【问题标题】:Vertical Hex Grid: Get x rings of tiles surrounding a specific coordinateVertical Hex Grid:获取围绕特定坐标的 x 圈瓷砖
【发布时间】:2016-04-24 20:49:33
【问题描述】:

问题

我要做的是从指定点获取 x 个环,并将这些环存储在 List<List<HexCoordinate>> 中,其中内部列表是该环中所有十六进制的列表,而 HexCoordinate 是下面定义的结构

理想情况下,我希望能够指定坐标,以及我想搜索多少环,并让算法为我抓取图块。

图像和尝试

我有一个看起来像这样的垂直(平顶)六角网格

在代码中,每个图块都由一个简单的HexCoordinate结构GitHub表示

public struct HexCoordinate : IEquatable<HexCoordinate>
{
    public int X { get; private set; }
    public int Y { get; private set; }

    public HexCoordinate(int x, int y)
    {
        X = x;
        Y = y;
    }

    public HexCoordinate North() => this + new HexCoordinate(0, -1);
    public HexCoordinate South() => this + new HexCoordinate(0, 1);

    public HexCoordinate West() => this + new HexCoordinate(-1, 0);
    public HexCoordinate East() => this + new HexCoordinate(1, 0);

    public HexCoordinate NorthWest() => this + new HexCoordinate(-1, -1);
    public HexCoordinate NorthEast() => this + new HexCoordinate(1, -1);

    public HexCoordinate SouthWest() => this + new HexCoordinate(-1, 1);
    public HexCoordinate SouthEast() => this + new HexCoordinate(1, 1);

    public bool Equals(HexCoordinate other)
    {
        return X == other.X &&
               Y == other.Y;
    }
    public static HexCoordinate operator +(HexCoordinate one, HexCoordinate two)
    {
        return new HexCoordinate(one.X + two.X, one.Y + two.Y);
    }
}

特别是对于我的示例,我绘制了一张图像,试图自己解决这个问题

我的尝试

因为我需要展示我已经尝试过的东西,所以这是我迄今为止尝试过的

public const int MAX_RINGS = 3;
public List<List<HexCoordinate> GetsRingsWithinHex(HexCoordinate coordinate, int maxRings = MAX_RINGS)
{
    // Attempt One Pseudocode
    // reference: http://gamedev.stackexchange.com/questions/51264/get-ring-of-tiles-in-hexagon-grid
    // int ring = 1
    //   Travel around the ring by traversing N,SE,S,SW,NW,N,NE multiplied by the ring number
    //   ring++
    //      Travel Around ring again
    //      cont until desired ring...

    var hexRings = new List<List<HexCoordinate>>();

    // Add in the current hex to the list
    var currentHex = new List<HexCoordinate>(); 
    currentHex.add(coordinate);
    hexRings.Add(currentHex);

    // Now go through and add the other rings
    while (hexRings.Count <= maxRings)
    {
        var ring = new List<HexCoordinate>();
        HexCoordinate tempCoordinate = coordinate;
        int currentRingNumber = hexRings.Count;

        // We start off by going north to the correct ring, and then adding it to our list
        for (int i = 0; i < currentRingNumber; i++)
        {
            tempCoordinate = tempCoordinate.North();
        }
        ring.add(tempCoordinate);

        // After that, we proceed to go clockwise around the ring until we come back to the start
        for (int i = 0; i < currentRingNumber; i++)
        {
            tempCoordinate = tempCoordinate.SouthEast();

            // If the ring is an odd number, you need to re-align the coordinates back to where whey should be
            if (IntExtensions.IsOdd(i)) tempCoordinate = tempCoordinate.North();

            ring.add(tempCoordinate);
        }

        // The rightmost segment is east because we can go straight down the required number of times
        for (int i = 0; i < currentRingNumber; i++)
        {
            tempCoordinate = tempCoordinate.South();
            ring.add(tempCoordinate);
        }

        // We utilize Current Ring - 1 because we only want this to run on rings that are greater than 1
        for (int i = 0; i < currentRingNumber - 1; i++)
        {
            tempCoordinate = tempCoordinate.SouthWest();
            ring.add(tempCoordinate);
        }

        // Coming into this statement, we are now at the bottom 3 coordinates.
        // Since our grid is laid out vertically, we can assume that these three hexes will be directly west of eachother
        // So we only have to go west twice to make our way to the next north segment 
        for (int i = 0; i < 2; i++)
        {
            tempCoordinate = tempCoordinate.West();
            ring.add(tempCoordinate);
        }

