How to calculate minimal distance in arrays? - c

I have problems calculating the minimal distance in arrays. Since I'm a newbie in using C, there could be a major obvious mistake I'm making but can't spot.
I will give an example. Delta_dis should be the calculated distance.
array[0][0]=9.317;
array[0][1]=3;
array[1][0]=3;
array[1][1]=3;
array[2][0]=10;
array[2][1]=10;
array[3][0]=3;
array[3][1]=10;
for (i=0; i<4; i++)
{
for (k=0; k<4; k++)
{
delta_dis[k] = sqrt(pow((array[i][0]-array[k][0]),2) + pow((array[i][1]-array[k][1]),2));
printf("\ndelta_dis_before[%i]: %f",i,delta_dis[k])
if (delta_dis[k-1]!=0)
{
if (delta_dis[k]>delta_dis[k-1])
{
delta_dis[k] = delta_dis[k-1];
}else if (delta_dis[k]<delta_dis[k-1] & i==k)
{
delta_dis[k] = delta_dis[k-1];
}else
{
delta_dis[k] = delta_dis[k];
}
}else
{
delta_dis[k]=delta_dis[k];
}
printf("\ndelta_dis_after[%i]: %f",i,delta_dis[k])
}
}
So in the first step I calculate the distance between these points. And the result (called delta_dis_before[k]) is correct. Then I want to check, if the previous calculated distance is smaller, if so, then delta_dis[k] should be exchanged with the smaller distance. However the minimal distance should not be changed if the selected arrays are the same i=k.
But as soon as I run this through, delta_dis_after is always 0.
For example what I get for i=0 is (I did round the numbers in the following)
delta_dis_before[0]: 0
delta_dis_after[0]: 0
delta_dis_before[1]: 6.317
delta_dis_after[1]: 0
delta_dis_before[2]: 7.033
delta_dis_after[2]: 0
delta_dis_before[3]: 9.429
delta_dis_after[3]: 0
but what I want is:
delta_dis_before[0]: 0
delta_dis_after[0]: 0
delta_dis_before[1]: 6.317
delta_dis_after[1]: 6.317
delta_dis_before[2]: 7.033
delta_dis_after[2]: 6.317
delta_dis_before[3]: 9.429
delta_dis_after[3]: 6.317
I hope I could explain properly what the problem is and what I want it to calculate.
Can someone please help me resolve this issue? Thank you!

Related

A rat in a maze using stacks

- mark[1,0]=1; top=1; stack[top].row=1; stack[top].col=0;
stack[top].dir=1; while (top>0) {
loc = pop();
cr=loc.row; cc=loc.col; cd=loc.dir;
while (cd <= 4) {
nr=cr+move[cd].r; nc=cc+move[cd].c;
if (nr==grow && nc==gcol) {
for(i=0; i<=top; i++) printLoc(stack[i]);
print(cr, cc); print(grow, gcol);
return;
}
if (maze[nr][nc]==0 && mark[nr][nc]==0) {
mark[nr][nc] = 1;
loc.row=cr; loc.col=cc; loc.dir=cd+1;
push(loc);
cr=nr; cc=nc; cd=1;
}
else cd=cd+1;
} } print(“no path in maze”);
I've been looking at this algorithm for about an hour but still can't understand it.
I understand that col/row/direction = column, row, direction, and mark means a 2-dimensional array. But I can't get how to implement this algorithm and turn it into code since I can't understand how it works.
Could someone explain this algorithm, please?
More info: 1 = 0,1(move right), 2 = 1,0(move down), 3 = 0, -1(move left), 4 = -1,0(move up). To store the path use stack. If you use stack choose depth-first search, if you use queue choose breadth-first search. 1 is the blocked path while 0 is the path you can walk on.

