POINPOLY - Editorial

PROBLEM LINK:

Practice
Contest

Author: Praveen Dhinwa
Tester: Hanlin Ren
Editorialist: Hanlin Ren

DIFFICULTY:

MEDIUM

PREREQUISITES:

Convex Polygon, constructive algorithms

PROBLEM:

Given an n-sided convex polygon where all vertices have integral coordinates. Output \lfloor n/10\rfloor distinct integral points strictly inside the polygon.

QUICK EXPLANATION:

This editorial contains 3 solutions: (Setter’s solution is the easiest among them)

Setter’s solution

  • For a polygon of 10 points, one can easily find one point by checking the midpoints of some chords.
  • Let P_1,P_2,\dots,P_n be the vertices.
  • For each k=10i+1, we find one integral point in the convex hull of P_k,P_{k+1},\dots,P_{k+9}.

A dfs/bfs-based solution

  • We need a data structure that given a point, tells in O(\log n) time if the point is strictly in the polygon.
  • First, we arbitrarily find one point in the polygon.
  • Then we check if its neighbors are in the polygon; check its neighbors’ neighbors; and so on.

Tester’s solution

  • Let U(x_0) be the intersection of x=x_0 with the upper hull. Similarly L(x_0) for lower hull.
  • Find (roughly) the smallest x_0 such that the open interval (L(x_0), U(x_0)) contains some integer.
  • Imagine a vertical line at x=x_0. Then we can output all integral points on the vertical line.
  • We repeatedly increase x_0, moving the vertical line rightwards. During this process we can maintain L(x_0),U(x_0) and iterate all integral points in the polygon.

Setter’s Solution

Click to view

A special case

Let’s first consider the case when n=10. Let the points be P_1,P_2,\dots,P_{10} in order. We want to find one integral point strictly inside the polygon. In fact, this can be done by the following: we simply look at every pair (i,j), and check if the midpoint of some P_i,P_j is integral. Of course, only pairs that are not on the edge of the polygon counts. Recall the midpoint of two points (x_1,y_1) and (x_2,y_2) is (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}).

Does this algorithm guarantee a solution? This can be proved by pigeonhole principle: imagine there are 4 holes: 00, 01, 10, 11. A point (x,y) is a pigeon, and it goes to the hole whose first bit is (x\bmod 2) and second bit is (y\bmod 2), e.g. (3,4) goes to 10 and (5,5) goes to 11. We have P_1,P_3,P_5,P_7,P_9 which are 5 pigeons, so 2 of them must be in the same hole, say P_i and P_j. The midpoint of P_i and P_j must be integral, and (i,j) is not on the edge.

General solution

Things become easy when we solved the n=10 case. Let the points be P_1,P_2,\dots,P_n. For any 0\le k<\lfloor\frac{n}{10}\rfloor, we regard P_{10k+1},P_{10k+2},\dots,P_{10k+9} as a polygon and find a point in it.

For example, in the polygon below, \lfloor n/10\rfloor=5, so we find a point Q_1 in the convex hull of \{P_1,P_2,\dots,P_{10}\}, find Q_2 in the convex hull of \{P_{11},\dots,P_{20}\}, and so on until Q_5. Then we output \{Q_1,Q_2,\dots,Q_{\lfloor n/10\rfloor}\}. We can guarantee that the points are integral, pairwise distinct, and strictly inside the polygon.

setter's solution

The total time complexity is O(n).

DFS/BFS Based Solution

Click to view

This problem can also be solved by something similar to DFS/BFS. Firstly, we can find an arbitrary point strictly inside the polygon, by what we did in Setter’s solution. (Or by whatever algorithm you like, if it’s correct and fast!) Then we start to search the neighbors of this point, until we find \lfloor n/10\rfloor points. One possible pseudocode is as follows:

dfs(p) //p is a point
  if num_visited == n / 10
    end the search process
  for q in p's neighbors //p has 4 neighbors
    if q is strictly inside the polygon
      if q is not visited
        visited[q] = 1 //we can use set<point> to maintain visited points
        num_visited += 1
        dfs(q)
//below is main procedure
p = arbitrary point strictly inside the polygon
num_visited = 1
visited[p] = 1
dfs(p)

A data structure

In the above pseudocode, we need a data structure that given a point (x,y), check if (x,y) is strictly inside the polygon. We can do it in O(\log n) time, and actually there are (at least) two simple ways to do it.

