I want to normalize the vectors, yet I could not understand how to do such a basic operation. There does not exist a direct method like Vector_3::normalize().
I call squared_length() and then CGAL::sqr_root() to find the vector length. Then when I want to divide my vector to its length, compiler does not permit since CGAL types are not compatible. Where am I wrong?
I'm expanding here the comment by #MarcGlisse.
There are two scenarios - you want exact calculations or you don't. In the first case you can use the Exact_predicates_exact_constructions_kernel_with_sqrt and the normalized vector will be exact.
In the second case you can use many kernels, but the normalized vector will probably have the length, slightly different from 1. Also some kernels don't have the function CGAL::sqrt at all - however you can always convert squared distances to the double type, using the function CGAL::to_double. Another way is to use a function CGAL::approximate_sqrt, which automatically does this conversion if it's necessary. These conversions make calculations not exact. Please see the example below:
#include <iostream>
#include <CGAL/Exact_predicates_exact_constructions_kernel_with_sqrt.h>
#include <CGAL/Exact_predicates_exact_constructions_kernel.h>
#include <CGAL/Simple_cartesian.h>
using KernelExactWithSqrt = CGAL::Exact_predicates_exact_constructions_kernel_with_sqrt;
using KernelExact = CGAL::Exact_predicates_exact_constructions_kernel;
using KernelInexact = CGAL::Simple_cartesian<double>;
template <typename T>
auto normalize(T const& V)
{
auto const slen = V.squared_length();
auto const d = CGAL::approximate_sqrt(slen);
return V / d;
}
template <typename T>
auto check()
{
typename T::Point_2 const p{0, 0}, q{1, 1};
auto const n = normalize(typename T::Vector_2{p, q});
return n.squared_length() == 1;
}
int main()
{
std::cout << "====== Exact kernel with square root ======" << std::endl;
std::cout << check<KernelExactWithSqrt>() << std::endl;
std::cout << "====== Inexact kernel ======" << std::endl;
std::cout << check<KernelInexact>() << std::endl;
std::cout << "====== Exact kernel with automatic conversion to double ======" << std::endl;
std::cout << check<KernelExact>() << std::endl;
}
The output:
====== Exact kernel with square root ======
1
====== Inexact kernel ======
0
====== Exact kernel with automatic conversion to double ======
0
So, this example shows that the Exact_predicates_exact_constructions_kernel_with_sqrt guarantees you that the normalized vector will be what you expect.
Related
is there a way how to compare each element of a vector with a constant? So far, I am comparing 2D vector Eigen::Vector2d with a constant double tolerance like this:
if (x(0) > tolerance && x(1) > tolerance)
{
...
}
I have found the function isApprox() but it did not worked somehow. Is there is a nicer or recommended way on how to do it?
One way to do this is to use the array method of the Vector class. Like this:
#include <Eigen/Dense>
#include <iostream>
int main(int argc, char * argv[]) {
Eigen::Vector2d A{ 7.5, 8.2 };
std::cout << A << '\n';
auto res = A.array() >= 8.0;
std::cout << res << '\n';
if (res.all()) {
std::cout << "True" << '\n';
}
else {
std::cout << "False" << '\n';
}
A(0) = 10.2;
auto res2 = A.array() >= 8.0;
std::cout << res2 << '\n';
if (res2.all()) {
std::cout << "True" << '\n';
}
else {
std::cout << "False" << '\n';
}
return 0;
}
In this case res and res2 are CwiseBinaryOp which contains booleans for each element in A. Use all to find when both are True.
I feel like a good call on this one is to write a simple function:
bool is_componentwise_greater_than(
Eigen::Ref<Eigen::VectorXd const> const &vector, double lower_bound) {
for (auto value : vector) {
if (value <= lower_bound)
return false;
}
return true;
}
The drawback of this way compared to the solution by #Matt is, that for more complicated use cases, using an Eigen expression can be more performant (no idea, if this applies here).
The (in my opinion huge) advantage of such an solution is, that you
can see exactly what it does from its usage.
Of course you can also pack Matts solution in an aptly named function to get this advantage. Another advantage is, that with the function I provided you know exactly what it does and never have to wonder, whether using auto with an Eigen type could bite you. I guess it won't and Matt probably knows exactly why. But I (and you?) do not and therefore wouldn't want to rely on it.
