Search: subworld:everything
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| BP929 |
| Bongard Problems about sequences of arbitrary length vs. Bongard Problems about sequences in which all examples have the same sequence length. |
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| BP930 |
| BP Pages on the OEBP where users are advised to upload examples that help people (by hinting at the solution) vs. other BP Pages. |
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COMMENTS
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Left examples have the keyword "help" on the OEBP.
BPs should be marked "help" when the OEBP wants most examples (at least on one side) to be helpful (not when just one or two uploaded examples are helpful).
Helpfulness can be a spectrum; most Bongard Problems are helpful to some degree just by not using the most convoluted unintelligible examples possible.
Examples that are helpful to people are often not particularly helpful to computers.
Any helpful Bongard Problem has a harder, not helpful version. For example, BP384 (square number of dots versus non-square number of dots) would be much harder if all examples had hundreds of dots that weren't arranged recognizably. Instead, the dots in the examples are always arranged in shapes that make the square-ness or non-square-ness of the numbers easy to check without brute counting.
When all examples in a Bongard Problem are helpful, it may become unclear whether the helpfulness is part of the Bongard Problem's solution.
E.g.: Is the left-hand side of BP384 "square number of dots", or is it "square number of dots that are arranged in a helpful way so as to communicate the square-ness"?
See seemslike, where examples being helpful is an irremovable aspect of the Bongard Problem's solution. |
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CROSSREFS
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Adjacent-numbered pages:
BP925 BP926 BP927 BP928 BP929 * BP931 BP932 BP933 BP934 BP935
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KEYWORD
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anticomputer, meta (see left/right), links, keyword, oebp, instruction
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WORLD
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bppage [smaller | same | bigger]
zoom in left (help_bp)
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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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| BP936 |
| Bongard Problem with solution relating to concept: area (geometry) vs. Bongard Problem unrelated to this concept. |
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