I found this test is the best one to spot the brainlets. They just can't solve it

I found this test is the best one to spot the brainlets. They just can't solve it.

The first one?

Here's a brainlet shooting blindly. Go on, explain your reasoning.

The 3rd from the right if i had to give the answer, but i think none of the alternatives seems correct.

The second from the left

>none of the alternatives seems correct.
Yeah, I'm starting to think the creator made a mistake.

Why?

Why?

I think its the first one, each row from top to bottom has in each set 3 arrows, looking just at the first one draw lines top to bottom and its always a different direction. down, downR left and down left right and down right down right

cuz it be the only one lookin shady muh nibba

Your "explanation" was less comprehensible than the matrix.

I think there's a mistake in that matrix, but if there isn't, the answer is (probably) the first from the left:

>going right to left, we see in the upper rows that left becomes up
>we also see that up becomes down
>so in the missing set, the middle arrow should be down, and both the lateral ones should be upward
>but we can see that the first column is asymmetrical, not like the other 2, that are fairly regular (two identical arrows on the sides and a different one in the center)
>since the first column is asymmetrical, the set should be: down arrow in the middle, one upward arrow on one side, and a different arrow on the other side.

The first from the left is the only one that KINDA makes sense.

This
Because the middle arrows has to be down. Why? The pattern for the middle arrows is down/up/left for the above 2 sequences.

Additionally, the first and third arrow for the 3rd sequence are identical in the two diagrams they’ve showed us, indicating that the correct answer must be up/down/up

>we also see that up becomes down
I should clarify: up becomes down only when it's in the middle.

Seems like the arrows in the middle and the ones on the sides follow different rules.

You're not considering that the first column is different than the other two: it's asymmetrical. The other two have the same arrow on the sides of each set, and a different one in the middle. The first column instead has:

>DOWN DOWN DOWNRIGHT
>DOWNRIGHT LEFT DOWN

It's only natural to assume that the last set of that column will be asymmetrical as well. Since it's clear that the middle arrow must be down, this leaves the first answer as the winner.

...

>while the brainlets try to choose between the offered alternatives, the superior mind makes up its own

...

It's the second from the left in accordance with sequential patterns. It was pretty easy.

t. brainlet

The first column is asymmetrical. The other two are symmetrical. The answer is the first from the left.

All instances of the left arrow follow an appearance of the up arrow.

No wait, you're probably right. I'm a moron.

Yes, that's obvious. But the second and third column are symmetrical (they follow an ABA pattern) while the first column is irregular (AAB and then ABC). It makes no sense to assume that the 3rd frame of the first column will suddenly be symmetrical.

Don't feel bad, this matrix is crazy.

middle object oscillate in a column the two on the side don't repeat in a column so first one fits

>column
mfw brainlet wiggers can't even speak their own language correctly

Second from the left.
The arrow in the middle should be down, as it should be the same as the top one.
Apart from the one in the middle of the middle row, all of the other arrows should appear twice, and that only happens in the second one.

OP you'd better come back and give the fucking answer. Its the first panel from the left

Nope. See:

Not gonna do your interview prep for you.

I don't know the answer. I wanted someone to point out the logic.

I think it's the first from the left, but the logic doesn't completely satisfy me.

Just admit you're a brainlet. Nobody will mock you.

Last row doesn't have to be assymetrical. The assymetry "rule" is exclusive to the first column and we don't know how it exactly works. I'd rather go for the 2 times rule, as it is broader, being applyable for all three columns.

There is no pattern to be seen, you cannot solve it. Even if the answer is listed, there isn't even data to solve for a pattern.

That'd be my second guess aside from this

The solution is obvious. I really want /pol/ gone from this board jesus fuck.
Sum the vectors in each block.
You come up with
(1,-3) (-2,-1) (1,0)
(0,-2) (2,-1) (0,3)
(?, ?) (-2, 1) (1,2)
Should be obvious from here.

>Should be obvious
no it isn't, and you don't know either

Middle arrows in respective top row and bottom rows are identical so we're left with the first or second option from the left.
The amount of identical vs different left and right arrows is constant in every column so the only solution which matches is the first one

sorry you're too retarded to understand simple shorthand, bet you couldn't even figure out the question lel

It's the third option, check the first sequence, you have 2 of the same arrows and one different for all the squares, in the second row the square #1 is different, since it has 3 different types, for which the third block in it needs to be the all the same, in the third row we see two blocks with 2 types, which means the first block needs to have 2 types too, but why is the second row different?
simple, the number of "types" of arrows or directions is not even, it follows a pattern, we start with 6 different types, since it's even it can distributed equally between the 3 squares, but the second row has only 5 different types of arrow, for which the 3 equals are needed, now the pattern is clear 6-5-4 for the directions of the arrows, we see that the row number 3 is already filled with 4 different types, so the answer needs to be a repetition of arrows we already have, the solution is number the option number 3.

There actually is a pattern surprisingly. The first number of the first vector in a row is always equal to the first number of the third vector in the same row. The second number of the first vector is equal to the sum of the components of the second vector, minus the second number of the third vector in the same row.

Following this logic, we should get (1,-3) as the vector. Unfortunately for vector poster up there none of the options fit that vector, so he's sadly just another brainlet.

I agree with this user

>He doesn't see the pattern
You need to go back

Not even sure if I should spoon-feed you niggers

>he's impotent
kek

all of you people are insufferable

your posturing is insufferable

Have you edited the pic to make it impossible?

The 45 degree vectors clearly have a length of 1, and not sqrt(2)

>le /pol/ boogeyman for no apparent reason

The more solid proof that the image is edited is that alternatives 4 and 5 are the same

>no apparent reason
You brainlets fucking up the board are a constant, everpresent reason. Sorry it's not PC to say it. Don't get mad just because I'm willing to speak the Great Truth.