Cells are 81 individual places in the 9 x 9 grid where numbers from 1 through 9 are either given, or must be found.
Rows are horizontal and contain nine cells. Numbers from 1 through 9 must only appear once in each of the nine rows, which are numbered R1 through R9 from top to bottom along the left edge of the grid.
Columns are vertical and contain nine cells. Numbers from 1 through 9 must only appear once in each of the nine columns, which are numbered C1 through C9 from left to right along the top edge of the grid.
Boxes are collections of nine cells defined by the intersection of three neighboring rows and three neighboring columns. They are called B1 through B9, and form the following grid-pattern:
B1 B2 B3
B4 B5 B6
B7 B8 B9
B1 is formed by the intersection of R1,R2,R3 and C1,C2,C3. The numbers 1 through 9 must only appear once in each box. So when the puzzle is solved, there will be nine 1's, with a single 1 in every Row, Column, and Box. Likewise for the other numbers, 2 through 9.
Box-Rows are rows of boxes, and there are three: B1,B2,B3 B4,B5,B6 B7,B8,B9
Box-Columns are columns of boxes, and there are three: B1,B4,B7 B2,B5,B8 B3,B6,B9
Every one of the nine boxes form a box-junction with one box-row and one box-column.
The box-junctions form nine patterns with five boxes in each pattern. They look like the following:
The junction-box is the single box at the intersection of a box-row and box-column.
The remaining four boxes of a box-junction are called companion-boxes.
For any junction-box, there are two box-column companions, and two box-row companions.
It's a good idea to try and solve for as many missing squares in a Junction box as possible.
For more about companions, click on Sudoku Companions.
When all four companion-boxes contain the same number, there's only one place to put that number in the junction-box. If only three companion-boxes contain a number, it may still be possible to determine where that number must appear in the junction-box, especially if some cells are already known. Even one number known in a single companion box can determine where that number must appear in the junction-box. It's a rare case, but it does happen.
It is very common to find a number by isolation. When a number appears in companion boxes, their appearance in those boxes excludes another occurrences in the same row or column containing that number. A pair of numbers, either intersecting within a junction-box, or running parallel through a junction-box, will exclude five or six cells within that junction-box. Here are examples of five-cell exclusions by intersection:
This is NOT a puzzle. It demonstrates how a pair of digits block out five cells in boxes where they intersect. If you draw a horizontal line through the "1" in the 2nd row, and a vertical line through the "1" in the 2nd column, they will intersect in the upper-left box, and block out five cells, one of which contains a "7". That means a "1" can only occur in one of the remaining four cells, and one of them is already occupied by "3". Likewise, if you draw a vertical line through the "1" in the 2nd row, and a horizontal line through the "1" in the 2nd column, they intersect in the center box, blocking out five cells, one of which contains a "4".
In the box where any number appears, all remaining cells in that box can't be that number; and in each of the other boxes through which the horizontal or vertical lines pass, there are three blocked cells.
When three numbers are either given, or have been found, within a box, and they are together in a single row or column, then the remaining six numbers must be in the same row or column in companion row-boxes or column-boxes, and must also be in the remaining six cells of this box. Such a three-number group is called a "wall".
What follows is a simple example of Isolation by Occupation. C3 in B4 contains 1,7,9 and therefore the remaining six cells of B4 must contain 2,3,4,5,6,8. That's also true of the five remaining cells in C3 (there's already a 5 in R1,C3).
Can you quickly find where to put another 2 in the pattern?
The previous example also illustrates something I call Phantom Finds. The 3rd column of B1 and the 2nd column of B4 are the only places a 2 can be placed in the B1,B4,B7 box-column. A 2 must go in B4's 2nd column because B4's 3rd column is occupied, and the 2 in B7 blocks B4's 1st column. That means BOTH the 1st and 2nd columns of B1 are blocked. Although you don't know where the 2 goes in B4's 2nd column, you know that 2 is limited to B1's 3rd column. There's already a number at (R1,C3) in B1's 3rd column, and the 2 in row R3 blocks (R3,C3), so (R2,C3) is where a 2 must go.
Here's another example where a "Phantom" occurs:
Can you find where to put another 5 in this pattern? Don't guess. Explain why it MUST go where you indicate.
What follows is an even more common form of "Phantom" finds. In this case there actually are two phantoms.
There are only three boxes shown, so there's another box above or below the left-most column-boxes, and another box to the left or right of the bottom-most row-boxes. This looks very much like the bottom-left corner of the previous example. Now look at the 3's, and you'll see they block out the bottom row and the 3rd column in the box with the 8 and 9. Therefore, a third 3 must be in the 1st column to the left of either the 8 or 9, but we don't know which of these two empty squares should get the 3. However, we know that this third 3 in the left-most column-box forces a fourth 3 in the 2nd column in the remaining box above or below this box-column. That 3 may be able to be found knowing the "Phantom" 3 in the 1st column blocks the entire column, as does the visible 3 in the 3rd column. The same logic applies to the 5's, with a "Phantom" third 5 next to the 8 or 9 in the 3rd column of the bottom-left box. Therefore, a fourth 5 MUST also be in the 2nd column of the remaining box, above or below this left-most box-column, because 5's are blocked in the 1st and 3rd columns of this box-column.
As you proceed to solve a puzzle, it is sometimes very handy to write what I call "What's Left" lists vertically above or below a column, or horizontally along side a row. There should be at least five (5) numbers already know in the row or column, so your list is only four (4) or less numbers, those which will fill in empty cells of that row or column. Rarely is a list of five (5) or more numbers very helpful.
These lists help you remember what's missing in a row or column, and can be used to isolate one or more of the missing numbers. As you find numbers in these lists, remember to cross them off your lists.
There is an exception to this general rule, when you know a "naked pair" exists in the row or column. A "naked pair" is a pair of numbers that are the ONLY two numbers that can exist in a pair of cells in the row or column. You may have two naked pairs (four numbers). In these cases, write the naked pair numbers together on the opposite end of the row or column, and construct your "What's Left" list on the other end from the remaining numbers that need to be found. This may make it easier to find those numbers because you can ignore the naked pair numbers and their associated cell pairs.
With a "what's left" list, you may be able to isolate an individual empty cell by two other forms of Exclusion. One form is when a missing number can only go into a single empty cell of a row or column because that number is blocked in all remaining empty cells. The other form is when a box has some (or none) of the missing numbers, and an empty cell in that box intersects with a row and/or column that contains the remaining missing numbers EXCEPT ONE. For example, if 1,2,4,7 are missing in a row that passes through an empty cell in a box that already contains a 1, and an intersecting column contains 4,7, then you know 2 must go in that cell. Some designers of these puzzles like to use "Isolation by Exclusion" at the very beginning. If you're having troubles finding anything, look for a single square taking a single digit.
Here's an example showing just the top two box-rows, but with a row that has a "what's left" list of 2,3,6,7.
Your challenge is to find all the missing numbers in that row (R3,Cx).
The 6 can be found by the first form of exclusion, and the 7 can be found by the second form of exclusion.
That is, 6 is blocked in C4 and C6 of B2, and in C9, so C2=6. 2,3,6 are excluded from C4, so C4=7.
Now 2 is still blocked in C6 of B2, so C9=2, and 3 is what's left: C6=3.
Answers in R3: C2=6, C4=7, C9=2, C6=3
Given that a "what's list" is associated with the empty cells of a row or column, these two forms of exclusion are:
A. One number in the list is excluded from all the empty cells, EXCEPT ONE. That number goes there.
B. All numbers in the list, EXCEPT ONE, are excluded from an empty cell. The missing number goes there.
There are other techniques that can assist in solving Sudoku Puzzles. Try Sudoku Walls and Sudoku Phantoms.
What follows is a more complete example of Isolation:
You should be able to quickly find: 4, 6, 7, 3, and then 1.
