 Solving the Evil Sudokus
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Posted by (+1293) one year ago
The newspaper Sudoku puzzles range from easy, medium, hard, very hard to evil (these ones have to be copied from the Sudoku web site.) Here is a link to an evil Sudoku that I will show how to solve: https://www.websudoku.com...4431612935 Go ahead and download the puzzle or copy it.

There are 81 small squares which I will simply call each individually as a "square." One should put small numbers in each square (1 to 81) to help follow easily. The first horizontal row would be squares 1-9. There are bigger squares which contain 9 squares each and I will refer to these larger squares as "blocks". They will be referred to as blocks 1-9. The other terminology I will use is horizontal row and vertical rows 1-9. I am not going to explain how Sudoku puzzles are filled out, this article is for those already filling out these puzzles.

For the evil Sudokus, there are generally 3 or 4 numbers everyone should get. In the above puzzle, one should be able to figure out that a "1" goes in square 69. Square 77 is easy, it is solved with a "5". Square 13 is a "6". Square 41 (center of the puzzle) is solved with a "7".

To solve the rest of the puzzle, I fill out all possible numbers for the empty squares. First I seek the three blocks (vertical and horizontal) which have the most squares solved. Blocks 1,2,3 have 9 solved squares. Blocks 4,5,6 have 11 solved squares. Blocks 7,8,9 have 10 solved. Vertical blocks 1,4,7 have 8 solved. Vertical blocks 2,5,8 have 14 solved. Blocks 3,6,9 have 8 solved, so I start with vertical blocks 2,5,8 since they have the most squares already solved.

In filling out the first empty square I come to, and filling out the possible numbers for that square (square 31), three numbers are possible for that square (2,8,9). For square 49, 2,8,9 are the only possibilities. Square 58 (2,3,8,9), and square 67 (2,8). Square 58 has the only "3" in this vertical row so eliminate the other nrs (2,8,9). Also, since there is now a "3" in horizontal rows 7 and 8, and vertical row 1, square 74 is now a 3.

Filling out vert row 5, square 5 is (2,4,8,9), 14 is (3,8,9), 23 is (2,3,4,9), 59 is (2,8,9) and 68 is (2,8). You will note in block 8 that squares 67 and 68 are both 2 or 8. That means that those two squares can only have a 2 or 8. . So, square 59 cannot have a 2 or 8 and has to be a 9. Now, go up vertically in row 5, block 2 and eliminate the three 9's.

Now, the possibilities for vertical row 6. Square 15: 8,9, 24: 2,9 square 33 (2,5,8,9), and 51 (2,5,8,9).

In looking at block 2, vertical row 6 has a "9" in square 15, and a "9" in square 24. The 9 has to be in the right vertical row of block 2. Therefore, there cannot be a 9 in square 33 or 51 and these two 9's should be eliminated. That is all for now in these three vertical blocks 2,5,8.

Now check the horizontal and vertical blocks and see which three have the most nrs solved. Blocks 7,8,9 have 15 solved if you count the two squares of 2,8 which are 80% solved. So, next I work on blocks 7,8,9.

Horizontal row 7: square 55 (1,6,7,8), 56 (1,6,8), 61 (1,2,6,7), 62 (1,2,7,8), 63 (1,2,6,7,8). Horizontal row 8: square 64 (6,7), 66 (4,6,7), 70 (4,6,7). Horizontal row 9: square 73 (1,8), 79 (1,4), 80 (1,4,8,9), 81 (1,4,8,9). Block 7 has only one (4), in 66. Eliminate the 6,7 in that square, leaving only the 4. In square 70, erase the 4.

Blocks 4,5,6 have the next most solved squares (11), they are next. Horizontal row 4 square 28 (2,5,8,9), 30 (8,9), 34 (1,2,4,5), 35 (1,2,48). Horizontal row 5: 37 (2,5,6,8), 38 (5,6,8), 44 (2,8), 45 (2,8). Horizontal row 6: 47 (4,5,8), 48 (8,9), 52 (4,5,7), and 54 (4,7,8).

