Of course if the point set is symmetrical, there can be fewer than 4 distinct rotations. On Thu, Oct 22, 2020 at 1:40 PM Tom Karzes <karzes@sonic.net> wrote:
No, it is independent of the point set. As long as there are at least two distinct points, there are exactly four minimal solutions. Look at the subject of the sentence you quoted:
The minimal solution is unique up to multiplication by a unit (1, i, -1, -i).
We are only talking about minimal solutions here. That means the GCD of all the points (when one is at the origin) is 1.
Your example is not minimal (the GCD is 5, which happens to be an integer, but in general is a Gaussian integer). Dividing by 5 produces a minimal reference solution. There are exactly 3 others, obtained by multiplying by the units, for a total of 4:
( 0, 0) ( 1, 0) ( 2, 0) ( 3, 0) ( 4, 0) ( 0, 0) ( 0, 1) ( 0, 2) ( 0, 3) ( 0, 4) ( 0, 0) (-1, 0) (-2, 0) (-3, 0) (-4, 0) ( 0, 0) ( 0, -1) ( 0, -2) ( 0, -3) ( 0, -4)
There are no more. Any other set of points with the same shape must be scaled by an amount greater than one. So my statement was entirely correct (except for the degenerate cases of only one point in the set, or no points in the set).
Once you start multiplying a minimal solution by Gaussian integers whose magnitude is greater than 1, then you can sometimes generate more than 4 solutions of that size, but that isn't what I was talking about. I was simply addressing the uniqeness of the desired solution.
Tom
Dan Asimov writes:
Tom Karzes wrote: ----- The minimal solution is unique up to multiplication by a unit (1, i, -1, -i). -----
This depends on the point set.
If say it's
X = {(0,0), (5,0), (10,0), (15,0), (20,0)}
Then there are 12 rotations about the origin that carry X into a set of integer points.
(By angles N*90º and N*90º ± atan2(3,4), N = 0, 1, 2, 3.)
—Dan
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