PnP 单目相机位姿估计（三）：二维码角点检测

---

IDE：visual studio 2013
使用库：Eigen opencv2.4.9
文档版本：1.0

---

解PnP问题时，用二维码的好处

1. 二维码由黑白两色组成，容易通过对图像阈值处理而发现
2. 二维码具有方向性（只要设计的不为中心对称），可以更好的辅助求解位姿
3. 二维码具有可识别性，即可解析二维码内部黑白块的排列顺序，从而确定该二维码是否为我们所要用的二维码

---

二维码识别的流程

该流程以该二维码为例

1. 阈值处理，二值化操作
2. 去掉二维码周围一圈，即保存为5*5的信息
3. 计算5*5图像内的海明距离，和程序所设定要查找的二维码的海明距离作比较
4. 若该二维码即为程序所要查找的二维码，则用opencv自带的findContours查找角点

---

代码

#include "Marker.h"

/************************************************构造函数****************************************************/
Marker::Marker(): id(-1)
{
}

/************************************************析构函数****************************************************/
Marker::~Marker() 
{
}

/*******************************************读取二维码内含信息**********************************************/
int Marker::getMarkerId(cv::Mat &markerImage, int &nRotations)
{
    assert(markerImage.rows == markerImage.cols);
    assert(markerImage.type() == CV_8UC1);

    cv::Mat grey = markerImage;
    //threshold image
    cv::threshold(grey, grey, 125, 255, cv::THRESH_BINARY | cv::THRESH_OTSU);

    //Markers  are divided in 7x7 regions, of which the inner 5x5 belongs to marker info
    //the external border should be entirely black
    //去掉周围的一圈黑色，提取出5x5的网格
    int cellSize = markerImage.rows / 7;

    for (int y = 0; y<7; ++y)
    {
        int inc = 6;
        //for first and last row, check the whole border
        if (y == 0 || y == 6) inc = 1;
        for (int x = 0; x<7; x += inc)
        {
            int cellX = x * cellSize;
            int cellY = y * cellSize;
            cv::Mat cell = grey(cv::Rect(cellX, cellY, cellSize, cellSize));
            //imshow("cell",cell);
            int nZ = cv::countNonZero(cell);
            //cout<<"nZ"<
            //cout<<"(cellSize*cellSize) / 2"<<" "<<(cellSize*cellSize) / 2<
            if (nZ >(cellSize*cellSize) / 2)
            {
                //can not be a marker because the border element is not black!
                return -1;
            }
        }
    }

    //将图像标记信息存放在一个 5x5 的 Mat 中
    cv::Mat bitMatrix = cv::Mat::zeros(5, 5, CV_8UC1);

    //get information(for each inner square, determine if it is  black or white)  
    for (int y = 0; y<5; ++y)
    {
        for (int x = 0; x<5; ++x)
        {
            int cellX = (x + 1)*cellSize;
            int cellY = (y + 1)*cellSize;
            cv::Mat cell = grey(cv::Rect(cellX, cellY, cellSize, cellSize));

            int nZ = cv::countNonZero(cell);
            if (nZ>(cellSize*cellSize) / 2)
                bitMatrix.at(y, x) = 1;
        }
    }

    //cout<<"bitMatrix:"<<" "<
    //check all possible rotations
    //因为会有4种放置方向
    cv::Mat rotations[4];
    //海明距离
    int distances[4];

    rotations[0] = bitMatrix;
    distances[0] = hammDistMarker(rotations[0]);

    std::pair<int, int> minDist(distances[0], 0);

    for (int i = 1; i<4; ++i)
    {
        //get the hamming distance to the nearest possible word
        rotations[i] = rotate(rotations[i - 1]);//逆时针转90度
        distances[i] = hammDistMarker(rotations[i]);

        if (distances[i] < minDist.first)
        {
            minDist.first = distances[i];
            minDist.second = i;
        }
    }
    //cout<<"minDist"<<" "<
    //cout<<"mat2id(rotations[minDist.second]):"<<" "<
    nRotations = minDist.second;
    //819//1100110011
    if (minDist.first == 0)
    {
        return mat2id(rotations[minDist.second]);
    }
    return -1;
}

/*****************************************Maker初始的坐标数据**********************************************/
int Marker::hammDistMarker(cv::Mat bits)
{
    //maker  1
    int ids[5][5] =
    {
        { 1, 1, 1, 1, 1 },
        { 1, 1, 0, 1, 1 },
        { 1, 0, 1, 0, 1 },
        { 1, 1, 1, 1, 1 },
        { 1, 1, 0, 1, 1 }
    };

    //maker 2
    int ids_1[5][5] =
    {
        { 1, 1, 1, 1, 1 },
        { 1, 1, 0, 1, 1 },
        { 1, 0, 1, 1, 0 },
        { 1, 1, 0, 1, 1 },
        { 1, 1, 1, 1, 1 }
    };

    int dist = 0;

    for (int y = 0; y<5; ++y)
    {
        float minSum = 1e5; //hamming distance to each possible word
        for (int p = 0; p<5; ++p)
        {
            float sum = 0;
            //now, count
            for (int x = 0; x<5; ++x)
            {
                sum += bits.at(y, x) == ids[p][x] ? 0 : 1;
            }
            if (minSum>sum)
                minSum = sum;
        }
        //do the and
        dist += minSum;
    }

    return dist;
}

/*********************************************计算二维码信息*************************************************/
int Marker::mat2id(const cv::Mat &bits)
{
    int val = 0;
    for (int y = 0; y<5; ++y)
    {
        val <<= 1;
        if (bits.at(y, 1)) val |= 1;
        val <<= 1;
        if (bits.at(y, 3)) val |= 1;
    }
    return val;
}

/********************************************CV图像旋转函数*************************************************/
cv::Mat Marker::rotate(cv::Mat in)
{
    cv::Mat out;
    in.copyTo(out);
    for (int i = 0; ifor (int j = 0; j(i, j) = in.at(in.cols - j - 1, i);
        }
    }
    return out;

}

/*****************************************Maker初始的坐标数据**********************************************/
bool operator<(const Marker &M1, const Marker&M2)
{
    return M1.id

---

最后

相关文章
solvepnp三维位姿估算
PnP 单目相机位姿估计（一）：初识PnP问题
PnP 单目相机位姿估计（二）：solvePnP利用二维码求解相机世界坐标