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Sudoku Third level game play 3

Sudoku level 3 Game 3: Learn to solve a high level hard Sudoku

Learn to solve Hard Sudoku level 3 puzzle game 3 by hard Sudoku techniques

Follow step by step solution to Sudoku level 3 game 3 to learn how to solve a hard Sudoku game. Hard Sudoku technique X wing provided the main breakthrough.

This Sudoku level 3 game happened to be not easy and needed use of advanced Sudoku technique X wing for the critical breakthrough.

You might try to solve it along an easier path.

Following is the Sudoku level 3 third game we'll solve after you solve it first.

How to solve Sudoku level 3 hard Sudoku Game 3

Though number of filled cells in the game is 28, whereas filled cells become as low as 22 in a very hard Sudoku game, this Sudoku level 3 puzzle game is not an easy one and is categorized as a 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 third 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 3 Stage 1: learn how to solve the hard Sudoku puzzle game

How to solve Sudoku level 3 hard Sudoku Game 3

First valid cell by row column scan: R3C4 4 by scan for 4 in R1, R2, C2 -- R4C5 4 by scan in R5, R6, C4, C6 -- R9C7 4 by scan for in R8, C8, C9 -- lastly, R7C3 4 by scan for 4 in R8, R9, C2. Digit 4 fully filled.

Scanning for next larger digit 5: R1C95 by scan in R2, C7, C8 -- R5C3 5 by scan for 5 in R6, C2 -- R4C4 5 by scan for 5 in R5, R6 -- R9C6 5 by scan for 5 in R7, R8, C4 -- lastly, R3C5 5 by scan for 5 in C4, C6. Digit 5 fully filled.

No valid digit by row column scan for 6.

Scanning for 7: R1C3 7 by scan for 7 in R2, C2 -- R3C7 7 by scan for 7 in R1, R2, C8. And that's all by row column scan.

R4C9 3 by DSA reduction of [2,6] in C9 from DS [2,3,6] in R4 Cycle (1,2,6) in cells R6C1, R6C3 and R6C8 by possible digit subset analysis. This helps to form a second Cycle (3,9) in R6C5, R6C6.

Reduced possible digit subset the rest three empty cells of central middle major square and in R5 is [2,6,7] -- R5C5 6 by reduction of [2,7] in C5 from DS [2,6,7] -- third Cycle (2,7) formed in R5C5, R5C7 -- R5C7 1 by reduction of [2,6].


A Note on Cycle (2,7)

In the Cycle (2,7) possible digit subset for both the cells R5C5 and R5C7 is [2,7].

That means in the final solution one of these two cells would have digit 2 and the other digit 7, but at this point of time we don't know for certain which cell will have which digit.

Effectively, these two digits Cycle between these two cells and won't appear in any other cell in the row R5.

The two digits are locked in these two cells only.

This is not as good as 100% certainty regarding digit occupancy of these two cells individually, but it certainly is as good if you consider the two cells as a group.

There may be Cycles involving three digits locked in 3 cells in a zone (row, column or major square).

Four or five digit Cycles also appear but are rare and not easy to identify.

Cycles always are valuable assets for simplification of a hard Sudoku game.


Finding a valid cell by Sudoku technique of possible digit subset analysis or DSA

We'll understand how we can place digit 3 in R4C9: the mechanism of DSA technique.

The cell R4C9 is at the junction of row R4 and column C9 as well as it is in parent right middle major square. These are the three parent zones of the cell R4C9.

The filled digit subset or filled DS in the major square is [1,4,5,7,8,9] which makes the possible digit subset for the three empty cells of the major square as possible DS [2,3,6]. This also is the set of possible digits in the cell R4C9 in the major square.

Question is, how many of these 3 possible missing digits are present in the other two parent zones of the cell R4C9! Observe that we have started the DSA analysis from the right middle major square and then analyzing the impact of the filled digits in two other parent zones on the set of possible digits in the specific cell of the major square.

This is a convenient way to carry out a DSA analysis.

Okay, coming back to track we find rather fortunately, that out ot the 3 missing digits [2,3,6] in R4C9, two of the set [2,6] are present in C9 and thus reduces the three possible digits [2,3,6] to the single digit 3 for the cell R4C9.

This is how a valid cell is identified by possible digit analysis or DSA technique.


Game status at this first stage shown below.

How to solve Sudoku level 3 hard Sudoku Game 3 Stage 1

Solution to the Sudoku level 3 game 3 Stage 2: Breakthrough by single digit lock and parallel scan

By scan for 2 in R8 and C9, digit 2 can appear in bottom right major square and in C8 in only 2 cells R7C8 and R9C8.

This is the special Sudoku digit pattern of single digit lock.

Note: A SINGLE DIGIT LOCK is a valuable digit pattern that usually provides a breakthrough, but for that the single digit lock is to be formed generally by a cross-scan of row and column.

This pattern on 2 implies that either of these two numbers of digit 2 must appear in one of these two cells in C8 so that no other empty cell in C8 can have digit 2.

Breakthrough by single digit lock on 2: R6C8 6 by reduction -- R4C7 2 by reduction -- R4C2 6 by reduction and exception in R4 -- R8C3 6 by scan for 6 in R9, C1, C2 -- R2C7 6 by scan for 6 in R1, C8, C9.

