic to a diffeomorphism''

I would know some references of this affirmation to see one proof.

-
Even more is true, any homotopy-equivalence is homotopic to a diffeomorphism. If the surface has boundary then you have to also require the homotopy-equivalence restricts to a homotopy-equivalence of the boundaries. A common reference is Zieschang, Vogt and Coldeway, Surfaces and planar discontinuous groups. – Ryan Budney Nov 28 2011 at 23:38
Statements like this abound in homotopy theory. For instance, any map $f$ is homotopic to an injection. Or a surjection. Or a smooth map. Or pretty much whatever you want (not all at the same time, though). The first two have to do with making the kernel or cokernel contractible. I'm not sure about the others, and I'd like to see an answer. – David White Nov 28 2011 at 23:40
Isn't this question almost the same as mathoverflow.net/questions/35198/… ? – Daniel Moskovich Nov 28 2011 at 23:52

3 Answers

You may find a proof of this using hierarchies in some notes of Lackenby. See Theorem 12.1. I think this sort of argument is probably due to Waldhausen, since his proof of homotopy rigidity of Haken 3-manifolds is dependent on this, so you could have a look at his paper too.

I think there are probably many other proofs of this fact. One possible proof is to endow $S$ and $R$ with hyperbolic metrics, and use the Douady-Earle map.

Edit: I was trying to answer the stronger question of whether a homotopy equivalence is homotopic to a homeomporphism (your use of the term homotopic threw me), which is answered in the above references. Another strengthening is to ask whether a homeomorphism is isotopic to a diffeomorphism? In other words, does a surface have a unique differential structure? In the context of PL structures, this uniqueness was answered by Rado (see Moise's book). I think it's also known that PL and differential structures are equivalent. This is discussed in Thurston's book (Theorem 3.10.9).

-
Didn't Munkres prove a homeomorphism is isotopic to a diffeomorphism? – Steve D Nov 29 2011 at 11:17
He proved uniqueness of differentiable structures on 3-manifolds. projecteuclid.org/… – Ian Agol Nov 29 2011 at 17:00

This fact is known as "the classification of surfaces". For a particularly non elementary view see http://www.math.uchicago.edu/~shmuel/tom-readings/ranicki-intro, chapter 4.

EDIT A particular elementary proof is here.

-
Well, "homeomorphic surfaces are diffeomorphic" is weaker than "homeomorphisms are homotopic to diffeomorphisms". But, then again, any two true statements are equivalent. ;) – Richard Kent Nov 29 2011 at 2:00
Any true statements may or may not be equivalent, but I was specifically answering the OP's statement. Look at the reference. – Igor Rivin Nov 29 2011 at 10:22
@Igor - I guess you are referring to Theorem 4.32, part (ii), on page 61 of the pdf file. But Ranicki doesn't give a proof. Instead he refers to chapter 9 of Hirsch's book "Differential Topology". Skimming chapter 9 it seems that the exact statement the OP wants is not there? – Sam Nead Nov 29 2011 at 10:55
@Sam: Ranicki doesn't give a full proof of classification of surfaces, that's true, but among other things, the subject of when homotopy equivalences are homotopic to diffeos is central to surgery theory, so once you understand classification of surfaces, the answer to the OP is immediate. The whole machine of surgery is a bit of an overkill, hence the "non elementary" comment. – Igor Rivin Nov 29 2011 at 12:32

A homeomorphism (of surfaces) is isotopic to a PL homeomorphism. See Theorem A4 of Epstein's "Curves on 2-manifolds and isotopies". He gives a proof. I haven't read the paper recently, but I recall that it is fairly self-contained.

-

format: plain


text:
The open source CFD toolbox
nullJoint.C
Go to the documentation of this file.
1/*---------------------------------------------------------------------------*\
2 ========= |
3 \\ / F ield | OpenFOAM: The Open Source CFD Toolbox
4 \\ / O peration |
5 \\ / A nd | www.openfoam.com
6 \\/ M anipulation |
7-------------------------------------------------------------------------------
8 Copyright (C) 2016 OpenFOAM Foundation
9-------------------------------------------------------------------------------
11 This file is part of OpenFOAM.
12
13 OpenFOAM is free software: you can redistribute it and/or modify it
15 the Free Software Foundation, either version 3 of the License, or
16 (at your option) any later version.
17
18 OpenFOAM is distributed in the hope that it will be useful, but WITHOUT
19 ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
20 FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
21 for more details.
22
23 You should have received a copy of the GNU General Public License
24 along with OpenFOAM. If not, see <http://www.gnu.org/licenses/>.
25
26\*---------------------------------------------------------------------------*/
27
28#include "nullJoint.H"
29#include "rigidBodyModel.H"
31
32// * * * * * * * * * * * * * * Static Data Members * * * * * * * * * * * * * //
33
34namespace Foam
35{
36namespace RBD
37{
38namespace joints
39{
41
43 (
44 joint,
45 null,
47 );
48}
49}
50}
51
52
53// * * * * * * * * * * * * * * * * Constructors * * * * * * * * * * * * * * //
54
56:
57 joint(0)
58{}
59
60
62:
63 joint(0)
64{}
65
66
68{
69 return autoPtr<joint>(new null(*this));
70}
71
72
73// * * * * * * * * * * * * * * * * Destructor * * * * * * * * * * * * * * * //
74
76{}
77
78
79// * * * * * * * * * * * * * * Member Functions * * * * * * * * * * * * * * //
80
82(
83 joint::XSvc& J,
84 const scalarField& q,
85 const scalarField& qDot
86) const
87{
89 << "Cannot calculate the state of a null-joint"
90 << abort(FatalError);
91}
92
93
94// ************************************************************************* //
Macros for easy insertion into run-time selection tables.
Add to construction table with typeName as the key.
Joint state returned by jcalc.
Definition: joint.H:125
Abstract base-class for all rigid-body joints.
Definition: joint.H:85
Null joint for the root-body.
Definition: nullJoint.H:68
virtual ~null()
Destructor.
Definition: nullJoint.C:75
virtual autoPtr< joint > clone() const
Clone this joint.
Definition: nullJoint.C:67
virtual void jcalc(joint::XSvc &J, const scalarField &q, const scalarField &qDot) const
Update the model state for this joint.
Definition: nullJoint.C:82
null()
Construct for given model.
Definition: nullJoint.C:55
Pointer management similar to std::unique_ptr, with some additional methods and type checking.
Definition: autoPtr.H:66
A list of keyword definitions, which are a keyword followed by a number of values (eg,...
Definition: dictionary.H:126
#define defineTypeNameAndDebug(Type, DebugSwitch)
Define the typeName and debug information.
Definition: className.H:121
#define FatalErrorInFunction
Report an error message using Foam::FatalError.
Definition: error.H:453
Namespace for OpenFOAM.
errorManip< error > abort(error &err)
Definition: errorManip.H:144
error FatalError
dictionary dict


text:
# Read reference data from external file

This is a function from a program I am working on for my internship, which involves research in computational science. I have some text files with reference data in a directory data/ with name e.g. AR00.1 that look like this:

90.000 73.629
120.00 98.338
150.00 123.00
180.00 147.64
210.00 172.28
...

The function works, with output as expected. I am fairly new to C++ and programming in general (had a little less than a year of C before I started with C++.) Am I doing things efficiently (i.e. loading all of the data into memory at once and calling it when needed) or is there a better way to access this data? The part that seems messiest to me is the nested vectors, though I can't think of another way to do them as I've no formal education in data structures.

In any case, here is a minimum example of the function with an arbitrary main function to show my usage. Thank you for taking the time to read and review this!

//includes go here

struct reference_data
{
double temperature,
volume_l_mol,
volume_m3,
compressibility;
};

using namespace std;

int main()
{
vector<string> species = {"AR","CH4","CO2","H2","HE","KR","N2","NE","XE"};
vector<string> pressures = {"00.1", "001", "005", "010", "020", "030"};

cout << NIST_data[0][0][0].temperature //90 for above data example
return 0;
}

{
vector<double> pressure_floats = {00.1, 001, 005, 010, 020, 030};
// ALL->SPECIES->PRES->STATE POINTS
vector<vector<vector<reference_data>>> NIST_data;
for(unsigned long species_ind = 0;species_ind<species.size();species_ind++)
{
//SPECIES->PRES->STATE POINTS
vector<vector<reference_data>> this_species;
string a_line;
for(auto pressure_ind= 0; pressure_ind < pressure_strings.size(); pressure_ind++)
{
string file_name = "data/";
file_name += species[species_ind];
file_name += pressure_strings[pressure_ind];
ifstream input(file_name);
if(input.is_open())
{
//PRES->STATE POINTS
vector<reference_data> this_pressure;
while (getline(input, a_line))
{
vector<string> this_line;
istringstream iss(a_line);
//copy the numbers of interest from the line into the vector this_line (thanks Doug / SO !)
copy(
istream_iterator<string>(iss),
istream_iterator<string>(),
back_inserter(this_line)
);
if(this_line.size())
{
reference_data this_point;
this_point.temperature = atof(this_line[0].c_str());
this_point.volume_l_mol = atof(this_line[1].c_str());
double liters = this_point.volume_l_mol*MOLES;
this_point.volume_m3 = liters / 1000.0;
this_point.compressibility = get_compressibility(this_point.temperature,
pressure_floats[pressure_ind],
this_point.volume_m3);
this_pressure.push_back(this_point);//add current temperature to the pressure it corresponds to
}
else
{
<< endl;
}
}
this_species.push_back(this_pressure);//add current pressure to species it corresponds to
}
else
{
cerr << "Error in opening file " << file_name
<< " in read_reference_data()" << endl;
}
input.close();
}
NIST_data.push_back(this_species);//add current species to all data
}
return NIST_data;
}

Instead of a 3D vector, I think you'll find a 2D map to be easier to use and maintain map<string,map<string,vector<reference_data>>>. This way the data is indexed by the actual strings you're using instead of generic index numbers where you'll need a legend to know which index applies to specific data.

Instead of getline when dealing with numeric data, it is much cleaner and simpler to use the extraction operator(>>), which will cast the data to the specific type of the variable you use:

double temperature = 0.0;
double volume = 0.0;
while(input >> temperature >> volume)
{
//process data
}

The loop will run until EOF.

