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I see the black dot move 1 to the left for each diagonal. Saw two videos about this online and can't see why all the shapes shift 1 spacs to the left. How does one ascertain this pattern?

enter image description here

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    $\begingroup$ It makes sense if you look at it starting from top row, going diagonally down (right-left diagonal (/)). All shapes identical and they all shift one step to the left. $\endgroup$ Commented Jan 16, 2023 at 13:57
  • $\begingroup$ I think the answer is based on the pattern applied to the block in column 1 row 2. But i dont know why its that block. It seems intuitive but I dont understand exactly why. $\endgroup$
    – Ra12
    Commented Jan 16, 2023 at 14:16
  • $\begingroup$ Which test is this from? $\endgroup$
    – bobble
    Commented Jan 16, 2023 at 15:13

4 Answers 4

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F

Because

There is a diagonal (SouthWest-NorthEast) pattern. All the squares with a same (i+j)mod 3 value (where i is the row number and j the column number) contain the same three items, once in each position.

In the empty (3,3) square, we want a white square, a black triangle and a black circle, just as in the (1,2) and (2,1) squares. The triangle will come first because it doesn't in either the (1,2) or the (2,1) squares, then come the circle and the square.

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  • $\begingroup$ imgur.com/a/zVmmBsT Please see the image I made (based on what Prim3numbah posted) . What I don't get is, for diagonal 1 - i see how it "jumps through to the other side". But how would I know diagonal 2 would do the same. Diagonal 1 was my only example of this "jump to the other side". I thought I'd need at least one other instance of this "jump" to confirm it. I looked up latin squares to try understand what you said about mod3 but struggling to understand. $\endgroup$
    – Ra12
    Commented Jan 17, 2023 at 3:43
  • $\begingroup$ @Ra12 (i+j)mod3 is just a way of identifying each "diagonal". Not sure what you mean about the "jump", but one way to describe what happens is: whent we go from one row to the next one within a given diagonal, the leftmost item "jumps" to the right. e.g., from (1,1) to (2,3) to (3,2), square-circle-triangle becomes circle-triangle-square and then triangle-square-circle... $\endgroup$
    – Evargalo
    Commented Jan 17, 2023 at 8:00
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The answer is

enter image description here

Because if you look at the

right-left diagonal (/), each shape moves one step to the left. When a shape is at the leftmost position it shifts to the rightmost after taking one step to the left. For example the string ABC, becomes BCA.

I've colored the shapes of diagonal 2 and named the three diagonals diagonal 1, diagonal 2 and diagonal 3.

enter image description here

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    $\begingroup$ I think I will delete mine as this is probably more along the lines of what OP is looking for. $\endgroup$
    – hexomino
    Commented Jan 16, 2023 at 15:07
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I came to the same conclusion as others but did it via process of elimination.

I see 3 different patterns in the 9 square grid.

When they are visually re-arranged, it becomes easy to identify which one is missing. Each shape occupies each location once (this is why it's easy to use the circle as the point of reference moving left to right). The option missing has a black circle and triangle, and a white square. By process of elimination, the black circle must be in the middle meaning it's option F. The next spoiler shows each of the patterns re-arranged visually to highlight the missing puzzle piece.

enter image description here

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Despite the approaches with the diagonal pattern you can also easily do this by combining column- and row-based patterns.

  1. Within each column from top to bottom the circle moves one space to the right.

  2. Within each row you have three combinations of black filled figures. Either only the circle, circle + triangle or circle + square.

Combining the two you know the circle has to be in the middle with triangle left and square right to it and circle and triangle have to be black. Correct answer therefore is F.

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