Adjacent-numbered pages:
BP389 BP390 BP391 BP392 BP393  *  BP395 BP396 BP397 BP398 BP399

hard, nice, solved, traditional, dithering, left-listable, right-listable

Jago Collins

If a "string" is wound tightly around the shape, does one of its segments lie directly on the shape?

All examples in this Problem are connected line segments or curves.

We are taking lines here to be infinitely thin, so that if the boundary of the convex hull intersects the endpoint of a line exactly it is understood that they meet at 1 point.

Adjacent-numbered pages:
BP559 BP560 BP561 BP562 BP563  *  BP565 BP566 BP567 BP568 BP569

Imagine wrapping a string around the pointed star. This string would take the shape of the boundary of the star's convex hull (a regular pentagon), and would only touch the star at the end of each of its 5 individual tips, therefore the star belongs on the left.

hard, nice, allsorted, solved, perfect

Leo Crabbe

This is a very inexact definition. Some left examples arguably should be placed on the right, since the particular way they are represented is arbitrary--the Platonic solids EX6730 and primes EX6734 especially, as these show arbitrary placement and arrangement of objects. Furthermore if arbitrary representations are allowed one cannot be sure for example the right hand drawing of random numbers EX6740 does not represent "numbers" in general. Still this Bongard Problem has been solved by people.

Adjacent-numbered pages:
BP808 BP809 BP810 BP811 BP812  *  BP814 BP815 BP816 BP817 BP818

fuzzy, abstract, stretch, math, solved, collective, experimental, dithering

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

Adjacent-numbered pages:
BP844 BP845 BP846 BP847 BP848  *  BP850 BP851 BP852 BP853 BP854

hard, convoluted, solved, rules

constant_change_seq_loop_right [smaller | same | bigger]

Aaron David Fairbanks

The rows in the panels on the right hand side show all the ways you can divide a certain number of dots between a certain number of bins, ignoring which bins they are placed in.

Adjacent-numbered pages:
BP899 BP900 BP901 BP902 BP903  *  BP905 BP906 BP907 BP908 BP909

solved, left-null, grid, left-listable, right-listable

Molly C Klenzak

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

This was an unintended solution for BP1130.

In category theory lingo, left examples are built by repeated horizontal composition and vertical composition. (Making horizontal lines as 0-ary vertical compositions is here forbidden.)

Anything fitting right in BP1130 fits right here.

Adjacent-numbered pages:
BP1124 BP1125 BP1126 BP1127 BP1128  *  BP1130 BP1131 BP1132 BP1133 BP1134

hard, less, convoluted, solved, inductivedefinition

[smaller | same | bigger]

Aaron David Fairbanks

The description in terms of rectangles was noted by Sridhar Ramesh when he solved this.

All examples in this Bongard Problem feature arced line segments connected at endpoints; these segments do not cross across one another and they are nowhere vertical; they never double back over themselves in the horizontal direction.

Furthermore, in each example, there is a single leftmost point and a single rightmost point, and every segment is part of a chain bridging between them. So, there is a topmost total chain of segments and bottommost total chain of segments.

Any picture on the left can be turned into a subdivided rectangle by the process of expanding points into vertical lines.

Here is another answer:

"Right examples: some junction point has a single line coming out from either the left or right side."

If there is some junction point with only a single line coming out from a particular side, the point cannot be expanded into a vertical segment with two horizontal segments bookending its top and bottom (as it would be if this were a subdivision of a rectangle).

And this was the original, more convoluted idea of the author:

"Start with a string along the top path. Sweep it down, region-by-region, until it lies along the bottom path. The string may only enter a region when it fully covers that region's top edge and likewise it must exit by fully covering the bottom edge. Only in left images can this process be done so that no segment of the string ever hesitates."

Quite convoluted when spelled out in detail. (Although it is not terribly complicated to imagine visually. See the keyword unwordable.)

The string-sweeping answer is the same as the rectangle answer because a rectangle represents the animation of a string throughout an interval of time. (A horizontal cross-section of the rectangle represents the string, and the vertical position is time.) Distorting the rectangle into a new shape is the same as animating a string sweeping across that new shape.

In particular, shrinking vertical lines of a rectangle into points means just those points of the string stay still as the string sweeps down.

The principle that horizontal lines subdividing the original rectangle become the segments in the final picture corresponds to the idea that the string must enter or exit a single region all at once.

BP1129 started as an incorrect solution for this Bongard Problem. Anything fitting right in BP1130 fits right in BP1129.

Adjacent-numbered pages:
BP1125 BP1126 BP1127 BP1128 BP1129  *  BP1131 BP1132 BP1133 BP1134 BP1135

hard, unwordable, solved

[smaller | same | bigger]

Aaron David Fairbanks