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HOW TO PLAY SU DOKU (unfinished)

|1 | | |

|6 |2 |4 |

|1 |7 |3 |

In the beginning

This is where the game might have started from. In the end diagram above, the numbers in this beginning position where shaded in light gray.

| |8 |5 |

|2 |1 | |

|8 |6 |3 |

The last number must be different than the rest:

|4 |7 |9 |

|2 |1 |5 |

|8 |6 |3 |

Similarly, if:

|2 |

|4 |

|1 |

|9 |

|5 |

|2 |

| |

| |

| |

|2 |

|1 |

|8 |

|7 |

| |

| |

| |

| | | |

| | | |

| | | |

| | | |

| | | |

| | | |

| | | |

| | |8 |

The 8 at the bottom tells us there is a 6 in the shaded square.

The same thing applies to a 3 × 3 box:

|1 |2 |3or8 |

|4 |5 |6 |

|7 |3or8 |9 |

And

|1 |2 | |

|7 | |9 |

Gives us an 8 in the top shaded square and a 3 in the bottom shaded square.

Square number 1, 2 or 3

Similarly, suppose we know the following:

| | | |

| | | |

| | |3or7 |

Tells us:

|5 | |3or7 |

| | | |

| | |3or7 |

Here are some variations of this.

|1or7 |3 |1or7 |

| |9 | |

| |6 |4 |

| | | |

| | | |

| | | |

| | | |

|2 | | |

| | | |

And a 2 goes in the shaded square.

With a pair of choices we can do the following:

| |

| |

| |

| |

| |

| |

|2 |

| |

From the information on the top row we know the shaded square has a 1 or a 2. But the information from the 2 in the column tells us it must be a 1.

More of the same:

| |

| |

|3or9 |

|3or9 |

|2or4 |

|1 |

|8 |

There is only one possible spot for the 4 in the first column:

|6 |

| |

|3or9 |

|3or9 |

|4 |

|1 |

|8 |

When a little bit of uncertainty doesn’t matter

To illustrate the idea, here is a simple situation. Suppose we know the following:

| |1 |6or8 |

|2 |3 |4 |

|5 |9 |6or8 |

Although we don’t know which of the shaded squares has a 6 and which has an 8, we know that the pair 6 and 8 are in the two squares. We conclude that the 7 is in the remaining square:

|7 |1 |6or8 |

|2 |3 |4 |

|5 |9 |6or8 |

Here’s an interesting situation:

| | | | | | |

|2 | | | | | |

The 9 can’t be in the top row so it must go in one of the dark shaded squares. So we’re uncertain about the exact location of the 9 but we know which row it is in: the middle one. This means that the 9 in the bottom left 3 × 3 box must be on the bottom row:

| |7 |6 | |3 |9 | | | | | | |2 | | | | | | | | | |8 |7 |5 | | | | |

Combining this with the 9 not being in either of the first two columns, we find where the 9 belongs:

| | | | | | | | | |9 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |9 | | | | | | | | | | |7 |6 | |3 |9 | | | | | | |2 | | | | | | | | |9 |8 |7 |5 | | | | |

Advanced techniques

When the remaining boxes are pairs of choices – the contradiction method

Consider the following Su Doku diagram. Most of it has been filled in.

|3 | |4 |1 |2 |5 |7 |9 | | |4 | |7 |5 |9 |6 |8 |3 | | | |7 |8 |6 |3 |1 |4 |2 | | |7 | |6 |3 |8 |9 |2 |4 | |3 | | |2 |4 |7 |8 |1 |5 | | | |4 |1 |9 |5 |3 |6 |7 | |4 | | |9 |7 |1 |2 |3 |6 | | | |3 |5 |8 |4 |7 |9 |1 | |7 | | |3 |2 |6 |4 |5 |8 | |

We have the following options. Note that in each undecided box there is exactly one pair of choices.

6 8 |3 |6 8 |4 |1 |2 |5 |7 |9 | |1 2 |4 |1 2 |7 |5 |9 |6 |8 |3 | |5 9 |5 9 |7 |8 |6 |3 |1 |4 |2 | |1 5 |7 |1 5 |6 |3 |8 |9 |2 |4 | |3 |6 9 |6 9 |2 |4 |7 |8 |1 |5 | |2 8 |2 8 |4 |1 |9 |5 |3 |6 |7 | |4 |5 8 |5 8 |9 |7 |1 |2 |3 |6 | |2 6 |2 6 |3 |5 |8 |4 |7 |9 |1 | |7 |1 9 |1 9 |3 |2 |6 |4 |5 |8 | |

Here’s a funny quote: ‘When you assume, you make an ass out of u and me.’ As we shall see below, the quote doesn’t apply to Su Doku!

Let’s assume that the number in the upper left box is an 8. Then we can start to fill in the rest of the boxes based on this assumption.

81 |3 |62 |4 |1 |2 |5 |7 |9 | |1 2 |4 |1 2 |7 |5 |9 |6 |8 |3 | |512 |911 |7 |8 |6 |3 |1 |4 |2 | |115 |7 |516 |6 |3 |8 |9 |2 |4 | |3 |64 |93 |2 |4 |7 |8 |1 |5 | |25 |86 |4 |1 |9 |5 |3 |6 |7 | |4 |57 |88 |9 |7 |1 |2 |3 |6 | |69 |210 |3 |5 |8 |4 |7 |9 |1 | |7 |113 |914 |3 |2 |6 |4 |5 |8 | |

We get a contradiction. Neither a 1 nor a 2 will work in the box below the upper left corner since there is already a 1 and a 2 in the first column. This means our assumption that the number in the upper left box is an 8 is false. Therefore the number in the upper left box is a 6. It is then easy to fill in the remaining numbers.

61 |3 |82 |4 |1 |2 |5 |7 |9 | |115 |4 |216 |7 |5 |9 |6 |8 |3 | |913 |514 |7 |8 |6 |3 |1 |4 |2 | |512 |7 |111 |6 |3 |8 |9 |2 |4 | |3 |99 |610 |2 |4 |7 |8 |1 |5 | |86 |25 |4 |1 |9 |5 |3 |6 |7 | |4 |84 |53 |9 |7 |1 |2 |3 |6 | |28 |67 |3 |5 |8 |4 |7 |9 |1 | |7 |117 |918 |3 |2 |6 |4 |5 |8 | |

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