Multi-byte vs. Unicode

Multi-byte	Unicode
char	TCHAR
strcat_s()	_tcscaft_s()
strcpy_s()	_tcscpy_s()
strncpy_s()	_tcsncpy_s()
strien()	_tcsien()
sprinif_s()	_stprintf_s()

How to get paths defined in the system environment

char path[_MAX_PATH];

const char* varName = "[system environment variable name]";
size_t len;
getenv_s(&len, path, 80, varName);

Print out a (decimal) number as a hexadecimal number

int decNum = 123456;
char hexNum[16];

// decimal number to hexadecimal number
sprintf_s(hexNum, 16, "%X", decNum);
printf("%d -> %s \n", decNum, hexNum);

// hexadecimal number to decimal number
sprintf_s(hexNum, 16, "FF");
decNum = (int)strtol(hexNum, nullptr, 16);
printf(" %s -> %d \n", hexNum, decNum);

Derivation of projective transformation based on shift, rotation and scale parameters

Reduced normal matrix for solving fast photogrammetric bundle adjustment

Reduced normatrix for photogrammetric bundle adjustment based on the collinearity equations

Note: There are $m_1$ photos and $m_2$ object points, and the number of EOPs is 6. If 6*$m_1$ is less than 3*$m_2$, solve unknowns related with EOPs. If not, solve object points coordinates, first. This post shows only the first case, 3*$m_2$ > 6*$m_1$.

[Fig. Normal matrix sample]

$$ \begin{pmatrix} N_1 & N_{12} \\ N_{12}^T & N_2 \end{pmatrix}\begin{pmatrix} X_1 \\ X_2 \end{pmatrix} = \begin{pmatrix} C_1 \\ C_2 \end{pmatrix} $$ $$ N_1 X_1 + N_{12} X_2 = C_1$$ $$ N_{12}^T X_1 + N_2 X_2 = C_2$$ In case $6m_1<3m_2$, let's slove $X_1$ first.
$$ X_2 = N_2^{-1} C_2 - N_2^{-1} N_{12}^T X_1$$, therefore
$$ (N_1 - N_{12} N_2^{-1} N_{12}^T) X_1 = C_1 - N_{12} N_2^{-1} C_2$$ $$ X_1 = (N_1 - N_{12} N_2^{-1} N_{12}^T)^{-1}(C_1 - N_{12} N_2^{-1} C_2)$$ And
$$ X_2 = N_2^{-1}(C_2 - N_{12}^T X_1) $$ For $X_2$, we can avoid solving the inverse matrix of $3m \times 3m$ .
For $i_{th}$ point,
$$ X_{2_i} = N_{2_{i}}^{-1}(C_{2_i} - N_{12_i}^T X_1) $$