The whole algorithm might look something like this:

map<string, map<string, vector<reference_data>>> data;
vector<string> species = {
"AR",
"CH4",
"CO2",
"H2",
"HE",
"KR",
"N2",
"NE",
"XE"
};
map<string, double> pressures ={
{"00.1", 0.1},
{"001", 1.0},
{"005", 5.0},
{"010", 10.0},
{"020", 20.0},
{"030", 30.0}
};
for (string s : species)
{
for (auto p : pressures)
{
string fileName = "data/" + s + p.first;
ifstream file(fileName);
if (file.is_open())
{
double temperature = 0.0;
double volume = 0.0;
while (file >> temperature >> volume)
{
reference_data this_point;
this_point.temperature = temperature;
this_point.volume_l_mol = volume;
double liters = this_point.volume_l_mol*MOLES;
this_point.volume_m3 = liters / 1000.0;
this_point.compressibility = get_compressibility(this_point.temperature,
p.second,
this_point.volume_m3);
data[s][p.first].emplace_back(this_point);
}
}

}
}

• I think that cast is slightly rough word to use in this case. May be convert would be better? – Incomputable Jun 15 '17 at 21:41
• In my experience cast generally applies more to operators, whereas convert applies more to specific functions. In this case since the conversion is done implicitly, I feel that cast is a more proper term for it. – tinstaafl Jun 15 '17 at 22:29
• good point. How about extract then? :) – Incomputable Jun 15 '17 at 22:31
• Seems kind of redundant to say that the extraction operator extracts data and doesn't make it plain that that data is actually changed by the operator, especially for someone who may not be familiar with how the extraction operator works. – tinstaafl Jun 15 '17 at 22:42
• fair enough, upvoted. – Incomputable Jun 15 '17 at 22:43


text:
# CBSE Sample Papers for Class 9 Maths Paper 4

CBSE Sample Papers for Class 9 Maths Paper 4 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 4

## CBSE Sample Papers for Class 9 Maths Paper 4

 Board CBSE Class IX Subject Maths Sample Paper Set Paper 4 Category CBSE Sample Papers

Students who are going to appear for CBSE Class 9 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 4 of Solved CBSE Sample Papers for Class 9 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

• All questions are compulsory.
• Questions 1-6 in Section-A are Very Short Answer Type Questions carrying 1 mark each.
• Questions 7-12 in Section-B are Short Answer (SA-I) Type Questions carrying 2 marks each.
• Questions 13-22 in Section-C are Short Answer (SA-II) Type Questions carrying 3 marks each.
• Questions 23 -30 in Section-D are Long Answer Type Questions carrying 4 marks each.

SECTION-A

Question 1.
Simplify: 32/3 x 72/3.

Question 2.
When P(x) = x3 – ax2 + x is divided by (x – a). Find the remainder.

Question 3.
On which axis does the point A(0, 4) lie?

Question 4.
Euclid divided his book “Elements” into how many chapters?

Question 5.
A coin is tossed 60 times and the tail appears 35 times. What is the probability of getting head?

Question 6.
What is the total surface area (T.S.A.) of a cone whose radius is $$\frac { r }{ 2 }$$ and slant height is 2l?

SECTION-B

Question 7.
Factorize: a(a – 1) – b(b – 1).

Question 8.
The angles of a triangle are in the ratio 2:3:7. Find the measure of each angle of triangle.

Question 9.
Find the angle at which the bisectors of any two adjacent angles of a parallelogram intersect.

Question 10.
Show that a median of a triangle divides it into two triangles of equal area.

Question 11.
Find the area of isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm.

Question 12.
The mean of 40 numbers was found to be 38. Later on, it was detected that a number 56 was misread as 36. Find the correct mean of the (data) given numbers.
OR
Following data gives a number of children in 40 families.
1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3,4, 2, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2,4, 3, 2, 1, 0, 5, 1, 2,4, 3, 4, 1, 6, 2, 2.
Represent it in the form of frequency distribution taking classes 0-2, 1-4, etc.

SECTION-C

Question 13.
Simplify

Question 14.
Find the product: (a-b-c) (a² + b² + c² + ab + ac – bc).
OR
The polynomial f(x) = x4 – 2x3 + 3x2 – ax + b when divided by (x – 1) and (x + 1) leaves the remainder 5 and 19 respectively. Find the value of a and b. Hence find the remainder when fix) is divided by (x – 2).

Question 15.
Fill in the blanks:
(i) For the line 4x + 3y = 12, x-intercept =___ and y-intercept = ___
(ii) If x = 4, y = 3, is a solution of 2x + ky = 14, then k =___
(iii) If the point P(P, 4) lies on the line 3x + y = 10, then P =___

Question 16.
In the adjoining figure, ABCD is a parallelogram in which E and F are the mid points of the sides AB and CD respectively. Prove that the line segment CE and AF trisect the diagonal BD.

Question 17.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that
ar(∆AOD) = ar(∆BOC)

Question 18.
Construct an isosceles triangle whose base is 5 cm and whose vertical angle is 72°

Question 19.
Plot the points A(-5, 2), B(3, 2), C(-4, -3) and D(6, 0), E(0, 2), F(4, -4) on the graph paper. Write the name of quadrant also.

Question 20.
The total surface area of a cylinder is 462 cm2. Its curved surface area is one-third of its total surface area. Find the volume of the cylinder.

Question 21.
A tent is in the form of a right circular cylinder, surmounted by a cone. The diameter of a cylinder is 24 m. The height of the cylindrical portion is 11m, while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.

Question 22.
Three coins are tossed simultaneously 150 times and it is found that 3 tails appeared 24 times; 2 tails appeared 45 times 1 tail appeared 72 times and no tail appear 9 times. If three coins are tossed simultaneously at random, find the probability of getting
(i) 3 tails,
(ii) 2 tails,
(iii) 1 tail.

SECTION-D

Question 23.

Question 24.
If the polynomials (2x3 + ax2 + 3x – 5) and (x3 + x2 – 2x +a) leave the same remainder when divided by (x – 2). Find the value of a. Also, find the remainder in each case.

Question 25.
Two men starts from point A and B respectively, 42 km apart. One walks from A to B at 4 km/hr and another walks from B to A at a certain uniform speed. They meet each other after 6 hours. Find the speed of the second man.

Question 26.
In a ∆ABC, ∠B > ∠C. If AM is the bisector of ∠BAC and AN ⊥ BC, prove that ∠MAN = $$\frac { 1 }{ 2 }$$ (∠B – ∠C).

Question 27.
In the given figure, ABCD is a square, M is the mid-point of AB and PQ ⊥ CM meets AD at P and CB produced at Q. Prove that
(i) PA = BQ,
(ii) CP = AB + PA

Question 28.
Prove that the sum of either pair of the opposite angles of a cyclic quadrilateral is 180°.
OR
Prove that the opposite angles of a cyclic quadrilateral are supplementary.

Question 29.
A copper wire of diameter 6 mm is evenly wrapped on a cylinder of length 15 cm and diameter 49 cm to cover its whole surface. Find the length and volume of the wire. If the specific gravity of copper be 9 g per cubic cm, find the weight of the wire.

Question 30.
15 students of Govt. School spend the following number of hours in a month for doing cleaning in their street 25, 15, 20, 20, 9, 20, 25, 15, 7, 13, 20, 12, 10, 15, 8.
(i) Find mean, median and mode from above data.
(ii) Which value is depicted from above data?

Solutions

Solution 1.
32/3 x 72/3 = (3 x 7)2/3 = (21)2/3

Solution 2.
P(x) = x3 – ax2 + x, x – a = 0 ⇒ x = a
When P(x) is divided by (x – a), we get
Remainder = P(+a) = a3 – a(+a)2 + (+a) = a3 – a3 + a = a
Remainder = P(a) = a
⇒ Remainder = a

Solution 3.
y-axis ⇒ coordinates of y-axis is (0, b).

Solutions 4.
13 chapters.

Solution 5.

Solution 6.

Solution 7.
a(a – 1) – b(b – 1) = a2 – a – b2 + b = a2 – b2 – (a – b) = (a – b)(a + b) – (a – b)
= (a – b) (a + b – 1)

Solution 8.
Let the angles of the given triangles measure (2x)°, (3x)° and (7x)°.
2x + 3x + 7x = 180° [By angle sum property of A]
12x = 180° ⇒ x = 15°
Hence angles are
2x = 30°
3x = 45°
7x =105°

Solution 9.

∠A + ∠B = 180°
⇒ $$\frac { 1 }{ 2 }$$∠A + $$\frac { 1 }{ 2 }$$∠B = 90°
In ∆AOB
$$\frac { 1 }{ 2 }$$∠A + $$\frac { 1 }{ 2 }$$∠B + ∠AOB = 180°
⇒ 90° + ∠AOB = 180°
⇒ ∠AOB = 90°

Solution 10.

Given: A ∆ABC in which AD is median.
Construction: Draw AL ⊥ BC
Proof: BD = DC
(AD is median so D is midpoint of BC)

=> ar(∆ABD) = ar(∆ADC) [∵ ar(∆) = $$\frac { 1 }{ 2 }$$ x base x height]
Hence, a median of a triangle divides it into two triangles of equal area.

Solution 11.
a = 13,b = 13,c = 24

Hence, area of the triangle (∆) = 60 cm²

Solution 12.
Calculated mean of 40 numbers = 38
∴ Calculated sum of these numbers = 38 x 40 = 1520
Correct sum of these numbers = [1520 – (wrong item) + correct item]
= 150 – 36 + 56 = 1540
The correct mean = $$\frac { 1540 }{ 40 }$$ = 38.5
Hence, correct mean = 38.5

Lowest data = 0, highest data = 6

Solution 13.

Solution 14.
(a – b – c) (a² + b² + c² + ab + ac – bc)
= [a + (-b) + (-c)] [a² + (-b)² + (-c)² – a(-b) – (-b) (-c) – (-c)a]
= a3 + (-b)3 + (-c)3 – 3a (-b)(-c)
= a3 – b3 – c3 – 3abc (a – b – c) (a² + b² + c² + ab + ac – bc)
= a3 – b3 – c3 – 3abc

OR
f(x) = x4 – 2x3 + 3x2 – ax + b
If f(x), divided by (x – 1), let remainder be R1 = 5.
x – 1 = 0, x = 1

Solution 15.
(i) For the line 4x + 3y = 12, x-intercept = 3 and y-intercept = 4, i.e., Ans = 3, 4
(ii) If x = 4, y = 3 is a solution of 2x + ky = 14, then k = 2.
=> Putting x = 4 and y = 3 in eqn. 2x + ky = 14
2 x 4 + k x 3 = 14 =>3k = 14 – 8 = 6
3k = 6 => k = 2
(iii) If the point P(P, 4) lies on the line 3x +y = 10, then P = 2
=> Putting x = P and y = 4 in eqn. 3x + y = 10
3P + 4 = 10 =>3P = 6 => P = 2.

Solution 16.

Let BD be intersected by CE and AF at P and Q respectively. AB || DC and AB = DC [opposite sides of || gm]
=> AE = FC and $$\frac { 1 }{ 2 }$$ AB = $$\frac { 1 }{ 2 }$$ DC => AE || FC and AE = FC
=> AECF is a || gm => AF || CE => EP || AQ and FQ || CP.
In ∆BAQ, E is the mid point of AB.
and EP || AQ so, P is the mid point of BQ
BP = PQ [By converse of midpoint theorem]
Again in ∆DPC, F is mid point of DC and FQ || CP. So Q is the mid point of DP.
=> PQ = QD [By converse of mid point theorem]
∴ BP = PQ = QD
Hence, CE and AF trisect the diagonal BD.

Solution 17.

∆ACD and ∆BCD being on the same base DC and between the same parallels DC and AB, then
ar(∆ACD) = ar(∆BCD)
Subtracting ar(∆DOC) on both sides
ar (∆ACD) – ar(∆DOC) = ar(∆BCD) – ar(∆DOC)
ar(∆AOD) = ar(∆BOC)

Solution 18.

Steps of Construction:
1. Draw a line segment BC = 5 cm
2. Make ∠CBX = 72°, below the line segment
3. Make ∠XBY = 90°
4. Draw the right bisector PQ of BC, intersecting BY at O.
5. With O as centre and radius OB, draw a circle, intersecting PQ at A.
6. Join AB and AC.
Then ∆ABC is the required isosceles triangle in which AB = AC.

Solution 19.
Point A(-5, 2) lies in => II quadrant
B(3, -2) lies in => I quadrant
C(-4, -3) lies in => III quadrant
D(6, 0) lies in => x-axis (abscissa)
E(0, 2) lies in => y-axis (ordinate)
F(4, -4) lies in => IV quadrant

Solution 20.
C.S.A. of cylinder = $$\frac { 1 }{ 3 }$$ x (T.S.A. of cylinder)
= $$\frac { 1 }{ 3 }$$ x 462 = 154 cm²
(T.S.A. of cylinder – C.S.A. of cylinder) = 462 – 154 = 308 cm²
2πr(h + r) – 2πrh = 308

Solution 21.

Radius of the cylinder = $$\frac { 24 }{ 2 }$$ = 12
m = R
Height, H = 11 m
Curved surface area of the cylindrical portion
= 2πRH sq. units
= (2π x 12 x 11) m²
= (264 π) m²
Radius of the cone = r = 12m,
height of the cone = (16 – 11) = 5m

Hence, the area of the canvas required for tent = 1320 m².

Solution 22.
Total number of trials = 150
Number of times 3 tails appeared = 24
Number of times 2 tails appeared = 45
Number of times 1 tail appeared = 72
Number of times 0 tail appeared = 9

Solution 23.

Solution 24.
Let f(x) = 2x3 + ax2 + 3x – 5 and g(x) = x3 + x2 – 2x + a
When f(x) is divided by (x – 2), then remainder = f(2) [x – 2 = 0 => x = 2]
f(2) = 2(2)3 + a(2)2 + 3(2) – 5 = 16 + 4a – 5 + 6 = 17 + 4a
When g(x) is divided by (x – 2), remainder = g(2) [x – 2 = 0 => x = 2]
g(2) = (2)3 + (2)2 -2 x 2 + a = 8 + a
But f(2) = g(2)
17 + 4a = 8 + a
3a = -9 => a = -3
Remainder in each case = 8 + (-3) = 8 – 3 = 5

Solution 25.
Let the speed of another man be x km/h.
Average speed of both men = (x + 4) km/h.

=> 6x = 18 =>x = $$\frac { 18 }{ 6 }$$ = 3
x = 3 km/h

Solution 26.
Given: In ∆ABC, in which ∠B > ∠C, AN ⊥ BC and AM is the bisector of ∠A. Also find the angle MAN if ∠B = 65° and ∠C = 30°.
To Prove: ∠MAN = $$\frac { 1 }{ 2 }$$ (∠B – ∠C)
Proof: Since AM is the bisector of ∠A

Solution 27.

In ∆PAM and ∆QBM
AM = BM (M is the mid point)
∠PAM = ∠QBM (each 90°)
∠PMA = ∠BMQ (V.O.A)
∴ ∆AMP ≅ ∆BMQ (ASA congruency)
∆AMP ≅ ∠BMQ (CPCT)
PA = BQ …(i)
and MP = MQ
Now join PC. Again in ∆CMP and ∆CMQ
PM = MQ (Proved)
∠CMP = ∠CMQ (each 90°)
CM = CM (Common)
∆CMP = ∆CMQ (RHS congruency)
CP = CQ (CPCT)
CP = CQ = BC + BQ = AB + PA[BC = AB, PA = BQ]
CP = AB + PA

Solution 28.
To Prove: ∠A + ∠C = 180°
∠B + ∠D = 180°
Construction: Join AC and BD

Solution 29.
Length of the cylinder = 15 cm
Radius = 24.5 cm, diameter of the wire = 0.6 cm

Solution 30.
Data in ascending order 7, 8, 9, 10, 12, 13, 15, 15, 15, 20, 20, 20, 20, 25, 25 => n = 15.

Median =15
Mode = Maximum number of observations = 20 (4 times)
(ii) Value => Social work.

We hope the CBSE Sample Papers for Class 9 Maths Paper 4 help you. If you have any query regarding CBSE Sample Papers for Class 9 Maths Paper 4, drop a comment below and we will get back to you at the earliest.


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# Installing Lua Modules for use in LuaLaTeX

I've done all of the searching that I can, and I think I've read the right answer, but am too dumb to know that I read it.

I can't seem to figure out the magic tricks to get Lua modules, installed by luarocks or by any other means, to show up in LuaLaTeX. So here is a minimal:

%!TEX TS-program = lualatex
\documentclass{article}
\usepackage{luatexbase}
\usepackage{luacode}
\begin{document}

\directlua{package.path = "~/.luarocks/share/lua/5.1/?.lua;~/.luarocks/share/lua/5.1/?/init.lua;" .. package.path }

\directlua{require('pl')}
\directlua{require('htk')}

\directlua{tex.sprint(package.path)}

the standard approximation $\pi = \directlua{tex.sprint(math.pi)}$

\end{document}

In installed htk using luarocks with the command luarocks --local install HTK . The Penlight module was installed the same way, except with penlight. Penlight uses the "pl/submodulename" scheme and htk uses just "htk.lua" in the ~/.luarocks/share/lua/5.1  directory.

The log output is:

./luamoduleload.tex:9: LuaTeX error <\directlua >:1: module 'pl' not found:
stack traceback:
[C]: in function 'require'
<\directlua >:1: in main chunk.
l.9     \directlua{require('pl')}

The lua interpreter ran into a problem, so the
remainder of this lua chunk will be ignored.

stack traceback:
[C]: in function 'require'
<\directlua >:1: in main chunk.
l.10    \directlua{require('htk')}

The lua interpreter ran into a problem, so the
remainder of this lua chunk will be ignored.

Would more of the log file be helpful?

In truth, if i just knew where to instal the lua modules that I want to use, I'd be a happy camper, especially if I don't need to do any magic tricks to get stand-alone lua scripts to work outside of LuaLaTeX.

Please let me know if I can be any clearer or if I'm not reading the right fine manuals. :)

-Andrew

-
Have you tried the CLUAINPUTS path? texmf.cnf says: % Lua needs to look for binary lua libraries distributed with packages. CLUAINPUTS = .;$SELFAUTOLOC/lib/{$progname,$engine,}/lua//. Perhaps you can add the location where luarocks adds libraries to CLUAINPUTS. – Aditya Mar 27 '11 at 20:34 So now that line looks like CLUAINPUTS = .;$SELFAUTOLOC/lib/{$progname,$engine,}/lua//;~/.luarocks/share/lua/5.1// and it appears to have no effect... Cry ;) – Andrew Starks Mar 28 '11 at 4:55
How about adding the .so file in the current directory to see if dynamic loading works at all. – Aditya Mar 28 '11 at 13:12
It does. If I drop the file into the project's directory it seems to work. – Andrew Starks Mar 28 '11 at 13:27
Could you please tell a bit more about this "pl"? What is the exact name (the extension) of the module? Where do you want to store it? A small complete example with a simple \directlua{require('pl')} and the log-file would be useful too. E.g. I can't understand where the reference to oberdiek.luatex.kpse_module_loader comes from. – Ulrike Fischer Mar 28 '11 at 14:34

 \directlua{package.path = "~/.luarocks/share/lua/5.1/?.lua;~/.luarocks/share/lua/5.1/?/init.lua;" .. package.path }

looks suspicious. I think you need to have \noexpands before the tilde characters, otherwise the expansion of the ~ macro ends up as part of package.path.
\directlua{package.path = "\noexpand~/.luarocks/share/lua/5.1/?.lua;\noexpand~/.luarocks/share/lua/5.1/?/init.lua;" .. package.path }


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## Cycling through SYCL

We recently gave an overview of recent history in the technical computing hardware market. In it, we mention the energy at Intel right now. The weight of Intel is behind the SYCL standard through its new software approach, oneAPI. SYCL is a cross-platform API that targets heterogeneous hardware, similar to OpenCL and CUDA. The SYCL standard was first introduced by Codeplay and is now being managed by the Khronos group. It allows single-source compilation in C++ to target multiple devices on a system, rather than using C++ for the host and domain specific kernel languages for the device. Furthermore, SYCL is fully C++ 17 standards compliant. You don’t have any extensions to the language that would prevent any standards compliant …

## ArrayFire v3.5 Official Release

Today we are pleased to announce the release of ArrayFire v3.5, our open source library of parallel computing functions supporting CUDA, OpenCL, and CPU devices. This new version of ArrayFire improves features and performance for applications in machine learning, computer vision, signal processing, statistics, finance, and more. This release focuses on thread-safety, support for simple sparse-dense arithmetic operations, canny edge detector function, and a genetic algorithm example. A complete list of ArrayFire v3.5 updates and new features are found in the product Release Notes. Thread Safety ArrayFire now supports threading programming models. This is not intended to improve the performance since most of the parallelism is happening on the device, but it does allow you to use multiple devices in …

## ArrayFire v3.0 is here!

Today we are pleased to announce the release of ArrayFire v3.0. This new version features major changes to ArrayFire’s visualization library, a new CPU backend, and dense linear algebra for OpenCL devices. It also includes improvements across the board for ArrayFire’s OpenCL backend. A complete list ArrayFire v3.0 updates and new features can be found in the product Release Notes. With over 8 years of continuous development, the open source ArrayFire library is the top CUDA and OpenCL software library. ArrayFire supports CUDA-capable GPUs, OpenCL devices, and other accelerators. With its easy-to-use API, this hardware-neutral software library is designed for maximum speed without the hassle of writing time-consuming CUDA and OpenCL device code. With ArrayFire’s library functions, developers can maximize …

## Intel OpenCL performance: 3rd generation hardware

Brian Kloppenborg 1 Comment

Introduction With Intel CPUs making up nearly 80% of the CPU market and 66% of computers using integrated graphics one can easily argue that integrated graphics devices represent one of the greatest markets for GPU-accelerated computing. Here at ArrayFire, we have long recognized the potential of these devices and offer built-in support for Intel CPUs, GPUs, and AMD APUs in the OpenCL backend of our ArrayFire GPU computing library. Yet one common theme for debate in the office has been how the hardware performs on different operating systems with different drivers across hardware revisions. To answer these questions (and, perhaps, to win some intra-office geek cred) I decided to write a series of blog posts about Intel’s GPU OpenCL performance. In this first installment I will compare the performance …

## Using zero-copy buffers on integrated GPUs

Brian Kloppenborg 1 Comment

One of the most powerful aspects of parallel program on integrated GPUs is taking advantage of shared memory and caches. The best example of this is sharing common data between the CPU and GPU via. zero-copy buffers. This technique permits your program to avoid the O(N) cost of copying data to/from the GPU. This feature is particularly useful for applications that deal with real-time data streams, like video processing.