Parallel scan for 8 on empty cells of C3: 8 in R9 eliminates R9C3, 8 in R6 eliminates R6C3 leaving only the cell R2C3 for 8: R2C3 8. This is an important breakthrough.


A note on parallel scan

Usual scan for a specific digit is done on the empty cells of a MAJOR SQUARE BY IDENTIFYING PRESENCE OF THE DIGIT IN SCANNED COLUMNS AND ROWS INTERSECTION THE MAJOR SQUARE.

But in this variation of parallel scan, scanning is done also for a single specific digit 8 in this case. But instead of examining the empty cells of the target major square, we examine the impact on EMPTY CELLS OF A ROW OR A COLUMN.

In this specific case, we are examining the three empty cells R9C3, R6C3 and R2C3. Because of presence of 8 in R9 and R6, the first two cells are prohibited for 8 leaving the single cell R2C3 for digit 8.

Usually a parallel scan is associated with formation of a Cycle in the scanned row or column and provides an important breakthrough.


The valid cell for 8 by parallel scan gives us a series of valid cells for digit 8: R8C2 8 by scan for 8 in R7, R9, C1 -- R3C8 8 by scan in R2, C7, C9 -- R1C6 8 by scan for 8 in R3C4. Digit 8 over.

This forms a Cycle of (6,9) in R3 and R3C2 1 by reduction of 9 by the Cycle (6,9).

Results of action taken at this stage shown below.

How to solve Sudoku level 3 hard Sudoku Game 3 Stage 2

Solution to Sudoku level 3 game 3 Stage 3: Another breakthrough by X wing of digit 7

By Cycle (2,3) in R1C4, R2C4, R5C4 7 by reduction of 2 -- R5C6 2 by reduction.

A single digit lock on 7 formed in R7C6 and R8C6 by scan for 7 in C4, C5.

A second single digit lock on same digit 7 is again created in same rows R7C9 and R8C9 by scan for 7 in C7, C8.

This formation of a single digit pairs sharing same two rows and columns R7, R8 and C6, C9 reserves both pairs of rows and columns for 7 only two of these diagonally opposite cells: either R7C6 and R8C9 or R7C9 and R8C6.

Thus digit 7 is prohibited any cells in these two pairs of rows and columns and we get a breakthrough valid digit in R9C1 7 by reduction of 7 in R8C1 -- R9C4 9 by DSA reduction.

This digit formation of two pairs of same digits sharing same pairs of columns and rows in possible digit subsets is called as X wing. The X wing is shown by black lines.

How to solve Sudoku level 3 hard Sudoku Game 3 Stage 3


A note on X wing digit pattern

This is one of the most valuable digit patterns you can identify and use for breakthrough in a hard Sudoku puzzle game.

In an X wing formation, the same possible digit appears in two pairs sharing a pair of rows and a pair of columns.

In the above Stage 3 figure, possible digit 7 appears in four cells: R7C6, R7C9, R8C6, R8C9. Shared rows are R7, R8 and shared columns are C6, C9.

This represents only the possible digit configuration.

But with this configuration in place we can be sure that if final solution has R7C6 7, then the diagonally opposite cell R8C9 must have 7 with no other cell in these two pairs of rows and columns having digit 7.

If instead, R7C9 has 7 in the final solution, then its diagonally opposite partner cell R8C6 must have 7 in it and no other cell in the two pairs of rows and columns having digit 7.

Effectively then, whichever be the case in the final solution, rows R7, R8 and columns C6, C9 cannot have any other empty cell with digit 7 in them. This is equivalent to the rows and columns having actually digit 7.

This property is used in scanning for the breakthrough.


Solution to the Sudoku level 3 game 3 final stage 4: Finishing off is a formality

With 9 in R9C4, R3C4 6 by reduction -- R3C6 9 by reduction -- R6C6 3 by reduction -- R6C5 9 by reduction -- R7C5 3 by exception in C5 and also by reduction.

R8C7 3 by scan for 3 in R7, R9, C9 -- R1C7 9 by reduction and exception in C7 -- R1C2 2 by reduction -- R2C1 9 by reduction -- R7C2 9 by exception in C2 and also by reduction -- R8C1 1 by reduction -- R6C1 2 by reduction and by exception in C1 -- R6C3 1 by reduction and by exception in R6 -- R9C3 2 by reduction and by exception in C3 -- R9C8 1 by reduction and by exception in R9 -- R7C8 2 by reduction -- R2C8 3 by reduction and by exception in C8 -- R2C9 1 by reduction by exception in top right major square.

R2C4 2 by reduction and by exception in R2 -- R1C4 3 by reduction and by exception in R1 -- R7C4 1 by exception in C4. Only 4 cells involved in X wing digit formation are left to be filled.

Break the formation by a simple scan for 6 in R8C3: R7C6 6 -- R7C9 7 by X wing property -- diagonally opposite R8C6 7 by X wing property and finally R8C9 9 by exception in the whole game.

This Sudoku level 3 is a hard Sudoku game to solve.

The final solution is shown below.

How to solve Sudoku level 3 hard Sudoku Game 3 Final Stage 4

A new game for you to solve at Sudoku level 3

We leave you here with a new Sudoku level 3 game to solve.

Enjoy.


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First and second level Sudoku games

You will get the earlier Sudoku game solutions at Beginner level Sudoku and Second level Sudoku.

Fourth level game plays

List of fourth level very hard Sudoku game plays are available at Fourth level Sudoku.