Printing number of items in a range from an array in Kotlin

Currently working on a little school assignment, trying to figure out if there is a specific way to print items out of a randomized array. Teacher wants us to generate 200 random numbers in a range of 1 through 100, store them, and then print them with the amount of each range(10s, 20s, 30s, etc). I've gotten the randomization and storage down, but I'm having a little issue trying to figure out how to properly display these.
We should have them separated by how many random numbers are 1-10, 11-20, etc etc, but also with a histograph to the side. I think I can figure out the histograph well enough, but I could use some help on how to display the numbers for each range! I hope this makes sense, any help is appreciated!
What I have so far is very simple, just having a rough time figuring out what goes in the println()
fun main() {
val rnds = (1..200).map { (1..100).random() }
println() // <-- ???
}
it should look something like:
1 - 10: # // <-- number of randomized numbers in that range here
11 - 20: # // <-- number of randomized numbers in that range here
21 - 30: # // <-- number of randomized numbers in that range here
etc etc, all the way until 91 - 100
I should also mention that I'm VERY new to kotlin, and still pretty green to programming in general. The easier the solution, the better, at least until I have a better understanding at what I'm seeing.
You can create your range buckets by using the map function on a range of each starting number of a bucket, and then associateWith to sort the original lists into these buckets:
val byRanges: Map<IntRange, List<Int>> = (1..100 step 10)
.map { it..(it + 9) }
.associateWith { range -> rnds.filter { it in range } }
Then when you want to convert this to text, you can use the joinToString() function on the map entries to format each item as text:
val asText: String = byRanges.entries.joinToString("\n") { (range, list) ->
"${range.start} - ${range.endInclusive}: # ${list.size}"
}
You can increase the "for" loop by range value and filter the list with desired condition to get the item count.
fun main() {
val randomRangeStart = 1
val randomRangeEnd = 100
val list = (1..200).map { (randomRangeStart..randomRangeEnd).random() }
for (i in randomRangeStart..randomRangeEnd step 10) {
println("Range $i - ${i+9}: ${list.filter { it >= i && it <= i+9 }.count()}")
}
}

