Part 2

After second part, our algorithm became something like this -

for each 0 <= i < n, initialize A = P

_{i}, B = P

_{(i + 1) % n}, C = P

_{(i + 2) % n}, then move C counter clockwise until area non-decreasing, move B to its adjacent point counter clockwise, if area non-decreasing, move C again until area non decreasing and so on. You will get the maximum area for each A and output the maximum of these areas. For a given point starting point P

_{i}, if the maximum area is ABC. We can say ABC is 'stable' triangle for A, in other word s(p

_{i}, P

_{i + 1}, P

_{i + 2}) = ABC, that means we cannot move B or C any further by not decreasing the area.

Now, after getting a maximum ABC for a fixed first point P

_{i}, we will start with P

_{i + 1}, P

_{i + 2}and P

_{i + 3}. We get A'B'C' as a stable triangle i.e. s(p

_{i}, p

_{i + 1}, p

_{i + 2}) = A'B'C'. Hence our algorithm is O(n ^ 2). But it can be proved that, rather than, start from P

_{i + 1}as a second point and P

_{i + 2}as a third point, we can start with B and C as a second and third point and get the same result, i.e. s(P

_{i + 1}, P

_{i + 2}, P

_{i + 3}) = s(P

_{i + 1}, B, C), where s(P

_{i}, P

_{i + 1}, P

_{i + 2}) = ABC.

Lets assume that we are wrong, and s(P

_{i + 1}, P

_{i + 2}, P

_{i + 3}) = A'B'C' > s(p

_{i}, P

_{i + 1}, P

_{i + 2}).

Case 1:

B' < B and C' = C

We know that,

ABC >= AB'C (otherwise largest triangle for A would be AB'C)

=> ABC + BB'A >= AB'C + BB'A

=> AB'BC >= AB'C + BB'A

=> AB'C + BB'C >= AB'C + BB'A

=> BB'C >= BB'A

=> BB'C >= BB'A' (for base BB' area is non decreasing from BB'C to BB'A, so it

will be non decreasing from BB'A to BB'C)

=> BB'C + A'B'C >= BB'A' + A'B'C

=> A'B'BC >= BB'A' + A'B'C

=> A'BC + BB'A' >= BB'A' + A'B'C

=> A'BC >= A'B'C

=> A'BC >= A'B'C' ...... [1] (as C = C')

So, for case 1, s(P

_{i + 1}, P

_{i + 2}, P

_{i + 3}) = s(P

_{i + 1}, B, C).

**Case 2:**

Let's say, A'B'C' is the largest triangle for first point A'.

Where, B' < B and C' > C. See the following image.

For first point A, ABC is the largest trianlge,

So, ABC >= AB'C

=> ABC + BB'C >= ABC + BB'C

=> ABC + BB'C >= AB'BC

=> ABC + BB'C >= ABC + BB'A

=> BB'C >= BB'A

i.e. BB'A <= BB'C Hence, BB'A <= BB'C' (otherwise, BB'A > BB'C and BB'A <= BB'C, which is not possible for a convex polygon) So, BB'A' <= BB'C' ... [2] (as the triangle area non increasing by moving C' to A, it will be non increasing from A to A') Now as we are considering, for first point A', the largest area triangle is A'B'C'. So, A'B'C' > A'BC'

=> A'B'C' + BB'A' > A'BC' + BB'A

=> A'B'C' + BB'A' > A'B'BC'

=> A'B'C' + BB'A' > A'B'C' + BB'C'

=> BB'A' > BB'C', this contradicts to [2].

So it is not possible to have, A'B'C' > A'BC' i.e. for case 2, s(P

_{i + 1}, P

_{i + 2}, P

_{i + 3}) = s(P

_{i + 1}, B, C).

**Case 3:**

Let's say, A'B'C' is the largest triangle for fixed first point A', where,

B' < B and C' < C

As ABC is the largest triangle having first point A,

either ABC' >= AB'C' (by moving B' to B, area increases) .. (3A)

or AB'C >= AB'C' (by moving C' to C, area increases) .. (3B)

**Case 3A**

ABC' >= AB'C'

=> ABC' + BB'A >= AB'C' + BB'A

=> AB'BC' >= AB'C' + BB'A

=> AB'C' + BB'C' >= AB'C' + BB'A

=> BB'C' >= BB'A

=> BB'C' >= BB'A' (As by moving C' to A area non increasing, moving A to A' will area be non increasing)

=> BB'C + A'B'C' >= BB'A' + A'B'C'

=> BB'A'C' >= BB'A' + A'B'C'

=> BB'A' + A'BC' >= BB'A' + A'B'C'

=> A'BC >= A'B'C'

So it is not possible to have, A'B'C' > A'BC' i.e. for case 3A, s(P

_{i + 1}, P

_{i + 2}, P

_{i + 3}) = s(P

_{i + 1}, B, C).

**Case 3B**

AB'C >= AB'C'

=> AB'C + CC'B' >= AB'C' + CC'B'

=> AB'C'C >= AB'C' + CC'B'

=> AB'C' + CC'A >= AB'C' + CC'B'

=> CC'A >= CC'B'

=> CC'A' >= CC'B' (As moving A to B', area decreases, it's not possible moving A' to B' area increases)

=> CC'A' + A'B'C' >= CC'B' + A'B'C'

=> A'B'C'C >= CC'B' + A'B'C'

=> CC'B' + A'B'C >= CC'B' + A'B'C'

=> A'B'C >= A'B'C'

So it is not possible to have, A'B'C' > A'B'C i.e. for case 3B, s(P

_{i + 1}, P

_{i + 2}, P

_{i + 3}) = s(P

_{i + 1}, B, C).