Note the (8,9), (8,9) in squares 30 and 48. Four or eight will end up in these two squares. So, eliminate the 8 and 9 in the other 4 squares in this block. Square 47 has the only 4 in the block, eliminate the 5 in that square.Now, eliminate the two 4's in squares 52 and 54.

Note that 34 and 35 contain the only (1,4) in the row. Eliminate the 2,5 and 2,8 in those two squares. Since 34 has lost its (5), check vertical row 7 and notice there is only now, one (5) in the row (square 52)/ So, eliminate the 7. Now, eliminate the 5 in square 51.

Now, notice the single 5 in square 33. Eliminate the 2,8. Now, eliminate the 5 in 28. Now this square has only a 2. Now, eliminate the 2 in 31 and 37. Squares 44 and 45 are going to be either 2 or 8. Therefore square 54 is 7. Now you can eliminate the 7 in 63.

Next I could go to either blocks 1,4,7 or 3,6,9. I will go to 1,4,7. Square 1 possible solutions are: (5,6,8,9), 19 (1,6,7,9), 2 (5,6,8), 20 (1,6). square 3 (3,6,7), 12 (3,7) 21 (3,6,7).

Squares 1 and 2 will have either a 5 or 8. Eliminate the 6,9 in square 1. Eliminate the 6 in square 2. Then eliminate the 8 in square 5. Square 19 is the only square in block 1 that has a 9. Eliminate the 1,6,7. Eliminate the 9 in square 24. Eliminate the 2 in 51. Eliminate the 8 in 31,48,49. Since 31 is 9, eliminate the 9 in 30.

Now, lets fill out the empty squares in Block 3. Square 16 is either 1 or 7. Square 8 is either 2,4,9. Square 26 is 1,4,7. Square 9 is (2,4,6,9), 18 (1,9) 27 (1,4,6).

Square 24 is now a 2. Eliminate the 2 in 5 and 23. Since 5 is now a 4, eliminate the 4 in 23. Eliminate the 3 in 14, eliminate the 8 in 15, eliminate the 9 in 18, eliminate the 1in 16,26,27,63, and 81. Squares 9 and 27 are the only 6's in block 3 so eliminate the 6 in 63.

Row 3 has a 3 in 23, so eliminate the 3 in 21. Block 1 has only the 1 in 20 so eliminate the 6. Now eliminate the 1 in 56. Horizontal row 2 has a 7 in 16, eliminate the 7 in 11. Eliminate the 3 in square 3. (Shouldn't have been in there anyhow. Horizontal row 1 has a 4, eliminate the 4 in 8 and 9.

Squares 8,9 can only have a 2 or 9. Eliminate the 6 in 9. Block 3 only has a 6in 27, eliminate the 4. Now there is only a 4 in 26 of block 3. Eliminate the 7. Eliminate the 6 in 21 and the 7 in square 3. Vertical row 5 has its 8, eliminate the 8 in 68. Eliminate the 2 in 67, the 9 in 49, the 7 in 61 and 70, and eliminate the 6 in 61. Block 9, square 62 has the only 7. Eliminate the 1,2, and 8. Eliminate the 7 in 55. That means 64 is a 7. Squares 34 and 79 in vertical row 7 are either 1 or 4. That means 61 cannot be a 1. Eliminate the 1.

That means 63 is an 8. Eliminate the 8 in 80,81, the 8 in 45, the 2 in 47, the 2 in 9, and eliminate the 9 in 8. Eliminate the 8 in 55, 56, the 6 in 55, the 1 in 73, the 6 in 38, the 5 in 37 and square 2, eliminate the 8 in 1, the 4 in 35, the 1 in 34, the 4 in 79, the 1 and 4 in 80,, and the 9 in 81. That should solve the puzzle.
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Posted by (+17053) one year ago
This guy was called a "killjoy" for publishing this.

A Pencil-and-Paper Algorithm for Solving Sudoku Puzzles by J.F. Crook