Perimeter is relatively easy and from Bernd Jähne book ( Digital Image Processing 5th: 508-511 ), it can be calculated as below code.
double CalculatePerimeter( const ChainCode& chains )
{
double p = 0;
double sqrt2 = sqrt( 2.0 );
for( size_t i=0; i < chains.size(); ++i )
{
if( chains[i]%2 == 0 )
{
p += 1.0;
}
else
{
p += sqrt2;
}
}
return p;
}
Bernd Jähne also showed how to get the area using a chain code but it doesn't work well when the chain code is 1 pixel width or height e.g. ( 0 0 4 4 ) will return 2 instead of 3. Refer below image how the area end up 1 pixel short.
Cris Luengo ( Cris's Image Analysis Blog : More chain code measures ) shows a matlab code that works well with 1 pixel wide chain code. In above case of 3 pixels, it adds two more addition when the direction changes which add up 1 more pixel. Refer below image.
// making a circular chain code by inserting the last code to the front
ChainCode cc;
cc.push_back( chains[ chains.size()-1 ] );
for( size_t i=0; i < chains.size(); ++i )
{
cc.push_back( chains[i] );
}
m_Area = CalculateArea( cc );
int CalculateArea( const ChainCode& cc ) // a circular chain
{
int (&M)[8][8]( TableAreaIncrement );
int B = 10, A = 0;
for( size_t i=1; i < cc.size(); ++i )
{
uint8_t c_i = cc[i];
uint8_t c_im1 = cc[i-1];
switch( M[c_im1][c_i] )
{
case 'X' :
throw Exception( StringFormat::As( _T("Invalid chain code found as %d->%d"),
c_im1, c_i ));
case 'A': A += -B+1; break;
case 'B': A += -B; break;
case 'C': A += B; break;
case 'D': A += B-1; break;
}
switch( c_i )
{
case 0: A+=B; break;
case 1: A+=(++B); break;
case 2: B++; break;
case 3: A+=-(++B)+1; break;
case 4: A+=-B+1; break;
case 5: A+=-(--B)+1; break;
case 6: B--; break;
case 7: A+=(--B); break;
}
}
return A;
}
int ChainCodeProperties::TableAreaIncrement[8][8] =
{
{'0','1','B','X','A','A','6','7'},
{'0','1','B','B','X','A','6','7'},
{'C','C','2','3','4','X','C','C'},
{'C','C','2','3','4','5','X','C'},
{'C','C','2','3','4','5','D','X'},
{'X','C','2','3','4','5','D','D'},
{'0','X','A','A','A','A','6','7'},
{'0','1','X','A','A','A','6','7'},
};
The center of mass can be calculated by summing up x and y coordinates. But just simply adding up does not cover 1 pixel wide chain code as below image. The (1,1) pixel added two times instead one time. The compensation can be done by adding more pixels when a direction changes in opposite way as the idea of Cris Luengo's area calculation.
void CalculateCentroid( const ChainCode& cc, double* cx, double *cy )
{
int (&O)[8][2]( TableOffsetIncrement );
int vpc = 0; // visitied pixel count
int sx = 0, sy = 0; // sum of x and y
int px = 0, py = 0;
for( size_t i=1; i < cc.size(); ++i )
{
uint8_t c_i = cc[i];
uint8_t c_im1 = cc[i-1];
if( c_i != c_im1 && (c_i & 0x3 ) == ( c_im1 & 0x3 ) )
{
// when direction changes in opposite direction
// e.g. 0 -> 4, 4 -> 0, 1 -> 5, ...
vpc++; sx += px; sy+=py;
}
px += O[ c_i ][0];
py += O[ c_i ][1];
vpc++; sx += px; sy+=py;
}
*cx = (double)sx / (double)vpc;
*cy = (double)sy / (double)vpc;
}
int ChainCodeProperties::TableOffsetIncrement[8][2] =
{
{ 1, 0 },
{ 1, -1 },
{ 0, -1 },
{ -1, -1 },
{ -1, 0 },
{ -1, 1 },
{ 0, 1 },
{ 1, 1 },
};
It works well most of the case but it fails if there is a pixel which is visited more than 2 times as below image. It has (1,1) pixel counted 3 times but it should be counted 2 times.
