OK, which of these figures belong in the same group? Please Register or Log in to view the hidden image!
Oh, don't get me started on Mensa tests. Figures so poorly drawn and small you couldn't make out what they were even if the mensan who drew them had an aptitude for such things. Sequences of numbers to complete that could be done 100 different EQUALLY CORRECT ways. For a background on how bad IQ tests are and where they came from, read Stephen Jay Gould's 'Mismeasure of Man'. If you are looking for one with an obtuse angle, the chevron shape is the odd one out. If you are looking for one in which lines cross, the two triangles would be the pick. Attributing high cognitive function ratings to those with better reasoning using only symbols like these is a huge mistake.
My answer to the last question is: the rectangle and rhombus have the same symmetries: two reflections and one rotation. The parallelogram has one rotation of 180°, the kite has one vertical reflection, so both isometries put them in the same order 2 group (I would say this is \( D_1 \), or \( \mathbb Z_2 \)). The rectangle and rhombus have order 4 groups, or Klein 4-groups (something I already knew), as \( D_2 \). The square is the only figure with \( D_4 \) symmetry. The trapezoid is also by itself. The question, though, is ambiguous: "belong in the same group" could mean "belong in a subgroup of". In that case the answer is they're all in the same group as the square.
Babylonian / Arabic numerals 0-6 are ordered according to the number of vertices when drawn using only straight line segments. That would have made a great geometrical test question, if you wished to do one with symbols. It also means something straightforward without resorting to obscure topological ideas no one is likely ever to noodle out on a test for the first time. Arabic numerals 1 and 2 are represented that way to represent male, female aspects respectively (sexy, eh?). Too bad, no transgender or gay numeric representations as far as I know. You can tell a lot about a culture simply by the way they count, evidently.
Well, let's look at what the Romans used. It had "composition rules" that were ambiguous when numbers got larger than thousands. Can you put a Mensa-style question together using Roman numerals for me?
Well, for starters, they had no symbol for zero, which means, unlike the Arabs, that they evidently believed that the chicken preceded the egg. It is one way to account for a numbering system that resembles hen scratches. Their numerals in some respects resemble their armies. It may have represented a cooperative effort to quickly count numbers of surviving troops in groups, simply by how they held their standards or pikes. They slew Archimedes over an argument about his math. Go figure. But they suffered a large number of losses from his war machines before they captured him. I have Mensa friends who will love to play around with this idea. I'll see if there are any other ideas.
Here's something I've worked out: The parallelogram has rotation symmetry but no reflection symmetry. With a rotation, you can label the vertices or the edges, a rotation means no element (edge or vertex) is fixed. The kite has reflection symmetry but no rotation symmetry and it does make a difference if you label the vertices (because two are fixed) or the edges, the edges under reflection is a double transposition, this baby is like a pair of twisted but uncrossed bands in my abstraction (of Dih(4)), the parallelogram's symmetry is like a pair of crossed but untwisted bands. So, a composition of these symmetries generates: a pair of crossed, twisted bands (and the identity). We have \( D_2 \).
Here's a bit more: if you've done a bit of graph theory, you know the figures are all the same cyclic (incomplete) graph, and it's a planar graph so embeds in any Euclidean domain. The identity transformation of a cyclic graph is equivalent to changing all the edges into vertices and vertices into edges. So some of the graphs are symmetric, but that's an additional structure (lengths of edges and angles). But, a bit of group theory tells you there must be some dual space thing here. The rectangle and rhombus are topological duals--one is the other with its vertices transformed into edges, and edges into vertices. The rectangle fixes a pair of edges under reflection, the rhombus fixes a pair of vertices, so the choice of labeling is a choice of a subset of permutations. Is the parallelogram dual to the kite? The square is its own dual graph.
So you see arfa context is the underlying principle of the comprehension debacle, what I have learned about test taking is you always get the mark if you are 100% confident you have chosen the right answer. This method helped me a lot. As for the question I will use the context "geometric shapes", and just say all the figures belong in the same group because they are all geometric shapes. In this question how I saw the context "group" and not "groups" if all the figures was in the same group, what singular group will that be? It would be the unifying group of geometric shapes.
