# VSN - Editorial

Practice

Contest

Tester: Misha Chorniy

Editorialist: Bhuvnesh Jain

EASY-MEDIUM

# Prerequisites

Cross Product, Binary Search

# Problem

Given a stationary point P and a point Q moving in straight line, indicate the time when Q is visible from P, given that there am opaque sphere between them.

# Explanation

We will try to find the solution in 2-D and then extend the idea to 3-D as well.  From the above figure, the first idea which can be clearly concluded is that, once point Q becomes visible from P, it will remain visible forever. Before that, it will always remain invisible. This, the function (whether Q is visible P at given time t) follows the idea that it is false initially and after a certain point, it always remains true. This hints that we can binary search on the answer and just need to check whether the point Q is visible from P at given time t or not. For more ideas on “Binary search” on such type of problem, refer to this awesome tutorial on topcoder. Hence, solution template looks like:


double low = 0, high = 10^18
for i in [1, 100]:
double mid = (low + high) / 2
if visible(P, Q, C, r, mid):
high = mid
else:
low = mid
double ans = (low + high) / 2 So, given a particular time t, we can first calculate the position of point Q. Join P and Q by a straight line. If the line doesn’t pass through the circle, the point Q is visible from P else it is not visible. To check if a given line passes through the circle or not, it is enough to check that the perpendicular distance of the line from the centre, C, of the circle is greater than the radius, r, of the circle. For this, we first complete the triangle PCQ and let the perpendicular distance of PQ from C be denoted by CO. Using the formula,

\text{Area of triangle} = \frac{1}{2} * \text{Base} * \text{Height} = \frac{1}{2} * CO * PQ

Since the area of the triangle can be found given 3 points in 2-D, and PQ is the Euclidean distance between P and Q, we can find the value of CO efficiently. Finally, we just need to compare it to r, the radius of the circle.

For extending the solution in 3-D, the idea is same and we just need to find the area of a triangle in 3-D. For details, one can refer here. It can be clearly seen that the above formula holds for the 2-D case as well.

\text{Area of triangle in 2/3-D} = \frac{|CP X CQ|}{2}

where CP and CQ are vectors, a X b denotes cross product of vectors and |a| denotes the magnitude of vector.

Thus, to find the length of CO, we have

|CO| = \frac{2 * \text{Area of triangle PCQ}}{|PQ|} = \frac{|CP X CQ|}{|PQ|}

For more details, you may refer to the editorialist solution which exactly follows the above idea.

### Extra Ideas/Caveats

The binary search implementation mentioned in the editorial is different from the one mentioned in Topcoder tutorial. Though both will give AC here, the one in Topcoder needs one to correctly set the Epsilon value for termination condition and sometimes can lead to wrong answers due to precision issues. The editorialist just prefers the above style to implement binary and ternary search involving doubles as calculation of epsilon is not required.

Also, for details regarding the exact number of iterations or complexity analysis for binary search problem on doubles, refer to this blog on codeforces.

### Note from the author

Finding the distance of a point from a line in 3-D is a generally common problem and also contains some edge cases where the point may not lie within the line segment region. But given the constraints of the problems, there are no edge cases but we should be aware of it in general. Feel free to share your approach, if it was somewhat different.

# Time Complexity

O(LogN) per test case (around 100 iterations for binary search).

# Space Complexity

O(1)

### AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here.

Tester’s solution can be found here.

Editorialist’s solution can be found here.

1 Like

I think the time complexity of O(1) per test case is a typo. Although the solutions arent available right now, I can make a guess that the implementation referred to is something like “Binary search at least 100-200 times.” to get the answer, that value of 100-200 is based on the value of N. We know the value of LogN and set a value higher than that as limit.

Can you please share the code snippet you’re talking about (I am assuming its not the topcoder one but the one in editorialist solution), if my above argument isnt applicable to your solution?

2 Likes

Just adding one more point ,we are basically finding the time at which line joining PQ’ is tangent to sphere,
where Q’=Q0+dt and solving for t.Final eqn is same as mentioned in the editorial

Want to mention two things:

1. Complexity is not O(1) per test case (As this will confuse ones who are new to discrete binary search) :The number of iterations (100 in this case) will need to be adjusted if tmax is larger, actual time complexity if O(log_2{t_{max})}, (per test case) in this case.

2. To get precision upto n decimal places, setting epilson as 10^{-n-1} will work fine as both low and high will be same upto n decimal places.

5 Likes

Alternative approach to checking whether Q(t) is visible from P or not at a given time t:

Obtain a parametric equation of the line joining P and Q(t) in terms of some parameter (say a) which varies from 0 to 1. Any point (x, y, z) lying on this line can be written in terms of a as:

x = Px.(1-a) + Qx(t).a

y = Py.(1-a) + Qy(t).a

z = Pz.(1-a) + Qz(t).a

At a = 0, you are at point P and at a = 1, you are at point Q(t). At any a between 0 and 1, you are at some point in the line between P and Q(t).

