Solving Sudoku Part 3

We left off last time finishing all the horizontal and vertical gimmies in the puzzle. These are the numbers that it takes little effort to find. Remember the technique (check rows and columns of containers for duplicates and then see if the remaining container has a unique position for the number in question) as you will use it throughout the puzzle solution. Every number you put into the puzzle should trigger another gimmie search  in both the horizontal and vertical directions for the container in question.

OK, now where to we go?

The easiest way I’ve found to proceed is to look at the rows, columns, and containers and select the one with the fewest empty spaces. It doesn’t always work out that this row, column, or container will be the best choice but it’s usually the best way to find another number and is easy to remember. Let’s see how it works for the puzzle we’re solving.

How to use row logic.

The row that’s highlighted as four empty spaces. The row above also has four spaces but yields no new numbers so we’ll stick with the one highlighted. The first step is to figure out what four numbers are missing. Those would be: 2, 4, 6, and 7. How do we know that? Simple. Just count across the row and make a note of the numbers between one and nine that are not there.Tongue Out

Now you have to use your memory a bit. Mentally put those four numbers into each of the empty squares and look at the combination of columns and containers for each square and see if existing numbers define a unique number for that square. For example, with a square being able to contain 2, 4, 6, & 7 sitting in a container having 2 & 4 and an intersecting column with 7 in them then you would know that the square must contain a 6 because none of the other numbers would be allowed in the square due to the no duplicates rule.

There are no unique numbers in any of the four squares above however look a little closer at the two red squares. Neither of those two squares can contain the 6 or 7. In the left red square the 6 is in the intersecting column and the 7 is in the container. In the right red square both the 6 and 7 are in the intersecting column. That means that the two red squares MUST contain either the 2 or the 4. That’s an important concept to grasp. Any time you have two squares that can only contain the same two numbers in any given row, column, or container then those two squares “capture” those numbers. You may not know in what order to use the two captured numbers but you do know that they can appear nowhere else in the row, column, or container in question. You might want to re-read this paragraph to make certain you understand it as this concept of capturing numbers in given squares is important to solving Sudoku puzzles.

In the case above you don’t know which of the two red squares has the 2 or the 4 but look what else that tells you. There are four numbers you have to place and two of them have now been captured. That means that the two green squares MUST contain either the 6 or the 7. Now, look at the columns for those two squares. One contains a 6 so you know the 7 has to go into the green square below it and the other contains a 7 so the 6 must go into the column under that one. Again, the no duplicate rules make this definition mandatory.

So, now the puzzle looks like this…

Two numbers from row logic.

Now let’s take a look at that lower left corner and see what we can do with that…

Logic used in a container.

This time the empty squares need the numbers 2, 4, 6, and 8. Once more there is no unique solution for one of the squares but, like above, we encounter the same situation. The red squares may not contain either the 6 or the 8 and therefore must contain either the 2 or 4. That means those two squares capture those numbers like they did above. (We knew from the row example above that the upper left square had already captured the 2 and 4.) Looking at the intersecting columns there is a 4 in the rightmost so the 2 has to go into the lower right square. That means that the 4 must go into the other red square as those two squares had captured those two numbers. The green squares have captured the 6 and 8 and, again like above, the intersecting numbers define which square each of those goes into (the 8 goes into the lower left and the 6 goes into the middle right).

The puzzle now looks like this…

Result of container logic.

And, since the two squares in the top row of the bottom containers had captured the 2 and 4 and one of those is filled with a 4 we now know that the green (remaining) square must have a 2. As you can see, we are sneaking up on the solution one number at a time.

We’ll consider the middle, bottom container next…

Next move toward solution.

There are three empty squares in this container. The numbers needed are 1, 4, and 5. You should by now be able to see exactly where each of those goes. Look at the left column for the container. There is a 1 and 4 in it so you know the 5 must go into that left-most square. That leaves the 1 and 4 and, you can see that there is a 4 in the middle row so the 4 can’t go into that square; it has to go into the bottom right square. That means the 1 has to go into the middle row square and that container is now full.

Where we stand now.

Excited to be solving the puzzle? I hope so because we’re going to leave it here for now and pick it up again in the next post.

The Series