        // We utilize Current Ring - 1 because we only want this to run on rings that are greater than 1
        for (int i = 0; i < currentRingNumber - 1; i++)
        {
            tempCoordinate = tempCoordinate.NorthWest();
            ring.add(tempCoordinate);
        }

        // The left most segment is easy because we can just go straight up
        for (int i = 0; i < currentRingNumber; i++)
        {
            tempCoordinate = tempCoordinate.North();
            ring.add(tempCoordinate);
        }

        // We utilize Current Ring - 1 because we only want this to run on rings that are greater than 1
        for (int i = 0; i < currentRingNumber - 1; i++)
        {
            tempCoordinate = tempCoordinate.NorthEast();

            // If the ring is an even number, you need to re-align the coordinates back to where whey should be
            if (IntExtensions.IsEven(i)) tempCoordinate = tempCoordinate.South();

            ring.add(tempCoordinate);
        }

        // Finally, we add the ring to our system rings and loop until we no longer fit the criteria
        hexRings.Add(ring);
    }

    return hexRings;
}

如果需要,这里是我的 IntExtensions

public static class IntExtensions
{
    public static bool IsBetween(this int num, int low, int high)
    {
        return num >= low && num <= high;
    }

    public static bool IsOdd(this int value)
    {
        return value % 2 != 0;
    }

    public static bool IsEven(this int value)
    {
        return value % 2 == 0;
    }
}

我目前的问题是该算法适用于第 1 和第 2 环,但是一旦到达第三环(如果我将其运行超过 3 环,可能会更远),沿底部和角落的坐标开始偏移 1 ...从下面我的控制台中的输出可以看出(应该手动手动编辑)

Ring 0 - System 5, 5

Ring 1 - System 5, 4
Ring 1 - System 6, 5
Ring 1 - System 6, 6
Ring 1 - System 5, 6
Ring 1 - System 4, 6
Ring 1 - System 4, 5

Ring 2 - System 5, 3
Ring 2 - System 6, 4
Ring 2 - System 7, 4
Ring 2 - System 7, 5
Ring 2 - System 7, 6
Ring 2 - System 6, 7
Ring 2 - System 5, 7
Ring 2 - System 4, 7
Ring 2 - System 3, 6
Ring 2 - System 3, 5
Ring 2 - System 3, 4
Ring 2 - System 4, 4

Ring 3 - System 5, 2
Ring 3 - System 6, 3
Ring 3 - System 7, 3
Ring 3 - System 8, 4
Ring 3 - System 8, 5
Ring 3 - System 8, 6
Ring 3 - System 8, 7
Ring 3 - System 7, 8 //(Should be 7, 7)
Ring 3 - System 6, 9 //(Should be 6, 8)
Ring 3 - System 5, 9 //(Should be 5, 8)
Ring 3 - System 4, 9 //(Should be 4, 8)
Ring 3 - System 3, 8 //(Should be 3, 7)
Ring 3 - System 2, 7 
Ring 3 - System 2, 6
Ring 3 - System 2, 5
Ring 3 - System 2, 4
Ring 3 - System 3, 4 //(Should be 3, 3)
Ring 3 - System 4, 3

有没有人可以帮助我找到正确的方向,或者为我提供一个算法来让我得到六角环?我个人在这个问题上被困了大约一天半,我似乎无法弄清楚这个问题。

【问题讨论】:

  • 我首先想到的是:为什么不对网格使用不同的编号方案?双分辨率。第一列的十六进制以 0 开头,步长为 2,第二列的十六进制以 1 开头,步长为 2。因此,您的十六进制位于具有双分辨率的规则方形网格上。然后,您也许可以使用 bresenham circle 算法来快速确定一个六边形是否与半径为 2 * ring 的圆相交。
  • 我的维基百科 bresenham C 例程的 F# 端口看起来不太好 :) 但从我目前的探索来看,我倾向于认为我上面描述的方法基本上是可行的。在这里,您预期的 ring3 导致了我的“双分辨率”方法。没有渲染六边形而是圆圈;)imgur.com/kZtWw8S
  • 您需要考虑xyLSB坐标并相应地修正位置。你不想要圆圈而是六边形,所以 Bresenham 是不可用的。相反,您正在搜索距离原点 1,2,3,4... 单元格的所有十六进制,因此A* 会更好。看看Improving performance of click detection on a staggered column isometric grid,尤其是cell2scrscr2cell,了解我所说的x,y 坐标校正的LSB(它是方形网格,但问题是一样的)。顺便说一句漂​​亮的草图
  • @BitTickler 我得到的印象是 OP 需要所有六边形层直到距离 N 而不仅仅是单个层。因此,从起点到 N 值的简单 A* 填充将简单快速地获得所需的所有内容。我的意思是基于网格的 A* 不是图形方法!!!
  • @killerrin 你只需要填充物......见Backtracking in A star

