Here's a grid of shapes:
Which of the following shapes should go in the missing cell, and why? There is only one correct answer, and the explanation should not be very complicated.
This is my own puzzle.
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1$\begingroup$ Is this your own puzzle, or did you get it from somewhere else? If it is not your own our policy is that you link the source in your post. $\endgroup$– bobbleCommented Jun 4, 2020 at 19:38
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3$\begingroup$ This is my own puzzle. $\endgroup$– David DimaCommented Jun 4, 2020 at 19:38
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$\begingroup$ rot13(Gur guerr va gur gbc ebj ner ubzrbzbecuvp nf cvrprjvfr yvarne 1-znavsbyqf. Gur guerr va gur frpbaq ebj ner nyfb (ohg ner abg ubzrbzbecuvp gb gur barf va gur svefg ebj). Gur obggbz yrsg vf ubzrbzbecuvp gb gur ebhaq pubvpr, ohg gur obggbz yrsg vfa'g ubzrbzbecuvp gb gur obggbz-zvqqyr, juvpu znxrf zr jbaqre vs lbh reerq ba gur ynggre.) $\endgroup$– msh210Commented Jun 4, 2020 at 21:05
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1$\begingroup$ @msh210 rot13(Lrf, lbh'er evtug... Fbeel nobhg gung, V'ir svkrq gur dhrfgvba.) $\endgroup$– David DimaCommented Jun 4, 2020 at 22:33
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$\begingroup$ I noticed that -- rot13(Gur funcrf va rnpu ebj unir gur fnzr ahzore bs nernf naq gur fnzr ahzore bs yvarf gbhpuvat gur rqtr. Naq juvyr nyy bs gur bcgvbaf sbe gur nafjre unq rvtug nernf, bayl bar bs gurz unq frira yvarf gbhpuvat gur rqtr: gur pvepyr. Vf gung n pbafrdhrapr bs gur gbcbybtl/qrsbezvat cnggrea lbh unir gnyxrq nobhg? Fb bar pna guvax nobhg vg nf fyvqvat gur yvarf nybat rnpu bgure orvat crezvggrq, ohg abg nyy gur jnl gb gur rqtr nf gung jbhyq ryvzvangr bar be zber nernf? Naq fvzvyneyl, gur rqtr yvarf pna fyvqr nyy gur jnl nebhaq gur rqtr, ohg abg yrnir gur rqtr.) $\endgroup$– SpiritFryerCommented Mar 1, 2023 at 12:53
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3 Answers
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The missing shape is
the circle.
Reasoning:
In each row the shapes are topologically equivalent.
More formally:
For any two shapes in a given row, there's a self-homeomorphism on the plane that maps one shape to the other.
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1$\begingroup$ Accepting this answer since it's the first to mention topological equivalence. @ferret: Basically, on each row, you can deform each shape into any other (mirroring is allowed, but not collapsing a line into a point or stretching a point into a line). $\endgroup$ Commented Jun 5, 2020 at 20:49
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From what I can see...
the circle. The first row has 4 outer intersection points and 2 inner intersection points each. The second row has 4 outer intersection points and 4 inner intersection poitns each. the two ont eh third row both have 7 outer intersection points. The first has 4 inner intersection points and the second 5, but that discontinuity goes away if we consider the X in the lower-right-hand corner to be passing by each other and not actually crossing. The circle has 7 outer intersections and 4 inner intersections as well... and is the only one of the options that has an odd number of outer intersection points.
and after the recent edit...
each row is topographically equivalent.
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$\begingroup$ The answer is right, but rows satisfy a much stronger condition than just having the same number of intersections. Can you find it? $\endgroup$ Commented Jun 4, 2020 at 22:37
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$\begingroup$ The image has been fixed now so you'll get a lot more consistencies. $\endgroup$– hexominoCommented Jun 4, 2020 at 23:11
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the circle
because
count the number of neighbours each region has. This is the same by row. Equivalently, the number and degree of each vertex, along with the degree of each adjacent vertex, are the same. Corners on the outer border aren't vertices, unless they are connected to a third edge.
