Adjacent-numbered pages:
BP811 BP812 BP813 BP814 BP815  *  BP817 BP818 BP819 BP820 BP821

precise, notso, stretch, unstable, perfect

fill_shapes [smaller | same | bigger]

Leo Crabbe

The conceptual limit of the sequence may not be the limit of the points in the image. For example in a sequence of halvings the limit value is never reached, so the bottom would never change color and thus its limit would not would not either.

Sequences progress from left to right (and there is not usually a way to intuitively extend the sequence in the other direction).

Adjacent-numbered pages:
BP847 BP848 BP849 BP850 BP851  *  BP853 BP854 BP855 BP856 BP857

notso, creativeexamples, perfect, infinitedetail, assumesfamiliarity, structure, contributepairs, rules

Aaron David Fairbanks

This is the "smallest" possible valid Bongard Problem on black and white bitmap images (up to switching sides).

Adjacent-numbered pages:
BP854 BP855 BP856 BP857 BP858  *  BP860 BP861 BP862 BP863 BP864

minimal, blackwhite, perfect, pixelperfect, experimental, funny, unstableworld

black_or_white_pixel [smaller | same | bigger]
zoom in left (black_pixel) | zoom in right (white_pixel)

Aaron David Fairbanks

This is a generalisation of BP861.

Adjacent-numbered pages:
BP855 BP856 BP857 BP858 BP859  *  BP861 BP862 BP863 BP864 BP865

hard, nice, stub, precise, stretch, unstable, hardsort, challenge, creativeexamples, perfect, pixelperfect

fill_shape [smaller | same | bigger]

Leo Crabbe

See BP860 for the more general version of this solution.

Adjacent-numbered pages:
BP856 BP857 BP858 BP859 BP860  *  BP862 BP863 BP864 BP865 BP866

nice, precise, unstable, perfect, pixelperfect

fill_shape [smaller | same | bigger]

Leo Crabbe

Or, "evokes idea of symmetry (see BP760) vs. not so," which is also a solution for BP847. - Aaron David Fairbanks, Jul 29 2020

Adjacent-numbered pages:
BP864 BP865 BP866 BP867 BP868  *  BP870 BP871 BP872 BP873 BP874

abstract, spectrum, anticomputer, subjective, concept, perfect

Leo Crabbe

There is a slight ambiguity here regarding whether a shape could be placed within another shape's hole. This is a question of how one perceives the Problem: are we sliding shapes around on a table in 2D or are we allowed to 'lift' them in 3D space?

Adjacent-numbered pages:
BP887 BP888 BP889 BP890 BP891  *  BP893 BP894 BP895 BP896 BP897

nice, precise, perfect, pixelperfect, help

Leo Crabbe

Adjacent-numbered pages:
BP907 BP908 BP909 BP910 BP911  *  BP913 BP914 BP915 BP916 BP917

perfect, contributepairs

zoom in right (shape_outline)

Leo Crabbe

A spot-the-difference exercise.

Arguably invalid (solution not simple).

Adjacent-numbered pages:
BP915 BP916 BP917 BP918 BP919  *  BP921 BP922 BP923 BP924 BP925

less, precise, convoluted, arbitrary, stretch, unstable, left-finite, left-full, perfect, pixelperfect, experimental, funny

bmp [smaller | same | bigger]

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