# Recent posts

## How to fix broken MathJax fonts on Linux

Posted by Diego Assencio on 2017.08.09 under Linux (General)

If you are using Linux and your MathJax fonts look ugly, and if you are certain that MathJax is correctly configured on the webpage you are accessing (e.g. by checking that things look fine on another device or browser), then your browser is probably selecting STIX fonts which are installed on your system instead of the ones offered by MathJax. The simplest way to solve this problem is by removing these STIX fonts. On Ubuntu/Debian, this can be done with the following command:

sudo apt remove fonts-stix


You can now see if this was the root cause of the problem by reloading the affected webpage with Ctrl+F5 (this will force your browser to bypass its cache).

## How std::move breaks return value optimization

Posted by Diego Assencio on 2017.08.07 under Programming (C/C++)

Consider the following program:

std::vector<int> random_numbers(const size_t n)
{
std::vector<int> numbers;

/* generate n random numbers, store them on numbers */

return numbers;
}

int main()
{
/* create an array with 100 random numbers */
std::vector<int> my_numbers = random_numbers(100);

...
}


Returning an std::vector<int> by value can make a developer nervous: will this cause the vector to be copied? That would be a costly performance penalty and therefore something to be avoided, if possible.

In this type of situation, a solid understanding of how return value optimization (or simply RVO, for short) works leads to inner peace. Any decent compiler will understand that numbers is a temporary variable whose value will be returned at the end of the random_numbers function, and since this returned value is an rvalue (a temporary object, see line 13), it can be used to initialize my_numbers directly, i.e., without going through any of the std::vector<int>'s constructors. In other words, no vectors should be copied or moved on the code above. How wonderful!

Let's take a closer look at how RVO removes the need for a copy operation. When random_numbers is called, memory on the call stack will be reserved for its return value (an std::vector<int>). Inside of random_numbers, numbers is clearly a temporary variable whose value will be returned at the end, so instead of copying this value to the memory location reserved for the function's return value at the very end, numbers is stored directly at that memory location. The elimination of such a copy operation when a function returns is what "return value optimization" stands for. In general, any type of optimization which causes copy operations to be eliminated is referred to as a copy elision optimization.

Despite all these facts, it is not uncommon for developers to be less confident than they should in this type of situation and prefer "being on the safe side" by writing this type of code:

std::vector<int> random_numbers(const size_t n)
{
std::vector<int> numbers;

/* generate n random numbers, store them on numbers */

return std::move(numbers);   /* don't do this! */
}


As innocent as this decision may be, it is severely flawed because of the way RVO works: only a local variable or a temporary object can be stored directly at the memory location for a function's return value (function parameters are not eligible for that), and this is only allowed if such an object is directly returned by the function and has the same type as the function's return type. By adding std::move on the return statement above, we converted the type of the returned object to std::vector<int>&& (an rvalue reference to an std::vector<int>), but random_numbers returns an std::vector<int>. This violates the conditions required for RVO, and the compiler will have no choice but to make numbers be constructed outside the memory area reserved for random_numbers's return value and then moved to that location when the function returns (a move operation is still possible here since std::move(numbers) is an rvalue and will therefore trigger the move constructor for the std::vector<int> which is constructed as the return value).

For many user-defined types, such an additionally incurred move operation will be cheap, but it will definitely not be cheaper than what RVO offers and therefore, by "playing safe", we ended up inevitably pessimizing our program. Also, notice that we were lucky to have a return type (std::vector<int>) with a move constructor; had this not been the case, the added std::move would have caused the return value to be initialized through a copy constructor, which is what we wanted to avoid in the first place. Ouch!

To finalize, here is an example which involves all concepts discussed so far:

#include <iostream>

class X
{
public:
/* default constructor */
X() { std::cout << "X::X()\n"; }

/* copy constructor */
X(const X&) { std::cout << "X::X(const X&)\n"; }

/* move constructor */
X(X&&) { std::cout << "X::X(X&&)\n"; }
};

X good()
{
std::cout << "good()\n";

X x;
return x;
}

{

X x;
return std::move(x);
}

int main()
{
X x1 = good();

return 0;
}


The program's output illustrates how the std::move on the bad function disables RVO and forces an unnecessary move operation (this will be the case even if you compile with lots of optimizations enabled, e.g. by using -O3 on gcc):

good()
X::X()
X::X()
X::X(X&&)


Try compiling and running this program, then remove the move constructor from X. The resulting output shows that std::move now causes X to be copied:

good()
X::X()
X::X()
X::X(const X&)


## The intersection area of two circles

Posted by Diego Assencio on 2017.07.12 under Mathematics (Geometry)

Let $C_1$ and $C_2$ be two circles of radii $r_1$ and $r_2$ respectively whose centers are at a distance $d$ from each other. Assume, without loss of generality, that $r_1 \geq r_2$. What is the intersection area of these two circles?

