Ghost Numbers

I introduced ghost numbers in triangulation and elimination so look at those before reading this discussion.

Ghost numbers are just possible numbers in a space. Some suggestions that help make ghost number work:

Ghost numbers are most useful, even essential, for difficult puzzles. The more difficult the puzzle, the more likely that you'll need to use ghost numbers. It may require a lot of patience and fail but it's the best I've come up with.

So let's look at an example suduko puzzle and solve it using triangulation, elimination, and ghost numbers. Every time we can pencil in ghost numbers based on the ghost rules, we will. Here is the puzzle:

If we triangulate 1, we can place four ghosts:

After one round of triangulation (numbers 1 through 9), we have the following information:

We know that the last three spaces in the first row must contain 5, 6, and 7. With 5 and 6 in the seventh column, we can place 7 and the last two spaces contain ghost 5s and 6s. We can remove the other ghost 7 in the third block. 2 and 9 must be in spaces six and seven in the block so 3 is in the fifth space (elimination) and 8 in the eighth space (triangulation or elimination, take your pick).

In the first block, 2 and 7 must be be spaces six and eight. Therefore, 1 must be in seven and 5 in four (elimination). With a 5 in the first block, we can now put a 5 in the second block (triangulation). Here's where we're at:

7, 8, and 9 are left in the second block and we can triangulate them in:

And by elimination, put 2, 7, and 9 in the first row of blocks:

A triangulation check puts 8s in the sixth block and 7s in the fouth and fifth blocks:

Now in the last column of blocks, we can triangulate a 9 in the ninth block and by elimination put in a 4 leaving 1|6 pairs in the ninth block and 4|8 pairs in the seventh column:

Knowing there must a 6 in the last column of the ninth block eliminates the 6 in the last space of the fist row, leaving a 5 with a 6 in the eighth space. In addition, we can now place 1|5 and 2|3 pairs in the sixth block:

The 1 and 2 in the fourth block eliminate the same numbers in the sixth block, leaving 3 and 5. By the same logic, we can place 2 and 6 in the fifth block:

We can place the last 2 in the seventh block by triangulation. 1, 3, and 4 are the remaining numbers in the fourth column. We can place them by elimination as well as the remaining 4 in the first row:

With the exception of the two blocks in which we already know number pairs but cannot eliminate anything, all the remaining rows, columns, and blocks have at least four empty spaces. Time to try a little triangulation. Going through the numbers, we get these results:

The 3 in the second space of the first row eliminates the 3 below it in the fourth block. By elimination we can place the remaining 3s in the puzzle. The 5 in the first row eliminates the 5 in the fourth block leaving a 5. And in the fifth block, 7 and 5 eliminate a 1 leaving 1 in the middle space:

The 5 in the sixth row eliminates the second 5 leaving 7 and in turn leaving 5 in the fifth block and 7 in the fourth block:

In the fourth row, the numbers 4, 8, and 9 are missing. 4 may not be in the first two empty spaces so it must be in the third and 8 goes in the sixth block as a result:

We can triangulate a 4 in block four leaving 8|9 pairs in the empty spaces in the middle row of blocks. The 8 in the first block allows us to place 8s and 9s in the middle blocks by elimination:

The rest of the puzzle can be completed using triangulation and elimination: