在X1 y1 X2 y2形式的10,000个点的文件中，如何检测至少四个正方形？ 爪哇

Question

I tried this example with only 4 points, it works with:

-1 -1
2 1
4 -2
1 -4

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;

public class CarreSimple {
    static int distance(Point point1, Point point2) {
        int dist = (int) (Math.pow(point1.getX() - point2.getX(), 2) + Math.pow(point1.getY() - point2.getY(), 2));
        dist = (int) Math.sqrt(dist);
        return dist;
    }

    public static void main(String[] args) {
        boolean EstCarre = false;
        Point[] tb = new Point[10000];
        int i = 0;
        BufferedReader br = null;
        try {
            br = new BufferedReader(new FileReader("C:/Users/walid/Downloads/points.txt"));
            String line;
            while ((line = br.readLine()) != null) {
                String[] parts = line.split(" ");
                int part1 = Integer.parseInt(parts[0]);
                int part2 = Integer.parseInt(parts[1]);

                Point p = new Point(part1, part2);
                tb[i] = p;
                System.out.println("X " + tb[i].getX() + " y " + tb[i].getY());
                i++;
            }
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                if (br != null) {
                    br.close();
                }
            } catch (IOException ex) {
                ex.printStackTrace();
            }
        }

        int[] distances = new int[3];
        distances[0] = distance(tb[0], tb[1]);
        distances[1] = distance(tb[0], tb[2]);
        distances[2] = distance(tb[0], tb[3]);
        System.out.println(distances[0]);

        int CoteEgal1 = -1;
        int CoteEgal2 = -1;
        int CotePasEgal = -1;


        if (distances[0] == distances[1]) {
            if (distances[0] != distances[2]) {
                CoteEgal1 = 0;
                CoteEgal2 = 1;
                CotePasEgal = 2;
            }
        } else if (distances[1] == distances[2]) {
            if (distances[1] != distances[0]) {
                CoteEgal1 = 1;
                CoteEgal2 = 2;
                CotePasEgal = 0;
            }
        } else if (distances[0] == distances[2]) {
            if (distances[0] != distances[1]) {
                CoteEgal1 = 0;
                CoteEgal2 = 2;
                CotePasEgal = 1;
            }
        }
        if (CoteEgal1 != -1) {
            int coinOpposÈ = 0;
            switch (CotePasEgal) {
                case 0:
                    coinOpposÈ = distance(tb[2], tb[3]);
                    break;
                case 1:
                    coinOpposÈ = distance(tb[1], tb[3]);
                    break;
                case 2:
                    coinOpposÈ = distance(tb[1], tb[2]);
                    break;
                default:
                    break;
            }

            if (coinOpposÈ == distances[CotePasEgal]) {
                int diagonal = coinOpposÈ;
                int adjacent = distances[CoteEgal1];
                boolean stillOK = true;
                for (int a = 0; a < 4; a++) {
                    int diagonalCount = 0;
                    int adjacentCount = 0;
                    for (int b = 0; b < 4; b++) {
                        if (a != b) {
                            int dist = distance(tb[a], tb[b]);
                            if (dist == diagonal) {
                                diagonalCount++;
                            } else if (dist == adjacent) {
                                adjacentCount++;
                            }
                        }
                    }
                    // est ce que on a 1 diagonal et 2 adjacents
                    if (!(diagonalCount == 1 && adjacentCount == 2)) {
                        stillOK = false;
                        break;
                    }
                }
                if (stillOK) {
                    EstCarre = true;
                }
            }
        }
        if (EstCarre) {
            System.out.println("C'est un carre");
        } else {
            System.out.println("Ce n'est pas un carre");
        }
    }
}

To make a loop on a file of 10,000 points and test on each combination of 4 I found too much difficult.

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;

public class Square {
    static int distance(Point point1, Point point2) {
        int dist = (int) (Math.pow(point1.getX() - point2.getX(), 2) + Math.pow(point1.getY() - point2.getY(), 2));
        dist = (int) Math.sqrt(dist);
        return dist;
    }

    public static void main(String[] args) {
        boolean EstCarre = false;
        Point[] tb = new Point[10000];
        int i = 0;
        BufferedReader br = null;
        try {
            br = new BufferedReader(new FileReader("C:/Users/walid/Downloads/exercice.txt"));
            String line;
            while ((line = br.readLine()) != null) {
                String[] parts = line.split(" ");
                int part1 = Integer.parseInt(parts[0]);
                int part2 = Integer.parseInt(parts[1]);

                Point p = new Point(part1, part2);
                tb[i] = p;
                System.out.println("X " + tb[i].getX() + " y " + tb[i].getY());
                i++;
            }
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                if (br != null) {
                    br.close();
                }
            } catch (IOException ex) {
                ex.printStackTrace();
            }
        }
        for (int i2 = 0; i2 < 10000; i2++) {
            for (int j = 0; j < 10000; j++) {
                for (int k = 0; k < 10000; k++) {
                    for (int l = 0; l < 10000; l++) {
                        if (i2 != j && i2 != k && i2 != l && j != k && j != l && k != l) {
                            int[] distances = new int[10000000];
                            distances[i2] = distance(tb[i2], tb[j]);
                            distances[j] = distance(tb[i2], tb[k]);
                            distances[k] = distance(tb[i2], tb[l]);

                            int CoteEgal1 = -1;
                            int CoteEgal2 = -1;
                            int CotePasEgal = -1;

                            if (distances[i2] == distances[j]) {
                                if (distances[i2] != distances[k]) {
                                    CoteEgal1 = i2;
                                    CoteEgal2 = j;
                                    CotePasEgal = k;
                                }
                            } else if (distances[i2] == distances[k]) {
                                if (distances[j] != distances[i2]) {
                                    CoteEgal1 = j;
                                    CoteEgal2 = k;
                                    CotePasEgal = i2;
                                }
                            } else if (distances[i2] == distances[k]) {
                                if (distances[i2] != distances[j]) {
                                    CoteEgal1 = i2;
                                    CoteEgal2 = k;
                                    CotePasEgal = j;
                                }
                            }

                            int coinOpposÈ = 0;
                            if (CoteEgal1 != -1) {
                                if (CotePasEgal == i2) {
                                    coinOpposÈ = distance(tb[k], tb[l]);
                                } else if (CotePasEgal == j) {
                                    coinOpposÈ = distance(tb[j], tb[l]);
                                } else if (CotePasEgal == k) {
                                    coinOpposÈ = distance(tb[j], tb[k]);
                                }

                                if (coinOpposÈ == distances[CotePasEgal]) {
                                    int diagonal = coinOpposÈ;
                                    int adjacent = distances[CoteEgal1];
                                    boolean stillOK = true;
                                    for (int a = 0; a < 10000000; a++) {
                                        int diagonalCount = 0;
                                        int adjacentCount = 0;
                                        for (int b = 0; b < 10000000; b++) {
                                            if (a != b) {
                                                int dist = distance(tb[a], tb[b]);
                                                if (dist == diagonal) {
                                                    diagonalCount++;
                                                } else if (dist == adjacent) {
                                                    adjacentCount++;
                                                }
                                            }
                                        }
                                        // est ce que on a 1 diagonal et 2 adjacents
                                        if (!(diagonalCount == 1 && adjacentCount == 2)) {
                                            stillOK = false;
                                            break;
                                        }
                                    }
                                    if (stillOK) {
                                        EstCarre = true;
                                    }
                                }
                            }
                        }
                        if (EstCarre) {
                            System.out.println("square found");
                        }
                    }
                }
            }
        }
    }
}

code 2: couple is a class that takes two points

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.util.Iterator;
import java.util.Map;
import java.util.HashMap;
import java.util.Set;

public class CarreP {
    public static void main(String[] args) {
        Point[] tb = new Point[10000];
        int i = 0;
        BufferedReader br = null;
        try {
            br = new BufferedReader(new FileReader("C:/Users/walid/Downloads/exercice.txt"));
            String line;
            while ((line = br.readLine()) != null) {
                String[] parts = line.split(" ");
                int part1 = Integer.parseInt(parts[0]);
                int part2 = Integer.parseInt(parts[1]);

                Point p = new Point(part1, part2);
                tb[i] = p;
                i++;
            }
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                if (br != null) {
                    br.close();
                }
            } catch (IOException ex) {
                ex.printStackTrace();
            }
        }