If $d \geq r_1 + r_2$, the circles intersect at most up to a point (when $d = r_1 + r_2$) and therefore the intersection area is zero. On the other extreme, if $d + r_2 \leq r_1$, circle $C_2$ is entirely contained within $C_1$ and the intersection area is the area of $C_2$ itself: $\pi r_2^2$. The challenging case happens when both $d \lt r_1 + r_2$ and $d + r_2 \gt r_1$ are satisfied, i.e., when the the circles intersect only partially but the intersection area is more than simply a point. Rearranging the second inequality, we obtain $r_1 - r_2 \lt d \lt r_1 + r_2$, so we will assume this to be the case from now on.

To solve this problem, we will make use of a Cartesian coordinate system with origin at the center of circle $C_1$ such that the center of $C_2$ is at $(d,0)$ as shown on figure 1.

 Fig. 1: Two intersecting circles $C_1$ (blue) and $C_2$ (red) of radii $r_1$ and $r_2$ respectively. The distance between the centers of the circles is $d = d_1 + d_2$, where $d_1$ is the $x$ coordinate of the intersection points and $d_2 = d - d_1$. Notice that $d_1 \geq 0$ since these points are always located to the right of the center of $C_1$, but $d_2$ may be negative when $r_2 \lt r_1$ since, in this case, the intersection points will eventually fall to the right of the center of $C_2$ as we move $C_2$ to the left, making $d \lt d_1$ and therefore $d_2 \lt 0$.

The circles $C_1$ and $C_2$ are described by the following equations respectively: $$\begin{eqnarray} x^2 + y^2 &=& r_1^2 \label{post_8d6ca3d82151bad815f78addf9b5c1c6_c1}\\[5pt] (x - d)^2 + y^2 &=& r_2^2 \\[5pt] \end{eqnarray}$$ At the intersection points, we have $x = d_1$. To determine $d_1$, we can replace $x$ with $d_1$ and isolate $y^2$ on both equations above to get: $$r_1^2 - d_1^2 = r_2^2 - (d_1 - d)^2$$ Solving for $d_1$ is a simple task: $$r_1^2 - d_1^2 = r_2^2 - d_1^2 + 2d_1d - d^2 \Longrightarrow d_1 = \displaystyle\frac{r_1^2 - r_2^2 + d^2}{2d} \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1}$$ From equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1}, we can see that $d_1 \geq 0$ since $r_1 \geq r_2$. The intersection area is the sum of the blue and red areas shown on figure 1, which we refer to as $A_1$ and $A_2$ respectively. We then have that: $$\begin{eqnarray} A_1 &=& 2\int_{d_1}^{r_1} \sqrt{r_1^2 - x^2}dx \label{%INDEX_eq_A1_def} \\[5pt] A_2 &=& 2\int_{d - r_2}^{d_1} \sqrt{r_2^2 - (x - d)^2}dx \end{eqnarray}$$ where the factors of $2$ come from the fact that each integral above accounts for only half of the area of the associated region (only points on and above the $x$ axis are taken into account); the results must then be multiplied by two so that the areas below the $x$ axis are taken into account as well.