Given an array, find combinations of n numbers that are less than c

This is a tough one, at least for my minimal c skills.
Basically, the user enters a list of prices into an array, and then the desired number of items he wants to purchase, and finally a maximum cost not to exceed.
I need to check how many combinations of the desired number of items are less than or equal to the cost given.
If the problem was a fixed number of items in the combination, say 3, it would be much easier with just three loops selecting each price and adding them to test.
Where I get stumped is the requirement that the user enter any number of items, up to the number of items in the array.
This is what I decided on at first, before realizing that the user could specify combinations of any number, not just three. It was created with help from a similar topic on here, but again it only works if the user specifies he wants 3 items per combination. Otherwise it doesn't work.
// test if any combinations of items can be made
for (one = 0; one < (count-2); one++) // count -2 to account for the two other variables
{
for (two = one + 1; two < (count-1); two++) // count -1 to account for the last variable
{
for (three = two + 1; three < count; three++)
{
total = itemCosts[one] + itemCosts[two] + itemCosts[three];
if (total <= funds)
{
// DEBUG printf("\nMatch found! %d + %d + %d, total: %d.", itemCosts[one], itemCosts[two], itemCosts[three], total);
combos++;
}
}
}
}
As far as I can tell there's no easy way to adapt this to be flexible based on the user's desired number of items per combination.
I would really appreciate any help given.
One trick to flattening nested iterations is to use recursion.
Make a function that takes an array of items that you have selected so far, and the number of items you've picked up to this point. The algorithm should go like this:
If you have picked the number of items equal to your target of N, compute the sum and check it against the limit
If you have not picked enough items, add one more item to your list, and make a recursive call.
To ensure that you do not pick the same item twice, pass the smallest index from which the function may pick. The declaration of the function may look like this:
int count_combinations(
int itemCosts[]
, size_t costCount
, int pickedItems[]
, size_t pickedCount
, size_t pickedTargetCount
, size_t minIndex
, int funds
) {
if (pickedCount == pickedTargetCount) {
// This is the base case. It has the code similar to
// the "if" statement from your code, but the number of items
// is not fixed.
int sum = 0;
for (size_t i = 0 ; i != pickedCount ; i++) {
sum += pickedItems[i];
}
// The following line will return 0 or 1,
// depending on the result of the comparison.
return sum <= funds;
} else {
// This is the recursive case. It is similar to one of your "for"
// loops, but instead of setting "one", "two", or "three"
// it sets pickedItems[0], pickedItems[1], etc.
int res = 0;
for (size_t i = minIndex ; i != costCount ; i++) {
pickedItems[pickedCount] = itemCosts[i];
res += count_combinations(
itemCosts
, costCount
, pickedItems
, pickedCount+1
, pickedTargetCount
, i+1
, funds
);
}
return res;
}
}
You call this function like this:
int itemCosts[C] = {...}; // The costs
int pickedItems[N]; // No need to initialize this array
int res = count_combinations(itemCosts, C, pickedItems, 0, N, 0, funds);
Demo.
This can be done by using a backtracking algorithm. This is equivalent to implementing a list of nested for loops. This can be better understood by trying to see the execution pattern of a sequence of nested for loops.
For example lets say you have, as you presented, a sequence of 3 fors and the code execution has reached the third level (the innermost). After this goes through all its iterations you return to the second level for where you go to the next iteration in which you jump again in third level for. Similarly, when the second level finishes all its iteration you jump back to the first level for which continues with the next iteration in which you jump in the second level and from there in the third.
So, in a given level you try go to the deeper one (if there is one) and if there are no more iterations you go back a level (back track).
Using the backtracking you represent the nested for by an array where each element is an index variable: array[0] is the index for for level 0, and so on.
Here is a sample implementation for your problem:
#define NUMBER_OF_OBJECTS 10
#define FORLOOP_DEPTH 4 // This is equivalent with the number of
// of nested fors and in the problem is
// the number of requested objects
#define FORLOOP_ARRAY_INIT -1 // This is a init value for each "forloop" variable
#define true 1
#define false 0
typedef int bool;
int main(void)
{
int object_prices[NUMBER_OF_OBJECTS];
int forLoopsArray[FORLOOP_DEPTH];
bool isLoopVariableValueUsed[NUMBER_OF_OBJECTS];
int forLoopLevel = 0;
for (int i = 0; i < FORLOOP_DEPTH; i++)
{
forLoopsArray[i] = FORLOOP_ARRAY_INIT;
}
for (int i = 0; i < NUMBER_OF_OBJECTS; i++)
{
isLoopVariableValueUsed[i] = false;
}
forLoopLevel = 0; // Start from level zero
while (forLoopLevel >= 0)
{
bool isOkVal = false;
if (forLoopsArray[forLoopLevel] != FORLOOP_ARRAY_INIT)
{
// We'll mark the loopvariable value from the last iterration unused
// since we'll use a new one (in this iterration)
isLoopVariableValueUsed[forLoopsArray[forLoopLevel]] = false;
}
/* All iterations (in all levels) start basically from zero
* Because of that here I check that the loop variable for this level
* is different than the previous ones or try the next value otherwise
*/
while ( isOkVal == false
&& forLoopsArray[forLoopLevel] < (NUMBER_OF_OBJECTS - 1))
{
forLoopsArray[forLoopLevel]++; // Try a new value
if (loopVariableValueUsed[forLoopsArray[forLoopLevel]] == false)
{
objectUsed[forLoopsArray[forLoopLevel]] = true;
isOkVal = true;
}
}
if (isOkVal == true) // Have we found in this level an different item?
{
if (forLoopLevel == FORLOOP_DEPTH - 1) // Is it the innermost?
{
/* Here is the innermost level where you can test
* if the sum of all selected items is smaller than
* the target
*/
}
else // Nope, go a level deeper
{
forLoopLevel++;
}
}
else // We've run out of values in this level, go back
{
forLoopsArray[forLoopLevel] = FORLOOP_ARRAY_INIT;
forLoopLevel--;
}
}
}

