Reverse this equation/function (2d to isometric)

Let your function be denoted by $A(x,y)$. Let $w_{\rm tile}, h_{\rm tile}$ be the tile width and height, $w_{\rm iso}, h_{\rm iso}$ be the isometric tile width and height, and $\Delta w, \Delta h$ be the offsets.

We have
$$A(x,y) = (\tfrac{1}{2}w_{\rm tile} x - w_{\rm iso} y + \Delta w , \tfrac{1}{4}h_{\rm tile} y + \tfrac{1}{2}h_{\rm iso}x + \Delta h)$$
This can be represented as the affine transformation $\mathbf{v} \mapsto T\mathbf{v} + \mathbf{v}_0$, where
$$T = \begin{bmatrix}\tfrac{1}{2}w_{\rm tile} & - w_{\rm iso} \\ \tfrac{1}{2}h_{\rm iso} & \tfrac{1}{4}h_{\rm tile}\end{bmatrix}$$
and $\mathbf{v}_0 = (\Delta w, \Delta h)$. The inverse $A^{-1}$ is given by subtracting $\mathbf{v_0}$ and then applying $T^{-1}$, which is the affine transformation $\mathbf{v} \mapsto T^{-1}(\mathbf{v} - \mathbf{v_0})$. The inverse of $T$ is given by (using the formula for the inverse of a $2\times2$ matrix)
$$T^{-1} = \frac{1}{\tfrac{1}{8}w_{\rm tile}h_{\rm tile} + \tfrac{1}{2}w_{\rm iso}h_{\rm iso}} \begin{bmatrix}\tfrac{1}{4}h_{\rm tile} & w_{\rm iso} \\ -\tfrac{1}{2}h_{\rm iso} & \tfrac{1}{2}w_{\rm tile}\end{bmatrix}.$$

Thus the inverse function would proceed in pseudocode as follows:


inv_det = 1 / (tileWidth * tileHeight / 8 + isoWidth * isoHeight / 2)

// subtracting v_0
u = x - isoXOffset
v = y - iosYOffset

// applying T^-1
s = (tileHeight / 4) * u + isoWidth * v
t = (-isoHeight / 2) * u + (tileWidth / 2) * v

return (inv_det * s, inv_det * t)