The computation of these integrals is straightforward. Before we proceed, notice first that: $$\begin{eqnarray} A_2 &=& 2\int_{d - r_2}^{d_1} \sqrt{r_2^2 - (x - d)^2}dx \nonumber \\[5pt] &=& 2\int_{- r_2}^{d_1 - d} \sqrt{r_2^2 - x^2}dx \nonumber \\[5pt] &=& 2\int_{d - d_1}^{r_2} \sqrt{r_2^2 - x^2}dx \nonumber \\[5pt] &=& 2\int_{d_2}^{r_2} \sqrt{r_2^2 - x^2}dx \label{%INDEX_eq_A2} \end{eqnarray}$$ where above we used the fact that $d_2 = d - d_1$. This is the same as equation \eqref{%INDEX_eq_A1_def} if we apply the substitutions $d_1 \rightarrow d_2$ and $r_1 \rightarrow r_2$. Therefore, by computing $A_1$, we will immediately obtain $A_2$ as well. Let's then compute $A_1$ first: $$\begin{eqnarray} A_1 &=& 2\int_{d_1}^{r_1} \sqrt{r_1^2 - x^2}dx \nonumber\\[5pt] &=& 2r_1 \int_{d_1}^{r_1} \sqrt{1 - \left(\frac{x}{r_1}\right)^2}dx \nonumber\\[5pt] &=& 2r_1^2 \int_{d_1/r_1}^{1} \sqrt{1 - x^2}dx \label{%INDEX_eq_A1} \end{eqnarray}$$ All we need to do now is to integrate $\sqrt{1 - x^2}$. The process is straightforward if we use integration by parts: $$\begin{eqnarray} \int \sqrt{1 - x^2}dx &=& x \sqrt{1 - x^2} - \int x \left(\frac{-x}{\sqrt{1 - x^2}}\right) dx \nonumber\\[5pt] &=& x \sqrt{1 - x^2} + \int \frac{x^2 - 1}{\sqrt{1 - x^2}} dx + \int \frac{1}{\sqrt{1 - x^2}} dx \nonumber\\[5pt] &=& x \sqrt{1 - x^2} - \int \sqrt{1 - x^2} dx + \sin^{-1}(x) \end{eqnarray}$$ Therefore: $$\int \sqrt{1 - x^2}dx = \frac{1}{2}\left( x \sqrt{1 - x^2} + \sin^{-1}(x) \right) \label{post_8d6ca3d82151bad815f78addf9b5c1c6_int_for_A1_A2}$$ Using equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_int_for_A1_A2} on equation \eqref{%INDEX_eq_A1} yields: $$\begin{eqnarray} A_1 &=& r_1^2 \left( \frac{\pi}{2} - \frac{d_1}{r_1}\sqrt{1 - \left(\frac{d_1}{r_1}\right)^2} - \sin^{-1}\left(\frac{d_1}{r_1}\right) \right) \nonumber\\[5pt] &=& r_1^2 \left( \cos^{-1}\left(\frac{d_1}{r_1}\right) - \frac{d_1}{r_1}\sqrt{1 - \left(\frac{d_1}{r_1}\right)^2} \right) \nonumber\\[5pt] &=& r_1^2 \cos^{-1}\left(\frac{d_1}{r_1}\right) - d_1 \sqrt{r_1^2 - d_1^2} \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_A1_final} \end{eqnarray}$$ where above we used the fact that $\pi/2 - \sin^{-1}(\alpha) = \cos^{-1}(\alpha)$ for any $\alpha$ in $[-1,1]$. This fact is easy to prove: $$\cos\left(\frac{\pi}{2} - \sin^{-1}(\alpha)\right) = \cos\left(\frac{\pi}{2}\right)\cos(\sin^{-1}(\alpha)) + \sin\left(\frac{\pi}{2}\right)\sin(\sin^{-1}(\alpha)) = \alpha$$ and therefore $\pi/2 - \sin^{-1}(\alpha) = \cos^{-1}(\alpha)$. As discussed above, we can now obtain $A_2$ directly by doing the substitutions $d_1 \rightarrow d_2$ and $r_1 \rightarrow r_2$ on the expression for $A_1$ on equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_A1_final}: $$A_2 = r_2^2 \cos^{-1}\left(\frac{d_2}{r_2}\right) - d_2 \sqrt{r_2^2 - d_2^2}$$ The sum of $A_1$ and $A_2$ is the intersection area of the circles: $$\boxed{ \begin{eqnarray} A_{\textrm{intersection}} &=& r_1^2 \cos^{-1}\left(\frac{d_1}{r_1}\right) - d_1\sqrt{r_1^2 - d_1^2} \nonumber \\[5pt] &+& r_2^2\cos^{-1}\left(\frac{d_2}{r_2}\right) - d_2\sqrt{r_2^2 - d_2^2} \nonumber \end{eqnarray} } \label{post_8d6ca3d82151bad815f78addf9b5c1c6_A_intersection}$$ where: $$\boxed{ d_1 = \displaystyle\frac{r_1^2 - r_2^2 + d^2}{2d} } \quad \textrm{ and } \quad \boxed{ d_2 = d - d_1 = \displaystyle\frac{r_2^2 - r_1^2 + d^2}{2d} } \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1_final}$$

### Summary

Given two circles $C_1$ and $C_2$ of radii $r_1$ and $r_2$ respectively (with $r_1 \geq r_2$) whose center points are at a distance $d$ from each other, the intersection area of the circles is:

 1 zero, if $d \geq r_1 + r_2$, since in this case the circles intersect at most up to a point. 2 $\pi r_2^2$, if $d \leq r_1 - r_2$, since in this case $C_2$ is entirely contained within $C_1$. 3 given by equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_A_intersection} in all other cases.