        HashMap<Couple, Integer> hmap = new HashMap<Couple, Integer>();
        int j = 0;
        while (j < tb.length) {
            for (int k = 0; k < tb.length; k++) {
                int xcarre = (int) Math.pow(tb[j].getX() - tb[k].getX(), 2);
                int ycarre = (int) Math.pow(tb[j].getY() - tb[k].getY(), 2);

                int distance = (int) Math.sqrt(xcarre + ycarre);
                Couple c = new Couple(tb[k], tb[k]);
                hmap.put(c, distance);
                Set set = hmap.entrySet();
                Iterator iterator = set.iterator();
                Iterator iterator2 = set.iterator();
                int[] distances = new int[hmap.size()];
                int s = 0;
                while (iterator.hasNext()) {
                    Map.Entry mentry = (Map.Entry) iterator.next();
                    distances[s] = (int) mentry.getValue();
                    System.out.println(distances[s]);
                    s++;
                }
                int CoteEgal1 = -1;
                int CoteEgal2 = -1;
                int CoteInegal = -1;
                for (int i1 = 0; i1 < distances.length; i1++) {
                    for (int i2 = 0; i2 < distances.length; i2++) {
                        for (int i3 = 0; i3 < distances.length; i3++) {
                            if (distances[i1] == distances[i2]) {
                                if (distances[i1] != distances[i3]) {
                                    CoteEgal1 = i1;
                                    CoteEgal2 = i2;
                                    CoteInegal = i3;
                                }
                            } else if (distances[i2] == distances[i3]) {
                                if (distances[i2] != distances[i1]) {
                                    CoteEgal1 = i2;
                                    CoteEgal2 = i3;
                                    CoteInegal = i1;
                                }
                            } else if (distances[i1] == distances[i3]) {
                                if (distances[i1] != distances[i2]) {
                                    CoteEgal1 = i1;
                                    CoteEgal2 = i3;
                                    CoteInegal = i2;
                                }
                            }
                        }
                    }
                }
            }
        }
    }
}

Answer 1

General idea is the following:

Suppose, we have two points: A and B. There are only two ways to build a square with this points: ABCD and ABEF (see picture below). Now, lets consider AB is a vector. We can rotate it 90 degrees clockwise and we will get point D. If we move it to point B, then we will get point C. Similarly we can rotate AB 90 degrees counterclockwise and get F and E.

Now we need only check if file contains points C and D, or E and F. If so, we have a square. We can do it quickly if we store all of points in hashmap.

Here is a sample implementation:

import java.util.HashSet;
import java.util.Set;

public class Main {
    public static Set<Point> pointsSet = new HashSet<>();

    public static void main(String[] args) {
        //Suppose, that points are already parsed from file
        Point[] points = new Point[]{
                new Point(-1, -1),
                new Point(2, 1),
                new Point(4, -2),
                new Point(1, -4)
        };

        //Fill hashset with points. Important: hashCode() and equals() methods are overrided in Point class
        for (Point point : points) {
            pointsSet.add(point);
        }

        Point point1;
        Point point2;
        Point suggested_point3;
        Point suggested_point4;

        //Consider every pair of points
        for (int i = 0; i < points.length - 1; i++) {
            point1 = points[i];
            for (int j = i + 1; j < points.length; j++) {
                point2 = points[j];

                //Calculate vector coordinates for pair of points
                Vector vector = new Vector(point2.getX() - point1.getX(), point2.getY() - point1.getY());

                //Rotate vector clockwise by 90 degrees and calculate coordinates,
                //where two more points should be to form a square with current pair of points
                Vector clockwise_rotated_vector = rotateVector(vector, 90);
                suggested_point3 = moveVectorToPoint(clockwise_rotated_vector, point1);
                suggested_point4 = moveVectorToPoint(clockwise_rotated_vector, point2);
                if (pointsSet.contains(suggested_point3) && pointsSet.contains(suggested_point4)) {
                    squareFound(point1, point2, suggested_point3, suggested_point4);
                }

