Learn to solve Hard Sudoku level 3 puzzle game 2 by a new Sudoku technique
Follow step by step solution to the Sudoku level 3 game 2 to learn how to solve a hard Sudoku game. A new Sudoku technique provided the main breakthrough.
This Sudoku level 3 game happened to be unexpectedly tough that needed identification, formation and application of a new Sudoku technique for the critical breakthrough.
Who knows, you may find it rather easy to solve!
Following is the Sudoku level 3 second game we'll solve after you solve it first.
Though number of filled cells in the game is 27, whereas filled cells become as low as 22 in a very hard Sudoku game, this Sudoku level 3 puzzle game certainly earns its place among the category of Hard Sudoku.
As a strategy, we'll always try to get a valid cell by row column scan first, by possible digit analysis or DSA second and by reduction caused by Cycles third. Cycles are valuable resources and we'll form a Cycle whenever we get the chance.
In addition we'll also use any other advanced Sudoku techniques of single digit lock, parallel digit scan or X wing if necessary.
Evaluation of possible digit subsets for all empty cells is often prescribed as a must-do activity. We strongly recommend to avoid this time-consuming activity as far as possible and instead go in for filling up the empty cells by unique valid digits using Sudoku techniques needed.
The Sudoku techniques for solving this second hard Sudoku level 3 puzzle game are briefly but clearly explained along with the step by step solution.
Let's go through the solution of the game.
Solution to Sudoku level 3 game 2 Stage 1: learn how to solve the hard Sudoku puzzle game
Opportunistic valid cell by row column scan: R9C6 7 by scan for 7 in R7, C4, C5. No more valid cell success by row column scan for 7, 8 or 9.
Going down to decreasing digits, next valid cell by scan is: R1C4 5 by scan in R3, C5, C6.
Trying on still lower digits, next valid cell by scan is: R4C9 3 by scan for 3 in R5, R6 -- R2C6 3 by scan for 3 in R1, R3, C5. And that's all by scan.
By next Sudoku technique of possible digit analysis DSA on empty cells of R4, instead of valid cell direct hit, Cycle (4,8) created in cells R4C2, R4C5 by digit subset reduction of [6,9]. Check for yourself.
Note: In the two digit Cycle of (4,8), possible digits are [4,8] in both the cells R4C2 and R4C5. If finally 4 happens to be there in R4C2, 8 must be there in R4C5 and vice versa. What does it mean? It means, the two digits [4,8] cannot appear in any empty cell of the PARENT ROW R4 as if the pair of digits are locked in these two cells and will Cycle within the two cells only.
By this Cycle (4,8), possible digit subset in rest of the two empty cells in R4 reduces to [6,9] -- R4C3 9 by reduction of 6 in left middle major square -- R4C6 6 by exception in R4. This is a good example of valid cell hit by forming Cycles.
Next valid cell hit by DSA: R8C3 5 by reduction of [1,6,8] from DS [1,5,6,8] in C3.
Finding a valid cell by Sudoku technique of possible digit subset analysis or DSA
We'll understand how we can place digit 5 in R8C3: the mechanism of DSA technique.
The cell R8C3 is at the junction of row R8 and column C3 as well as it is in parent bottom left major square. These are the three parent zones of the cell R8C3.
The filled digit subset or filled DS in C3 is [2,3,4,7,9] which makes the possible digit subset for the four empty cells of C3 as possible DS [1,5,6,8]. This also is the set of possible digits in the cell R8C3 in C3.
Question is, how many of these 4 possible missing digits are present in the other two parent zones of the cell R8C3. Observe that we have started the DSA analysis from C3 and then analyzing the impact of the filled digits in two other parent zones on the set of possible digits. This is a convenient way to carry out a DSA analysis.
Okay, coming back to track we find, rather fortunately, that out ot the 4 missing digits [1,5,6,8] in R8C3, two of the set [6,8] are present in R8 and the third digit 1 is present in parent bottom left major square. This reduces the three digits [1,6,8] from the possible digit subset [1,5,6,8] leaving just the single digit 5 for the cell R8C3.
This is how a valid cell is identified by possible digit analysis or DSA technique.
This helps to form the Cycle (1,8) in R1C3, R5C3 producing valid cell hit R7C3 6 by reduction in C3 -- R9C4 6 by scan for 6 in R7, R8.
R9C9 5 by scan for 5 in R7, R8, C7, C8.
Cycle (2,8) formed in R9C2, R9C7 by possible digit analysis -- R9C8 3 by reduction.
Game status at this first stage shown below.
Solution to the Sudoku level 3 game 2 Stage 2: Breakthrough by new Sudoku technique Trapezium digits
Without scope for any more valid cell by row column scan, Cycle or DSA, we must look for a new Sudoku digit pattern for a breakthrough.
And the new digit combination is provided by the digit 2 occurring in two pairs in R8C2, R9C2 and R8C8, R9C7 of possible digits.
Two conditions are satisfied by these four cells:
- First, each pair of digit 2 appears in only the two cells in the parent major square. That means, for any pair, say R8C2, R9C2, digit 2 can be placed in only one of these two cells in bottom left major square. In the same way digit 2 in R8C8, R9C7 can occur in either of these two cells only.
