NTHCIR - Editorial
#PROBLEM LINK:
[Practice][111]
[Contest][222]
**Author:** [Adury Surya Kiran][4444]
**Tester:** [Mugurel Ionut Andreica][5555]
**Editorialist:** [Lalit Kundu][6666]
#DIFFICULTY:
Medium-Hard
#PREREQUISITES:
advanced maths, geometry, recurrences
#PROBLEM:
We are given radius of circles $C_1$, $C_2$, $C_3$ and $C_4$ in the following image:
<img src="https://s3.amazonaws.com/codechef_shared/download/JULY15/NTHCIR1.png" height="20%" width="20%">
Now, we are going to draw circles $C_5$, $C_6$ and so on in such a way that $C_n$ touches circles $C_1$, $C_2$ and $C_{n-1}$. Like as show in shown in the following image:
<img src="https://s3.amazonaws.com/codechef_shared/download/JULY15/NTHCIR2.png" height="20%" width="20%">
Given upto $10^7$ values of $n$, find sum of radius of circle $C_n$ for all $n$.
#QUICK EXPLANATION:
================ ========================
Using Descartes Descartes' Theorem, if we have $3$ existing circles of curvature(inverse of radius) $a_1$, $a_2$, $a_3$, we can draw a $4^{th}$ circle of curvature $a_4$, touching all three circles such that $a_4 = a_1 + a_2 + a_3 \pm 2 \sqrt{a_1*a_2 + a_2*a_3 + a_1*a_3}$, where $a_4$ is negative if it represents a circle that circumscribes the first three.
Now, $n^{th}$ circle in our problem is formed by finding the $4$th $4^{th}$ circle touching circles $C_1$, $C_2$ and $C_{n-1}$. So, we can define our recurrence, solving which we get general term for the radius of $n^{th}$ circle.
#EXPLANATION:
================
A person should know Descartes Theorem One of the ways to solve this problem. problem is via Descartes' Theorem. First, lets define what curvature of a circle is.
The curvature of a circle with radius $r$ is defined as $\pm \frac{1}{r}$. The plus sign here applies to a circle that is externally tangent to the other circles, like the red circle is externally tangent to the three black circles in the image. For an internally tangent circle like the big red circle, that circumscribes the other circles, the minus sign applies.
<img src="https://upload.wikimedia.org/wikipedia/commons/thumb/8/87/Descartes_Circles.svg/500px-Descartes_Circles.svg.png" height="20%" width="20%">
Descartes Descartes' Theorem states that if four circles are tangent to each other at six distinct points, and the circles have curvatures $a_i$ (for $i = 1, ..., 4$), then:
$(a_1+a_2+a_3+a_4)^2=2(a_1^2+a_2^2+a_3^2+a_4^2)$.
When trying to find the radius of a fourth circle tangent to three given kissing circles(circles that touch each other mutually), the equation is best rewritten as:
$a_4 = a_1 + a_2 + a_3 \pm 2 \sqrt{a_1*a_2 + a_2*a_3 + a_1*a_3}$
The $\pm$ sign reflects the fact that there are in general two solutions. One solution is positive and the other is either positive or negative; if negative, it represents a circle that circumscribes the first three (as shown in the diagram above).
Now, let's see how we employ this theorem to find solution to the problem.
We know that $C_n$(for $n \ge 4$) is formed as a $4^{th}$ circle tangent to three kissing circles $C_1$, $C_2$ and $C_{n-1}$, where $C_1$ is being touched by all circles internally. And also, the circle $C_n$ will touch all other circles externally, so we are interested in positive curvature of $C_n$. And, curvature of $C_1$ is $-\frac{1}{r_1}$.
So, we can write using Descartes' Theorem that $a_n = a_1 + a_2 + a_{n-1} + 2 \sqrt{a_1*a_2 + (a_2 + a_1)*a_{n-1}}$. Now, we can employ this recurrence to find $n^{th}$ circle, in $O(n)$. This will be enough to pass **subtask 1**.