And yet again, the top left figure is the only one with an interior angle of greater than 180 degrees.
I suppose the question as posed: "which figures belong in the same group", would probably be interpreted differently by people with different levels of math education, say. Group usually refers to a symmetry group; the average joe might think the question is about sets. And so, the square has \( D_4 \) symmetry, all the others have symmetries which are subgroups of \( D_4 \). That might be one answer. The thing is these questions are usually about testing the way you think things through. It isn't multiple choice, you're supposed to give an answer and a reason you think it's an answer. If you think there's more than one answer, I didn't see anything about not being allowed multiple answers. Maybe that's what the testers are looking for: people who can see the questions are ambiguous.
Roman numerals. I remember writing the composition rules down after I did a formal languages course; that is, the Roman number system is a language with grammar and syntax, composition of strings and so on, and you can construct a FSA that accepts the strings if they're valid. But there is that ambiguity problem, what happens with that is you can't write a valid string or construct an FSA, the rules break down. The Romans weren't dummies, but they used a number system that they knew was flawed, and not just for a while, for like 2000 years. We can look back and laugh, I suppose, but how do you explain its success? Perhaps because it does have a regular, entirely predictable behaviour up until numbers reach a certain size, a size that must not have been a great concern to the Romans.
I have an excellent proposition for you arfa, when you get the results you should post them in as much detail you will have available. Then hopefully from that point all inferences to the op and other questions can be dissected hench forth. If you have the results already you should post them.
Sorry, no results yet. No results are expected, either. I'll leave you to interpret that however you will.
One further observation someone with good visualisation might see in the piccy: Please Register or Log in to view the hidden image! is that a parallelogram can fit inside a rhombus such that each vertex of the first figure is on each edge of the latter. The rhombus embeds the same way--vertex to edge--in the rectangle, and the rectangle inside the square. The trapezoid seems to be excluded, but the kite should fit inside a square, right? On second thought, the kite should fit in a rectangle and square if you can change its internal angles, but not a parallelogram. Oops, I guess it does fit in the parallelogram too. I'm pretty sure you can't draw a square around a parallelogram or rhombus. Aha, if you're allowed to change angles, the kite embeds in all the other figures. If you're allowed to change the edges as well then a kite can become any of the other figures. A rhombus is a square with a change of angles, a rectangle and parallelogram have this relation too. Angles and lengths of sides is where be symmetries (arrr).
It's ok thanks I just like to get as much certainty as possible in this universe, but the reason I made my previous statement was, once I arrived at the correct answer and recieved the mark I knew for sure that my answer was correct, when compared to other equally valid options. But I do agree with what you said about the question bieng ambiguous the term group is not specified so it would always remain just an inference as to what context the author was implying.
I'm a little late to this party and haven't read the other responses so please forgive if this has been mentioned. The first three are simple closed curves and the fourth one's not. Note that the complement of the first three consists of two connected components; and the complement of the fourth figure consists of three connected components. https://en.wikipedia.org/wiki/Jordan_curve_theorem
Further to note arfa, the implication of context in the question used the word "figures", so that's as specific as the question indicates what context of the word "group" in the singular, the author implied. So the English definition of the word "figure" is basically shape or form in this context. So the question can be rephrased in context by saying " which of these shapes or forms are in the same group" . The only singular common group that's these shapes or forms can belong in is the group of geometric shapes. If you have many different groups to allocate these shapes then you will not be able to single out any group as more special compared to the other groups. This is ambiguous, this question seems to be more about the comprehesion of English rather than pure mathematical theory. If all the other shapes was allocated in distinct seperate groups but only 2 shapes shared a common group then the method of multpile groups would make more sense. Since you could single out one of the groups as bieng more special than the rest, it would be the only group that would contain "figures" plural.
I think you are correct you also picked the same shape as arfa brane, so I think the both of you are correct.