Some rows and columns now have five or more numbers found.
Write what is missing above or below such columns, or next to rows.
If you did what was described above, you should have arrived at what is shown below:
The numbers found were the following:
(R8,C4) = 4 in B8
(R1,C3) = 3 in B1
(R4,C2) = 7 in B4
(R9,C7) = 6 in B9
(R7,C7) = 1 in B9
(R9,C8) = 7 in B9 by isolation
(R7,C9) = 8 in B9 by isolation
I placed minus (-), plus (+), and colon (:) symbols after the numbers found to indicate how they were found. The 4+ in B8 was found by Isolation from the 4 in B2. The numbers with the minus were found by pairs of numbers in boxes of the same box-row or box-column, along with an intersection of that same number in a row or column of an interescting box-column or box-row.
A good example is the 3- in B1 found by the pair of 3's in B4 and B7 with the intersection by the 3 in B2.
The "7" in B1 then isolates the 3- at (R1,C3).
The colon-numbers were found in B9 by Isolation after finding the 6- in B9. The 1: and 7: are easy to find, then the 8: is found. This leads to another set of "what's left" next to R7.
Because B8 is only missing three numbers, which are 1,3,9, you can write "what's left" above C6 for the four cells in the two boxes above B8, namely B5 and B2. The trailing periods in the list indicate it is a short list.
Solving the rest of the puzzle goes something like this:
(R5,C6) = 7 found because the 7 in B1 forces 7 to be only in C4 of B2 (Phantom).
Along with the 7's in R4, R6, and C5, you find this 7 in B5.
With four numbers remaining in R5 (2,4,5,8), we can see 2,4,5 in C4 of B8, so ...
(R5,C4) = 8 by Exclusion.
This leads immediately to another 8 ...
(R2,C6) = 8 in B2.
Because we've found both 7 and 8 in C6, we're left with 2 and 5 in B5.
So, because of the 5 in R4 of B6, we get ...
(R4,C6) = 2 and
(R6,C6) = 5
This leads to several more finds:
(R2,C5) = 5 by Isolation from 5 in B8
(R5,C3) = 5 by Isolation from 5's in B5 and B6.
(R5,C7) = 2 because of the 2 in B5
(R5,C5) = 4 because of the 2 in B5 (what's left in R5)
This completes R5.
Let's look at the puzzle with all these finds filled in, and updated "what's left" lists:
(R4,C7) = 4 by Isolation using the 4's in B4 and B5.
This leads to a new what's left list for R4 of 1,3,6,9.
(R4,C4) = 3 bacause of the phantom 3 in B6 and 3's in B2 and B4. (Exclusion)
Now, because of phantom 9 in B6 and 9 in B4, we find ...
(R4,C5) = 9 in B5. (Exclusion)
What's left in C5 is ...
(R6,C5) = 1 which completes C5, also ...
(R6,C4) = 6 what's left in B5, which completes B5.
You may have noticed a new form of what's left for R6. There's a vertical bar (|) between two pairs of numbers. Each pair is isolated to occur in separate boxes. R6 of B4 has 2 and 8, and B6 has 3 and 9. These pairs are sometimes called "naked pairs".
Now let's look at the what's left list in R8, which is 1,6,8,9. B9 already has 1,6,8, so the only empty cell in B9 for R8 must be ...
(R8,C8) = 9 by Exclusion. (Remember to cross out found numbers in what's left lists.)
We are now left with 1,6,8 in R8, but notice that B8 already has 6 and 8, so ...
(R8,C6) = 1 by Exclusion.
We can now fill in the 3,9 in B6 (above B9) because of the 9 in B9:
(R6,C7) = 9
(R6,C8) = 3 (what's left), which completes B6.
We can now complete B9, which has only two empty cells.
(R7,C8) = 4 because of the 3 in C8.
(R9,C9) = 3 (what's left), which completes B9.
This now allows us to complete B8 ...
