Search: -meta:BP501
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Displaying 11-20 of 26 results found.
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BP849 |
| Loop is time-symmetrical (loop is like ABCBABCBA... instead of like ABCABCABC...) up to rotation of object vs. not so (neither up to rotation nor reflection). |
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BP860 |
| Finitely many copies of the shape can be arranged such that they are locked together vs. not so. |
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CROSSREFS
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This is a generalisation of BP861.
Adjacent-numbered pages:
BP855 BP856 BP857 BP858 BP859  *  BP861 BP862 BP863 BP864 BP865
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KEYWORD
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hard, nice, stub, precise, stretch, unstable, hardsort, challenge, creativeexamples, perfect, pixelperfect
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CONCEPT
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tiling (info | search)
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WORLD
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fill_shape [smaller | same | bigger]
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AUTHOR
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Leo Crabbe
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BP898 |
| Can fold into tetragonal disphenoid ("isosceles tetrahedron") vs. cannot. |
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COMMENTS
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Which two sides are the long sides and which side is the short side, or equivalently which angles are the wider angles and which angle is the narrower angle, is the only relevant information to consider for each triangle. Triangles are all assumed isosceles and congruent to one another.
All examples in this Problem feature four of these triangles connected by corners and/or edges. |
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CROSSREFS
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BP897 was conceived as a false solution for this.
Adjacent-numbered pages:
BP893 BP894 BP895 BP896 BP897  *  BP899 BP900 BP901 BP902 BP903
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KEYWORD
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hard, precise, allsorted, notso, math, preciseworld
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CONCEPT
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triangle (info | search)
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WORLD
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[smaller | same | bigger]
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AUTHOR
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Molly C Klenzak
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BP899 |
| Regions in drawing (ignore background) can be coloured using three or fewer colours such that no adjacent regions are coloured the same colour vs. four colours are required. |
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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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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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BP944 |
| Image of Bongard Problem that would sort ANY image of a valid Bongard Problem on one of its sides vs. image of Bongard Problem whose categorization of a BP image would depend on the solution or examples in it. |
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COMMENTS
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"Any" here means any image of a Bongard Problem in the relevant format, i.e. with white background, black vertical dividing line, and examples in boxes on either side.
All examples shown in this Problem clearly sort themselves on the left or right.
A self-referential but maybe simpler solution is "would sort all examples in this whole Bongard Problem on one of its sides vs. not so." Users adding examples please try to maintain this: for any example you add to the right of this Bongard Problem, make sure it does not sort all the other examples in this Bongard Problem on just one of its sides. - Aaron David Fairbanks, Aug 26 2020 |
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CROSSREFS
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Adjacent-numbered pages:
BP939 BP940 BP941 BP942 BP943  *  BP945 BP946 BP947 BP948 BP949
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KEYWORD
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hard, challenge, presentationinvariant
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WORLD
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boxes_bpimage_sorts_self [smaller | same | bigger] zoom in left | zoom in right
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AUTHOR
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Jago Collins
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BP965 |
| If you place the image on top of itself so that it lines up with itself exactly within a small region, it also lines up everywhere else vs. not so. |
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COMMENTS
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Rotations are allowed. To avoid confusion about whether reflections are allowed, no examples are included on the right that require reflections to match up with themselves locally but not globally; no examples are included on the left that can match up with themselves locally but not globally using a reflection.
Only parts of ellipses are used, and only one type of ellipse per image, to make everything easier to read and reason about. |
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CROSSREFS
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See BP1246 for a variation on this idea where instead of lining the image up with itself along arbitrarily small regions, you line the image up with itself along individual separate objects.
Adjacent-numbered pages:
BP960 BP961 BP962 BP963 BP964  *  BP966 BP967 BP968 BP969 BP970
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KEYWORD
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hard, precise, distractingworld, perfect
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CONCEPT
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local_global (info | search)
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AUTHOR
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Aaron David Fairbanks
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BP998 |
| X "X _" vs. all are "X _"; X Y. |
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COMMENTS
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Right:
All are "all but one are ___"; all but one are black.
All are "every other is ___"; every other is solid polygons.
All are "gradually becoming ___"; gradually becoming thickly outlined.
Left:
All but one are "all but one are ___".
Every other is "every other is ___".
Gradually becoming "gradually becoming ___".
Here is another way of putting it:
Call it "meta" when the whole imitates its parts, and call it "doubly-meta" when the whole imitates its parts with respect to the way it imitates its parts. Left are doubly-meta, while right are just meta.
Here is a more belabored way of putting it:
Call something like "is star-shaped" a "rule". An object can satisfy a rule.
Call something like "all but one are ___" a "rule-parametrized rule". A collection of objects can satisfy a rule-parametrized rule with respect to a particular rule.
On the right: every collection fits the same rule-parametrized rule (with respect to various rules); furthermore the collection of collections fits that same rule-parametrized rule (with respect to some unrelated rule that collections can satisfy).
On the left: The collection of collections fits a rule-parametrized rule with respect to the rule of fitting that rule-parametrized rule (with respect to various rules).
Previously, an unintended solution to this BP was "not all groups share some noticeable property vs. all do." It is hard to come up with examples foiling this alternative solution because the rule-parametrized rule (see explanation above) usually has to do with not all objects in the collection fitting the rule. (See BP568, which is about BP ideas that are always overridden by a simpler solution.) The example EX10108 "all five are 'all five are ___'" was added, foiling the alternative solution. |
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CROSSREFS
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The right side of this Problem is a subset of BP999left.
Adjacent-numbered pages:
BP993 BP994 BP995 BP996 BP997  *  BP999 BP1000 BP1001 BP1002 BP1003
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EXAMPLE
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"Odd one out with respect to what property is the odd one out" would not fit left: even though this example does seem doubly-meta, it is not doubly-meta in the right way. There is no odd one out with respect to the property of having an odd one out.
Similarly, consider "gradual transition with respect to what the gradual transition is between", etc. Instead of having the form "X 'X __' ", this is more like "X [the __ appearing in 'X __']". Examples like these two could make for a different Bongard Problem. |
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KEYWORD
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hard, unwordable, challenge, overriddensolution, infodense, contributepairs, funny, rules, miniworlds
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CONCEPT
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self-reference (info | search)
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WORLD
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zoom in right
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AUTHOR
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Aaron David Fairbanks
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BP1011 |
| Polygon can be inscribed in a circle vs. not so. |
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