Way 1: Finding upper/lower hulls

Click to view

First, let’s find the upper and lower hull. Let L=\arg\min_{i=1}^nx*, R=\arg\max_{i=1}^nx*. Since the polygon is given in counter-clockwise order, the upper hull is given by P_R, P_{R+1}, \dots, P_L and the lower hull is P_L, P_{L+1}, \dots, P_R.

finding upper & lower hull

Consider a query point Q=(x_0,y_0). If it’s strictly below the upper hull, and strictly above the lower hull, then it’s inside the polygon; otherwise it’s not. Take the upper hull as an example. By binary search we can find the smallest i s.t. x_i\ge x_0. We find the y coordinate of the intersection of the segment (x_{i+1},x_i) and the vertical line x=x_0. If y>y_0, then Q is strictly below the upper hull. For example, in the following picture, Q_1 is strictly below the upper hull, but Q_2 is not.

binary search on upper hull

We use the same procedure to check if Q is above the lower hull. Then we can decide whether a point is strictly inside the polygon in two binary searches.

Way 2: Sorting around the leftmost point

Click to view

This method works for any point but we consider the leftmost point for convenience. The leftmost point of a convex polygon is defined as the point with smallest x. In case a tie happens, we arbitrarily choose one. Suppose this polygon is P_1P_2\dots P_n, and the leftmost point is P_1. We define the angle of a point (x,y) is \mathrm{atan2}(y-P_1.y,x-P_1.x)(See atan2). To give a first impression, consider the following figure:

leftmost point; angles

  • In the left polygon, the point P is the only lowest point;
  • In the right figure, the angle of A is \alpha, while the angle of B is -\beta.
  • It’s easy to see that, for any 2\le i\le n, the angle of P_i is in [-\pi/2,\pi/2].

In the preprocessing phase, we just compute the angles of all points \alpha_2,\alpha_3,\dots,\alpha_n. We have \alpha_2<\alpha_3<\dots<\alpha_n.

Consider a query point Q(x,y) whose angle is \alpha. If \alpha\le \alpha_2 or \alpha\ge\alpha_n, it’s definitely not in the polygon(strictly); otherwise we find the largest i such that \alpha_i < \alpha. If the segment (P_1,Q) intersects with (P_i,P_{i+1}), then Q is not in the polygon(see Q_2 of the following figure); otherwise Q is in the polygon. This takes O(\log n) time.

angular binary search

Tester’s Solution

Click to view
  • Let’s define L(t) be the y coordinate of the intersection point of the lower hull and vertical line x=t.
  • Similarly define U(t) be that of upper hull and vertical line x=t.
  • Find the smallest x_0 such that the open interval (L(x_0),U(x_0)) contains integer.
  • For all integer y\in (L(x_0),U(x_0)), we output (x_0,y_0). That is, the blue points below.
  • Imagine we have a vertical line at x=x_0. Now we move it to the right, i.e. x_0\gets x_0+1.
  • In this process, we can maintain L(x_0) and U(x_0) efficiently by maintaining the intersected segment of polygon and vertical line, i.e. the red segments below.

tester's solution

  • We repeatedly move the vertical line rightwards, i.e. x_0\gets x_0+1. When we increase x_0, we maintain the two red segments, and L(x_0),U(x_0) can be computed in O(1) time. Then we spend O(U(x_0)-L(x_0)) time to output all integral points on the vertical line. Remember to quit as soon as we have \lfloor n/10\rfloor points.
  • The time complexity is O(n).

How to find the first x_0?

We don’t need to find x_0 accurately; an estimate x'\in [x_0-5,x_0] is okay. (In fact this 5 is arbitrary; it means we scan 5 useless points.) Let L be the leftmost point. To find x_0, we consider three cases:

  • Case I: y[L+1] e y[L] and y[L-1] e y[L]. In this case, x_0=x[L]+1.
  • Case II: y[L-1]=y[L]. Since L-1,L,L+1 are not collinear, y[L+1] e y[L]. Since \cot heta=\frac{x[L+1]-x[L]}{y[L+1]-y[L]}, the answer is x[L]+\cot heta=x[L]+\frac{x[L+1]-x[L]}{y[L+1]-y[L]}.
  • Case III: y[L+1]=y[L]. This is symmetric to Case II, and the answer is x[L]+\frac{x[L-1]-x[L]}{y[L-1]-y[L]}.

finding first $x_0$

Bonus: how to generate a convex polygon?

