Shorthand method of Huarong Road formula
In the first line, move 1, 2, 3 to the target position one by one, regardless of other numbers. When moving the number 4, first move it below the target bit. At this time, it is found that 4 cannot be moved down to the target bit. You can move the two digits on the left of the same row of 4 to the right by one space, and then move the three digits of 1, 2 and 3 in the first row counterclockwise to the left by one space.
At this point, you can move 4 to the target position in the upper right corner. After moving, remove the irrelevant digits to the left of 4, and reverse the whole 1, 2, 3 clockwise to the right. The first line is right. The second solution is the same as the first line.
In the third line, if 10 is below 9, you need to hide 10 in the right square, and then let 9 move freely in the left square. At this point, you can move 10 to the left by 9. This step only looks at 9 and 10, without considering other numbers.
For 1 1, 12, if 1 1 is in the same square, 1 1 is below 10. So the third row is right.
In the last line, all aligned 9, 10,1,12 are moved counterclockwise to the left by two spaces, and 13, 14, 15 are moved counterclockwise to the right by two spaces, which is/. Then move 9 clockwise by two squares to 15, and it's all right.