- The second important characteristic is: both pairs of occurrences of 2 SHARE SAME TWO ROWS R8 and R9 - 2 in R8C2 and 2 in R8C8 share same row R8 and 2 in R9C2 and R9C7 share same row R9.
Now let's consider the effect of this special digit formation of TRAPEZIUM DIGITS.
For each pair of digit 2 in R8 or R9, as one of the two will certainly appear in the row in the final solution, digit 2 cannot be placed in any other empty cell of these two rows. Row sharing effectively blocks or reserves the two shared rows for placing digit 2.
This is the general effect of this special configuration.
Specific to our problem, 2 is reduced from possible digit subset [1,2,4] in R8C5 and that creates the all important breakthrough by Cycle (1,4,8) in C5.
Result is R2C5 2.
This is the most critical breakthrough.
Results of action taken at this stage shown below.
Formation of the digit 2 in the four cells as shown above share two rows BUT NOT TWO COLUMNS.
If you imagine the four cells to be the four vertices of a quadrangle, two sides will be parallel to each other but the other two sides won't be parallel to each other. Such a geometric figure shown below is called as a Trapezium.
Digit 2 can be placed in only two cells R8C2 and R9C2 of bottom left major square. Same way 2 can be placed in only two cells R8C8 and R9C7 both in bottom right major square.
These 4 numbers of digit 2 in possible digit subsets for the four cells share the two common rows R8 and R9 but not the columns. That's why it is always ensured that if any of the four, say 2 in R8C2, does not belong to the final solution, the 2 in R8C8 and 2 in R9C2 must be the valid digits in the cells.
In the same way, if 2 in R8C8 does not belong to the final solution, its row partner 2 in R8C2 and major square partner R9C7 must be the valid digits in the cells.
It means, in the final solution digit 2 must appear either at the end of diagonal AC OR at the end of diagonal BD ensuring that a 2 IS IN ANY CASE PRESENT IN BOTH SHARED ROWS R8 AND R9.
Effect of this is REDUCTION OF ANY INSTANCE OF DIGIT 2 IN THE DSs OF THE OTHER EMPTY CELLS OF THE TWO COMMON ROWS R8 AND R9.
The intersecting diagonals look like an X and that's why it is easier to remember this especially effective digit pattern as an X Wing.
Note: If instead of R9C7, the digit 2 were in R9C8, we would have both pairs of rows and columns shared and the diagonals would have been equal in length. That is the case of proper and perfectly formed X Wing. In this puzzle we have instead a deformed X wing.
Solution to sudoku level 3 game 2 Stage 3: End game holds no tough barriers
By Cycle (1,4,8) in C5, R2C5 2 by reduction of [1,4] -- R1C8 2 by scan for 2 in R2, R3 -- R9C7 2 by scan for 2 in C8, C9 -- Cycle (4,9) formed in R8C8, R8C9 -- R7C9 8 by reduction of 4 -- R9C2 8 by reduction of 2 -- R8C2 2 by reduction -- R7C1 3 by reduction -- R8C1 7 by reduction -- R8C5 1 by reduction of 4 in Cycle (4,9) in R8 -- R8C4 3 by exception in R8.
With 8 in C2, R4C2 4 by reduction -- R4C5 8 by reduction -- R6C5 4 by exception in C5.
Turning attention to favorable zone C2, R1C2 1 by DSA reduction of [5,7] in R1 from DS[1,5,7] in C2 and hence in R1C2 -- R2C2 5 by reduction -- R5C2 7 by exception in C2 R5C1 5 by scan for 5 in R6, C3 -- R6C1 8 by reduction -- R5C3 1 by exception in left major square -- R1C3 8 by reduction.
We could have finished the game at this stage itself, but instead will close the stage here for clarity of understanding from the less cluttered game status figure below.
Solution to the Sudoku level 3 game 2 final stage 4: Finishing off is a formality
R6C7 7 by scan for 7 on empty cells of R6 by 7 in C8, and 7 in central middle major square -- R6C4 1 by scan in R5 -- R6C8 9 by exception in R6. R3C6 1 by scan for 1 in R1, C4 -- R2C9 1 by scan for 1 in R3, C7, C8.
R3C7 8 by scan for 8 in R2, C9 -- R5C8 8 by scan in C7, C9 -- R5C9 6 by scan for 6 in C7 -- R5C7 4 by exception in right middle major square.
R2C8 6 by scan for 6 in C7, C9 -- R3C1 6 by scan for 6 in R1, R2.
R2C7 9 by exception in C7 -- R2C1 4 by reduction -- R1C1 9 by exception in C1 -- R1C6 4 by exception in R1 -- R7C6 2 by reduction -- R7C4 4 by exception in R7 -- R5C6 9 by exception in C6 -- R5C4 2 by exception in R5 -- R3C4 9 by exception in C4 -- R3C9 4 by exception in R3 -- R8C9 9 by exception in C9 -- R8C8 4 by exception in whole game.
The final solution is shown below.
A new game for you to solve at Sudoku level 3
We leave you here with a new Sudoku level 3 game to solve.
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