XOR. Consider each box as a 4-bit bitstring specifying which corners are "on" and "off". Then the bottom cell in each column is the XOR of the two cells above it, and the right cell in each row is the XOR of the two cells to the left of it.

If you were to lay the top row on the 2nd row, any line that is drawn twice disappears for the 3rd row, all lines drawn once are present. So top left plus middle left = bottom left

On each row, when the first square is placed on top of the second square, overlapping curves/quarter circles disappear. On the third row when the first square is placed on top of the second square, there'll be no overlapping curves, which means all curves in the 1st and 2nd square will appear on the third square! :)

in the pattern option B ,C ,D have already been used in the pattern. so the only one left is option A. thus the answer is A

• Superimpose the first two column.
• Delete what's common.

I thought it was one of those moveable puzzles and all that was missing was the middle piece.. one containing all 4 quarters of a circle...

In first row the first box(i.e fiorst corner) there are three quarter circles, in 3rd box(i.e at the second corner) is one quarter circle), in last row the first box contains two quarter circles so eventually there must be four quartter circles i.e A

If you go vertically (top to bottom), and trace Pattern in Square 1 over Pattern in Square 2, the parts which overlap is removed to obtain Pattern in Square 3.

Look at one row. Look at each corner Of the squares. They have a sequence of on, on, off.

Solution using Set Intersection, Set Difference, and Set Union.

Label each corner 1 through 4 starting from the top left corner as 1 labeling clockwise with the bottom left corner as 4. If we make each cell a set that contains a labeled corner only if it has a semi-circle then the labelling would look like so:

Notice that for rows 1 and 2 that $C = A \cup B - (A \cap B)$ where $A,B,C$ are the columns. Thus we should assume the pattern follows for row 3 and gives the answer A. or $\{1,2,3,4\}$ .

Col 1 NAND col 2 = col 3

You can see from row 1 upper right that it's not XOR.

I looked at it in a row-wise mannner. U add up all the blocks in the first 2 to get the third one.. And if any bpock is overlapping it gets cancelled out.

In these type of sums the thing is that there can be a variety of solutions and each of them have to be considered.

First let's divide each square in four, keeping this in mind we can see that each row follows the same rule, if the 2 first little squares are differents, then the last little square should have a semi circle, otherwise the last little square should be empty, follow this rule in the last row we can see that every 2 first little squares are differents, so every last little squares should have a semicircle. That is the A answer.

Divide each row into 2 equal rows. For each quarter circle facing left or right, there needs to be a corresponding quarter circle facing left or right respectively. This applies for columns too, where "left" and "right" is to be replaced with Up and down.

I just know that A is right because I see patterns

Simple: A is the only block that is unique.

Nice mind work...

Patterns B, C and D are already used. So I guessed it's A.