## Machine Learning with ArrayFire: Linear Classifiers

Linear classifiers perform classification based on the linear combinition of the component features. Some examples of Linear Classifiers include: Naive Bayes Classifier, Linear Discriminant Analysis, Logistic Regression and Perceptrons. ArrayFire’s easy to use API enables users to write such classifiers from scratch fairly easily. In this post, we show how you can map mathematical equations to ArrayFire code and implement them from scratch. Naive Bayes Classifier Perceptron Naive Bayes Classifier Naive bayes classifier is a probabilistic classifier that assumes all the features in a feature vector are independent of each other. This assumption simplifies the bayes rule to a simple multiplication of probabilities as show below. First we start with the simple Baye’s rule.  p(C_k | x) = \frac{p(C_k)}{p(x)} …

## OpenCL on Intel HD / Iris graphics on Linux

Under Windows and Mac the Intel GPU drivers include OpenCL support; however, on Linux OpenCL on Intel GPUs is implemented through an open source project called Beignet (pronnounced like “ben-yay”, a type of French pastry akin to a what we would call a “fritter” in English). Below I have written a step-by-step guide on how you can get Beignet running on an Ubuntu 14.10 system which has an Intel 3rd, 4th, or 5th generation Intel processor. Instructions for other variants of Linux will be similar, except for the commands to install the prerequisite packages. There are several little caveats which need to be discussed up front. Foremost, the Beignet project supports the following hardware: 3rd Generation Intel Core Processors Intel …

## Getting Started with the Intel Xeon Phi on Ubuntu 14.04/Linux Kernel 3.13.0

You may already know that the Intel MPSS (Manycore Platform Software Stack) officially only supports the RedHat and SUSE Linux distros. Using an enterprise distro might be very interesting if your company is running a large server environment or short on specialized people and you rely on the distro official support for more complicated tasks. Not all companies use enterprise Linux distributions. Ubuntu has a large share of the Linux distro market (if not the largest). A while back, I needed to setup a machine running Ubuntu 14.04 and MPSS 3.4.x and could not find any documentation running the newest versions of Ubuntu/Linux Kernel/MPSS. In this blog, I will try to document how to get the Intel Xeon Phi running …


text:
TCGAvisualize_EAbarplot Function Issue
1
0
Entering edit mode
pgreylin • 0
@pgreylin-9908
Last seen 5.9 years ago

Hello all,

I am trying to run the TCGAvisualize_EAbarplot function, but instead of producing the expected charts it produced nothing, specifically it runs the function, then after the last step says null device. Any ideas why it didn't run and just produced the null device message? Please help, my friend and I's project depends on it. thank you!

tcga • 907 views
0
Entering edit mode

This is the exact message it produces:

TCGAvisualize_EAbarplot(tf = rownames(ansEA$ResBP), + GOBPTab = ansEA$ResBP,
+                         GOCCTab = ansEA$ResCC, + GOMFTab = ansEA$ResMF,
+                         PathTab = ansEA\$ResPat,
+                         nBar = 20)
null device
1

0
Entering edit mode

Hi,

I am facing the same problem. Did you fix it? If yes, can you please share how?

0
Entering edit mode
eshelden ▴ 20
@eshelden-14132
Last seen 2.6 years ago

R did this to me too. Turns out, it created the graph in a PDF file in the current folder.


text:
# Downsapled and blurred image

I have an image with size $256*256$ then i get the Fourier transform of the image, If i only take the central region of the Fourier transform whose size is $64*64$ and get the inverse FT ,What is the effect on the output image$(64*64)$?

Down-sampling, Blurring, or both.

## 3 Answers

Have you fftshifted your results or not? The answers will be different:

If you fftshift your $256$ x $256$ 2-dimensional Fourier transform result, retain only the $64$ x $64$ central co-efficients, inverse fftshift, and then inverse fourier transform the result, the effect will be a blurring of your original image.

This is because you have culled our your images' high frequency components, hence, the effect will be a blurring.

If however you have not fftshifted, then you are doing the opposite, and will actually be doing a high pass filtering. (Your image will certainly not be blurred).

By the way, hard thresholding fourier co-efficients like this is certainly used, but will give you ringing effects. However, those effects might be negligible and/or removable based on your application.

• so i won't get aliased image, only blurring. do you have any idea how to remove this blurring by wiener filtering? – HforHesham Dec 13 '13 at 19:03
• @HforHesham If you only remove fourier co-efficients, you will not be causing aliasing, just a filtering operation. If you have a question about Weiner filtering the best thing to do would be to make a new question about it. – Tarin Ziyaee Dec 13 '13 at 19:15

By taking an FFT, retaining a sub-spectrum, then taking the inverse FFT you do actually perform two operations: low-pass filtering then downsampling. Reducing the size of tje spectrum can be seen as first setting to 0 everything outside the 64-by-64 area, then cropping out this 64-by-64 piece.

Let's detail a bit what happens:

1. First FFT: you compute $\hat{I} = \mathcal{F}(I)$, where $I$ is the original image (256-by-256).
2. Set to 0 the coefficients outside the 64-by-64 area: $LPF(\hat{I})(f) = \hat{I}(f)$ if $f$ is in the 64-by-64 region, 0 otherwise. This is a low-pass filter (LPF) since you are actually removing high frequencies.
3. Downsampling: crop out the frequency region of interest: $\hat{X} = LPF(\hat{I})_{64 \times 64}$.
4. Invert FFT: you obtain a smaller image $X = \mathcal{F}^{-1}(\hat{X})$.

This downsampling procedure is however not common because low-pass filters with smoother decays are preferred.

It is blurring. You eliminated the high frequency components.

Downsampling is implemented in spatial domain. You can try imagesc(abs(fftshift(fft2(img(1:2:end,1:2:end))))) and compared the result figure with the one you created with the method described in your question.

Wiener filter is a low pass filter, it is used to restored the image blurred with white noise (includes high frequency). You can try wiener2 in Matlab with your image.


text:
# Base

of a topological space $X$, base of a topology, basis of a topology, open base

A family $\mathfrak{B}$ of open subsets of $X$ such that each open subset $G \subseteq X$ is a union of subcollections $U \subseteq \mathfrak{B}$. The concept of a base is a fundamental concept in topology: in many problems concerned with open sets of some space it is sufficient to restrict the considerations to its base. A space can have many bases, the largest one of which is the family of all open sets. The minimum of the cardinalities of all bases is called the weight of the topological space $X$. In a space of weight $\tau$ there exists an everywhere-dense set of cardinality $\le \tau$. Spaces with a countable base are also referred to as spaces satisfying the second axiom of countability. The dual concept of a closed base, formed by the complements of the elements of a base, is used in compactification theory.

A local base of a space $X$ at a point $x \in X$ (a base of the point $x$) is a family $\mathfrak{B}(x)$ of open sets of $X$ with the following property: For any neighbourhood $O_x$ of $x$ it is possible to find an element $V \in \mathfrak{B}(x)$ such that $x \in V \subseteq O_x$. Spaces with a countable local base at every point are also referred to as spaces satisfying the first axiom of countability. A family $\mathfrak{B}$ of open sets in $X$ is a base if and only if it is a local base of each one of its points $x \in X$.

Let $\mathfrak{m}, \mathfrak{n}$ be cardinal numbers. A base $\mathfrak{B}$ of the space $X$ is called an $\mathfrak{m}$-point base if each point $x \in X$ belongs to at most $\mathfrak{m}$ elements of the family $\mathfrak{B}$; in particular, if $\mathfrak{m} = 1$, the base is called disjoint; if $\mathfrak{m}$ is finite, it is called bounded point finite; and if $\mathfrak{m} = \aleph_0$, it is called point countable.

A base of the space is called -local if each point has a neighbourhood intersecting with at most elements of the family ; in particular, if , the base is referred to as discrete; if is finite, it is called bounded locally finite; and if , it is called locally countable. A base is called an -point base (or an -local base) if it is a union of a set of cardinality of -point (-local) bases; examples are, for , -disjoint, -point finite, -discrete and -locally finite bases.

These concepts are used mainly in the criteria of metrizable spaces. Thus, a regular space with a countable base, or satisfying the first axiom of countability and with a point countable base, is metrizable; a regular space with a -discrete or -locally finite base is metrizable (the converse proposition is true in the former case only).

A base of the space is called uniform (-uniform) if for each point (each compact subset ) and for each one of the neighbourhoods () only a finite number of elements of the base contain (intersect with ) and at the same time intersect with the complement (). A space is metrizable if and only if it is paracompact with a uniform base (a Kolmogorov or -space with a -uniform base).

A base of the space is called regular if for each point and an arbitrary neighbourhood of it there exists a neighbourhood such that the set of all the elements of the base which intersect both with and is finite. An accessible or -space is metrizable if and only if it has a regular base.

A generalization of the concept of a base is the so-called -base (lattice base), which is a family of open sets in the space such that each non-empty open set in contains a non-empty set from , i.e. is dense in according to Hausdorff. All bases are -bases, but the converse is not true; thus, the set in the Stone–Čech compactification of the set of natural numbers in forms only a -base.

#### References

 [1] P.S. Aleksandrov, "Einführung in die Mengenlehre und die Theorie der reellen Funktionen" , Deutsch. Verlag Wissenschaft. (1956) (Translated from Russian) [2] P.S. [P.S. Uryson] Urysohn, , Works on topology and other fields of mathematics , 1–2 , Leningrad (1951) (In Russian) [3] P.S. Aleksandrov, B.A. Pasynkov, "An introduction to the theory of topological spaces and general dimension theory" , Moscow (1973) (In Russian) [4] A.V. Arkhangel'skii, V.I. Ponomarev, "Fundamentals of general topology: problems and exercises" , Reidel (1984) (Translated from Russian) [5] N. Bourbaki, "Elements of mathematics. General topology" , Addison-Wesley (1966) (Translated from French)


text:
the learning never stops!

# Instrument Procedures

I shot the approach shown below in 1992, back in the bad old USSR, during a shooting war between Armenia and Azerbaijan. The airport sits just north of Mount Ararat, the place where Noah's Ark is supposed to be, hence the very high MSAs. Oh yes, I was getting a check ride. The weather was above minimums, but not by much. The good news is that the approach was well designed and when we broke out there was runway in front of us. It doesn't always work out that way.

Over the years I've seen more than my share of instrument approaches that left me wondering what the approach designers were thinking, if air traffic control was trying to set me up, or if I had simply dreamt up my instrument rating and really didn't know what I was doing.

So here we go, a few examples that I hope will help you figure out when either the approach designers or air traffic controllers do not have your health and well being in mind. If you are here without an instrument rating, well you can figure the solution to that.

All of this is hard-learned technique. (No references provided.)

### Poor Approach Design — Example: Kingston, St. Vincent TVSV

If you ever get a chance to turn down a trip to E. T. Joshua Airport (TVSV) in St. Vincent, you should jump on that. I've been there once and wish I had never seen the place. It is a short runway (4,649') that sits at the edge of a mountain. The only instrument approach is an NDB that makes no sense whatsoever. Unfortunately, I never figured that out until after I flew there. Once I figured it out, I said no to further trips. But it remains as a good example of how a poorly designed approach can fool you. (It fooled me.)

#### First Look

Figure: TVSV NDB Rwy 07 planview, profile, and minimums, from Jeppesen Airway Manual, pg. TVSV 16-1, 2 Aug 13.

At first glance it looks okay. You head outbound from the IAF for 3-1/2 minutes and if you fly that at the same speed you make the turn, your offset should be good enough. I thought the GV was the perfect airplane for the job. Not only did we have a very low approach speed for the runway, that approach speed would give us the tightest possible turn radius. We flew this thing perfectly in our GV, 120 KCAS outbound, in the turn, and intercepting the final approach. But we didn't realize there was a problem.

Before you scroll down, look at the approach plate and try to see if you can spot the problem.

#### Planview

Figure: TVSV NDB Rwy 07 planview, with drawn triangle, from Jeppesen Airway Manual, pg. TVSV 16-1, 2 Aug 13.

If you make a standard rate turn your turn radius should be equal to:

$\mathrm{turn radius}=\frac{\mathrm{\left(nm/min\right)}}{3}$

In the case of a GV traveling at 120 KCAS (2 nm/min):

$\mathrm{turn radius}=\frac{2}{3}=\mathrm{2/3 nm}$

Which of course means your turn diameter is twice that, so the distance a in the drawing is 4/3 nm. The hypotenuse of that triangle is what we are interested in. How much distance do we have to make our descent from the MDA to the runway?

$\mathrm{distance to descend \left(c\right)}=\frac{a}{sin40}$

In the case of our GV doing 120 KCAS:

$\mathrm{distance to descend \left(c\right)}=\frac{2\mathrm{\left(2/3\right)}}{sin40}=2.07\mathrm{nm}$

That seems like enough, but is it?

#### Profile View

Figure: TVSV NDB Rwy 07 profile view, with drawn triangle, from Jeppesen Airway Manual, pg. TVSV 16-1, 2 Aug 13.

The approach appears to keep you at the MDA of 1,500' until making the 063° turn to the runway. Since we want a 3° glide path whenever possible, we need to begin the descent:

$\mathrm{distance to descend \left(b\right)}=\frac{a}{tan3}$

In the case of an MDA of 1,500':

$\mathrm{distance to descend \left(b\right)}=\frac{1500}{tan3}=\mathrm{28,622 feet}=4.71\mathrm{nm}$

No wonder we could never make the approach work! You cannot fly it with a hard MDA, it has to be flown as a visual. There is another problem with using this approach under instrument conditions. The visibility required is 3,200 meters, which comes to 1.7 nautical miles. You could find yourself shooting this approach with legal visibility with no possibility of landing in a controlled manner. (It would take an 8 degree descent rate from the MDA.)

#### Conclusion & Lesson Learned

This is not a circling approach but it is not a properly designed straight-in approach either. If you had to fly to this runway, I would insist on VMC conditions with at least 5 nm (5.75 sm) visibility.

The lesson here is that if the approach looks unusual, it might be. If you've never seen the design before you should consider a "chair flying" session where you imagine your airplane in each phase of the approach and consider the problems with safely getting to the next phase.

### Legal (but impractical) Approaches — Example: Amsterdam, Netherlands EHAM

Figure: EHAM ILS DME Rwy 27 Planview, Initial Clearance, from Jeppesen Airway Manual, page EHAM 11-7, 16 Sep 11

#### Approach as depicted

An ICAO Course Reversal must be entered in a specific corridor and if outside that corridor there may be a holding pattern that gives you the opportunity to reverse course twice. Here is an example I've flown recently.

The ATIS was one line long with winds, temperature, ceiling, visibility and instructions to expect the ILS to Runway 06. After we checked in with approach control we were cleared "ARTIP ILS DME RWY 27" and that was it.

Visual approaches are not allowed. What would you do?

#### Approach as intended (and expected)

Figure: EHAM ILS DME Rwy 27 Planview, Initial Clearance, from Jeppesen Airway Manual, page EHAM 11-7, 16 Sep 11

If you are not within a specific entry sector under ICAO Course Reversal rules, you must proceed to the holding pattern, execute one turn, then execute the course reversal as depicted. Yes, you are reversing course twice. More about this: Course Reversals.

The EHAM Airport Briefing pretty much spells this out in the case of lost communications, but it also says "navigation in the initial and intermediate approach segment is primarily based on radar vectors by ATC."

You must reduce to 220 Knots within 15 nm of SPL and be prepared to fly the entire procedure.

#### Approach as usually given

Figure: EHAM ILS DME Rwy 27 Planview, Initial Clearance, from Jeppesen Airway Manual, page EHAM 11-7, 16 Sep 11

While you must be prepared to fly the entire procedure do not delay other cockpit tasks thinking you have ten minutes to spare. You will very likely be given vectors or clearance "Direct Papa Alpha Mike, cleared the ILS DME Runway 27." You need to be ready to descend and configure quickly.

#### Conclusion & Lesson Learned

There is nothing wrong with this approach, other than it wastes time and congests the skies over an already congested airport. Knowing the proper procedure prepares you if the controller gets too busy or forgets the last minute change. But you need to be ready for that change as late as it might come.

The lesson here is to understand the operating rules of your environment; you should not fly any instrument procedure outside the United States until you understand ICAO course reversal rules. Once you've done that, you need to be able to predict "what is expected" versus "what is depicted."

### Circling (just about all of them) Approaches — Example: JFK, New York, KJFK

Just about every circling approach at minimums is impossible to complete with a stable approach on final. More about this: (U.S.) Circling Approach Visibility Minimums (Are Too Low!).

Ever wonder why you normally demonstrate your circling abilities in a simulator wired to KMEM LOC 27 circle Runway 18R or to KJFK Loc 4R circle Runway 31R? It is because they both have a runway perpendicular to the landing runway which is offset, giving you extra distance and improving your odds. The chances of having to really circle at either of these airports is quite low. But even with the help of another runway to help keep sight of the airport environment, you will still not be able to roll out at 500' above the runway on a 3° glidepath if the visibility is at minimums. For example . . .

#### The Normal Quick Look

Figure: KJFK ILS or LOC Rwy 4 planview and minimums, circling to Rwy 31R depicted, from Jeppesen Airway Manual, page KJFK 21-2, 22 Nov 13

The minimum Category D obstruction clearance radius for circling approaches under older TERPS criteria is 2.3 nautical miles. The minimum visibility allowed for a Category D approach under U.S. rules is 2 statute miles. Keep in mind that comes to only 1.74 nautical miles.

If you typically brief "I will commence my circle at 2.3 miles" (the green circle on the accompanying approach plate) you shouldn't be surprised if it takes you a bit longer to spot the runway. How much longer? It will be 2.3 - 1.74 = 0.56 nm. Yes, we just robbed you of more than a half a mile.

This might explain why this circling approach seems more difficult in actual practice than it does in theory. But so what, you say, you will just have to keep the airplane a half a mile tighter. Not a big deal, right?

#### Stable Approach Impossible

Figure: KJFK airport diagram extract, circling to Rwy 31R depicted, from Jeppesen Airway Manual, page KJFK 20-9, 14 Aug 15

If you want to land in the touchdown zone your aim point will be 750' from the threshold.

A 3° glide path requires 318' per nautical mile, which means you need to be wings level no later than 500 / 318 = 1.57 nm (9,553') from touchdown. The usual way to measure that is from the runway threshold, which makes it 9,553 - 750 = 8,803' from the threshold. Of course this isn't a normal runway layout, we have the added benefit of Runway 4R/22L to help keep sight of the airport. The edge of that runway is 1,900' from the threshold of Runway 31R, meaning we only have to roll out 8,803 - 1,900 = 6,903' from that runway.

A Category D aircraft flying at the maximum category speed will be doing 166 knots and will have a turn radius of 5,164' under no-wind conditions at sea level. That means the maximum distance from the airport environment (Runway 4R/22L) will be 12,068' and that comes to 2 nautical miles.

But the minimum visibility for the approach is 2.0 statute miles, which is only 1.74 nautical miles!

The numbers are just as bleak for a Category C aircraft which can maneuver in only 1.7 nautical miles but is only given 1.75 statute miles (1.49 nautical miles) visibility.

#### Conclusion & Lesson Learned

Circling at minimums is demanding even under ideal conditions. But for most conditions where a 90° or greater turn involved, it can be impossible. Newer TERPS criteria that bestow the "inverse C" on an approach give you more maneuvering area but do nothing about the minimum visibility criteria.

The lesson here is a simple one: don't circle at minimums. How low can you go? See (U.S.) Circling Approach Visibility Minimums (Are Too Low!) for a technique to determine realistic minimums.

### Mountainous Straight-In (many of them) Approach — Example: Eagle, Colorado KEGE

The first time I flew into Eagle, Colorado was in an Air Force GIII that did not have an EGPWS, GPS, or just about any other of the niceties we take for granted these days. We briefed that if we didn't see the runway at the point now called POWRS, we would go someplace else. If we did spot the runway, we would fly down the valley.