For **subtask 2**, we need a constant time solution for each $n$, that means we need to arrive at a closed form of the $n^{th}$ term of recurrence. Let's see how we can solve this recurrence. Basically, our recurrence is:
$a_n = c_1 + a_{n-1} + 2 \sqrt{c_2 + c_1 a_{n-1}}$,
where $c_1 = a_1 + a_2$ and
$c_2 = a_1 * a_2$
Here, we substitute $x_n$(a new recurrence) as $\sqrt{c_2 + c_1 a_{n}}$, which implies that
$a_n = \frac{x_n^2 - c_2}{c_1}$ --------Eqn (1)
So, our recurrence is now $a_n = c_1 + a_{n-1} + 2 x_{n-1}$. In this equation, substitute $a_{n-1}$ and $a_n$ using equation $1$ to get,
$x_n^2 = x_{n-1}^2 + 2c_1x_{n-1} + c_1^2$, which is basically $x_n^2 = (x_{n-1}+ c_1)^2$. Now, this has greatly been simplified to $x_n = x_{n-1} + c_1$, whose general term can be written as $x_n = nc_1 + C$, where $C$ is an arbitrary constant, which depends on $x_1$.
Now, to find $a_n$ we substitute general term for $x_n$ in equation $1$, which yields $a_n = \frac{(nc_1 + C)^2 - c_2}{c_1}$.
Now, we know the value of $a_3$ as $\frac{1}{r_3}$, so we can find value of $C$ by substituting $n$ as $3$ in the general form of $a_n$. As you might note, it will give us two values for $C$ because its a quadratic in $C$, so, we take the value which gives us correct value for $a_4$ whose value we already know as $\frac{1}{r_4}$.
Now, our solution for a single $n$ is constant time.
#ALTERNATE SOLUTION:
SOLUTION:
========================
There is another solution using the concept of inversion of a plane. Let's see this concept in detail.
We are going to transform all points in an $x$-$y$ plane(say plane 1) to another $x$-$y$ plane(say plane 2). For each point $(x, y)$ in plane 1, we map it to $\frac{x}{r^2}, \frac{y}{r^2}$, where $r$ is distance of point $(x, y)$ from origin *i.e.* $\sqrt{x^2 + y^2}$. If we consider point $(x, y)$ in polar coordinates as $r e^{i \theta}$, then we map it to $\frac{1}{r} e^{i \theta}$ in the plane 2. Transformation from plane 2 to plane 1 is simple, we just have to apply the same transformation again for all points in plane 2.
Now, let's see how it affects simple figures: circle and lines. Here I am going to summarise the results, you can easily derive these results by considering equations of lines and circles(in circles(doing in polar coordinates).
coordinates will be easier).
- All the circles which pass through origin in plane 1 will be mapped to straight lines in the second. plane 2. A circle with center at $(r, 0)$ and radius $r$ will be mapped to line $x = \frac{1}{2 r}$ in plane 2.
- All the straight lines which do not pass through origin in the first plane will be mapped to circles which pass through origin in the second.
- All the circles which do not pass through origin in the first plane will be mapped to some other circles in the second plane.
- All the straight lines which pass through origin in the first plane will be mapped to the same lines in the second. Also, note that if there are two points on such a line where one point is at a distance $r_1$ from origin and another at distance $r_2$, where $r_2 \gt r_1$, then distance between them in plane 1 is $r_2 - r_1$ and distance between these points in transformed plane will be $\frac{1}{r_1} - \frac{1}{r_2}$.
Now, let's see how we employe this transformation to solve our problem. First, we'll choose a suitable origin in plane 1(plane where all our circles are present). Now, we see that all circles are tangent to circles $C_1$ and $C_2$ and we remember that we can transform circles passing through origin to straight lines, so, we choose the point where $C_1$ and $C_2$ touch as the origin. Now, we also assume that their centers lie on $y$-axis at points $R_1$ and $R_2$ respectively.
