Adjacent-numbered pages:
BP785 BP786 BP787 BP788 BP789  *  BP791 BP792 BP793 BP794 BP795

math, structure, orderedtriplet, traditional

Aaron David Fairbanks

This answer is independent of whether angles are measured clockwise or counter-clockwise; only a consistent choice must be made.

Adjacent-numbered pages:
BP786 BP787 BP788 BP789 BP790  *  BP792 BP793 BP794 BP795 BP796

math, structure, orderedtriplet, traditional

Aaron David Fairbanks

This is the "harder version" of BP505.

Adjacent-numbered pages:
BP796 BP797 BP798 BP799 BP800  *  BP802 BP803 BP804 BP805 BP806

hard, less, abstract, math, subjective, challenge, right-unknowable, collective, experimental, finishedexamples

Aaron David Fairbanks

See BP372 for the version of this with BPs with black shapes as examples instead of boxes.

Adjacent-numbered pages:
BP800 BP801 BP802 BP803 BP804  *  BP806 BP807 BP808 BP809 BP810

nice, dual, handed, leftright, creativeexamples, structure, contributepairs, rules

boxes_bdimage_six_per_side_extra_below_clear_sort [smaller | same | bigger]

Aaron David Fairbanks

This is solvable; it was solved by Sridhar Ramesh.

A full turn is considered "the same angle" as no turns; likewise for adding and subtracting full turns from any angle. All sequences of angles shown start at the rightmost tick.

It doesn't matter whether the angle is measured clockwise or counterclockwise, as long as the choice is consistent.

Adjacent-numbered pages:
BP820 BP821 BP822 BP823 BP824  *  BP826 BP827 BP828 BP829 BP830

hard, convoluted, notso, math, solved

Aaron David Fairbanks

The original pre-transformed square is the same across all examples, however it is not shown in most examples; what the pre-transformed square looks like must be deduced by the solver.

BP1260 is about applying the same transformation to different objects.

Adjacent-numbered pages:
BP834 BP835 BP836 BP837 BP838  *  BP840 BP841 BP842 BP843 BP844

easy, abstract, arbitrary, anticomputer, traditional, rules

Aaron David Fairbanks

https://en.wikipedia.org/wiki/Moving_sofa_problem

Adjacent-numbered pages:
BP845 BP846 BP847 BP848 BP849  *  BP851 BP852 BP853 BP854 BP855

nice, precise, physics, creativeexamples, proofsrequired, left-narrow, right-narrow, dithering

Leo Crabbe

All examples are Bongard Problems fitting left or right in BP793.

All examples here are in the conventional format, i.e. white background, black vertical dividing line, and examples in boxes on either side.

Border cases are Bongard Problems that always self-sort one way given their particular visual format (e.g. fixed number of boxes), but self-sort a different way in another slightly different format.

Meta Bongard Problems appearing in BP793 that are presentationinvariant necessarily fit right here.

It is interesting to think about how this Bongard Problem sorts itself. The only self-consistent answer is that it fits right.

See BP793 "sorts self left vs. sorts self right".

See BP944 "sorts every BP on one side vs. doesn't".

Adjacent-numbered pages:
BP922 BP923 BP924 BP925 BP926  *  BP928 BP929 BP930 BP931 BP932

hard, solved, presentationinvariant, visualimagination

boxes_bpimage_sorts_self [smaller | same | bigger]
zoom in left (boxes_bpimage_sorts_self_incarnation_dependent) | zoom in right

Aaron David Fairbanks

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.

Adjacent-numbered pages:
BP929 BP930 BP931 BP932 BP933  *  BP935 BP936 BP937 BP938 BP939

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

3_dots_on_square_grid [smaller | same | bigger]

Leo Crabbe

"Any" here means any image of a Bongard Problem in the relevant format, i.e. with white background, black vertical dividing line, and examples in boxes on either side.

All examples shown in this Problem clearly sort themselves on the left or right.

A self-referential but maybe simpler solution is "would sort all examples in this whole Bongard Problem on one of its sides vs. not so." Users adding examples please try to maintain this: for any example you add to the right of this Bongard Problem, make sure it does not sort all the other examples in this Bongard Problem on just one of its sides. - Aaron David Fairbanks, Aug 26 2020

Adjacent-numbered pages:
BP939 BP940 BP941 BP942 BP943  *  BP945 BP946 BP947 BP948 BP949

hard, challenge, presentationinvariant

boxes_bpimage_sorts_self [smaller | same | bigger]
zoom in left | zoom in right

Jago Collins