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Sudoku Third level game play 5

Sudoku level 3 Game 5: Breakthroughs by Hard Sudoku Techniques

Solve Sudoku level 3 game 5 by hard Sudoku techniques

How to make breakthroughs by hard Sudoku techniques of single digit lock and parallel scan explained in this step by step solution to Sudoku level 3 game 5.

First solve this fifth game of Sudoku level 3 and then if you need, go through the solutions.

How to solve Sudoku level 3 hard Sudoku Game 5

This Sudoku level 3 puzzle game 5 is a hard one and needed use of hard Sudoku techniques for breakthroughs.

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 fifth 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 5 Stage 1: First breakthrough by single digit lock

How to solve Sudoku level 3 hard Sudoku Game 5

First valid cell by row column scan is for 6: R4C1 6 by scan for 6 in R5, C3 -- R8C2 6 by scan for 6 in R9, C1, C3 -- R6C5 6 by scan for 6 in R4, R5, C6 -- lastly, R7C4 6 by scan for 6 in R8, R9, C5, C6. Digit 6 fully filled.

Scanning next for 7: R4C3 7 by scan in R5, R6 -- R8C1 7 by scan for 7 in R9, C3. An opportunistic valid cell hit in the neighborhood: R9C2 8 by scan in C3 -- R7C3 9 by scan for 9 in R9 -- R8C3 2 by exception in bottom left major square.

Again an opportunistic valid cell for 9: R5C1 9 by scan for 9 in R6, C2 -- R5C5 3 by scan for 3 in R4, R6 -- R2C9 9 by scan for 9 in C7, C8 -- R3C6 9 by scan in R1, R2, C4, C5. Digit 9 chosen for scan as it has maximum filled digits 7 out of 9.

With no easy valid cell hits by scan or DSA, short length possible digits in empty cells of the favorable central zones are filled.

The breakthrough Sudoku digit pattern identified among these DSs that we'll explain next stage.

Just observe the single digit lock on digit 4 caused by a simple scan for 4 in R6 that allows 4 only in two cells R4C5, R4C6 of R4. This is the single digit lock on 4 and it prohibits digit 4 to be placed in any other cell in R4.

Game status at this first stage shown below.

How to solve Sudoku level 3 hard Sudoku Game 5 Stage 1


A note on breakthrough by Single digit lock

The more frequent form of single digit lock is shown above in R4C4, R4C5. Digit 4 appears only in the possible digit subsets of these two cells in the major square and row R4.

In the final solution, one of these two must be present ensuring that digit 4 is prohibited in all other possible digit subsets of empty cells in R4.

As a result, digit 4 is reduced from the possible digit subset [4,8] in R4C7 to make a valid cell hit of R4C7 8. This is one of the most important breakthrough in the game.

Such an effective single digit lock can happen either by scan of a single row or column or by cross-scan of a row and a column.

It follows that by scan of a digit in two columns or two rows cannot form a single digit lock that can be directly effective (it can yet become effective in forming an X wing though).


Solution to the Sudoku level 3 game 5 Stage 2: Breakthrough by single digit lock and parallel scan

Breakthrough by single digit lock on 4 in R4C4, R4C5: R4C7 8 by reduction of 4 -- R6C6 8 by scan for 8 in R4.

Now is the time for a breakthrough by DSA: R9C9 [1,4] by reduction of [3] from possible digit subset [1,3,4] in R9 -- Cycle (1,4) formed in R5C9 and R9C9 -- R6C9 2 by reduction of 1 by the Cycle (1,4) -- R6C7 5 by reduction -- R6C3 1 by exception in R6 -- R5C2 5 by exception in left middle major square.

With possible digit subset [7,8] in R7C9, R8C9 and digit 7 in R8, R8C9 8 by reduction -- R7C9 7 by exception in C9.

In neighborhood cell, R7C5 8 by scan for 8 in R8 and C6.

Cycle (2,4) in R7C6, R7C7 by reduction of 5 from DS [2,4,5] in R7 by parallel scan of 5 in C6, C7 -- R7C8 5 by exception.

R3C7 7 by scan for 7 in R1, C8 -- R2C2 7 by scan for 7 in R1, R3, C1 -- R3C2 1 by exception in C2. DS [3,5] in C3 and 5 in R1 -- R1C3 3 by reduction of 5 -- R3C3 5 by exception in C3.

R3C4 3 by elimination of two of three empty cells of R3 by parallel scan for 3 on these 3 cells: 3 in C1 eliminates R3C1 and 3 in C5 eliminates R3C5 for 3. Only R3C4 is left for 3. R1C4 1 by scan for 1 in R2, R3.

Result of actions taken at this stage shown below.

How to solve Sudoku level 3 hard Sudoku Game 5 Stage 2


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 3 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 R3C1, R3C4 and R3C5 of row R3.

3 in C1 eliminates empty cell R3C1 for 3 and 3 in C5 eliminates R3C5 for 3. Only the third cell R3C4 is left where 3 can be placed.

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


Solution to Sudoku level 3 game 5 Final Stage 3: No more challenges left

R4C4 4 by reduction -- R4C5 1 by reduction. R2C4 8 by scan for 8 in C5, C6 -- R8C4 5 by exception in C4 -- R1C1 8 by scan for 8 in R2, R3 -- R2C8 3 by scan for 3 in R1 -- R9C7 3 by scan for 3 in R7, C8, C9.

R2C5 5 by scan R3, C6.

R8C5 2 by DSA reduction of [1,3] from DS [1,2,3] in three empty cells of R8 -- R3C5 4 by exception in C5 -- R2C6 2 by exception in top middle major square -- R2C1 4 by exception in R2 -- R3C1 2 by exception in C1.

With 2 in R2C6, R7C6 4 by reduction -- R9C6 1 -- R8C6 3 by exception in C6 -- R8C8 1 by exception in R8 -- R9C9 4 by exception in R9 -- R7C7 2 by exception in R7. R5C8 4 by reduction -- R5C9 1 by exception in C9 -- R1C8 2 by exception in C8 -- R1C7 4 by exception in whole game.

Final solution shown below.

How to solve Sudoku level 3 hard Sudoku Game 5 final Stage 3

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.

how to play Sudoku level 3 game 5 exercise


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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.