# Revision history [back]

for both L2 and HAMMING norm, the position of the values is irrelevant, as long as you stay consistant with the ordering

(xx + yy + zz) == (yy + xx + zz) == (zz + yy + x*x) == ...

your example above will "fail" for both norms, since you arbitrarily change the order in one of the vectors. again, if you use the same permutation for all of your vectors, the results will be the same.

for both L2 and HAMMING norm, the position of the values is irrelevant, as long as you stay consistant with the ordering

(xx

 (x*x + yy y*y + zz) z*z) == (yy (y*y + xx x*x + zz) z*z) == (zz (z*z + yy y*y + x*x) == ......


your example above will "fail" for both norms, since you arbitrarily change the order in one of the vectors. again, if you use the same permutation for all of your vectors, the results will be the same.

for both L2 and HAMMING norm, the position of the values is irrelevant, as long as you stay consistant with the ordering

 (x*x (x1*x2 + y*y y1*y2 + z*z) z1*z2) == (y*y (y1*y2 + x*x x1*x2 + z*z) z1*z2) == (z*z (z1*z2 + y*y y1*y2 + x*x) x1*x2) == ...


your example above will "fail" for both norms, since you arbitrarily change the order in one of the vectors. again, if you use the same permutation for all of your vectors, the results will be the same.

for both L2 and HAMMING norm, the position of the values is irrelevant, as long as you stay consistant with the ordering

 (x1*x2 + y1*y2 + z1*z2) == (y1*y2 + x1*x2 + z1*z2) == (z1*z2 + y1*y2 + x1*x2) == ...


let me try another drawing:

// ok.
(x1 y1 z1)
(x2 y2 z2)

// still ok (same result as above):
(z1 y1 x1)
(z2 y2 x2)

// not ok:
(x1 y1 z1)
(z2 y2 x2)


your example above will "fail" for both norms, since you arbitrarily change the order in one of the vectors. again, if you use the same permutation for all of your vectors, the results will be the same.