标签: c# algorithm hexagonal-tiles


【解决方案1】:

好的,所以我想我可能已经想出了解决问题的方法。我已经测试了多达 4 个环,它在相应的环中为我提供了所有正确的十六进制。

public List<List<HexCoordinate>> GetsRingsSurroundingHex(HexCoordinate coordinate, int maxRings)
    {
        // idea reference: http://gamedev.stackexchange.com/questions/51264/get-ring-of-tiles-in-hexagon-grid
        // int ring = 1
        //   Travel around the ring by traversing N,SE,S,SW,NW,N,NE multiplied by the ring number
        //   ring++
        //      Travel Around ring again
        //      cont until desired ring...

        var hexRings = new List<List<HexCoordinate>>();

        // Add in the current hex to the list
        var currentHex = new List<HexCoordinate>();
        currentHex.Add(coordinate);
        hexRings.Add(currentHex);

        // Now go through and add the other rings
        while (hexRings.Count <= maxRings)
        {
            var ring = new List<HexCoordinate>();
            HexCoordinate tempCoordinate = coordinate;
            int currentRingNumber = hexRings.Count;

            // We start off by going north to the correct ring, and then adding it to our list
            for (int i = 0; i < currentRingNumber; i++)
            {
                tempCoordinate = tempCoordinate.North();
            }
            ring.Add(tempCoordinate);

            // After that, we proceed to go clockwise around the ring until we come back to the start
            for (int i = 0; i < currentRingNumber; i++)
            {
                tempCoordinate = tempCoordinate.SouthEast();

                // If the ring is an odd number, you need to re-align the coordinates back to where whey should be
                if (IntExtensions.IsOdd(i)) tempCoordinate = tempCoordinate.North();

                ring.Add(tempCoordinate);
            }

            // The rightmost segment is east because we can go straight down the required number of times
            for (int i = 0; i < currentRingNumber; i++)
            {
                tempCoordinate = tempCoordinate.South();
                ring.Add(tempCoordinate);
            }

            // We utilize Current Ring - 1 because we only want this to run on rings that are greater than 1
            for (int i = 0; i < currentRingNumber - 1; i++)
            {
                if (currentRingNumber.IsEven())
                {
                    if (i.IsEven())
                        tempCoordinate = tempCoordinate.SouthWest();
                    else
                        tempCoordinate = tempCoordinate.West();
                }
                else
                {
                    if (i.IsEven())
                        tempCoordinate = tempCoordinate.West();
                    else
                        tempCoordinate = tempCoordinate.SouthWest();
                }

                ring.Add(tempCoordinate);
            }

            // Coming into this statement, we are now at the bottom 3 coordinates.
            // Since our grid is laid out vertically, we can assume that these three hexes will be directly west of each other
            // So we only have to go west twice to make our way to the next north segment 
            for (int i = 0; i < 2; i++)
            {
                tempCoordinate = tempCoordinate.West();
                ring.Add(tempCoordinate);
            }

            // We utilize Current Ring - 1 because we only want this to run on rings that are greater than 1
            for (int i = 0; i < currentRingNumber - 1; i++)
            {
                if (i.IsEven())
                    tempCoordinate = tempCoordinate.NorthWest();
                else
                    tempCoordinate = tempCoordinate.West();

                ring.Add(tempCoordinate);
            }

            // The left most segment is easy because we can just go straight up
            for (int i = 0; i < currentRingNumber; i++)
            {
                tempCoordinate = tempCoordinate.North();
                ring.Add(tempCoordinate);
            }

            // We utilize Current Ring - 1 because we only want this to run on rings that are greater than 1
            for (int i = 0; i < currentRingNumber - 1; i++)
            {
                if (currentRingNumber.IsEven())
                {
                    if (i.IsEven())
                        tempCoordinate = tempCoordinate.East();
                    else
                        tempCoordinate = tempCoordinate.NorthEast();
                }
                else
                {
                    if (i.IsEven())
                        tempCoordinate = tempCoordinate.NorthEast();
                    else
                        tempCoordinate = tempCoordinate.East();
                }

                ring.Add(tempCoordinate);
            }

            // Finally, we add the ring to our system rings and loop until we no longer fit the criteria
            hexRings.Add(ring);
        }

        return hexRings;
    }

【讨论】:

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