There are many ambiguities here. The solver is expected to determine what things are "allowed" to be inputs for each process. To avoid confusion examples should not be sorted differently if you consider inputting nothing.

In each example there is at least some overlap between the set of possible inputs and the set of possible outputs for each process. If we did not apply this constraint, an easy example to be sorted right would be a process that turns blue shapes red.

A harder-to-read but more clearly defined version of this Problem could include within each example a mini Bongard Problem sorting left all allowed inputs for the process.

https://en.wikipedia.org/wiki/Fixed_point_(mathematics)

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structure, rules, miniworlds

Leo Crabbe

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nice, precise, allsorted, stretch, perfect, traditional, preciseworld

3_or_4_points [smaller | same | bigger]

Leo Crabbe

"Full overlapping not allowed" means you cannot overlay an image onto itself without moving it; if this were allowed all images would be sorted on the left. The copies can be moved around (translated) in 2D but can not be flipped or rotated.

There are examples on the right drawn with thick lines, and these could be created by copying an image with slightly thinner lines and moving it over a tiny amount. If you fix this issue by saying "the copy has to be moved over more than a tiny amount" then the Bongard Problem is perfect but not precise, but if you fix this issue by saying "interpret the figures as made up of (infinitesimally) thin lines" then it's precise but not perfect. - Aaron David Fairbanks, Jun 17 2023

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nice, notso, creativeexamples

Leo Crabbe

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precise, allsorted, number, left-null, help, preciseworld

dots [smaller | same | bigger]

Leo Crabbe

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precise, stretch, boundingbox, invalid, experimental, preciseworld

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zoom in left (square_arrangement_of_white_pixels) | zoom in right (oblong_rectangular_arrangement_of_white_pixels)

Leo Crabbe

Evidently this solution is stretching the idea of what makes images "different". Each corresponding pair of panels would be parsed as identical in any other case. It is worth noting that if you view this Problem on a template, the solution no longer applies at all.

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less, precise, dual, antihuman, contributepairs, invalid, experimental, funny

Leo Crabbe

Panels on the left hand side contain geometrical objects that appear distorted (to a human) due to surrounding information.

Vicente Sierra-Vázquez & Ignacio Serrano-Pedraza, Application of Riesz transforms to the isotropic AM-PM decomposition of geometrical-optical illusion images, April 2010, Figure 1.

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fuzzy, anticomputer, subjective, contributepairs, invalid, experimental

Leo Crabbe

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precise, allsorted, unstable, left-narrow, perfect, unorderedpair

2_fill_shapes [smaller | same | bigger]

Leo Crabbe

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nice, precise, allsorted, unstable, left-narrow, perfect, pixelperfect, unorderedpair

2_fill_shapes [smaller | same | bigger]

Leo Crabbe

In other words, we take the distance between points (a,b) and (c,d) to be equal to |c-a| + |d-b|, or, in other words, the distance of the shortest path between points that travels along grid lines. In mathematics, this way of measuring distance is called the 'taxicab' or 'Manhattan' metric. The points on the left hand side form equilateral triangles in this metric.

⠀

An alternate (albeit more convoluted) solution that someone may arrive at for this Problem is as follows: The triangles formed by the points on the left have some two points diagonal to each other (in the sense of bishops in chess), and considering the corresponding edge as their base, they also have an equal height. However, this was proven to be equivalent to the Manhattan distance answer by Sridhar Ramesh. Here is the proof:

⠀

An equilateral triangle amounts to points A, B, and C such that B and C lie on a circle of some radius centered at A, and the chord from B to C is as long as this radius.

A Manhattan circle of radius R is a turned square, ♢, where the Manhattan distance between any two points on opposite sides is 2R, and the Manhattan distance between any two points on adjacent sides is the larger distance from one of those points to the corner connecting those sides. Thus, to get two of these points to have Manhattan distance R, one of them must be a midpoint of one side of the ♢ (thus, bishop-diagonal from its center) and the other can then be any point on an adjacent side of the ♢ making an acute triangle with the aforementioned midpoint and center.

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hard, allsorted, solved, left-finite, right-finite, perfect, pixelperfect, unorderedtriplet, finishedexamples

3_dots_on_square_grid [smaller | same | bigger]

Leo Crabbe