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如何写一个简单的手写识别算法,可以精准快速的识别出自定义的简单图形: 实现代码: 实现代码: 第四步把小片段的移动略去,最后就能得出其实是画了一个凹的形状。 实现代码: 这个算法的厉害之处是可以实时识别,画到一半也能判断出来。
把所有的笔画定义了个8个方向,然后将B的笔画可以分解成一个字符串。然后当人在触摸屏上画出一个符号时,也将它分解成8个方向的字符串,最后比较两个字符串的距离就能判断出和不同符号的近似度。
实现起来也很简单,第一步去噪,因为不同触摸屏的采样频率不同。
实现代码:void GestureAlgorithm::addPoint(int x, int y)
{
int d_x, d_y;
d_x = x-positions.back().x;
d_y = y-positions.back().y;
if( d_x*d_x + d_y*d_y >= MIN_MOVEMENT)
{
updateStatistic(x, y);
recognizeGesture();
}
}
void GestureAlgorithm::updateStatistic(int x, int y)
{
positions.push_back(Point(x, y));
point_num = positions.size();
if(point_num >1)
{
// For Point Recognization
dist_sum += positions.begin()->dist(x,y);
dist_average =dist_sum/(point_num - 1);
// For Line Recognization
// Need a patch for the V0 calculation.
Point v0 = Point(positions[1].x - positions[0].x, positions[0].y );
Point v1 = Point(x - positions[0].x, y -positions[0].y);
if(normalize(v0) && normalize( v1))
{
float theta = acos(dot(v0, v1));
theta_sum += theta;
theta_sqsum += sq(theta);
theta_average = theta_sum / (float)(point_num - 1);
theta_factor = sqrt((float)(point_num - 1)*theta_sqsum - sq(theta_sum))/(point_num-1);
}
}
mainDirections = detectDirection(positions);
//Statistic Update
pos_x_sum += x;
pos_y_sum += y;
pos_xx_sum += sq(x);
pos_xy_sum += x * y;
midPoint = Point(pos_x_sum/point_num, pos_y_sum/point_num);
curGestureRender->render_bbox->addPoint(x, y);
}
第二步把去噪后的数据转换成方向序列,把之前得到的点换成方向序列,并把方向序列归纳到之前定义的8个方向中去。PosList GestureAlgorithm::limitDirections(const PosList &positions)
{
PosList res;
int lastx, lasty;
bool firstTime = true;
for( PosList::const_iterator ii = positions.begin(); ii != positions.end(); ++ii )
{
if( firstTime )
{
lastx = ii->x;
lasty = ii->y;
firstTime = false;
}
else
{
int dx, dy;
dx = ii->x – lastx;
dy = ii->y – lasty;
if( dy > 0 )
{
if( dx > dy || -dx > dy )
dy = 0;
else
dx = 0;
}
else
{
if( dx > -dy || -dx > -dy )
dy = 0;
else
dx = 0;
}
res.push_back( Point( dx, dy ) );
lastx = ii->x;
lasty = ii->y;
}
}
return res;
}
第三步把连续一致的方向合并。
Position Num: 141
X= 113 Y= 0
X= 0 Y= -15
X= 0 Y= 179
X= 13 Y= 0
X= -110 Y= 0
X= 0 Y= 6
X= 0 Y= -101
X= 3 Y= 0
Directions Number: 8
Directions Length:540
UP Number: 3 Down Number: 2 Left: 1 right 2
Position Num: 142
PosList GestureAlgorithm::simplify(const PosList &positions)
{
PosList res;
int lastdx = 0, lastdy = 0;
bool firstTime = true;
int index=0;
for( PosList::const_iterator ii = positions.begin(); ii != positions.end(); ++ii )
{
if( firstTime )
{
lastdx = ii->x;
lastdy = ii->y;
firstTime = false;
}
else
{
bool joined = false;
if( (lastdx > 0 && ii->x > 0) || (lastdx < 0 && ii->x < 0) )
{
lastdx += ii->x;
joined = true;
}
if( (lastdy > 0 && ii->y > 0) || (lastdy < 0 && ii->y < 0) )
{
lastdy += ii->y;
joined = true;
}
if( !joined )
{
res.push_back( Point( lastdx, lastdy ) );
lastdx = ii->x;
lastdy = ii->y;
}
}
}
if( lastdx != 0 || lastdy != 0 )
{
res.push_back( Point( lastdx, lastdy ) );
}
return res;
}
PosList GestureAlgorithm::removeShortestNoise(const PosList &positions)
{
PosList res;
int shortestSoFar;
PosList::const_iterator shortest;
bool firstTime = true;
for( PosList::const_iterator ii = positions.begin(); ii != positions.end(); ++ii )
{
if( firstTime )
{
shortestSoFar = ii->x*ii->x + ii->y*ii->y;
shortest = ii;
firstTime = false;
}
else
{
if( (ii->x*ii->x + ii->y*ii->y) < shortestSoFar ) { shortestSoFar = ii->x*ii->x + ii->y*ii->y;
shortest = ii;
}
}
}
for( PosList::const_iterator ii = positions.begin(); ii != positions.end(); ++ii )
{
if( ii != shortest)
res.push_back( *ii );
}
return res;
}
PosList GestureAlgorithm::detectDirection(const PosList &positions)
{
PosList directions = simplify(limitDirections(positions));
double minLength = calcLength(directions) *minMatch;
while(directions.size() > 0 && calcLength(removeShortestNoise(directions)) > minLength)
{
directions = simplify(removeShortestNoise(directions));
}
upNum = 0; downNum = 0; leftNum = 0; rightNum =0;
for(int i = 0; i< directions.size(); i++) { if(directions[i].y >= 0 && directions[i].x ==0)
upNum++;
else if(directions[i].y < 0 && directions[i].x ==0) downNum++; else if(directions[i].x >= 0 && directions[i].y ==0 )
leftNum++;
else if(directions[i].x < 0 && directions[i].y ==0 )
rightNum++;
}
return directions;
}
GameRes游资网 2015-08-23 08:48:18
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