How to draw a polygon from a set of unordered points

Currently, I am using a convex hull algorithm to get the outer most points from a set of points randomly placed. What I aim to do is draw a polygon from the set of points returned by the convex hull however, when I try to draw the polygon it looks quite strange.
My question, how do I order the points so the polygon draws correctly?
Thanks.
EDIT:
Also, I have tried sorting using orderby(...).ThenBy(...) and I cant seem to get it working.
Have you tried the gift wrapping algorithm ( http://en.wikipedia.org/wiki/Gift_wrapping_algorithm)? This should return points in the correct order.
I had an issue where a random set of points were generated from which a wrapped elevation vector needed a base contour. Having read the link supplied by #user1149913 and found a sample of gift-wrapping a hull, the following is a sample of my implementation:
private static PointCollection CalculateContour (List<Point> points) {
// locate lower-leftmost point
int hull = 0;
int i;
for (i = 1 ; i < points.Count ; i++) {
if (ComparePoint(points[i], points[hull])) {
hull = i;
}
}
// wrap contour
var outIndices = new int[points.Count];
int endPt;
i = 0;
do {
outIndices[i++] = hull;
endPt = 0;
for (int j = 1 ; j < points.Count ; j++)
if (hull == endPt || IsLeft(points[hull], points[endPt], points[j]))
endPt = j;
hull = endPt;
} while (endPt != outIndices[0]);
// build countour points
var contourPoints = new PointCollection(points.Capacity);
int results = i;
for (i = 0 ; i < results ; i++)
contourPoints.Add(points[outIndices[i]]);
return contourPoints;
}
This is not a full solution but a guide in the right direction. I faced a very similar problem just recently and I found a reddit post with an answer (https://www.reddit.com/r/DnDBehindTheScreen/comments/8efeta/a_random_star_chart_generator/dxvlsyt/) suggesting to use Delaunay triangulation which basically returns a solution with all possible triangles made within the data points you have. Once you have all possible triangles, which by definition you know won't result on any overlapped lines, you can chose which lines you use which result on all nodes being connected.
I was coding my solution on python and fortunately there's lots of scientific libraries on python. I was working on a random sky chart generator which would draw constellations out of those stars. In order to get all possible triangles (and draw them, just for fun), before going into the algorithm to draw the actual constellations, all I had to do was this:
# 2D array of the coordinates of every star generated randomly before
points = list(points_dict.keys())
from scipy.spatial import Delaunay
tri = Delaunay(points)
# Draw the debug constellation with the full array of lines
debug_constellation = Constellation(quadrants = quadrants, name_display_style = config.constellation_name_display_style)
for star in available_stars:
debug_constellation.add_star(star)
for triangle in tri.simplices:
star_ids = []
for index in triangle:
star_ids.append(points_dict[points[index]].id)
debug_constellation.draw_segment(star_ids, is_closed = True)
# Code to generate the image follows below
You can see the full implementation here: fake_sky_chart_generator/fake_libs/constellation_algorithms/delaunay.py
This is the result:

Point in Polygon Algorithm

I saw the below algorithm works to check if a point is in a given polygon from this link:
int pnpoly(int nvert, float *vertx, float *verty, float testx, float testy)
{
int i, j, c = 0;
for (i = 0, j = nvert-1; i < nvert; j = i++) {
if ( ((verty[i]>testy) != (verty[j]>testy)) &&
(testx < (vertx[j]-vertx[i]) * (testy-verty[i]) / (verty[j]-verty[i]) + vertx[i]) )
c = !c;
}
return c;
}
I tried this algorithm and it actually works just perfect. But sadly I cannot understand it well after spending some time trying to get the idea of it.
So if someone is able to understand this algorithm, please explain it to me a little.
Thank you.
The algorithm is ray-casting to the right. Each iteration of the loop, the test point is checked against one of the polygon's edges. The first line of the if-test succeeds if the point's y-coord is within the edge's scope. The second line checks whether the test point is to the left of the line (I think - I haven't got any scrap paper to hand to check). If that is true the line drawn rightwards from the test point crosses that edge.
By repeatedly inverting the value of c, the algorithm counts how many times the rightward line crosses the polygon. If it crosses an odd number of times, then the point is inside; if an even number, the point is outside.
I would have concerns with a) the accuracy of floating-point arithmetic, and b) the effects of having a horizontal edge, or a test point with the same y-coord as a vertex, though.
Edit 1/30/2022: I wrote this answer 9 years ago when I was in college. People in the chat conversation are indicating it's not accurate. You should probably look elsewhere. 🤷‍♂️
Chowlett is correct in every way, shape, and form.
The algorithm assumes that if your point is on the line of the polygon, then that is outside - for some cases, this is false. Changing the two '>' operators to '>=' and changing '<' to '<=' will fix that.
bool PointInPolygon(Point point, Polygon polygon) {
vector<Point> points = polygon.getPoints();
int i, j, nvert = points.size();
bool c = false;
for(i = 0, j = nvert - 1; i < nvert; j = i++) {
if( ( (points[i].y >= point.y ) != (points[j].y >= point.y) ) &&
(point.x <= (points[j].x - points[i].x) * (point.y - points[i].y) / (points[j].y - points[i].y) + points[i].x)
)
c = !c;
}
return c;
}
I changed the original code to make it a little more readable (also this uses Eigen). The algorithm is identical.
// This uses the ray-casting algorithm to decide whether the point is inside
// the given polygon. See https://en.wikipedia.org/wiki/Point_in_polygon#Ray_casting_algorithm
bool pnpoly(const Eigen::MatrixX2d &poly, float x, float y)
{
// If we never cross any lines we're inside.
bool inside = false;
// Loop through all the edges.
for (int i = 0; i < poly.rows(); ++i)
{
// i is the index of the first vertex, j is the next one.
// The original code uses a too-clever trick for this.
int j = (i + 1) % poly.rows();
// The vertices of the edge we are checking.
double xp0 = poly(i, 0);
double yp0 = poly(i, 1);
double xp1 = poly(j, 0);
double yp1 = poly(j, 1);
// Check whether the edge intersects a line from (-inf,y) to (x,y).
// First check if the line crosses the horizontal line at y in either direction.
if ((yp0 <= y) && (yp1 > y) || (yp1 <= y) && (yp0 > y))
{
// If so, get the point where it crosses that line. This is a simple solution
// to a linear equation. Note that we can't get a division by zero here -
// if yp1 == yp0 then the above if will be false.
double cross = (xp1 - xp0) * (y - yp0) / (yp1 - yp0) + xp0;
// Finally check if it crosses to the left of our test point. You could equally
// do right and it should give the same result.
if (cross < x)
inside = !inside;
}
}
return inside;
}
To expand on the "too-clever trick". We want to iterate over all adjacent vertices, like this (imagine there are 4 vertices):
i
j
0
1
1
2
2
3
3
0
My code above does it the simple obvious way - j = (i + 1) % num_vertices. However this uses integer division which is much much slower than all other operations. So if this is performance critical (e.g. in an AAA game) you want to avoid it.
The original code changes the order of iteration a bit:
i
j
0
3
1
0
2
1
3
2
This is still totally valid since we're still iterating over every vertex pair and it doesn't really matter whether you go clockwise or anticlockwise, or where you start. However now it lets us avoid the integer division. In easy-to-understand form:
int i = 0;
int j = num_vertices - 1; // 3
while (i < num_vertices) { // 4
{body}
j = i;
++i;
}
Or in very terse C style:
for (int i = 0, j = num_vertices - 1; i < num_vertices; j = i++) {
{body}
}
This might be as detailed as it might get for explaining the ray-tracing algorithm in actual code. It might not be optimized but that must always come after a complete grasp of the system.
//method to check if a Coordinate is located in a polygon
public boolean checkIsInPolygon(ArrayList<Coordinate> poly){
//this method uses the ray tracing algorithm to determine if the point is in the polygon
int nPoints=poly.size();
int j=-999;
int i=-999;
boolean locatedInPolygon=false;
for(i=0;i<(nPoints);i++){
//repeat loop for all sets of points
if(i==(nPoints-1)){
//if i is the last vertex, let j be the first vertex
j= 0;
}else{
//for all-else, let j=(i+1)th vertex
j=i+1;
}
float vertY_i= (float)poly.get(i).getY();
float vertX_i= (float)poly.get(i).getX();
float vertY_j= (float)poly.get(j).getY();
float vertX_j= (float)poly.get(j).getX();
float testX = (float)this.getX();
float testY = (float)this.getY();
// following statement checks if testPoint.Y is below Y-coord of i-th vertex
boolean belowLowY=vertY_i>testY;
// following statement checks if testPoint.Y is below Y-coord of i+1-th vertex
boolean belowHighY=vertY_j>testY;
/* following statement is true if testPoint.Y satisfies either (only one is possible)
-->(i).Y < testPoint.Y < (i+1).Y OR
-->(i).Y > testPoint.Y > (i+1).Y
(Note)
Both of the conditions indicate that a point is located within the edges of the Y-th coordinate
of the (i)-th and the (i+1)- th vertices of the polygon. If neither of the above
conditions is satisfied, then it is assured that a semi-infinite horizontal line draw
to the right from the testpoint will NOT cross the line that connects vertices i and i+1
of the polygon
*/
boolean withinYsEdges= belowLowY != belowHighY;
if( withinYsEdges){
// this is the slope of the line that connects vertices i and i+1 of the polygon
float slopeOfLine = ( vertX_j-vertX_i )/ (vertY_j-vertY_i) ;
// this looks up the x-coord of a point lying on the above line, given its y-coord
float pointOnLine = ( slopeOfLine* (testY - vertY_i) )+vertX_i;
//checks to see if x-coord of testPoint is smaller than the point on the line with the same y-coord
boolean isLeftToLine= testX < pointOnLine;
if(isLeftToLine){
//this statement changes true to false (and vice-versa)
locatedInPolygon= !locatedInPolygon;
}//end if (isLeftToLine)
}//end if (withinYsEdges
}
return locatedInPolygon;
}
Just one word about optimization: It isn't true that the shortest (and/or the tersest) code is the fastest implemented. It is a much faster process to read and store an element from an array and use it (possibly) many times within the execution of the block of code than to access the array each time it is required. This is especially significant if the array is extremely large. In my opinion, by storing each term of an array in a well-named variable, it is also easier to assess its purpose and thus form a much more readable code. Just my two cents...
The algorithm is stripped down to the most necessary elements. After it was developed and tested all unnecessary stuff has been removed. As result you can't undertand it easily but it does the job and also in very good performance.
I took the liberty to translate it to ActionScript-3:
// not optimized yet (nvert could be left out)
public static function pnpoly(nvert: int, vertx: Array, verty: Array, x: Number, y: Number): Boolean
{
var i: int, j: int;
var c: Boolean = false;
for (i = 0, j = nvert - 1; i < nvert; j = i++)
{
if (((verty[i] > y) != (verty[j] > y)) && (x < (vertx[j] - vertx[i]) * (y - verty[i]) / (verty[j] - verty[i]) + vertx[i]))
c = !c;
}
return c;
}
This algorithm works in any closed polygon as long as the polygon's sides don't cross. Triangle, pentagon, square, even a very curvy piecewise-linear rubber band that doesn't cross itself.
1) Define your polygon as a directed group of vectors. By this it is meant that every side of the polygon is described by a vector that goes from vertex an to vertex an+1. The vectors are so directed so that the head of one touches the tail of the next until the last vector touches the tail of the first.
2) Select the point to test inside or outside of the polygon.
3) For each vector Vn along the perimeter of the polygon find vector Dn that starts on the test point and ends at the tail of Vn. Calculate the vector Cn defined as DnXVn/DN*VN (X indicates cross product; * indicates dot product). Call the magnitude of Cn by the name Mn.
4) Add all Mn and call this quantity K.
5) If K is zero, the point is outside the polygon.
6) If K is not zero, the point is inside the polygon.
Theoretically, a point lying ON the edge of the polygon will produce an undefined result.
The geometrical meaning of K is the total angle that the flea sitting on our test point "saw" the ant walking at the edge of the polygon walk to the left minus the angle walked to the right. In a closed circuit, the ant ends where it started.
Outside of the polygon, regardless of location, the answer is zero.
Inside of the polygon, regardless of location, the answer is "one time around the point".
This method check whether the ray from the point (testx, testy) to O (0,0) cut the sides of the polygon or not .
There's a well-known conclusion here: if a ray from 1 point and cut the sides of a polygon for a odd time, that point will belong to the polygon, otherwise that point will be outside the polygon.
