# How do people crack SUDOKU

## 9 professional tips to solve difficult Sudokus faster

Difficult Sudokus can be solved with pure logic, but that often takes far too long. Methods like Swordfish, 3D Medusa, Sue de Coq or Death Blossom are far too complicated and time-consuming.

If you want to be fast, you have to go the short ways. Correct digits have to be found, and as quickly as possible, because every new number entered makes the puzzle easier. Sudoku master Stefan Heine reveals his tricks with which even the most complicated Sudokus can be cracked.

Sudoku specialist Stefan Heine regularly supplies the German Sudoku Championship and - as the only German - even Sudoku World Championships. Stefan Heine developed the following tips together with Sebastian Matschke, one of the best German puzzle solvers. Year after year he qualifies as one of the fastest in Germany for the German team at the World Cup.

### 9 professional tips from Stefan Heine for extra difficult Sudokus

Using the following nine situations, you will learn how to solve even the most complicated Sudokus quickly. The rest is practice, practice, practice. So you too can become a Sudoku master!

1 We look for the number in the box with the star. In the same horizontal line there are already 1, 5 and 2, in the same vertical column 3, 7 and 9, and in the same 3x3 field 6 and the 8. Neither 1, 2, 3, 5, 6, 7 are allowed in this box , 8 or 9 can be entered. This leaves only a 4 for the star box.

2 The three boxes marked with the small numbers 1 and 2 can only be 1 or 2. However, if there was a 1 in the box with the asterisk, the other two boxes should each contain a 2. As a result, however, there would be no more space for a 2nd in the right middle 3x3 field. This is why the box with the star cannot contain the 1, but only the 2.

3 In both lines 5 and 7, the number 4 can only be placed in boxes A and B, because in all other places there is already a 4 in the same 3x3 field or in the same vertical column. So the 4 goes either in the two A-boxes or in the two B-boxes.
The 4 must not be in any other box in column 7, not even in the box marked with an asterisk. However, 1, 2, 3, 5, 7, 8 and 9 are not allowed here, as these digits already appear in the same line, in the same column or in the same 3x3 field. Therefore a 6 must be entered in the asterisk box.

4 In the middle 3x3 field there are two possible places for the 4, A and B. If you write the 4 in the A box, it must also be in the A box in the middle, lower 3x3 field. If you put the 4 in one of the B-boxes, it must also be written in the other B-box. In both cases, stay in the lower right 3x3 field only one possible box left for the 4 - the one with the star.

5 In the middle 3x3 field, the 8 can only be in the two A boxes, which means that the 8 is in column 6 and can no longer be in the two boxes below the 4. Now let's look at the three boxes in the middle, lower 3x3 field, which are marked with the numbers 357, 35 and 57: They may only contain the 3, 5 and 7. This means that the 3, 5 and 7 must not be placed anywhere else in the middle, lower 3x3 field - three boxes, three digits.

It follows that the box with the asterisk must not contain the 3, 5 and 7 in addition to the digits already in the same row and column (1, 4, 6, 8 and 9). This leaves only a 2 for the box with the star.

6 In line 1, the 2 can only be placed in the box with A or the one with an asterisk. If it were in A, there would also have to be a 2 in one of the two
Box with B. In addition, the 2 in line 7 would only have room in the C box.

Now try to find a place for the 2 in column 1. The 2 in B blocks the top 3x3 field and the 2 in C blocks the bottom 3x3 field. The middle 3x3 field cannot have a 2 in column 1 from the start (horizontal arrows). From this it follows that there cannot be a 2 in A and so we found a 2 in the box with a star!

7 In the upper left 3x3 field there are two possible positions for the 4, the box marked with A and the box marked with an asterisk. If the 4 were in A, there would only be space for the 4 in the upper right 3x3 field in the box also marked with an A. Also in the lower left 3x3 field only the A box would remain. If we now look at the lower right 3x3 field, we would no longer have any space there for the 4. It follows: At the top left, the 4 cannot be in position A - but only in the asterisk field!

8 In the lower left 3x3 field, the 4 can only become an A or a star. If we try position A, the other A's definitely also contain 4s. Can you follow? In the upper middle 3x3 field, there would only be box B for a 1. The second B would then also be a 1. Now look down to the left again. With the 4 at A, you could no longer enter a 1 there. As a result, our assumption is wrong from the start and the 4 must go to the star and not to position A.

9 In column 1, the 4 can only go to the two positions B. In the 3x3 field to the right, we know where 4 and 9 are going. We don't yet know which one is above or below.

If we now assume that there is also a 9 at position B, this puzzle would always have at least two solutions. Because even if everything else were filled in, the 4/9 pairs would remain in columns 1 and 4. That must not be. So you can conclude that the 9 must not be at B, but must be higher up in one of the two C boxes.

Against this background, if we look at the box with an asterisk and count through (1, 2, 3, 4, 5 in the 3x3 field, the 9 at C and 6 and 7 in column 3), we are happy that an 8 is found on the Position with the star.

Have fun trying out the tips!