                //Same for counterclockwise rotated vector
                Vector counterclockwise_rotated_vector = rotateVector(vector, -90);
                suggested_point3 = moveVectorToPoint(counterclockwise_rotated_vector, point1);
                suggested_point4 = moveVectorToPoint(counterclockwise_rotated_vector, point2);
                if (pointsSet.contains(suggested_point3) && pointsSet.contains(suggested_point4)) {
                    squareFound(point1, point2, suggested_point3, suggested_point4);
                }
            }
        }
    }

    private static void squareFound(Point point1, Point point2, Point point3, Point point4) {
        System.out.println(String.format("%s, %s, %s, %s", point1, point2, point3, point4));
    }

    private static Point moveVectorToPoint(Vector vector, Point point) {
        double x = point.getX() + vector.getX();
        double y = point.getY() + vector.getY();
        return new Point(x, y);
    }

    private static Vector rotateVector(Vector original_vector, double degrees) {
        double radians = Math.toRadians(degrees);
        double x = original_vector.getX() * Math.cos(radians) - original_vector.getY() * Math.sin(radians);
        double y = original_vector.getX() * Math.sin(radians) + original_vector.getY() * Math.cos(radians);
        return new Vector(x, y);
    }
}

Class Point :

import java.math.BigDecimal;
import java.util.Locale;

public class Point {
    public Point(double x, double y) {
        this.x = x;
        this.y = y;

        //Store decimal values with certain scale, which would later be used for hashCode() calculation
        this.decimalX = new BigDecimal(x).setScale(DECIMAL_SCALE, BigDecimal.ROUND_HALF_EVEN);
        this.decimalY = new BigDecimal(y).setScale(DECIMAL_SCALE, BigDecimal.ROUND_HALF_EVEN);
    }

    private static final int DECIMAL_SCALE = 3;

    private double x;
    public double getX() {
        return x;
    }

    private double y;
    public double getY() {
        return y;
    }

    private BigDecimal decimalX;
    private BigDecimal decimalY;

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        Point point = (Point) o;

        if (!decimalX.equals(point.decimalX)) return false;
        return decimalY.equals(point.decimalY);
    }

    @Override
    public int hashCode() {
        //calculate for decimal values, so two very close enough points will have equal hash codes
        int result = decimalX.hashCode();
        result = 31 * result + decimalY.hashCode();
        return result;
    }

    @Override
    public String toString() {
        return String.format(Locale.ENGLISH, "(%2f,%2f)", x, y);
    }
}

Class Vector :

public class Vector {
    public Vector(double x, double y) {
        this.x = x;
        this.y = y;
    }

    private double x;
    public double getX() {
        return x;
    }

    private double y;
    public double getY() {
        return y;
    }
}

UPDATE: As mentioned in comments, there were few moments to improve, so I updated my code.

Answer 2

Finding all possible combinations of squares from 10000 points would be too much work doing it brute-force.

So, you need a way to find points that can be combined to form a square, ie points that share X or Y coordinate.

So, load all points into memory, "indexing" them by building a Map<Integer, Set<Integer>> , mapping X coordinates to collection of Y coordinates. The value is a Set so duplicate points are eliminated, and to improve performance of code below.

Now comes the more brute-force part, but it's greatly reduced, so shouldn't be too bad. Search like this (pseudo-code so you have to do part of this too) :

for eRight in xMap:
    xRight = eKey.key
    continue if eRight.value.size() == 1
    for eLeft in xMap:
        xLeft = eLeft.key
        continue if xLeft >= xRight
        continue if eLeft.value.size() == 1
        // at this point we now have two distinct X coordinates
        // with their corresponding collection of Y coordinates
        for yTop : eLeft.value:
            continue if ! eRight.value.contains(yTop)
            for yBottom : eLeft.value:
                continue if yBottom >= yTop // assuming (0,0) is lower-left
                continue if ! eRight.value.contains(yBottom)
                // at this point we now have two distinct Y coordinates
                // that fit both X coordinates, i.e we have a rectangle
                new Rectangle(xLeft, yTop, xRight, yBottom)

The above code finds rectangles , not squares . I'll leave it to you to changing it to find squares.
Hint : The yBottom (or yTop ) loop is replaced with a calculation.

Answer 3

I have an answer that is similar to Marat Safin's above, except that I think we must be careful about the fact that floating point arithmetic is imprecise and we must be careful that after some steps of calculation, a floating point number like 3.1415926 may become 3.1415927 (just differ by 1E-7). The solution to this is that we must also declare an argument epsilon (tolerance) that allows us to do "approximately equal" of two floating point numbers (such as double ).

My answer also has test input which you can use to verify your program.

You can find my answer here in another SO post of Detect square in a List of Points