相關閱讀 |
>>> 創業先鋒 眾人拾柴火焰高 >>> | 簡體 傳統 |
如何寫一個簡單的手寫識別算法,可以精準快速的識別出自定義的簡單圖形: 實現代碼: 實現代碼: 第四步把小片段的移動略去,最后就能得出其實是畫了一個凹的形狀。 實現代碼: 這個算法的厲害之處是可以實時識別,畫到一半也能判斷出來。
把所有的筆畫定義了個8個方向,然后將B的筆畫可以分解成一個字符串。然后當人在觸摸屏上畫出一個符號時,也將它分解成8個方向的字符串,最后比較兩個字符串的距離就能判斷出和不同符號的近似度。
實現起來也很簡單,第一步去噪,因為不同觸摸屏的采樣頻率不同。
實現代碼: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);
}
第二步把去噪后的數據轉換成方向序列,把之前得到的點換成方向序列,并把方向序列歸納到之前定義的8個方向中去。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;
}
第三步把連續一致的方向合并。
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游資網 2015-08-23 08:48:18
稱謂:
内容: