Search: author:Leo Crabbe
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BP943 |
| Visual Bongard Problems whose solutions cannot be deduced when viewed in template form vs. not so. |
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COMMENTS
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Left examples are not required to be valid, as long as their solution doesn't apply in the traditional Bongard Problem format (6 panels vs. 6 panels, all one image). Additionally, they do not necessarily have to be rendered invalid by being viewed in the template format, but their solution does have to be altered. In some cases left examples are simply Problems whose solution is specific to the computer medium (BP941), however some examples have more profound solutions that the pen-and-paper template medium is too restrictive to represent (BP854). |
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CROSSREFS
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See BP568.
Adjacent-numbered pages:
BP938 BP939 BP940 BP941 BP942  *  BP944 BP945 BP946 BP947 BP948
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KEYWORD
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meta (see left/right), links, oebp, time
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WORLD
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bp [smaller | same | bigger]
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AUTHOR
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Leo Crabbe
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BP942 |
| Square bounding box vs. oblong rectangular bounding box. |
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BP941 |
| JPEG image vs. PNG image. |
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BP940 |
| Bongard Problems such that there is a way of making an infinite list of all relevant possible right-sorted examples vs. Bongard Problems where there is no such way of listing all right-sorted examples. |
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BP939 |
| Optical illusions vs. not so. |
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BP938 |
| Bongard Problem with solution relating to concept: shape perimeter vs. Bongard Problem unrelated to this concept. |
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BP937 |
| Shapes have equal perimeter 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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BP935 |
| Shapes have equal area vs. not so. |
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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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