Now we substitute (x, y, z) in terms of the parameter a in the equation of the sphere. We will obtain an equation in terms of t and a, which will be a quadratic in the parameter a. The point of tangency is when this quadratic has exactly one root, so we binary search on t, searching for that t when the quadratic’s discriminant becomes zero.

Link to my solution: CodeChef: Practical coding for everyone

5 Likes

I used simple 3D Coordinate Geometry.

Let the general point on which point Q is traveling be Q’.

Therefore,

Q’x=Qx + Dxt

Q’y=Qy + Dyt

Q’z=Qz + Dzt

Now, the equation of line PQ’ in parametric form(with parameter k) is,

X = Px + (Q’x-Px)k

Y = Py + (Q’y-Py)k

Z = Pz + (Q’z-Pz)k

Now, equation of sphere is,

(X-Cx)2 + (Y-Cy)2 +(Z-Cz)2 = R2

Now solve Line PQ’ with equation of sphere, and we will get a quadratic equation in parameter k.

For condition of tangency(single solution), delta must be equal to zero.

So calculate delta and make it equal to zero, and here we get another quadratic in t.

Solve for t and consider the postitive value ,since it is time.

4 Likes

If D is the point upto which Q goes, then

|(PCXPD)/|PD||=r

If k is the time, a quadratic in k can easily be solved.

My solution:CodeChef: Practical coding for everyone

3 Likes

Can somebody tell why my solution was not working ? Still not getting it. CodeChef: Practical coding for everyone exact same method and took care of out of bound situations as well.

Binary search for time t and calculate Q’(x,y,z) = Q(x,y,z) + d(x,y,z) * t

Distance of point C(center) from line \overrightarrow{PQ'} from Idea -> \frac{|\overrightarrow{PC} \ X \ \overrightarrow{PQ'}|}{|\overrightarrow{PQ'}|}\ ==\ r

EDIT
Float comparision is wrong here.
Instead search ends when t(low) >= t(high) each being added or subtracted by \epsilon respectively in each iteration

Slightly different approach - use the vector equation of the cone: F(u) = u.d - |u||d|cos(theta) = 0,
where the axis of the cone is given by the vector d and the origin is the vertex of the cone.
Shift origins to P, then PC is d, and PQ is u.
Then, you only have to implement the dot product and find the smallest t when q = q0 + d*t satisfies F(q) = 1 with a binary search.

Solution here.

Or Take the Reference of this site to solve this.

http://www.ambrsoft.com/TrigoCalc/Sphere/SpherLineIntersection_.htm

Condition on :B * B-4 * A * C

SInce it is Monotonic.We can use Binary Search

1 Like

Can you prove that binary search will give the correct answer in max 100 iterations?

My submission: TLE submission

Source of condition: Line Sphere intersection

Hi. Both Author’s and Editorialist’s solution fail in this test case:

1
4 0 0 -10 6 0 1 0 0 0 0 0 3

I mailed the testcase to Misha 2 days before the contest ended but it wasn’t updated. Any reason why?

Edit:He mailed . He was busy for his exams so didn’t see my mail before.

3 Likes

I think the below link is enough to solve the Question in just half an hour.
Here,they provided full proof of of dot product as well as cross product formula…its became really helpful to me I think test cases were weak during contest as per given case by @soham1234

1
4 0 0 -10 6 0 1 0 0 0 0 0 3

this solution gives output 19.2915 which is wrong (this is my code accepted during contest)
many submissions giving different outputs passed during contest.

this solution gives output 8.7085 which is correct (i submitted after contest)

Mistake in my code was to find minimum positive t.

1 Like

I did not use binary search.

Point P will make an enveloping cone (cone by tangents) C to sphere. Find equation of C.

Point Q with vector d, will make a straight line (ray) L. Find equation of L.

Now the point of intersection of C and L is the required point, which will give minimum time t after solving.

Per test case O(1) solution:
https://www.codechef.com/viewsolution/18861170

my solution
Could anyone please tell me what is the mistake in my code?
I’m getting the most precise answer;

Anyone having solution without binary search ??? my solution

Anyone with a similar approach and working well please post your solution in the comment section I solved it in O(1) for every test case.
My approach was initializing P to [x1,y1,z1], Q to [x2 + tdx, y2 + tdy, z2 + tdz] and C to [x3,y3,z3].

And applying formula |PC X PQ| / |PQ| (considering vectors).

This formula will give us shortest distance between line PQ and centre of Circle (c). Since for time t to be minimum, this distance should be equal to R (radius).

SO, r = |PC X PQ| / |PQ|.

Substitute values for P, Q and C and at end you will get an quadratic equation of the form b^2 - 4ac = 0,

Solve it and take the minimum positive root(incase we get both positive roots).

Easy solution : CodeChef: Practical coding for everyone

1 Like

This approach is different from binary search approach.
And it works! it is based on forming a quadratic equation based on distance formula and requires very few lines of code :
Here is the link to the code:
https://www.codechef.com/viewsolution/18842935