Hackerrank Solution: Find the Point

Original Problem

Consider two points, $p=(p_x, p_y)$ and $q=(q_x, q_y)$. We consider the inversion or point reflection, $r=(r_x, r_y)$, of point $p$ across point $q$ to be a $180^{\circ}$ rotation of point $p$ around $q$.

Given $n$ sets of points $p$ and $q$, find $r$ for each pair of points and print two space-separated integers denoting the respective values of $r_x$ and $r_y$ on a new line.

Input Format

The first line contains an integer, $n$, denoting the number of sets of points.
Each of the $n$ subsequent lines contains four space-separated integers describing the respective values of $p_x$, $p_y$, $q_x$, and $q_y$ defining points $p=(p_x,p_y)$ and $q=(q_x,q_y)$.

Constraints

• $1\leq n\leq 15$
• $-100\leq p_x,p_y,q_x,q_y\leq 100$

Output Format

For each pair of points $p$ and $q$, print the corresponding respective values of $r_x$ and $r_y$ as two space-separated integers on a new line.

Sample Input

2
0 0 1 1
1 1 2 2

Sample Output

2 2
3 3

Explanation

The graphs below depict points $p$, $q$, and $r$ for the $n=2$ points given as Sample Input:

1. 
   Thus, we print $r_x$ and $r_y$ as 2 2 on a new line.
2. 
   Thus, we print $r_x$ and $r_y$ as 3 3 on a new line.

Solution

The rotation matrix around $180^\circ$ is:

$$R(180^{\circ })={\begin{bmatrix}\cos 180^{\circ } &-\sin 180^{\circ } \\\sin 180^{\circ } &\cos 180^{\circ } \\\end{bmatrix}}={\begin{bmatrix}-1&0\\[3pt]0&-1\\\end{bmatrix}}$$

Now it's quite easy to translate by point $q$, rotate with matrix $R$ and translate back to $q$:

$$r = q + R(180^{\circ }) \cdot (p - q) = 2q - p$$

The final implementation then looks like this in Ruby:

gets.to_i.times{
    x = gets.split.map(&:to_i)
    print 2 * x[2] - x[0], ' ', 2 * x[3] - x[1], "\n"
}