Now if we transform the plane, then circles $C_1$ and $C_2$ will become straight lines which are parallel to each other. All the circles $C_3$, $C_4$, $C_5$ $\cdots$ will become some other circles in the transformed plane. But the fact that they all are touching $C_1$ and $C_2$ before transformation means that they all touch the two straight lines which are parallel to each other after transformation, which makes all their radii equal in the transformed plane(this radius is equal to $\frac{1}{4R_1} - \frac{1}{4R_2}$. \frac{1}{4R_2}$). See the following figure.
<img src="http://i.imgur.com/tJG1Cux.png" height="100%" width="100%">
So, the equations of lines $C_1^{\prime}$ and $C_2^{\prime}$ are $y = \frac{1}{2 R_1}$ and $y = \frac{1}{2 R_2}$ respectively.
Now, to find the radius of any $n^{th}$ circle in plane 1, we first find coordinates(in plane 2) of two points on the circle and we transform these points back to plane 2 and get the distance between them as the diameter.
We already know the $y$-coordinate of the centre of circles $C_3^{\prime}$, $C_4^{\prime}$ and so on. We need to find their $x$ coordinate now. If we have $x$-coordinate of the center of circle $C_3^{\prime}$, then we can easily find centre of any circle and we already know the radius of circles in plane 2, so we can also find the coordinates of a point on the circumference, which we'll transform back.
Let's see how we find $x$-coordinate of center of $C_3^{\prime}$ in plane 2. Keep referring to image below while reading this for more clarity. We know that a line passing through origin in plane 1 remains same in plane 2, so we draw a line $Z_1$ in plane 1 passing from origin and center of circle $C_3$. Now, this line in plane represented as $Z_1^{\prime}$ will also pass through center of circle $C_3^{\prime}$ and origin of plane 2. If we know distance of center of $C_3^{\prime}$ from origin somehow, we can calculate its $x$-coordinate also, since we already have $y$-coordinate.
We now, try to relate distances in plane 2 with distances in plane 1. Keep referring to image below while reading this for more clarity. Say, the distance $d$ = distance between origin and center of $C_3^{\prime}$. So, radius circle $C_3^{\prime}$ cuts line $Z_1^{\prime}$ at two points, distance between these points in plane 1 must be the diameter of $C_3$ *i.e.* $2R_3$ because the same line intersects circle $C_3$ at same two points. Distance of these two points from origin in plane 2 can be written as $d - r_3^{\prime}$ and $d - + r_3^{\prime}$. Now, if we transform them to plane 1 and see distance between them, we can say that:
$2 r_3 = \frac{1}{d - r_3^{\prime}} - \frac{1}{d - + r_3^{\prime}}$
<img src="http://i.imgur.com/URwsTar.png" height="100%" width="100%">
In above equation, we can find the $d$ as all other variables are known. And using value of $d$ we can find $x$-coordinate of center of $x_3^{\prime}$ $=$ $C_3^{\prime}$ as $\pm \sqrt{d^2 - y^2}$. So, center of $n^{th}$ circle will be $x_3^{\prime} + 2(n-3)R_3$ or $x_3^{\prime} - 2(n-3)R_3$, we can now check which one to use by finding radius of $C_4$ and match with given value in input.
Now, our problem is solved. You can do the calculations yourself to find the closed formula.
#COMPLEXITY:
Constant per query.
#AUTHOR'S, TESTER'S SOLUTIONS:
[setter][333]
[tester][444]
[123]: https://s3.amazonaws.com/codechef_shared/download/JULY15/NTHCIR1.png
[124]: https://s3.amazonaws.com/codechef_shared/download/JULY15/NTHCIR2.png
[125]: https://upload.wikimedia.org/wikipedia/commons/thumb/8/87/Descartes_Circles.svg/500px-Descartes_Circles.svg.png
[111]: http://www.codechef.com/problems/NTHCIR
[222]: http://www.codechef.com/JULY15/problems/NTHCIR
[4444]: http://www.codechef.com/users/adurysk
[5555]: http://www.codechef.com/users/mugurelionut
[6666]: http://www.codechef.com/users/darkshadows
[333]: http://www.codechef.com/download/Solutions/JULY15/Setter/NTHCIR.cpp
[444]: http://www.codechef.com/download/Solutions/JULY15/Tester/NTHCIR.cpp