Click to view

There is an easy way to generate convex polygons:

  • Take n-1 random vectors v_1,v_2,\dots,v_{n-1}, and compute v_n=-\sum_{i=1}^{n-1}v_i. This is to ensure that \sum_{i=1}^nv_i is the zero vector.
  • Then sort v_1,v_2,\dots,v_n by their angle(the angle of vector (x,y) is defined as \mathrm{atan2}(y,x), see atan2).
  • Let’s walk around the vectors. Choose a start point P, and the vertices of the polygon are P,P+v_1,P+v_1+v_2,P+v_1+v_2+v_3,\dots,P+\sum_{i=1}^{n-1}v_i.
  • See the figure below for better understanding.

generate random polygons

Basically we use this method to construct test data. We also have some handmade data to hack flawed solutions.

ALTERNATIVE SOLUTION:

There are many solutions for this problem. Please, feel free to share your approaches!

AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here.
Tester’s solution 1 can be found here.
Tester’s solution 2 can be found here.
Tester’s solution 3 can be found here.

8 Likes

The entire editorial dosen’t even talk about Pick’s Theorm, which is required to even check whether or not it is possible to find that many integer points.

1 Like

Another approach :

1)Find centroid of Polygon. The polygon centroid must lie inside the polygon as it is convex.

2) Join each vertex and the polygon centroid, thus forming ‘N’ triangles.

3) Find centroid for each of ‘N’ triangles. (Centroid of all triangles will also lie inside each respective triangle and obviously the polygon)

4) Now you have total ‘N+1’ points inside polygon.

5) Check for each point whether it lies inside polygon. Break after you find floor(N/10) such points.

6) Time complexity = O(n) + O(n*log(n)) = O(nlog(n)).

[Note:

a) This approach surely works of N>50 I guess. Probably requires some fine tunning for 10<=N<=50.

b) Approach fits in time limit probably because the first floor(N/10) points we calculate always lie inside the polygon, so for(N>50) no need even to check whether they lie inside the polygon or not]

1 Like

From the given n points- Form 4 groups like even_even, even_odd, odd_even and odd_odd(where x_y are x and y cordinates respectively).
Find the mid points of of all elements in a group.
They will have integer cordinates and will be inside the polygon.

I did this problem by finding the centroid of the polygon. Since the polygon is Convex, Centroid is surely strictly inside the polygon.Then find the mid pts of each of the N vertices and the centroid and stop when the number of points exceed floor(n/10). This gave me 70 pts…probably some careful implementation was required for low N. For small N I did horizontol scanline. (Drawing horizontol lines and checking the intersection points).

since we only have to find n/10 points we can iterate over all given points of the polygon and look for the point in the immediate neighbour i.e(left, right, up, down) and add them to the set.
the point checking part could be done with area of triangle approach where area of all 3 subtriangle == area of main triangle(3 consecutive points of the polygon).

I followed the tester’s approach by scanning a line from left to right. I found the intersection points by maintaining a list of lower and upper parts of polygon and calculating intersection points from the equation of a line. I got 70 points and couldn’t pass the first test case in first subtask. Here is a link to my solution.
Can anybody point out to me which test case I am missing?

I used the bfs approach, but to check the point is inside or outside, I used a different approach, I basically triangulated the polygon and in each triangle I used the bfs approach to find the points. Checking if a point lies inside a triangle or not takes O(1).

My approach was very different and very easy too. We just use one property of a polygon: If A and B are distinct vertices, then every point on segment AB will be inside it.

There are N vertices. We can divide them into 4 disjoint groups based on the parities of X and Y coordinates. (first group: all vertices (x,y) where x is odd and y is odd etc.)

It’s clear that inside each group, if we take any 2 points A and B, the midpoint of segment AB will have integer coordinates, since both X and Y coordinates have same parities. There are 4 groups, so the largest group will have at least [N/4] points. We can simply choose one vertex and connect it to every other in this group, giving us [N/4] - 1 distinct points, which is more than [N/10] already.

After some pondering I realized that the constraints (convexity and non-collinearity) seems to force that there must exist a lot of internal points (and hence enough points will always exist).

I tried a rough solution that I did not expect to work at the time, randomly pick two non-adjacent corner points A=(x_1,y_1) and B=(x_2,y_2) and if g=\gcd(x_1-x_2,y_1-y_2) eq 1 then there exist g-1 integer points along line AB which can easily be found. Repeat until enough unique points are found.