**Case 4:**

C' < C and B' = B

Here,

ABC >= ABC'

=> ABC + CC'A >= ABC' + CC'A

=> ABC + CC'A >= ABC'C

=> ABC + CC'A >= ABC + CC'B

=> CC'A >= CC'B

=> CC'A' >= CC'B

=> CC'A' + A'BC' >= CC'B + A'BC'

=> A'BC'C >= CC'B + A'BC'

=> A'BC + CC'B >= CC'B + A'BC'

=> A'BC >= A'BC'

=> A'BC >= A'B'C' (as B' = B)

So it is not possible to have, A'B'C' > A'BC i.e. for case 3B, s(P

_{i + 1}, P

_{i + 2}, P

_{i + 3}) = s(P

_{i + 1}, B, C).

**Case 5:**

B' > B and C' < C

Now,

A'BC >= A'BC' (From case 4)

=> A'BC + CC'B >= A'BC' + CC'B

=> A'BC'C >= A'BC' + CC'B

=> A'BC' + CC'A' >= A'BC' + CC'B

=> CC'A' >= CC'B

=> CC'A' >= CC'B'

=> CC'A' + A'B'C' >= CC'B' + A'B'C'

=> CC'B'A' >= CC'B' + A'B'C'

=> CC'B' + A'B'C >= CC'B' + A'B'C'

=> A'B'C >= A'B'C'

So it is not possible to have, A'B'C' > A'B'C i.e. for case 5, s(P

_{i + 1}, P

_{i + 2}, P

_{i + 3}) = s(P

_{i + 1}, B, C).

So, from all of the 5 possible cases, we found that, s(P

_{i + 1}, P

_{i + 2}, P

_{i + 3}) = s(P

_{i + 1}, B, C). So our algorithm will still work, if we start our iteration for first point i = i + 1, without changing the position 2nd and 3rd point. We will move first point from 0 to n - 1, and 2nd and 3rd point will also not move more than n times. So our algorithm in worst case is O (3 * n) = O(n)

My simple implementation in C++.

#include<cstdio>

#include<cstring>

#include<cmath>

#include<cstdlib>

#include<vector>

#include<string>

#include<algorithm>

#include<set>

#include<map>

#include<cassert>

#include<sstream>

#include<queue>

#include<stack>

#include<iostream>

using namespace std;

struct Point {

int x;

int y;

Point(int _x, int _y) : x(_x), y(_y) {}

Point() : x(0), y(0) {}

bool operator <(const Point& p) const {

return x < p.x || (x == p.x && y < p.y);

}

};

bool left_turn(const Point& p1, const Point& p2, const Point& p3) {

return (p2.y - p1.y) * (p3.x - p1.x) - (p2.x - p1.x) * (p3.y - p1.y) > 0;

}

// Returns list of points of convex hull in counter clockwise

// The last point and first point will be equal

vector<Point> convex_hull(vector<Point> p) {

vector<Point> ret;

sort(p.begin(), p.end());

// build lower hull

for (int i = 0; i < p.size(); ++i) {

while (ret.size() >= 2 &&

left_turn(ret[ret.size() - 2], ret[ret.size() - 1], p[i])) {

ret.pop_back();

}

ret.push_back(p[i]);

}

int lower_hull_size = ret.size();

// build upper hull

for (int i = p.size() - 2; i >= 0; --i) {

while (ret.size() - lower_hull_size >= 1 &&

left_turn(ret[ret.size() - 2], ret[ret.size() - 1], p[i])) {

ret.pop_back();

}

ret.push_back(p[i]);

}

return ret;

}

double triangle_area (const Point& p1, const Point& p2, const Point& p3) {

return abs((double) p1.x * p2.y +

(double) p2.x * p3.y +

(double) p3.x * p1.y -

(double) p1.y * p2.x -

(double) p2.y * p3.x -

(double) p3.y * p1.x) / 2.0;

}

int main (void)

{

int n;

while (scanf("%d", &n) && n != -1) {

vector<Point> p;

for (int i = 0; i < n; ++i) {

int x, y;

scanf("%d%d", &x, &y);

p.push_back(Point(x, y));

}

p = convex_hull(p); p.pop_back();

int i = 0;

int j = i + 1;

int k = j + 1;

double res = 0.;

while (true) {

double area = triangle_area(p[i], p[j], p[k]);

while (true) {

while (true) {

int nk = (k + 1) % n;

double narea = triangle_area(p[i], p[j], p[nk]);

if (narea >= area) {

k = nk;

area = narea;

} else {

break;

}

}

int nj = (j + 1) % n;

double narea = triangle_area(p[i], p[nj], p[k]);

if (narea >= area) {

j = nj;

area = narea;

} else {

break;

}

}

if (area > res) res = area;

i++;

if (i == j) j = (j + 1) % n;

if (j == k) k = (k + 1) % n;

if (i == n) break;

}

printf("%.2lf\n", res);

}

return 0;

}

very no good

ReplyDeleteThis paper disproves the O(n) alg. https://arxiv.org/pdf/1705.11035.pdf

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