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penpot-exporter-figma-plugin/plugin-src/utils/matrixInvert.ts

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2024-04-15 02:48:03 -05:00
export const matrixInvert = (M: number[][]): number[][] | undefined => {
// if the matrix isn't square: exit (error)
if (M.length !== M[0].length) {
return;
}
// create the identity matrix (I), and a copy (C) of the original
const dim = M.length;
let i = 0,
ii = 0,
j = 0,
e = 0;
const I: number[][] = [],
C: number[][] = [];
for (i = 0; i < dim; i += 1) {
// Create the row
I[i] = [];
C[i] = [];
for (j = 0; j < dim; j += 1) {
// if we're on the diagonal, put a 1 (for identity)
if (i === j) {
I[i][j] = 1;
} else {
I[i][j] = 0;
}
// Also, make the copy of the original
C[i][j] = M[i][j];
}
}
// Perform elementary row operations
for (i = 0; i < dim; i += 1) {
// get the element e on the diagonal
e = C[i][i];
// if we have a 0 on the diagonal (we'll need to swap with a lower row)
if (e === 0) {
// look through every row below the i'th row
for (ii = i + 1; ii < dim; ii += 1) {
// if the ii'th row has a non-0 in the i'th col
if (C[ii][i] !== 0) {
// it would make the diagonal have a non-0 so swap it
for (j = 0; j < dim; j++) {
e = C[i][j]; // temp store i'th row
C[i][j] = C[ii][j]; // replace i'th row by ii'th
C[ii][j] = e; // replace ii'th by temp
e = I[i][j]; // temp store i'th row
I[i][j] = I[ii][j]; // replace i'th row by ii'th
I[ii][j] = e; // replace ii'th by temp
}
// don't bother checking other rows since we've swapped
break;
}
}
// get the new diagonal
e = C[i][i];
// if it's still 0, not invertable (error)
if (e === 0) {
return;
}
}
// Scale this row down by e (so we have a 1 on the diagonal)
for (j = 0; j < dim; j++) {
C[i][j] = C[i][j] / e; // apply to original matrix
I[i][j] = I[i][j] / e; // apply to identity
}
// Subtract this row (scaled appropriately for each row) from ALL of
// the other rows so that there will be 0's in this column in the
// rows above and below this one
for (ii = 0; ii < dim; ii++) {
// Only apply to other rows (we want a 1 on the diagonal)
if (ii === i) {
continue;
}
// We want to change this element to 0
e = C[ii][i];
// Subtract (the row above(or below) scaled by e) from (the
// current row) but start at the i'th column and assume all the
// stuff left of diagonal is 0 (which it should be if we made this
// algorithm correctly)
for (j = 0; j < dim; j++) {
C[ii][j] -= e * C[i][j]; // apply to original matrix
I[ii][j] -= e * I[i][j]; // apply to identity
}
}
}
// we've done all operations, C should be the identity
// matrix I should be the inverse:
return I;
};