After getting AC with this the best rationale I can give for this working is that for this not to work you would have to pick a set of points where all pairwise differences in x and y are coprime, which seems quite unlikely for sets with a large number (\geq10) of points.

3 Likes

centroid must lie inside the triangle;

  • c=((x1+x2+x3/3),(y1+y2+y3)/3)
  • (x,y) is congruent to (0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2) mod 3
  • so in this only few combinations will have ( c=((x1+x2+x3),(y1+y2+y3))) congruent (0,0) mod 3
  • for the above combinations c=((x1+x2+x3/3),(y1+y2+y3)/3) is integer always;
  • and by php(pigeon hole principle) we know that atleast X=floor(n/9) will have modulo 3,so we can find (X)C3 centroids which are integers;
  • but this might not always give floor(n/10) points due to some centroids might overlap
  • so we are finding all centroids which are integers if it crosses floor(n/10) we stop
  • we use a set too avoid repetition.
  • time complextity is O(nlogn)
  • view my soln here- https://www.codechef.com/viewsolution/17348337

@lokesh2002
Even I did follow a similar approach First I chose 3 points from the polygon and checked if the centroid of the triangle (Since It always lies inside the polygon) has integer coordinates but I did not check the condition “if my centroids obtained of two different triangles will coincide or not”. My code seems to pass even without checking this condition . So, Is it like two centroids of two different triangles in a convex polygon never coincide (or) any specific reason ??

My code - https://www.codechef.com/viewsolution/17364307

I took a random triangle and used binary search to find the smallest triangle in which the point lies. Then i checked if the point lies inside the triangle by finding the areas of the three trianglesand the entire triangle.
Also i didnt check for all the vertices. Checking the adjacent point of each vertex will suffice.
Complexity -O(nlogn) as we only need to check the 4 adjacent points of each vertex of the polygon.

I just checked the 2 adjacent points of each vertex. Yikessss!!
Checked whether inside by checking if that point lies on the left of 2 edges meeting at that vertex

@r_64 @vijju123

what i did was find the leftmost x and rightmost x

loop through i = lx+1 to rx-1

then for each i , find where the vertical line intersects lower_y coordinate and upper_y coordinate.

then from floor(lower_y)+1 to ceil(higher_y)-1 store all the

values until it reaches our desired result.

is this method wrong ??

https://www.codechef.com/viewsolution/17432565

@admin I think there could have been a better question,by giving the required no of points as input itself, rather than setting it to (n/10). The solution for (n/10) is pretty simpler . But then a case could have been asked for
to print all the points inside the polygon. My approach would handle such cases as well.
I divided the n sided polygon into (n-2) triangles. Now for each triangle, starting from its centroid I ran a bfs to cover all points lying inside it. If there are k points, Complexity is O(klog(k)) since I used set to avoid duplicates.

My code : https://www.codechef.com/viewsolution/17322525 In this code, set the req variable to as desired, to get that many points as output.

1 Like

I used line sweeping algorithm to print the points. Link to my solution is as follow:
https://www.codechef.com/viewsolution/17365491

i have tried both approach dividing into triangles and simply using whole polygon:

  1. Approach 1:simply using polygon

in simply using whole polygon approach first i checked it contain sufficient point or not by using pick’s theorem then calculated maxY and minY and then start sweeping line to adjacent point on boundary of polygon and calculating all points on line by using bresenham line points generating algorithm and checking inclusion also…by this i passed top 6 subcases passed in less time limit and got TLE for last 2.link text

  1. Approach 2:dividing whole polygon into triangle

as said in editorial i divided polygon into triangle and used triangle inclusion test not that Area test for inclusion given in editorial because i feel it will give answer true(for inside) when point is in boundary which will give wrong answer.and similarly find points inside triangle as in previous approach(rotating line point to next point)…but this give large TLE on face only top 2 test case passed.and after testing time taken by traingle inclusion i found out it is taking too much time than fast polygon inclusion test.link text

**After contest over i saw some solutions and find that this question did a comedy with me. it is not a medium problem as tags are saying(binary search…etc.) and can be bypassed easily with some observations.link text
**

Here, in the convex polygon, all interior angles are less than or equal to 180 degrees, OR strictly less than 180 degrees.

Weak Test case Have a look at this solution of Poinpoly problem (it’s not mine) https://www.codechef.com/viewsolution/17306405

1 Like