Let's play the third Sudoku game at second level hardness
This is the third game play at second level Sudoku. You should find it a bit harder to solve. The most basic technique of row column scan for finding a valid cell won't be enough any more. After proceeding a few steps by using row column scan, you will find suddenly that you are no longer able to proceed further.
This is the point where we have to devise new techniques for finding a valid cell based on new logic of pattern analysis.
This is the hallmark of Sudoku game.
As you try to solve increasingly difficult Sudoku problems, you need to create and use more sophisticated pattern analysis techniques.
We will keep open the option for you to devise your own techniques to solve the game. You may solve the game on your own and check the solution given at the end of the session. If you are stuck, you may refer to our step by step solution process.
Or, you may decide to learn the new techniques straightaway from our detailed solution.
The Sudoku second level game 3
A rough criterion to judge the difficulty level of a Sudoku game is the number of empty cells in the game board. In all our earlier beginner level games, if you had noticed, the number of empty cells was 45 to 47 at most. In this second hardness level game, the number of empty cells is 49, a larger number. On top of it, the filled up digit clustering adds to its difficulty.
Together this simply means, there won't be many friendly digits waiting for you to easily form a scan pattern and pinpoint a valid cell. You have to work more intelligently and use more complex techniques to find a valid cell.
Here we will leave you with the game to solve yourself or go forward with our solution.
Solution process of the 3rd Sudoku game at hardness level 2
To start with we will repeat the very brief Sudoku concepts under Aside 1 and Aside 2. If you are already comfortable in playing easy Sudoku, you may skip these two sections.
Furthermore, If you feel the need to brush up the basics of how to play Sudoku in more details, you may go back to our session 1 and session 2 of beginner level Sudoku game play or session 1, and session 2 of second level Sudoku game play.
A Sudoku game board of 81 square cells comes with a few of the cells already filled up with valid digits 1 to 9. The objective of the game is to fill up rest of the empty cells following the basic Sudoku rule of NOT using any of the digits 1 to 9 more than once in any row, column or 9-cell square (identified by thick borders, there are 9 such 9-cell squares).
At each step when we locate a valid cell where we can put one and only one digit, we will identify the cell by its row number (R1 to R9) suffixed by column number (C1 to C9), mention the digit, and mention the column row scan information ( or nothing if the cell happens to be only one empty in the containing row, column or 9-cell square).
If we use a new technique to find the valid cell, we will mention the type of technique and if needed will explain it in more details. Sometimes we will add a remark.
The only objective at each step is to find a cell in which one and only one digit can be placed. This is what we call a valid cell.
Valid cell and Valid digit:
We would repeat the definition of valid digit and the valid cell.
We define a VALID digit as the digit you put in an empty cell so that,
It is the only possible digit which you can put in the cell following the rules of the game.
As before, the most important objective is, how to find out the cell in which you can put one and only one digit out of the digits 1 to 9.
We have used the most basic technique of row column sweep or horizontal-vertical cross-scanning mainly. We have used Exclusion technique wherever we found it possible.
We used the concept of a favorable zone as a column or a row or a 9-cell square with high digit occupancy so the number of digits possible to fill up the empty cells in the zone is much less and so by eliminating the few other contenders getting only a single digit for a single cell is that much higher. We have also used digits with high occurrence in the board, that is, the favorable digits to our advantage.
For a favorable digit that has appeared, say, in 7 cells out of maximum 9 possible all over the board, finding the rest of the two valid cells for this favorable digit by row-column sweep is generally very easy because of elimination of a large part of the board by its occurrence in most zones.
Let's find the valid cells - the stages
We will show the game board again for convenience. The following is the Sudoku game we have to solve.
Our strategy will generally be to find a valid cell by row column scan and if it doesn't work then use more powerful techniques. We will stop and explain how a new technique works when we meet it for the first time.
The first step is easy and we get our first valid cell at R5C8 for digit 5 by scan of C7 and C9 and the second at R1C1 for 5 by scan of R2, R3. But after these two easy successes we were stuck and after checking for any more scan we went for DSA. Let's see the details.
Starting valid cell R5C8 5, scan C7, C9 -- R1C1 5, scan, R2, R3 -- R3C6 3 DSA [1,2,3,4] in 9 cell square -- R1C6 2, DSA [1,2,4] in 9 cell square -- R2C5 1, DSA [1,4] in 9 cell square -- R1C5 4. Then we formed the digit subsets for the three cells, R2C2, R3C2 and R3C3.
We'll explain how DSA technique worked.
The technique Digit Subset Analysis or DSA in short
For quick success by this technique, we identify a most favorable zone (row, column or a 9-cell square) with highest number of cells filled up so that the possibilities of digits occupying the few empty cells are minimum.
Usually, we get the opportunity to apply this powerful technique quickly on a row or column with 7 cells filled up and only two cells remaining empty. If the empty cells are more than 2, chances of success is less.
In this game such a favorable zone is the top middle 9-cell square. In its four empty cells the four digits [1,2,3,4] are the only possibilities. These four digits form the Digit Subset.
We represent this digit subset as [1,2,3,4].
Now we will analyze this digit subset by cross checking each of these four empty cells with the intersecting rows and columns to see whether any three out of the four digits appear in an intersecting row and column. If so, we can put the fourth digit in the corresponding cell. This is usual DSA.
In this case we get such a break by DSA in cell R3C6 where the DSA set was [1,2,3,4]. Scanning the intersecting row R3 and column C6 we find these two together have the three digits 1,2, and 4. So these three digits are eliminated as prospective occupants for this cell R3C6 leaving the single digit 3. This is the condition of a valid digit. So we put 3 in cell R3C6.
Digit subset analysis is a general concept based on which you may devise your own technique.
Continuing in the next stage we get the valid cells,
R2C2 8, DSA cancel -- Cycle (1,7) formed in 9 cell square -- formation of a second cycle (4,8) in two remaining cells R3C8, R3C9 in row R3.
Usefulness of a cycle
You may get an idea of the usefulness of the cycle (1,7) by observing its ability of restricting validity of the two digits 1 and 7 only to these two cells, R3C2 and R3C3.
A cycle of length 2 restricts validity of the two digits to these two cells only, reducing the size of the digit subset for other cells in the row, column or 9-cell square where the cycle has been formed.
Just observe even though there were 4 empty cells in R3 and four possible digits, because of formation of the cycle (1,7) another 2 length cycle has also been formed in the two remaining cells R3C8 and R3C9 in the row R3. The cycle of (1,7) works as if these two cells have already been filled up by 1 and 7.
Let's proceed to next stage.
R3C9 4, DSA cancel by scan C9-- R3C8 8, DSA cancel -- R7C7 8, scan C8, C9 -- R5C7 3 DSA [6,2,3,7,9] in 9 cell square and scan of C7 and R5. This is a large DSA, but still successful, you never know how a DSA may produce results. All the rest four digits 6,2,7 and 9 appear in intersecting row R5 and column C7, leaving the lone digit 3 for this cell.
This breakthrough allowed formation of a cycle of (6,2) in cells R4C9 and R6C9 in column C9 of 9-cell square. This is by scan of C7, C8 for 6 and C7 for 2. This produces further positive results.
R4C9 for 6 by DSA cancel -- R6C9 2 by DSA cancel -- R9C9 1, scan C8 -- R8C8 9, DSA [3,9] scan R8 -- R7C8 3. -- R2C8 2 DSA [2,7] in C8 scan R1 -- R1C8 7.
Let us produce the results achieved till now.
Proceeding further, R2C9 9 -- R1C9 3 -- R1C7 1 -- R5C5 2, scan R4, R6 -- R5C3 6, DSA [4,6] in R5, scan C3 -- R5C2 4 -- R9C1 4, scan C2, C3, R7, R8 -- R4C4 4, scan R5, R6, C5, C6 -- R4C5 3, scan R6, C6 -- R9C4 3, scan R7, C5. The situation till now is as below.
Proceeding further, R6C4 9, DSA [6,9] in C4, scan bottom middle 9-cell square -- R7C4 6 -- R7C1 1, DSA [1,9] in R7, scan C1 -- R7C2 9 -- R4C7 9, scan R6 -- R6C7 7 -- R6C3 1, scan R4, C1 -- R3C3 7, DSA cancel -- R3C2 1 -- R8C2 7, DSA [5,7], scan R8 -- R9C2 5 -- R9C3 8, DSA [5,8] in C3, scan R9 -- R4C3 5 -- R8C1 6 -- R9C5 7 -- R8C5 8 -- R6C5 6 -- R4C1 7, scan R6 -- R6C1 8 -- R6C6 5 -- R4C6 8. End. Game solved.
A special note for you
We haven't tried to optimize. So we feel that this more complex game is solvable by following an easier path.
You should try to reach one of the simpler solutions.
Lastly, we leave a game for you to solve.
A game for you to solve
We leave you here with a new game for you to solve. In our next session we will present its solution and another new game.