To expand on #chowlette's answer where the second line checks if the point is to the left of the line,
No derivation is given but this is what I worked out:
First it helps to imagine 2 basic cases:
the point is left of the line . / or
the point is right of the line / .
If our point were to shoot a ray out horizontally where would it strike the line segment. Is our point to the left or right of it? Inside or out? We know its y coordinate because it's by definition the same as the point. What would the x coordinate be?
Take your traditional line formula y = mx + b. m is the rise over the run. Here, instead we are trying to find the x coordinate of the point on that line segment that has the same height (y) as our point.
So we solve for x: x = (y - b)/m. m is rise over run, so this becomes run over rise or (yj - yi)/(xj - xi) becomes (xj - xi)/(yj - yi). b is the offset from origin. If we assume yi as the base for our coordinate system, b becomes yi. Our point testy is our input, subtracting yi turns the whole formula into an offset from yi.
We now have (xj - xi)/(yj - yi) or 1/m times y or (testy - yi): (xj - xi)(testy - yi)/(yj - yi) but testx isn't based to yi so we add it back in order to compare the two ( or zero testx as well )
I think the basic idea is to calculate vectors from the point, one per edge of the polygon. If vector crosses one edge, then the point is within the polygon. By concave polygons if it crosses an odd number of edges it is inside as well (disclaimer: although not sure if it works for all concave polygons).
This is the algorithm I use, but I added a bit of preprocessing trickery to speed it up. My polygons have ~1000 edges and they don't change, but I need to look up whether the cursor is inside one on every mouse move.
I basically split the height of the bounding rectangle to equal length intervals and for each of these intervals I compile the list of edges that lie within/intersect with it.
When I need to look up a point, I can calculate - in O(1) time - which interval it is in and then I only need to test those edges that are in the interval's list.
I used 256 intervals and this reduced the number of edges I need to test to 2-10 instead of ~1000.
Here's a php implementation of this:
<?php
class Point2D {
public $x;
public $y;
function __construct($x, $y) {
$this->x = $x;
$this->y = $y;
}
function x() {
return $this->x;
}
function y() {
return $this->y;
}
}
class Point {
protected $vertices;
function __construct($vertices) {
$this->vertices = $vertices;
}
//Determines if the specified point is within the polygon.
function pointInPolygon($point) {
/* #var $point Point2D */
$poly_vertices = $this->vertices;
$num_of_vertices = count($poly_vertices);
$edge_error = 1.192092896e-07;
$r = false;
for ($i = 0, $j = $num_of_vertices - 1; $i < $num_of_vertices; $j = $i++) {
/* #var $current_vertex_i Point2D */
/* #var $current_vertex_j Point2D */
$current_vertex_i = $poly_vertices[$i];
$current_vertex_j = $poly_vertices[$j];
if (abs($current_vertex_i->y - $current_vertex_j->y) <= $edge_error && abs($current_vertex_j->y - $point->y) <= $edge_error && ($current_vertex_i->x >= $point->x) != ($current_vertex_j->x >= $point->x)) {
return true;
}
if ($current_vertex_i->y > $point->y != $current_vertex_j->y > $point->y) {
$c = ($current_vertex_j->x - $current_vertex_i->x) * ($point->y - $current_vertex_i->y) / ($current_vertex_j->y - $current_vertex_i->y) + $current_vertex_i->x;
if (abs($point->x - $c) <= $edge_error) {
return true;
}
if ($point->x < $c) {
$r = !$r;
}
}
}
return $r;
}
Test Run:
<?php
$vertices = array();
array_push($vertices, new Point2D(120, 40));
array_push($vertices, new Point2D(260, 40));
array_push($vertices, new Point2D(45, 170));
array_push($vertices, new Point2D(335, 170));
array_push($vertices, new Point2D(120, 300));
array_push($vertices, new Point2D(260, 300));
$Point = new Point($vertices);
$point_to_find = new Point2D(190, 170);
$isPointInPolygon = $Point->pointInPolygon($point_to_find);
echo $isPointInPolygon;
var_dump($isPointInPolygon);
I modified the code to check whether the point is in a polygon, including the point is on an edge.
bool point_in_polygon_check_edge(const vec<double, 2>& v, vec<double, 2> polygon[], int point_count, double edge_error = 1.192092896e-07f)
{
const static int x = 0;
const static int y = 1;
int i, j;
bool r = false;
for (i = 0, j = point_count - 1; i < point_count; j = i++)
{
const vec<double, 2>& pi = polygon[i);
const vec<double, 2>& pj = polygon[j];
if (fabs(pi[y] - pj[y]) <= edge_error && fabs(pj[y] - v[y]) <= edge_error && (pi[x] >= v[x]) != (pj[x] >= v[x]))
{
return true;
}
if ((pi[y] > v[y]) != (pj[y] > v[y]))
{
double c = (pj[x] - pi[x]) * (v[y] - pi[y]) / (pj[y] - pi[y]) + pi[x];
if (fabs(v[x] - c) <= edge_error)
{
return true;
}
if (v[x] < c)
{
r = !r;
}
}
}
return r;
}

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