Left examples are quantity-comparison Problems in which the dividing point on the spectrum separating the sides lands between them. Right examples are quantity-comparison Problems in which the dividing point is the entirety of one of the sides (always the left side, for clarity).

Only less-than vs. greater-than Problems may have imprecise (BP876^right) dividing point between sides.

Adjacent-numbered pages:
BP872 BP873 BP874 BP875 BP876  *  BP878 BP879 BP880 BP881 BP882

hard, challenge, meta (see left/right), miniproblems, creativeexamples, assumesfamiliarity, structure, presentationinvariant

bpimage_shapes_quantity_soln [smaller | same | bigger]

Aaron David Fairbanks

"Allsorted" Bongard Problems (BP875^left) always fit as right examples.

Left examples are never "allsorted".

Adjacent-numbered pages:
BP873 BP874 BP875 BP876 BP877  *  BP879 BP880 BP881 BP882 BP883

hard, challenge, meta (see left/right), miniproblems, creativeexamples, assumesfamiliarity, structure, presentationinvariant

bpimage_shapes_exact_sort [smaller | same | bigger]
zoom in left | zoom in right (bpimage_shapes_exact_sort_no_midway)

Aaron David Fairbanks

Which two sides are the long sides and which side is the short side, or equivalently which angles are the wider angles and which angle is the narrower angle, is the only relevant information to consider for each triangle. Triangles are all assumed isosceles and congruent to one another.

All examples in this Problem feature four of these triangles connected by corners and/or edges.

BP897 was conceived as a false solution for this.

Adjacent-numbered pages:
BP893 BP894 BP895 BP896 BP897  *  BP899 BP900 BP901 BP902 BP903

hard, precise, allsorted, notso, math, preciseworld

[smaller | same | bigger]

Molly C Klenzak

This Problem is related to the four colour theorem.

Adjacent-numbered pages:
BP894 BP895 BP896 BP897 BP898  *  BP900 BP901 BP902 BP903 BP904

hard, nice, math

[smaller | same | bigger]

Jago Collins

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-categorize one way given their particular visual format (e.g. fixed number of boxes), but self-categorize a different way in another slightly different format.

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

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

Loosely speaking, examples on the left are "Bongard Problems that can be self-similar". However, Bongard Problems with images of themselves deeply nested in boxes or rotated/flipped are not considered "self-similar"; the Bongard Problem must use itself, as-is (allowing downward scaling and ignoring pixelation), as a panel.

All examples here are in the conventional format, i.e. white background, black vertical dividing line, and examples in boxes on either side. (A more general version of this Bongard Problem might allow many formats of Bongard Problems, sorting an image left if a self-similar version is possible having the same solution and format. This more general version would no longer be tagged presentationinvariant, since sorting would not only depend on solution, but also format.)

It would hint at the solution (keyword help) to only include images of Bongard Problems that, as it stands, are already clearly categorized on one side by themselves. (That is, images of Bongard Problems that belong on one of the two sides of BP793.) It is tricky to come up with images that are categorized by themselves as it stands but that could NOT be recursively included within themselves. EX7967, EX7999, EX7995, and EX6574 are some examples.

See BP987 which narrows down the left-hand side of this BP further based on whether or not the BP could contain itself as a panel on both sides.

Adjacent-numbered pages:
BP949 BP950 BP951 BP952 BP953  *  BP955 BP956 BP957 BP958 BP959

hard, stub, abstract, challenge, meta (see left/right), miniproblems, infinitedetail, presentationinvariant, visualimagination

Leo Crabbe

Rotations are allowed. To avoid confusion about whether reflections are allowed, no examples are included on the right that require reflections to match up with themselves locally but not globally; no examples are included on the left that can match up with themselves locally but not globally using a reflection.

Only parts of ellipses are used, and only one type of ellipse per image, to make everything easier to read and reason about.

See BP1246 for a variation on this idea where instead of lining the image up with itself along arbitrarily small regions, you line the image up with itself along individual separate objects.

Adjacent-numbered pages:
BP960 BP961 BP962 BP963 BP964  *  BP966 BP967 BP968 BP969 BP970

hard, precise, distractingworld, perfect

Aaron David Fairbanks

Right:

All are "all but one are ___"; all but one are black.

All are "every other is ___"; every other is solid polygons.

All are "gradually becoming ___"; gradually becoming thickly outlined.

Left:

All but one are "all but one are ___".

Every other is "every other is ___".

Gradually becoming "gradually becoming ___".

Here is another way of putting it:

Call it "meta" when the whole imitates its parts, and call it "doubly-meta" when the whole imitates its parts with respect to the way it imitates its parts. Left are doubly-meta, while right are just meta.

Here is a more belabored way of putting it:

Call something like "is star-shaped" a "rule". An object can fit a rule.

Call something like "all but one are ___" a "rule-parametrized rule". A collection of objects, with respect to a particular rule, can fit a rule-parametrized rule.

A drawing on the right shows many collections. Every collection fits the same rule-parametrized rule (with respect to various rules); furthermore the collection of collections fits that same rule-parametrized rule (with respect to some rule collections can fit).

Likewise a drawing on the left shows a collection of collections, with some noticeable recurring rule-parametrized rule. The collection of collections must fit that rule-parametrized rule with respect to the rule of fitting that rule-parametrized rule (with respect to various rule).

An unintended solution to this BP is "not all groups share some noticeable property vs. all do." It is hard to come up with examples foiling this alternative solution because (what was above called) the rule-parametrized rule usually has to do with not all objects in the collection fitting the rule. (See BP568, which is about BP ideas that are always overridden by a simpler solution.)

Some examples would fit left under a certain interpretation: EX8220 "all are 'all are ___' " and EX8222 "palindrome with respect to being a palindrome with respect to ___" (every shown collection is a palindrome with respect to some property, and all things in a list being the same is a palindrome). But those rules are not necessarily the most obvious ways of interpreting these pictures, so they have been marked as ambiguous. Either of these placed on the left would prevent the intended solution being overridden (see the previous paragraph).

Here is a list of left example ideas that would be impossible to make:

- Exhaustive list of all exhaustive lists of all ____.

The right side of this Problem is a subset of BP999^left.

Adjacent-numbered pages:
BP993 BP994 BP995 BP996 BP997  *  BP999 BP1000 BP1001 BP1002 BP1003

"Odd one out with respect to what property is the odd one out" would not fit left in this Problem: even though this example does seem doubly-meta, it is not doubly-meta in the right way. There is no odd one out with respect to the property of having an odd one out.

Similarly, consider "gradual transition with respect to what the gradual transition is between", etc. Instead of having the form "X 'X __' ", this is more like "X [the __ appearing in 'X __'] ". Examples like these two could make for a different Bongard Problem.

hard, unwordable, challenge, overriddensolution, infodense, contributepairs, funny, rules, miniworlds

zoom in right

Aaron David Fairbanks