Step 1

The objective is to get the digits 0 through 9 in every row, every column and every 3x3 square, which are annotated by the thicker border lines.

For each digit, 0 through 9, when you find one in the grid, imagine that all cells in the containing row, column and 3x3 square change color, indicating that the digit of interest cannot go in those cells.

Step 2

A 1 can be placed in the lower right side square, because of the 1 digits in the 2 neighboring squares.

Imagine a red line through the 1 digits in the neighboring squares, passing through the row, column and square they inhabit.

This leaves only one open cell in the lower right square, so a 1 digit must go there.

Step 3

The lower left square can now determine where it’s 1 digit should go because of the two 1 digits in the other squares of that row, and the 1 digit in the square above it.

Step 4

Some times it is best to temporarily not sequence through all the digits from 0 to 9, but instead take advantage of recently placed digits.

We placed a 1 digit in the lower left square. Now it is possible to determine the placement of the 3 digit in that square.

Two vertical rays and two horizontal rays from existing equal 3 digits (in this case, number 3) will always determine the placement of a digit.

Step 5

Starting over with the 2 digit, there may be an opportunity in the lower left square again. It doesn’t pan out, because there are two empty cells once the neighboring 2 digit rays are drawn through the lower left square.

BUT the two empty cells are in the same row, so we know that the bottom cell row in the 2 neighboring squares to the right cannot have a 2 digit – depicted as a purple ray.

So this leaves only one empty cell in the lower center square, so a 2 digit must go there.

Step 6

The middle center square has three neighboring squares that contain a 3 digit. Drawing the rays into the middle center square leaves only one empty cell – where a 3 digit must go.

Step 7

The upper right square has four 3 digit rays passing through it, determining which cell MUST have a 3 digit.

Step 8

Noticing an opportunity, although out of sequence, I saw two 6 digit rays passing through the upper left square,
Leaving only one empty cell which must contain a 6 digit.

It is fruitful to look at the rows, columns and squares that are getting full. As they near completion, it is easier to find new digit placements.

Step 9

A similar opportunity to place a 6 digit is in the lower right square. Two neighboring squares create 6 digit rays that leave only one cell empty, which must therefore have a 6 digit in it.

Step 10

The last two 6 digits placed on the grid now add enough constraints to place another 6 digit in the upper right square.

Step 11

Moving on to the 7 digit, there is an opportunity in the middle center square, imposing the constraints of the 7 digits in 2 neighboring squares

Step 12

A 7 digit can also be determined in the upper right square, using three 7 digit vectors (2 horizontal – which determines the cell row, and 1 vertical).

Step 13

The same situation exists in the middle left square, which determines another 7 digit placement.

Step 14

There is only one 7 digit left to place,
In the lower left square. Two vertical 7 vectors determine where the last 7 digit is placed on the grid.

Step 15

There are only two 8 digits on the gird, which isn’t enough to place any more 8 digits.

The lower left square is able to place a 9 digit. From now on, this tutorial will not depict the vectors which disqualify cells from receiving a digit. You will have to visualize them yourself.

But the diagram does highlight the 9 digits in neighboring squares which constrain the placement of the 9 digit, circled in blue.

Step 16

Another 9 digit is placed.
The choice of what square to look at when placing digits is an art.
You can do it in a methodical way, or a random way – being guided by instinct. Look where you think there may be a possibility.

If a square, row or column has more than 5 cells filled, it is a reasonable candidate to investigate.

Step 17

Another 9 digit is placed, this time using the vector from just one neighboring cell.

This was possible because the square already had 6 cells filled, so there are fewer possibilities for the other 3 cells.

Step 18

In the last diagram, we placed the seventh digit in the upper left square, leaving 2 empty cells.

Once you determine the eight cell, you automatically have determined the ninth cell also.

The last 2 digits to place in the upper left square is 4 and 8. A 4 digit in the square below determines where the last two digits are placed.

Step 19

Column one now has eight digits in it.
The two digits not yet in the column are 2 and 5.

Figure out why they are placed as shown (one of them has a constraint).

Step 20

Now there are eight digits in the lower left square. Placing the last digit is easy.

Step 21

Now row seven has eight digits in it, so placing the ninth digit is easy.

Step 22

Row eight also has eight digits, so placing the last one is easy.

Notice how fast things are going now that the grid is filling up, as you go it gets easier and easier.

Step 23

Eight digits are in column two, so place the ninth.

Step 24

There are now seven digits in row six.
The remaining digits to place are 4 and 8. We are able to determine where the 4 goes – how come?

(hint: look below it)

Step 25

Placing the 4 digit in row 6 automatically determines where the 8 digit is placed, since it is last.

Step 26

Placing last digit needed in column 5

Step 27

Row 2 has seven digits, needs 4 and 5.
We are able to determine where to place the 5. Why?

Step 28

Place the 4 to finish row two.

Step 29

That leaves seven digits in the upper right square. It needs 1 and 2.

We are able to place the 1 because the 2 cannot be placed in that row (there is a 2 in that row already)

Step 30

And we finish off the square by placing the 2.

Step 31

There are now eight digits in column eight.
Place the 5 digit to complete the column

Step 32

Column nine needs a 5 and an 8.
It is easy to place the 8.

Step 33

We finish off the middle right square by placing the final 2.

Step 34

And then we finish off column nine by placing the 5.

Step 35

Finish off row one by placing the 1 digit.

Step 36

Row four only needs a 1 and 2.
It is easy to place the 1 digit.

Step 37

Then place the final 2 digit in that row

Step 38

Then place the final digit in the middle left square.

Step 39

Most sudoku puzzles end with the scenario now shown. Empty cells in squares that line up, in parallel.

Middle Center needs a 5 and a 6.
It is easy to place the 6 (because the 5 cannot go in that position).

Step 40

Finish off row five.

Step 41

Upper center square needs a 4 and 6.
Now that we placed the 6 in the square below, we know where the 4 and 6 must go in this square.

Step 42

Place the final 6 in row three.

Step 43

Place the final 6 in row three.

Step 44

The lower center square needs 4 and an 8. It is easy to place the 8, because the 4 cannot go there.

Step 45

Place the final 4, and you are done.