Years later, in a high tech CL-604, I was in the right seat and watched with dismay as the pilot (also the chief pilot and my boss) flew right down the centerline of the approach. Our first EGPWS warning was over the 9,036' peak. Then we got a few more as his VVI exceeded 1,000 fpm one or two times. After that, I convinced him and our other pilots that the valley was the way to go. (We also got very familiar with an excellent airport just a few minutes to the west, Rifle, Colorado.)

There is much to be learned about how to diagram an approach with this example . . .

#### A Lot of Mountains

Figure: KEGE RNAV(GPS)-D, from Jeppesen Airway Manual, page KEGE 12-1, 22 May 15

Looking at the approach plate there are obviously a lot of mountains but one can be forgiven for thinking this approach meets TERPS criteria so it can't be too bad. Right?

The descent rates from POWRS to AWACC and from AWACC to NEPRY are not too bad, certainly within all guidance. But what about the rate from AWACC at 11,100' down to the runway at 6,547'? The distance between those two points is 7.5 nm so the descent rate is (11,100 - 6,547) / 7.5 = 607' / nm. That's almost double what you need for a 3° glide path.

#### A look at the elevation profile

Figure: KEGE RNAV(GPS)-D elevation profile, flown as published, from Google Earth, 13 Sep 2015

You can input the relevant points into Google Earth and come up with an elevation profile that shows the magnitude of the problem. As you can see in the chart below map, the highest obstacle comes about a third of the way, shortening your distance to descend. There is a perfectly good valley to the north, why not use it?

#### A saner approach

Figure: KEGE RNAV(GPS)-D elevation profile, flown along valley, from Google Earth, 13 Sep 2015

By approximating the ground track of the valley you can significantly reduce your obstacle clearance problem. At this point, your descent gradient starts at 12,000' and ends at the runway elevation of 6,547', which is 12.5 as the crow flies but significantly longer via the ground track. The Google Earth chart says "16.1 mi," but it is up to you to specify nautical or statute miles. I've specified nautical, which means our new descent rate is (12,000 - 6,547) / 16.1 = 339' / nm. That is quite a bit more manageable than the straight down the middle requirement.

#### Conclusion & Lesson Learned

The example approach cannot be flown along course centerline without exercising the EGPWS, scaring the passengers, and completely ignoring stabilized approach criteria. By raising your approach minimums to require a visual sighting of the runway at 12,000', you can fly along the nearby valley and eliminate all the previous concerns.

The lesson learned is this: just because an approach passes TERPS criteria doesn't mean it is safe!

### Instrument Approach Math

You don't need an engineering degree to diagram an approach as shown in these examples. If you can plug and chug with a scientific calculator, all you need are some formulas. Here are some handy ones. (Click on the figure to download a PDF version.)

### References

Jeppesen Airway Manuals

Google Earth

Top


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# Win That Prize

Logic Level 2

You are on a game show and is playing in the jackpot round.

In jackpot round, the host says you have 3 doors to choose from, one of these doors contain the prize, a new car, and the other 2 contain rocks.

You are given 2 chances to open 2 doors. You nervously open door 2, and it turns out to contain rocks, you are given 1 chance left, but the round gets more intense as the host adds 5 more doors and changes the room where the prize is, but the host adds 2 consolation prizes, $5,000 and$10,000, into random rooms, and gives you an additional chance.

What would be the percentage that you will win one of three prizes that lies in one of the rooms?

×


text:
# Fast training for large-scale one-versus-all linear classifiers using tree-structured initialization

H. Fang, M. Cheng, C.-J. Hsieh, M. P. Friedlander
SIAM International Conference on Data Mining, 2019

## Abstract

We consider the problem of training one-versus-all (OVA) linear classifiers for multiclass or multilabel classification when the number of labels is large. A naive extension of OVA to this problem, even with hundreds of cores, usually requires hours for training on large real world datasets. We propose a novel algorithm called OVA-Primal++ that speeds up the training of OVA by using a tree-structured training order, where each classifier is trained using its parent’s classifier as initialization. OVA-Primal++ is both theoretically and empirically faster than the naive OVA algorithm, and yet still enjoys the same highly parallelizability and small memory footprint. Extensive experiments on multiclass and multilabel classification datasets validate the effectiveness of our method.

## BiBTeX

@InProceedings{2019HuangFast,
author =       {H. Fang and M. Cheng and C.-J. Hsieh and M. P. Friedlander},
title =        {Fast training for large-scale one-versus-all linear
classifiers using tree-structured initialization},
year =         2019,
booktitle =    {Proc. SIAM Inter. Conf. Data Mining (SDM19)}
}


text:
This is a continuation of my last post on this subject. As Gordon remarked in one of his posts, you may need to refresh your browser if some of the embedded gifs do not appear as they should.

### Dualities and isomorphisms of classical groups

Four of the five families of classical generalised quadrangles come in dual pairs: (i) $Q(4,q)$ and $W(3,q)$; (ii) $H(3,q^2)$ and $Q^-(5,q)$. Both can be demonstrated by the Klein correspondence. Recall from the last post that the Klein correspondence maps a line of $\mathsf{PG}(3,q)$ represented as the row space of

$M_{u,v}:=\begin{bmatrix}u_0&u_1&u_2&u_3\\v_0&v_1&v_2&v_3\end{bmatrix}$

to the point $(p_{01}:p_{02}:p_{03}:p_{12}:p_{31}:p_{23})$ where

$p_{ij}=\begin{vmatrix}u_i&u_j\\v_i&v_j\end{vmatrix}$.

Now consider the symplectic generalised quadrangle $W(3,q)$ defined by the bilinear alternating form

$B(x,y):=x_0y_1 - x_1y_0 + x_2y_3 - x_3y_2.$

A totally isotropic line $M_{u,v}$ must then satisfy

$0 = B(u,v) = p_{01} + p_{23}$.

Therefore, the lines of $W(3,q)$ are mapped to points of $Q^+(5,q)$ lying in the hyperplane $\pi:X_0+X_5=0$. Now the quadratic form defining $Q^+(5,q)$ is $Q(x)=x_0x_5+x_1x_4+x_2x_3$ and $\pi$ is the tangent hyperplane at the projective point $(1:0:0:0:0:1)$, which does not lie in the quadric. Hence the hyperplane $\pi$ is non-degenerate and so we see that $W(3,q)$ maps to points of $Q(4,q)$. That this mapping is bijective follows from noting that the number of lines of $W(3,q)$ is equal to the number of points of $Q(4,q)$ (namely, $(q+1)(q^2+1)$).

Hence $\mathsf{P\Gamma Sp}(4,q)\cong \mathsf{P\Gamma O}(5,q)$.

Now we will consider a more difficult situation which reveals that the generalised quadrangles $H(3,q^2)$ and $Q^-(5,q)$ are also formally dual to one another. We could take the approach above, but instead, we will use a cunning way (told to me by Tim Penttila) to produce an Hermitian form on $\mathsf{PG}(3,q^2)$. First we consider $Q^-(5,q)$ as the subspaces of $Q^+(5,q^2)$ that are fixed by a particular involution known as a Baer involution. In order for this to look nice, I will change the form that defines $Q^+(5,q^2)$:

$Q(x) = x_0x_5 + x_1x_4 + f(x_2,x_3)$

where $f$ is an irreducible homogeneous quadratic in two variables over $GF(q)$. So when we restrict this form to the points with entries only in $GF(q)$, we have a form of minus type; a polar space isomorphic to $Q^-(5,q)$. The Baer involution $\tau$ is simply the map which raises each coordinate to its q-th power, and so our $Q^-(5,q)$ is exactly the fixed singular elements of this involution. Moreover, $\tau$ switches the latins and greeks. To see this, note that the latins and greeks form a block-system for the full orthogonal group. Now consider the following singular plane:

$\pi = \{ (a:b: c: d: 0: 0) \mid (a,b,c,d) \ne 0, f(c,d)=0\}$.

Then applying $\tau$ to each point comprising $\pi$ will give us a different plane which intersects $\pi$ in the line

$L = \{(a:b:0:0:0:0)\mid (a,b)\ne 0\}$.

Therefore, $\pi^\tau$ is in a different class to $\pi$. Furthermore, if we take a line of $Q^-(5,q)$, it is fixed by $\tau$ and the two singular planes incident with this line are interchanged by $\tau$. If we relate what $\tau$ is doing back to $\mathsf{PG}(3,q^2)$, we see that it is inducing a polarity of $\mathsf{PG}(3,q^2)$ having $(q^2+1)(q^3+1)$ absolute lines. By the classification of polarities, this is a unitary polarity. So we see that $Q^-(5,q)$ corresponds to the hermitian surface $H(3,q^2)$. Hence $P\Gamma O^-(5,q)\cong P\Gamma U(4,q)$.

An ovoid of $Q^+(5,q)$ is a set of $q^2+1$ points such that no two are collinear in a line of $Q^+(5,q)$. So under the Klein correspondence, we have a set of $q^2+1$ lines of $\mathsf{PG}(3,q)$ such that no two are concurrent in a point of $\mathsf{PG}(3,q)$ (and equivalently, they do not span a plane of $\mathsf{PG}(3,q)$). So an ovoid of $Q^+(5,q)$ is equivalent to a spread of $\mathsf{PG}(3,q)$.

We also mention here a beautiful observation of Shult and Thas:

any 1-system of $Q^+(5,q)$, q odd, is equivalent to the classical 1-system (arising from the Klein image of an elliptic quadric).

A 1-system of $Q^+(5,q)$ is a set of $q^2+1$ lines of $Q^+(5,q)$ that are pairwise opposite: there is no singular plane on one of the lines meeting the other in a point. Under the Klein correspondence, we end up with a set of point/plane incident pairs, such that no two will intersect in a line that passes through their given points. Such a configuration defines an ovoid of $\mathsf{PG}(3,q)$: a set of $q^2+1$ points with no 3 collinear. By a result attributed to the independent work of Barlotti and Panella (1955), this ovoid is projectively equivalent to en elliptic quadric. So the 1-system is the Klein image of an elliptic quadric.

### Isomorphisms of simple gorups

The Klein Correspondence yields a simple way to explain why $A_8 \cong \mathsf{PSL}(4,2)$ and $\mathsf{PSL}(3,2)\cong \mathsf{PSL}(2,7)$. First, consider an eight dimensional vector space over $\mathbb{F}_2$ equipped with the quadratic form

$Q(x):=x_0x_5+x_1x_4+x_2x_3+x_6x_7$.

