Search: subworld:everything
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| BP927 |
| Image of Bongard Problem whose self-sorting depends on examples in it vs. image of Bongard Problem that will sort any Bongard Problem with its solution on either its left or right regardless of examples chosen. |
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COMMENTS
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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. |
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CROSSREFS
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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
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KEYWORD
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hard, solved, presentationinvariant, visualimagination
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WORLD
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boxes_bpimage_sorts_self [smaller | same | bigger]
zoom in left (boxes_bpimage_sorts_self_incarnation_dependent) | zoom in right
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AUTHOR
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Aaron David Fairbanks
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| BP931 |
| Some number labels its own position in the sequence from left to right vs. not so. |
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| BP932 |
| Every vertex is connected to every other vs. vertices are connected in a cycle (no other connections). |
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COMMENTS
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Complete graphs with zero, one, two, or three vertices would be ambiguously categorized (fit in overlap of both sides).
Left examples are called "fully connected graphs." Right examples are called "cycle graphs." |
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CROSSREFS
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Adjacent-numbered pages:
BP927 BP928 BP929 BP930 BP931 * BP933 BP934 BP935 BP936 BP937
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KEYWORD
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precise, left-narrow, right-narrow, both, preciseworld
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CONCEPT
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graph (info | search),
distinguishing_crossing_curves (info | search),
all (info | search),
loop (info | search)
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WORLD
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connected_graph [smaller | same | bigger]
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AUTHOR
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Aaron David Fairbanks
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| BP933 |
| Ball will reach edge of bounding box under gravity vs. not so. |
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COMMENTS
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Strictly this Problem's solution is not actually about gravity, it is about a constant downwards force (the ball's time-independent path does not depend on the magnitude of the force, only direction). The phrasing for the solution is a shorthand that takes advantage of human physical intuition. |
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CROSSREFS
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Adjacent-numbered pages:
BP928 BP929 BP930 BP931 BP932 * BP934 BP935 BP936 BP937 BP938
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KEYWORD
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physics
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CONCEPT
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bounding_box (info | search),
imagined_motion (info | search),
gravity (info | search)
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WORLD
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dot_with_lines_or_curves [smaller | same | bigger]
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AUTHOR
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Leo Crabbe
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| BP934 |
| If "distance" is taken to be the sum of horizontal and vertical distances between points, the 3 points are equidistant from each other vs. not so. |
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COMMENTS
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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.
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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:
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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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CROSSREFS
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Adjacent-numbered pages:
BP929 BP930 BP931 BP932 BP933 * BP935 BP936 BP937 BP938 BP939
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KEYWORD
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hard, allsorted, solved, left-finite, right-finite, perfect, pixelperfect, unorderedtriplet, finishedexamples
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CONCEPT
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triangle (info | search)
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WORLD
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3_dots_on_square_grid [smaller | same | bigger]
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AUTHOR
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Leo Crabbe
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| BP935 |
| Shapes have equal area vs. not so. |
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| BP937 |
| Shapes have equal perimeter vs. not so. |
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| BP939 |
| Optical illusions vs. not so. |
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