?【Tips】?

Use the × mark frequently! Use the × mark frequently! Use the × mark frequently!

?Simple mode 10*10? [Must click grid]

6: The middle two squares must be clicked □□□□■ ■□□□□

7: The four squares in the middle must be clicked □□□■■ ■■□□□

8: Two spaces before and after □□■■■ ■■■□□

9: Leave one space empty before and after □■■■■ ■■■■□

10: All points ■■■■■ ■■■■■

There is only one solution: the sum of black squares and spaces equals 10:

Such as 1 8: ■□■■■ ■■■■■(1 1 8=10)

2 7：■■□■■ ■■■■■(2 1 7=10)

And so on 3 6, 4 5, 5 4, 6 3, 7 2, 8 1 Same as above

?Difficulty mode 15*15? [Must point grid]

8: The middle square must be clicked

□□□□□ □□■□□ □□□□□

9: The three squares in the middle must be clicked

□□□□□ □■■■□ □□□□□

10: The five squares in the middle must be clicked

□□□□□ ■■■■■ □□□□□

11: Leave four spaces on the left and right

□□□□■ ■■■■■ ■□□□□

12: Three empty spaces on the left and right

□□□■■ ■■■■■ ■■□□□

13: Leave two spaces on the left and right

□□■■■ ■■■■■ ■■■□□

14: Leave one space on the left and right

□■■■■ ■■■■■ ■■■■□

15: All points

The sum of black squares and spaces is exactly equal to 15. There is only one solution:

Such as 2 5 6 (2 1 5 1 6=15):

■■□■■ ■■■□■ ■■■■■

By analogy, there is no need to go into details about numbers and spaces.

[Advanced version must click the grid]

When the sum of the black and white squares does not equal 10 but the number is very large (1 1 7 = 9) like 1 7, you can also click on the necessary squares.

First assume that 1 7 is from left to right or top to bottom (just determine the direction).

We consider the leftmost situation first, that is, assuming that this 1 is the first one on the left, then the second one must be a space, and the situation becomes ■□□□□ □□□□□. [Actually, don’t mark this 1 in black! All you need to do is remember where the dark spots are! 】

So now that we have determined that only the two on the left are black and white, what we have to do is to select seven of the eight spaces on the right and mark them black. At this time, the intersection principle is used, and the intersection is an inevitable part. Starting from the third grid, counting seven to the right, and marking the seventh grid in black, the situation becomes ■□□□□ □□□■□.

Then count from the first square on the right to the seventh on the left and mark it black, and the situation becomes ■□□■□ □□□■□.

Then mark the middle of the two black grids just black, and the situation becomes ■□□■■ ■■■■□.

It can be seen that the current 1 7 already has 1 6, but our 1 is assumed mentally and is not actually marked black, so the actual final situation should be □□□■■ ■■■■□.

1 cannot be determined, just leave it empty. The above six black squares are the must-dot squares for 1 and 7.

If it is 2, mentally assume that the two outermost squares are black, and the other numbers are the same as above.

And so on:

1 6：□□□□■ ■■■□□

2 4：□□□□□ □■□□□

5 1：□□□■■ □□□□□

1 1 5：□□□□□ ■■■■□

1 4 1：□□□□■ ■□□□□