Think of the vectors as eight-tuples, so that we have a natural action of the symmetric group $S_8$ on the vectors: simply by permutation of coordinates. Then $S_8$ fixes $(1,1,1,1,1,1,1,1)$ and the subspace $W$ of vectors whose coordinates sum to $0$. Then $S_8$ induces an action on the quotient six-dimensional vector space

$W/\langle(1,1,1,1,1,1,1,1)\rangle$.

We should also observe that $S_8$ preserves the quadratic form $Q$, and the inherited quadratic form on $W/\langle(1,1,1,1,1,1,1,1)\rangle$. So $S_8$ has a representation into $\mathsf{O}^+(6,2)$, and then by orders, we have an isomorphism. The Klein correspondence then yields $A_8 \cong \mathsf{PSL}(4,2)$.

Now for $\mathsf{PSL}(3,2)\cong \mathsf{PSL}(2,7)$, we observe that we can use the same setting since $\mathsf{PSL}(3,2)$ can be realised inside the stabiliser of a point or a plane of $\mathsf{PG}(3,2)$. Take the following singular plane of $Q^+(5,2)$:

$\pi:=\{(a+b+c,a+b,a+c,a,b+c,b,c,0)\mid 0\ne(a,b,c)\in\mathbb{F}_2^3\}$

Now $S_8$ does not fix this plane by acting on coordinates, but we do have a subgroup of $S_8$, namely $\mathsf{PSL}(2,7)$ that acts 2-transitively on these coordinates if we think of its natural action on the projective line labelled in the order $(0,1), (1,0),(1,1), (1,3), (1,2), (1,6), (1,4), (1,5)$. Then by comparing orders, we see that $\mathsf{PSL}(2,7)$ is isomorphic to $\mathsf{PSL}(3,2)$.


text:
Someone today asked me: why graphs? “Why are graphs so widely used in computer science, and in related fields such as software engineering?” Here's my glib response:

The short answer is that graphs can be used to reason symbolically about any kind of pairwise relationship between any kind of entity, and that we like to think about pairwise relationships because unary relationships aren't powerful enough and $$k$$-way relationships for $$k\gt 2$$ add extra complication without adding any real power.

So already in 1736 Euler was using graphs to model transportation connections between pairs of places (the bridges of Königsberg; more modernly this idea shows up in airline ticket planning, where the vertices represent airports and the edges represent flights, or in mapquest-like route planning, where the vertices represent road intersections and the edges represent intersection-free segments of roads). We have graphs representing people and social networks connecting them (online friendships, sexual contacts, parenthood, coauthorship, etc). We have graphs representing subroutines in a computer program and caller-callee relations between them. We have graphs representing web pages and html links between them. We have graphs representing proteins in your body and the chemical interactions they participate in. Etc etc.

Graphs are powerful because the same kinds of problems and algorithms turn out to be important in many of these different applications. So by taking away the application-specific features of all of those different problems and turning them into something as abstract as a graph, we only have to solve these problems once instead of repeatedly solving the same problems in different disguises.

None:
2009-02-12T08:18:53Z
that's the explanation I give as well. Very well written.
11011110:
2009-02-12T15:11:43Z
Thanks!
None: k-relationships
2009-02-12T08:33:46Z
$$k$$-way relationships for $$k\gt 2$$ add extra complication without adding any real power
Is there any way of understanding why this is the case, and in what way? What would be the formally equivalent statement?
11011110: Re: k-relationships
2009-02-12T15:10:39Z
I meant something mildly technical by this: one can define k-ary hypergraphs for any k, of course, but they're just the same things as bipartite graphs. So the hypergraph point of view doesn't really get you away from graphs.
None: Re: k-relationships
2009-02-13T00:43:05Z
Yes, but the language of hypergraphs can be very convenient. Just as an example (but many others could be provided) see Ronald Fagin, Degrees of acyclicity for hypergraphs and relational database schemes JACM 1983, and G. Gallo et al. Directed hypergraphs and applications Discrete Applied Mathematics 1993
leonardo_m:
2009-02-12T11:21:22Z
11011110:
2009-02-12T15:17:44Z
Yeah, I see that it has only a brief and easily missed line about this before going into the more dry technical material. I'll have to see what I can dig up in the way of reliable sources that would let me add it there; I don't think it should be too difficult.
None: Agreed!
2009-02-12T14:36:32Z

Wow, I just did my first graph lecture for Algs and Data Structures Tuesday, and you said very nicely what I HOPE the students took away from the first part of the lecture -- where I ask them to think of examples of data that can be modelled as graphs. But you said it very clearly.

Michael Mitzenmacher

11011110: Re: Agreed!
2009-02-12T15:18:34Z
Thanks! That's where I usually put it in my algorithms classes, too.
2009-02-12T22:48:50Z

This is great! I hope Graph theory community also sees this comment (and perhaps it helps for grant proposal as well)..

Also, I hope TCS realizes how important Graph Theory is. (Seriously! Graph Theory creates a lot of "exciting" and "deep" theories in the last two decades or so! If one only knows Graph Theory 30 years, he or she may object this point, though..)

Thanks.

Ken-ichi Kawarabayashi

None: Energy Landscape?
2009-02-13T07:25:00Z

Hi,

I am a biologist, and I encounter the following graph-theoretic(I think) problem:

We have a big graph with billions of of nodes, and for each node we can assign a real value-"height" to it. Then we want to obtain a global view of this landscape and try to study some property of this graph.

Does this kind of situation sounds familiar to you?

Thank a lot!!!


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## Dialect in Nice

Visible to anyone in the world
Edited by Patrick Andrews, Saturday, 3 Aug 2019, 11:25

We recently visited Nice and were struck by many signs written in the local dialect.  As someone who understands French quite well, I was struck how different it was from standard French.  For example, the badge of the football team has “Despi 1904”, which I assume is the equivalent of “depuis 1904”.  The picture below shows a bilingual sign with French at the top and the dialect underneath.

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# Handling timing+physics properly

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## Recommended Posts

Hi guys, I've been starting a small game, kind of old 2D space shooter. My problem is that I'm currently using a 15ms delay between each update of the game state, but I'd like to move to a more precise timing. So far, I have a spaceship that can move (method update), this method is called every 15ms and checks if the user is using throttle and other inputs like turning left, update the position of the spaceship and go on. I'd like to know how to change my code to use the: speed = distance/time equation. I can compute the time between each frame, but what about the speed? How could I know it? Currently 3 forces are applied to the spaceship: gravity, frictions and a force in the direction of the ship if the user is using throttle. This give me a resulting force, but how can I translate this into a "speed"? I don't know if I'm explaining well, so anyway I uploaded my code (C# MDX, only the starship class will be useful http://xboxemul.free.fr/dotnet/spaceship.zip) so maybe one of you can take a look. Cheers, Tony

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Simply stated, a = F/m, when means acceleration equals force divided by mass.

Acceleration is the derivative of velocity, so to find the velocity, you integrate the acceleration. Now, it probably isn't possible to integrate the acceleration analytically in your case, so you must do it numerically. Numerical integration isn't exact, but it is usually close enough. There are several numerical integration algorithms. The simplest (and least accurate) is called Euler integration. Every frame, you simply do this: v = v + at, where v is the current velocity, a is the current acceleration, and t is the amount of time since the last update. v and a can be scalars or vectors.

Since velocity is the derivative of distance, you can do the same thing to find the position every frame: s = s + vt, where s is the current position, v is the current velocity, and t is the amount of time since the last update. Again, s and v can be scalars or vectors.

If you have problems with the accuracy, you might consider using a better (and more complicated) integration algorithm such as Verlet or Runge-Kutta.

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I recommend not letting the passage of real time affect your game logic. If your game is being updated at close enough to every 15 ms, that is good enough. Old style games did not use variable update times, and even new games usually don't.

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Quote:
 Original post by JohnBolton

Great answer. I'm gonna give this a try.

Vorpy> Well 15ms is a fairly good timing of course, but I'm not happy with it. The game still computes 5/6 frames between those timings, and well if you compile the code, you can see it's a little bit choppy when you move fast enough. I'm not happy with it.

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Quote:
 Original post by tonyp7I'm not happy with it.

15ms resolution for Physics calculations should be fine.
Perhaps the choppiness you see, is not the Physics itself, but an issue with the Graphics and rendering updates?

Generally, while you keep your physics calcs at a fixed interval, the graphics will occure at their own different rate as well; either locked to the monitor refresh rate, or operating as fast as possible

If those rates and your physics rate happen to have a 'beat frequency' you can get some very ugly choppiness problems. Happened to me once...

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Here's the code that's currently handling the timing:
            // Update time elapsed since last frame            m_timeElapsed = (float)(m_currentTime - m_lastTime) / (float)m_ticksPerSecond;            m_runningTime += m_timeElapsed;            //if 15ms is reached, game is ready for a new turn            if (game_ticks_counter >= 0.015)            {                game_ready = true;                game_ticks_counter = 0;            }            else            {                game_ready = false;                game_ticks_counter += m_timeElapsed;            }

Instead of game_ticks_counter = 0, you should use game_ticks_counter -= .015. That will get you closer to your target number of updates per second.

You can also smooth out the graphics without affecting the game logic/physics. I think your current code always draws the same thing between updates. There's no point in drawing the scene multiple times if it looks the same each time. You're getting several redraws per second because your game isn't doing much work right now. You also might be getting some ugly screen tearing if your redraws aren't v-synched, and no update method can fix that (besides v-synching).

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Quote:
 Original post by VorpyInstead of game_ticks_counter = 0, you should use game_ticks_counter -= .015. That will get you closer to your target number of updates per second.

Good point. I don't know why I haven't thought of it.

Quote:
 Original post by VorpyYou can also smooth out the graphics without affecting the game logic/physics. I think your current code always draws the same thing between updates. There's no point in drawing the scene multiple times if it looks the same each time.

Well that's the problem. Each 15ms, the scene moves, then several draws are being done but there are indeed the same thing until another 15ms is reached.
Between those 15ms, if a vertex moves from point A to point B, it would be of course nice to use the time between the 15ms to smooth the movement, instead of brutally moving from A to B, but I don't know how to achieve that.

You are really of a great help. I wouldn't expect some people to actually take the time to explain me those basic things (I have read some timing articles though...). Thanks a lot.

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you want to look at linear interpolation. A big word, but basically, if you want to move from A to B, with a fraction u,

P(u) = A + u * (B - A), 0 <= u <= 1.

The trick is to calculate 'u'.

Say A was computed at frame 'ta', and B at frame 'tb', and you are at time 't', t in range [ta, tb].

u = (t - ta) / (tb - ta);

however... Doing it like that, you would require to run the rendering 'behind' a little bit, so that when you render, you always have a 'B' position for your object, else you have to extrapolate.

u = ((t - frame_time) - ta) / (tb - ta);
u = max(min(u, 1.0f), 0.0f);

You can use other types of interpolation, such as cubic, or quadratic. But linear should do fine.

That's kind of how interpolation works for some network games (frame_time being replaced by the send_rate), and how player positions are represented on client machines. Running behind also introduces 'lag', which depends on your target framerate.

You can also have less restriction, and use a exponential decay, such as this, to smooth out the transition between the latest calculated physics position and your last rendered position. That also does not require you to cache your physics positions from frame to frame.

RenderPos += (PhysicsPos - RenderPos) * 0.5f;

So every time it is called, the 'gap' between physics and reder positions will be reduced by 2. This is not frame rate independent though. This would be the easiest thing to do to smooth out the rendered position.

It's not very hard to make it framerate independent as well.


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# Alternately Multiplying and Dividing Within Pascal's Triangle

I recently noticed that, if you take the $$(2n)$$th row of Pascal's triangle, and alternately multiply and divide the numbers that crop up, almost everything cancels. For example, the 4th row (1 4 6 4 1) becomes $$1/4 \times 6/4 \times 1$$, which simplifies to $$3/8$$. I performed the same procedure for some later rows, and got $$5/16$$, $$35/128$$, $$63/256$$, $$(3 \times 7 \times 11)/(2^{10})$$, and $$(3 \times 11 \times 13)/(2^{11})$$.

All of the denominators seem to be powers of 2, which I don't know how to prove. Also, the fractions seem to me to be (very) slowly approaching 0, which I also don't know how to prove (if it's even true). Any assistance with either question would be greatly appreciated- I'm not very well versed in methods of proof.

• 4th row the way mathematicians count them.
– user645636
Jun 12, 2019 at 23:14
• Okay. I'll switch that, then. Jun 12, 2019 at 23:15

$$\frac{\displaystyle \binom{2n}{0}}{\displaystyle \binom{2n}{1}}\frac{\displaystyle \binom{2n}{2}}{\displaystyle \binom{2n}{3}}\cdots\frac{\displaystyle\binom{2n}{2n-2}}{\displaystyle \binom{2n}{2n-1}} = \frac{1}{2n}\frac{3}{2n-2}\cdots\frac{2n-1}{2}$$
$$\frac{1\cdot2\cdot3\cdot4\cdots(2n-1)(2n)}{2^2\cdot 4^2\cdot6^2\cdots (2n)^2} =\frac{(2n)!}{\big[2^n n!\big]^2}=\frac{1}{4^n}\binom{2n}{n}.$$
• And $\frac{1}{4^n}\binom{2n}{n}$ approaches 0 as n approaches infinity? Jun 12, 2019 at 23:21
• Yes, see central binomial coefficient for estimates and elementary inequalities. In particular, $$\frac{1}{4^n}\binom{2n}{n}\sim \frac{1}{\sqrt{\pi n}}.$$ Not sure off the top of my head of an elementary proof of a sufficient inequality. Jun 12, 2019 at 23:27


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# zbMATH — the first resource for mathematics

Invariant of the hypergeometric group associated to the quantum cohomology of the projective space. (English) Zbl 1073.32014
Summary: We present a simple method to calculate the Stokes matrix for the quantum cohomology of the projective spaces $$\mathbb C \mathbb P^{k-1}$$ in terms of certain hypergeometric group. We present also an algebraic variety whose fibre integrals are solutions to the given hypergeometric equation.

##### MSC:
 32S40 Monodromy; relations with differential equations and $$D$$-modules (complex-analytic aspects) 14N35 Gromov-Witten invariants, quantum cohomology, Gopakumar-Vafa invariants, Donaldson-Thomas invariants (algebro-geometric aspects)
Full Text:
##### References:
 [1] Balser, W.; Jurkat, W.B.; Lutz, D.A.; Balser, W.; Jurkat, W.B.; Lutz, D.A., On the reduction of connection problems for differential equations with an irregular singular point to ones with only regular singularities, I, II, SIAM J. math. anal, SIAM J. math. anal, 19, 2, 398-443, (1988) · Zbl 0639.34034 [2] Barannikov, S., Semi-infinite Hodge structures and mirror symmetry for projective spaces · Zbl 1304.58003 [3] Beilinson, A.A., Coherent sheaves on cpn and problems of linear algebra, Funct. anal. appl, 13, 2, 68-69, (1978) · Zbl 0402.14006 [4] Beukers, F.; Heckman, G., Monodromy for the hypergeometric function _nfn−1, Invent. math, 95, 325-354, (1989) · Zbl 0663.30044 [5] Dubrovin, B., Painlevé transcendents in two dimensional topological field theory, (), 287-412 · Zbl 1026.34095 [6] Dubrovin, B., Geometry and analytic theory of Frobenius manifolds, (), 315-326 · Zbl 0916.32018 [7] Givental’, A., Equivariant Gromov-Witten invariants, Intern. math. res. notices, 13, 613-663, (1996) · Zbl 0881.55006 [8] Golyshev, V.V., Riemann-Roch variations, Izv. math, 65, 5, 853-887, (2001) · Zbl 1068.14047 [9] A.L. Gorodentsev, S.A. Kuleshov, Helix theory, MPIM preprint series 97, 2001 [10] Greuel, G.-M., Der Gauss-Manin zusammenhang isolierter singularitäten von vollständigen durchschnitten, Math. ann, 214, 3, 235-266, (1975) · Zbl 0285.14002 [11] Guzzetti, D., Stokes matrices and monodromy of the quantum cohomology of projective spaces, Comm. math. phys, 207, 2, 341-383, (1999) · Zbl 0976.53094 [12] Kontsevich, M., Homological algebra of mirror symmetry, (), 120-139 · Zbl 0846.53021 [13] Levelt, A.H.M., Hypergeometric functions, Indag. math, 23, 361-403, (1961) [14] Okubo, K.; Okubo, K., Connection problems for systems of linear differential equations, (), 238-248, (1987), Tokyo Metropolitan University · Zbl 0633.15013 [15] S. Tanabé, Transformée de Mellin des intégrales - fibres associées à l’intersection complète non-dégénérée, Math. AG/0405399 [16] Zaslow, E., Solitons and helices: the search for a math-phys bridge, Comm. math. phys, 175, 337-375, (1996) · Zbl 0851.53055
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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## Lakshya Education MCQs

Question: Two - thirds of a consignment was sold at a profit of 6% and the rest at a loss of 3%. If however there was an overall profit of Rs. 540, the value of consignment was -
Options:
 A. Rs. 15000 B. Rs. 16000 C. Rs. 18000 D. None of these

Earn Reward Points by submitting Detailed Explaination for this Question

## More Questions on This Topic :

Question 1. Krishna purchased a number of articles at Rs. 10 for each and the same number for Rs. 14 each. He mixed them together and sold them for Rs. 13 each. Then his gain or loss percent is = ?
1.    Loss $$8\frac{1}{3}\%$$
2.    Gain $$8\frac{2}{3}\%$$
3.    Loss $$8\frac{2}{3}\%$$
4.    Gain $$8\frac{1}{3}\%$$
Question 2. A man purchases some oranges at the rate of 3 for Rs. 40 and the same quantity at 5 for Rs. 60. If he sells all the oranges at the rate of 3 for Rs. 50, find his gain or loss percent ( to the nearest integer ) = ?
1.    34% loss
2.    31% profit
3.    31% loss
4.    32% profit
Question 3. Albert buys 4 horses and 9 cows for Rs. 13400. If he sells the horses at 10% profit and the cows at 20% profit, then he earns a total profit of Rs. 1880. The cost of a horse is -
1.    Rs. 1000
2.    Rs. 2000
3.    Rs. 2500
4.    Rs. 3000
Question 4. A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity (in kg) sold at 18% profit is.
1.    400
2.    560
3.    600
4.    640
Question 5. A man bought oranges at the rate of 8 for Rs. 34 and sold them at the rate of 12 for Rs. 57. How many oranges should be sold to earn a net profit of Rs. 45 ?
1.    90
2.    100
3.    135
4.    150


text:
Question

# The ionic product of water changes when:A. An acid is added to itB. A base is added to itC. Either a base or an acid is added to it.D. The temperature has changed.

Hint: The ionic product of water is the product of concentrations of ${{\text{H}}^{\text{ + }}}$ and ${\text{O}}{{\text{H}}^ - }$ ions in water at a particular temperature. It is denoted by ${{\text{K}}_{\text{w}}}$

Step-by-Step Explanation: Water is a weak electrolyte and ionizes only to a small extent. ‘The ionic product of the water is the product of concentrations of ions in which it is dissociated at a particular temperature’. The water is dissociated in the following manner-
$\Rightarrow {{\text{H}}_2}{\text{O}} \rightleftharpoons {{\text{H}}^{\text{ + }}} + {\text{O}}{{\text{H}}^ - }$
Then ${{\text{K}}_{\text{w}}}$=$\dfrac{{\left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{{\text{H}}_2}{\text{O}}} \right]}}$
$\Rightarrow {{\text{K}}_{\text{w}}}\left[ {{{\text{H}}_2}{\text{O}}} \right] = \left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]$
Because $\left[ {{{\text{H}}_2}{\text{O}}} \right] = 1$ then,
$\Rightarrow {{\text{K}}_{\text{w}}} = \left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]$
Ionic products are temperature dependent because the Ionic product is directly proportional to the temperature.
By Le-Chatelier’s principle, as the temperature increases, the position of equilibrium shifts to the right of the equation to minimize the effect of increased temperature.
This means with the increase in temperature the concentration of$\left[ {{{\text{H}}^{\text{ + }}}} \right]$ and$\left[ {{\text{O}}{{\text{H}}^ - }} \right]$ ions will increase which will result in increased value of${{\text{K}}_{\text{w}}}$ as forward reaction is favoured.
This means that the ionic product depends only on temperature and is unaffected by any other changes.

Hence correct answer is ‘D’.

Note: Since concentration of hydrogen ion increases when temperature is increased and we know that pH is inversely proportional to concentration of hydrogen ion then the pH of water will decrease with increase in temperature. Also,
When,$\left[ {{{\text{H}}^{\text{ + }}}} \right] = \left[ {{\text{O}}{{\text{H}}^ - }} \right]$ then the solution is neutral [pure water is neutral].
When $\left[ {{{\text{H}}^{\text{ + }}}} \right] > \left[ {{\text{O}}{{\text{H}}^ - }} \right]$ then the solution becomes acidic.
When $\left[ {{{\text{H}}^{\text{ + }}}} \right] < \left[ {{\text{O}}{{\text{H}}^ - }} \right]$ then the solution becomes basic.