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It could definitely not be anything.
It's 5.
If you can proove it's some other kind of patterns, do it.

It could be absolutely anything. There is no way to guess the pattern given a finite number of examples

What the fuck are your talking about. If you can't see the pattern, maybe it's because your a brainlet and a literal faggot

Confirmed. It's 5.

I already said that it's supposed to be 5. However there's no way to justify that answer. I got the answer just like you but you're the one who fails to understand what I'm saying about "guess the pattern" questions, brainlet.

5

they add like logic both in rows and columns.

winrar. similar lines cancel; different lines add
in both the columns and the rows. XOR

any given drawing in composed of the two other drawings in it's column or row. sticks that overlap between two drawings are removed, but if they don't overlap they are added.

>It could be absolutely anything. There is no way to guess the pattern given a finite number of examples
WHAT THE FUCK?

Have you never taken a math course? Have you never seen even the simplest series?
>1 + 2 + ... + n
Rings any bells?

You wouldn't expect n+1 to be 50 if n is 30.

what does this have to do with series?

Please rigorously prove to me that the answer is 5.

it's a series, you just follow the obvious pattern to find the next item

OPs problem is not a series and I'd like to know what you think an "obvious pattern" is, and how you rigorously mathematically prove that a sequence of objects obeys such a pattern given only a finite number of terms.

number the vertices starting from the top going clockwise V = {1, 2, 3, 4}
Graph = (Vertices, Edges) Edges are a set of ordered pairs
Let X be an operation on two graphs
G1 = (V, {(1,2)})
G2 = (V, {(1, 2)})
G1 X G2 = (V, NULL)

G3 = (V, {(1, 2), (1, 4)})
G4 = (V, {(2,3), (3, 4), (1, 4)})
G3 X G4 = (V, {(2,3), (3, 4), (1, 2)})

G5 = (V, {(1, 4)})
G6 = (V, {(1, 2), (2, 3), (3, 4), (1, 4)})
G5 X G6 = (V, E)

By intuition, it can be shown that X represents taking the complement of the intersection of both edge sets.
It does not show that it is the only answer, only that it is a valid one.

>It does not show that it is the only answer
So we agree. The answer could be anything.

Well, I haven't taken algebra yet, so I don't have the experience to prove set theory rigorously. But common sense tells me there must be a way to prove at least by contradiction that the other possible answers aren't valid.

Can someone please explain in lay terms how this works

what's the next number in my secret sequence?
1, 2, 3, 4, _

>X represents taking the complement of the intersection of both edge sets
That works going horizontally. Other people ITT have observed a different pattern which works both horizontally and vertically.

The pattern is that the third graph's edge set is the symmetric difference (xor) of the previous two graphs' edge sets.

Shit like this is why "guess the pattern" questions are flawed.

Yes, I agree that figuring out in which way the pattern goes is a problem with these tests. You could also argue that it moves diagonally or even in a spiral or whatever, and find a sequence that fits that direction.

My answer was the same as XOR anyway so it works vertically too. Basically taking the complement of the intersection, in other words all the edges that are in either one of the sets but not in both. But I agree with you.

pick a symbol, overlay it on a symbol to its right or a symbol below it. overlapping sticks cancel

The complement of the intersection is not the same as the symmetric difference.

if some element x is in neither A nor B, then x is in the complement of the intersection, but x is not in the symmetric difference

>if some element x is in neither A nor B, then x is in the complement of the intersection
Absolutely not if the Universe is the union of both A and B

Yes but in the vertical colums in OPs picture we see sets A and B whose union is not the whole "universe" (I'm assuming by "universe" you mean the set of 4 edges forming a square)

In the horizontal rows you always have sets whose union is the univers

That's wrong. I'm not going to explain what the universe is, it's high school set theory. But it works perfectly horizontally and vertically.

math.stackexchange.com/questions/84184/relation-between-xor-and-symmetric-difference

The "universe" is whatever we agree it is, my friend. Why your continued petulance and insistence that other people ought to figure out things that are arbitrary?
Going across any horizontal row, the union of the first two sets turns out to be the set of all 4 edges forming a square.

Going down some of the vertical columns (the left one and the right one), it's different.

BTW the link you posted explains why xor and symmetric difference are the same. They are NOT the same thing as the complement of the intersection though.

>X represents taking the complement of the intersection of both edge sets

Here is what the left vertical column would look like if the operation were taking the complement of the intersection.

XOR
O--O
ROX

God damn it do you need every single detail spelled out you autists?

(A union B) ∖ (A intersection B)

>"The set of elements in (A union B) but not in (A intersection B)"

I obviously meant the relative complement of only the present edges, not of the cyclic graph.

If that's what you meant, then perhaps you should have said so. Don't expect me to translate your incorrect ramblings into something intelligible and correct.

You said the complement of the intersection. That's wrong, my friend. You're welcome.

You can't expect these linear minded brainlets to understand. What you said is more towards logic/philosophy than actual math, these brainlets only know math.

Holy shit yo, I think the answer is 5. But my brain is doing so many complex maths right now that I don't even consciously comprehend. Feels god man.

5
The 1st and 2nd image in a row are overridden and if they have a line in the same position then it's deleted in the 3rd

In the first row, there is a line in the same position in both the first and second image. When merged, they result into empty dots

In the second row, the only common line between first and second image are the one that goes from top to left. The third image results in only that line missing

In the third row, the only common line would be again top to left

The pattern also works in columns

5

Its 5

Line + line = 0
Line + 0 = line
0 + line = 0+line

>>>print("hello world")
hello world

...

upvoted

You can come up with adequate numerobabble to explain all the options.

The "correct" one is the one the maker of this problem had in mind.

Five right? If you labelled columns a to c and rows 1 to 3 then a1+a3=a2. If the same repeats for the other columns then and since c1 is empty then c3 is the same as c2

The correct one is the simplest pattern. All other solutions are probably overfitted.

>simplest aka the solution I like :^)
fuck off

Come up with a valid rationalization for 3 that works horizontally and vertically.

No, the one with least parameters. Tests like this are designed such that there's only one simplest solution.

Here's an extremely simple explanation why 3 is right, that involves 0 parameters: the pattern is always that figure 3 goes in the bottom right corner regardless of the other figures shown.

Its 5