I want to create an Eigen::Replicate object that can be accessed like a vector, i.e. with a single index. I got that to work with the fixed-size replicate<Index,Index>(), which I can't use in reality, the non-one factor is not a compile-time constant. It also works when manually creating a Replicate object, but I feel like I'm just overlooking the obvious way of using a replicate function to achieve this:
#include <Eigen/Dense>
#include <iostream>
using namespace Eigen;
int main(){
Vector3i v (3);
v << 0,1,2;
constexpr int nReplications {2};
auto replDynamic { v.replicate(nReplications, 1) };
/* with a dynamic replication, two indexes are required to access a coeff */
std::cout << "5th entry: " << replDynamic(4,0) << '\n';
auto replFixed { v.replicate<nReplications, 1>() };
/* I want to use only one index, but I require the number of replications
* in one dimension to be dynamic */
std::cout << "5th entry: " << replFixed(4) << '\n';
/* don't know how to access the VectorwiseOp variant */
// auto replVector { v.replicate(nReplications) };
// std::cout << "5th entry: " << replVector(4) << '\n';
/* this function doesn't exist */
// auto replDefined { v.replicate<Dynamic,1>(nReplications, 1) };
// std::cout << "5th entry: " << replDefined(4) << '\n';
/* I'd rather not define it manually (it's not the intended way), but it works */
Replicate<Vector3i,Dynamic,1> replManual { v, nReplications, 1 };
std::cout << "5th entry: " << replManual(4) << '\n';
return 0;
}
The source code shows VectorwiseOp<...>::replicate(Index factor) in line 134, which sounds like what I need, but I don't seem to be able to access it.
And a function such as replicate<Index,Index>(Index,Index) doesn't exist.
Assuming I understand what you are asking, since a Vector3i is a one column Eigen::Matrix, you can get a VectorwiseOp<...> expression template from a Vector3i (say) by using the colwise() function and then call the one argument replicate with that.
That is,
#include <Eigen/Dense>
#include <iostream>
using namespace Eigen;
int main() {
Vector3i v(3);
v << 0, 1, 2;
auto foo = v.colwise().replicate(2);
std::cout << "5th entry: " << foo(4) << '\n';
return 0;
}
Note though that using type deduction on a expression template, or "pseudo expression" as they are called in the Eigen documentation, is generally a bad idea i.e. writing Eigen::Matrix<int, 6, 1> foo = v.colwise().replicate(2) is safer; the Eigen documentation mentions the issue here.
By adding a .reshaped() after the replicate(...) call, the ColsAtCompileTime are set to 1, and therefore, the resulting object can be accessed like a vector:
#include <Eigen/Dense>
#include <iostream>
using namespace Eigen;
int main(){
Vector3i v (3);
v << 0,1,2;
constexpr int nReplications {2};
auto replReshaped { v.replicate(nReplications, 1).reshaped() };
std::cout << "5th entry: " << replReshaped(4) << '\n';
return 0;
}
What is the correct way to determine if a number (in my case it is a value of power of two calculated by pow(2,n)) is within the limits of values that one variable type can take? I'm doing it like this: if(pow (2,128)>std::numeric_limits<float>::max()), but this is evaluated as true although it is expected that float's maximum value is 2^128 or something more. Is there any better way to do this comparison?
For these kinds of limit checking, you can move the terms around to stay within the limits of the type.
In this case, pow(2,n) == exp(ln(2)*n) mathematically, so, rearranging terms, you can use n > ln(maxval)/ln(2)
You can take the base 2 logarithm of the maximum limit for the type of variable and compare it to n. For example: if(n > std::log2(std::numeric_limits<float>::max()). You probably don't want n to be exactly on the limit though, since I think stuff like floating point error might cause some problems.
First of all can you answer what is the result of pow(2, 128)?
The real question is what is the type for this expression?
The second question is do you know how floating point numbers work?
Take a look on this code to give you a hints:
#include <cmath>
#include <iostream>
#include <limits>
template<class T>
void printInfo(const std::string& desc, T x)
{
std::cout << desc << ' ' << typeid(x).name() << ' ' << x << std::endl;
}
int main()
{
printInfo("A", std::pow(2, 128));
printInfo("B", std::pow(2.0f, 128));
printInfo("A", std::pow(2, 128.0f));
auto c = std::pow(2.0f, 128.0f);
printInfo("C", c);
std::cout << (c > std::numeric_limits<float>::max()) << std::endl;
std::cout << (c == std::numeric_limits<float>::infinity()) << std::endl;
return 0;
}
https://wandbox.org/permlink/bHdKqToDKdC0hSvW
I recommend review documentation of numeric_limits.
And analyze this code:
#include <cmath>
#include <iostream>
#include <limits>
template<class T>
void print2exp()
{
std::cout << typeid(T).name() << '\n';
std::cout << "Radix = " << std::numeric_limits<T>::radix << '\n';
auto maxExp = std::numeric_limits<T>::max_exponent;
std::cout << "Max exp = " << maxExp << '\n';
std::cout << "2^maxExp = " << std::pow(static_cast<T>(2), static_cast<T>(maxExp)) << '\n';
std::cout << "2^(maxExp - 1) = " << std::pow(static_cast<T>(2), static_cast<T>(maxExp - 1)) << '\n';
}
int main()
{
print2exp<float>();
print2exp<double>();
print2exp<long double>();
return 0;
}
https://wandbox.org/permlink/J0hACKUKvKlV8lYK
So proper approach to this is (assuming that radix is 2):
if (x < std::numeric_limits<T>::max_exponent) {
return std::pow(static_cast<T>(2), static_cast<T>(x));
} else {
throw invalid_argument("x is to big to be use as 2^x");
}
There is this:
https://codeyarns.com/2016/02/16/how-to-compare-eigen-matrices-for-equality/
But there is no isApprox for tensors.
