Search: -meta:BP501
|
Displaying 31-39 of 39 results found.
|
( prev ) page 1 2 3 4
|
|
Sort:
id
Format:
long
Filter:
(all | no meta | meta)
Mode:
(words | no words)
|
|
|
|
|
BP1011 |
| Polygon can be inscribed in a circle vs. not so. |
|
| |
|
|
|
|
|
BP1038 |
| Reciprocals of right numbers summed in infinite series converge vs. diverge. |
|
| |
|
|
CROSSREFS
|
Adjacent-numbered pages:
BP1033 BP1034 BP1035 BP1036 BP1037  *  BP1039 BP1040 BP1041 BP1042 BP1043
|
|
KEYWORD
|
hard, precise, allsorted, handed, leftright, math, challenge, meta (see left/right), miniproblems, assumesfamiliarity, structure, preciseworld, presentationinvariant
|
|
WORLD
|
boxes_dots_bpimage_clear_set_of_numbers [smaller | same | bigger]
|
|
AUTHOR
|
Aaron David Fairbanks
|
|
|
|
|
BP1040 |
| Left is union of (non-constant) arithmetic progressions vs. not so. |
|
| |
|
|
CROSSREFS
|
Adjacent-numbered pages:
BP1035 BP1036 BP1037 BP1038 BP1039  *  BP1041 BP1042 BP1043 BP1044 BP1045
|
|
KEYWORD
|
hard, precise, allsorted, convoluted, notso, handed, leftright, math, challenge, meta (see left/right), miniproblems, assumesfamiliarity, structure, preciseworld, presentationinvariant
|
|
WORLD
|
boxes_dots_bpimage_clear_set_of_numbers [smaller | same | bigger]
|
|
AUTHOR
|
Aaron David Fairbanks
|
|
|
|
|
BP1055 |
| Equidiagonal quadrilaterals vs. non-equidiagonal quadrilaterals |
|
| |
|
|
|
|
|
BP1123 |
| Can be cut into tiles forming a checkerboard pattern vs. not so. |
|
| ?
|
|
|
|
|
|
|
BP1129 |
| An oval is sorted left; shapes are sorted left when they can be built out of others sorted left by A) joining side by side (at a point) or B) joining one on top of the other (joining one's entire bottom edge to the other's entire top edge). |
|
| |
|
|
|
|
|
BP1130 |
| Start with a rectangle subdivided further into rectangles and shrink the vertical lines into points vs. the shape does not result from this process. |
|
| ?
|
|
|
|
COMMENTS
|
The description in terms of rectangles was noted by Sridhar Ramesh when he solved this.
All examples in this Bongard Problem feature arced line segments connected at endpoints; these segments do not cross across one another and they are nowhere vertical; they never double back over themselves in the horizontal direction.
Furthermore, in each example, there is a single leftmost point and a single rightmost point, and every segment is part of a path bridging between them. So, there is a topmost total path of segments and bottommost total chain of segments.
Any picture on the left can be turned into a subdivided rectangle by the process of expanding points into vertical lines.
Here is another answer:
"Right examples: some junction point has a single line coming out from either the left or right side."
If there is some junction point with only a single line coming out from a particular side, the point cannot be expanded into a vertical segment with two horizontal segments bookending its top and bottom (as it would be if this were a subdivision of a rectangle).
And this was the original, more convoluted idea of the author:
"Start with a string along the top path. Sweep it down, region-by-region, until it lies along the bottom path. The string may only enter a region when it fully covers that region's top edge and likewise it must exit by fully covering the bottom edge. Only in left images can this process be done so that no segment of the string ever hesitates."
Quite convoluted when spelled out in detail, but not terribly complicated to imagine visually. (See the keyword unwordable.)
The string-sweeping answer is the same as the rectangle answer because a rectangle represents the animation of a string throughout an interval of time. (A horizontal cross-section of the rectangle represents the string, and the vertical position is time.) Distorting the rectangle into a new shape is the same as animating a string sweeping across that new shape.
In particular, shrinking vertical lines of a rectangle into points means just those points of the string stay still as the string sweeps down.
The principle that horizontal lines subdividing the original rectangle become the segments in the final picture corresponds to the idea that the string must enter or exit a single region all at once. |
|
CROSSREFS
|
BP1129 started as an incorrect solution for this Bongard Problem. Anything fitting right in BP1130 fits right in BP1129.
Adjacent-numbered pages:
BP1125 BP1126 BP1127 BP1128 BP1129  *  BP1131 BP1132 BP1133 BP1134 BP1135
|
|
KEYWORD
|
hard, unwordable, solved
|
|
CONCEPT
|
topological_transformation (info | search), imagined_motion (info | search)
|
|
WORLD
|
[smaller | same | bigger]
|
|
AUTHOR
|
Aaron David Fairbanks
|
|
|
|
|
BP1200 |
| The whole rectangle can be filled in by successively replacing pairs of adjacent rectangles with one vs. not so. |
|
| |
|
|
COMMENTS
|
Another wording: "can be repeatedly broken along 'fault lines' to yield individual pieces vs not." |
|
REFERENCE
|
Robert Dawson, A forbidden suborder characterization of binarily composable diagrams in double categories, Theory and Applications of Categories, Vol. 1, No. 7, p. 146-145, 1995. |
|
CROSSREFS
|
All of the examples fitting left here would fit right in BP1199 except for (1) a single rectangle, (2) two rectangles stacked vertically, or (3) two rectangles side by side horizontally.
All of the examples fitting right in in BP1097 (re-styled) would fit right here (besides a single solid block, but that isn't shown there).
Adjacent-numbered pages:
BP1195 BP1196 BP1197 BP1198 BP1199  *  BP1201 BP1202 BP1203 BP1204 BP1205
|
|
KEYWORD
|
hard, precise, challenge, proofsrequired, inductivedefinition, left-listable, right-listable
|
|
AUTHOR
|
Aaron David Fairbanks
|
|
|
|
|
BP1245 |
| When two players alternate coloring regions, either can force connection from top edge to bottom edge vs. either can force connection from left edge to right edge. |
|
| |
|
|
|
|
Welcome |
Solve |
Browse |
Lookup |
Recent |
Links |
Register |
Contact
Contribute |
Keywords |
Concepts |
Worlds |
Ambiguities |
Transformations |
Invalid Problems |
Style Guide |
Goals |
Glossary
|
|
|
|
|
|
|
|
|