(R7,C6) = 3 because of the 3's in B7 and B9.
(R9,C6) = 9 (what's left), which completes B8.
And several more fall into place just as easily ...
(R7,C2) = 2 because of 9 in C2.
(R7,C3) = 9 completes R7.
(R6,C2) = 8 because of 2 in C2.
(R6,C3) = 2 completes R6.
(R2,C7) = 3 by Isolation from 3's in B1, B2, B6, B9.
Let's now find as many empty cells as quickly as possible.
(R1,C8) = 5 because of the 2 in R1.
(R3,C8) = 2 what's left in C8.
(R2,C1) = 2 by Isolation from 2's in B2, B3, B7.
The next one is found by Exclusion, but using intersecting rows.
Since there are four values left in C2, we look for any of those
values in Rows that intersect C2 that are the still blank cells. Thus:
(R3,C2) = 6 because of the 6's in R1, R2, and R9.
That leaves us with 1,4,5, and we find 5's in R1, R2 ...
(R9,C2) = 5 because 5's are excluded elsewhere in C2.
We're now left with only 1,4 in C2 of B1, but we're also left with just 1,4 in B7.
Since there's a 4 in C1, one of the empty cells in B7, we get this ...
(R9,C1) = 1 and
(R9,C3) = 4
We can now resolve the missing pair in B4, row R4, because of the 1 in C1 ...
(R4,C1) = 6 and
(R4,C3) = 1
And we can finish both C3 and B7 as follows:
(R3,C3) = 8 because there's already a 6 in B1, and
(R8,C3) = 6 completing C3 AND
(R8,C1) = 8 completing B7 and R8.
We're left with 5,9 in C1 of B1, and there's already a 5 in R1, so ...
(R1,C1) = 9 and
(R3,C1) = 5 completing C1.
All that's left are empty cells in the top box-row.
We can quickly find them using the what's left lists.
Using the 7,9 list for R3, we resolve R3 as follows:
(R3,C7) = 7 because 9 is blocked in C7,
(R3,C4) = 9 what's left in R3, and
(R1,C7) = 8 what's left in C7.
Now all that's left in C4 is 1,7, but there's a 7 in R2, so ...
(R2,C4) = 1 and
(R1,C4) = 7 completing C4 and B2.
We are left with 4,9 in R2, AND C9 of B3; and 1,4 in C2 of B1, thus ...
(R2,C2) = 4 because 9 is blocked in C2, and therefore ...
(R2,C9) = 9 what's left in R2, and
(R1,C2) = 1 what's left in C2, and
(R1,C9) = 4 AND WE ARE DONE!!
The puzzle is now solved, and looks like this:
Having gone through a complete puzzle, step-by-step, here's another puzzle we can try in class, or you can do on your own. It has a very interesting feature... there's one diagonal that is completely filled in, from the upper-left corner to the lower-right corner. In class, each participant has a copy of this puzzle on paper, with room to write "what's left" lists above the columns or to the right of the rows. If there's not enough time to do this in class, then take the puzzle home and try it yourself. For those who didn't attend class, I'm presenting the puzzle below, AND I've got a link to a PDF of this same puzzle you can download. The link appears at the end of this section.
If you'd like, you can download or view a PDF of this Sudoku Class Puzzle,
just click on this reference link. Here's a link to the Sudoku Class Solved".
There's also a frame-by-frame html version you can navigate at Sudoku-class.
For those of you who would like to try an interesting puzzle, click on this Interesting Sudoku Puzzle link.
The solution is partially worked out, but you are encouraged to finish it.
For those of you who would like to try a harder puzzle, click on this Sudoku Hard Puzzle link.
The solution is available at Sudoku Hard Solution.
For those of you who would like to try a VERY HARD puzzle, click on this Sudoku Very Hard Puzzle link.
The solution is available at Sudoku Very Hard Solution.
I believe in "guessing" when the puzzle is so hard, you're stuck. I've added another Subpage which discusses my technique.