The following doesn't do what I want:
#include <Eigen/Core>
#include <unsupported/Eigen/CXX11/Tensor>
#include <array>
#include <iostream>
using namespace Eigen;
using namespace std;
int main()
{
// Create 2 matrices using tensors of rank 2
Eigen::Tensor<int, 2> a(2, 3);
Eigen::Tensor<int, 2>* b = &a;
cerr<<(*b==*b)<<endl;
}
because it does coordinate wise comparison and returns a tensor of the same dimension instead of a true/false vale.
How do I check if two tensors are identical? No isApprox for tensors.
I could write my own function, but I want to be able to use GPU power when available, and it seems like Eigen has built-in GPU support.
For an exact comparison of 2 tensors A and B, you can use the comparison operator followed by a boolean reduction:
Tensor<bool, 0> eq = (A==B).all();
This will return a tensor of rank 0 (i.e. a scalar) that contains a boolean value that's true iff each coefficient of A is equal to the corresponding coefficient of B.
There is no approximate comparison at the moment, although it wouldn't be difficult to add.
You can always use a couple of Eigen::Maps to do the isApprox checks.
#include <iostream>
#include <unsupported/Eigen/CXX11/Tensor>
using namespace Eigen;
int main()
{
Tensor<double, 3> t(2, 3, 4);
Tensor<double, 3> r(2, 3, 4);
t.setConstant(2.1);
r.setConstant(2.1);
t(1, 2, 3) = 2.2;
std::cout << "Size: " << r.size() << "\n";
std::cout << "t: " << t << "\n";
std::cout << "r: " << r << "\n";
Map<VectorXd> mt(t.data(), t.size());
Map<VectorXd> mr(r.data(), r.size());
std::cout << "Default isApprox: " << mt.isApprox(mr) << "\n";
std::cout << "Coarse isApprox: " << mt.isApprox(mr, 0.11) << "\n";
return 0;
}
P.S./N.B. Regarding Eigen's built in GPU support... Last I checked it is fairly limited and with good reason. It is/was limited to fixed size matrices as dynamic allocation on a GPU is really something you want to avoid like the common cold (if not like the plague). I take it back. It looks like the Tensor module supports GPUs pretty well.
I'm trying to use cxx-11's std::unique() to find
the unique elements in an array:
#include <iostream>
#include <algorithm>
#include <vector>
#include <typeinfo>
int main(){
const int n=11;
double x[n],a3[n],a1[n];
x[0]=-0.717778;
x[1]=-0.496843;
x[2]=-0.429063;
x[3]=-0.3596;
x[4]=-0.205607;
x[5]=0.0730536;
x[6]=0.138018;
x[7]=0.585526;
x[8]=2.40104;
x[9]=3.75268;
x[10]=4.55704;
a3[0]=0.790832;
a3[1]=0.569896;
a3[2]=0.502116;
a3[3]=0.432653;
a3[4]=0.343625;
a3[5]=0.512472;
a3[6]=0.56708;
a3[7]=1.01459;
a3[8]=2.32799;
a3[9]=3.67962;
a3[10]=4.48398;
std::cout.precision(10);
std::copy(a3,a3+n,a1);
for(int i=0;i<n;i++) a1[i]+=x[i];
std::sort(a1,a1+n);
for(int i=0;i<n;i++) std::cout << a1[i] << std::endl;
std::cout << "---" << std::endl;
int n_1=std::unique(a1,a1+n)-a1;
std::cout << "length of unique subvector " << n_1 << std::endl;
std::cout << "---" << std::endl;
for(int i=0;i<n_1;i++) std::cout << a1[i] << std::endl;
std::cout << "---" << std::endl;
}
but when I'm running this code (link to coliru)
it returns:
original array
0.073053
0.073053
0.073053
0.073054
0.138018
0.5855256
0.705098
1.600116
4.72903
7.4323
9.04102
---
length of unique subarray 10
---
unique array
0.073053
0.073053
0.073054
0.138018
0.5855256
0.705098
1.600116
4.72903
7.4323
9.04102
---
the unique array still contains a duplicate (and so is wrong)!
what am I doing wrong?
Let's try with a bit more precision, std::cout.precision(20):
0.073052999999999979064
0.073053000000000034575
0.073053999999999952308
0.13801800000000000179
0.58552559999999997942
0.70509800000000000253
1.6001160000000000938
4.7290299999999998448
7.4322999999999996845
9.0410199999999996123
Since most decimal fractions can't be represented exactly by a binary floating point format, slightly different rounding errors cause slightly different results.
In general, you can't expect the results of different floating point calculations to be exactly equal, even if the corresponding calculations applied to mathematical real numbers would be.
You could instead test for "almost equality", carefully choosing a tolerance that's appropriate for your numerical domain. unique allows you to specify your own predicate, instead of a simple equality test:
std::unique(a1,a1+n,[](double x, double y){return std::abs(x-y) < tolerance;});
How about:
int n_1 = std::unique(a1,a1+n,
[](float a, float b)
{
return std::fabs(a-b) < 10e-9;
}
) - a1;
?
Live demo link