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Sudoku Third level game play 8

Sudoku level 3 Game 8: Step by Step Easy to Understand Solution

Step by step easy and quick solution to Sudoku level 3 game 8

Solution to the Sudoku level 3 game 8 is explained step by step in an easy to understand manner. Each breakthrough by Sudoku technique explained.

We'll solve the following Sudoku level 3 puzzle. First try to solve the puzzle before going through the solution.

How to solve Sudoku level 3 hard Sudoku Game 8

Strategy of solving a Sudoku hard puzzle

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, even if it doesn't result in an immediate valid cell hit.

In addition we'll also use other advanced Sudoku techniques of single digit lock, parallel digit scan or X wing whenever we ge the chance.

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 eighth 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 8 Stage 1: Breakthrough by possible digit subset analysis or DSA technique

How to solve Sudoku level 3 hard Sudoku Game 8

First valid cell by row column scan for digit 2: R5C9 2 by scan R4, R6, C7 -- R2C8 2 by scan in R1, R3 -- R8C6 2 by scan in R7, C4, C5 -- R9C2 2 by scan in R7, R8, C1, C3. Digit 2 fully filled.

Next success by row column scan is for 6: R4C8 6 by scan in R5, R6 -- R9C1 6 by scan for 6 in R7, C2, C3 -- and lastly R8C9 6 by scan in R7, R9, C7. All 6s are filled.

For next larger digit 7: R5C7 7 by scan in R6.

Cycle (4,5,9) in R6C7, R6C8, R6C9 and Cycle (1,8) by reduction in R6C2, R6C4.

By going back to scanning for lower incomplete digits, success for 3: R7C1 3 by scan C2, C3 -- R9C9 3 by scan in R7, R8 -- R7C9 8 by scan for 8 in R8, C8.

R9C5 8 by DSA reduction of [5,9] from DS [5,8,9] in C5 -- R1C5 5 by DS reduction -- R5C5 9 by exception in C5. R7C3 9 by scan in R9, C2 -- R4C1 9 by scan in R5, C2, C3.

Breakthrough Cycle (1,4) formed in R5C3, R9C3 by DSA. We'll see its effect in next stage.

Game status at this first stage shown below.

How to solve Sudoku level 3 hard Sudoku Game 8 Stage 1

Solution to the Sudoku level 3 game 8 Stage 2: Breakthroughs By DSA and Cycles

Breakthrough R2C3 8 by reduction of 1 because of the Cycle (1,4) in C3 -- R4C3 7 by exception in C3.

R2C6 4 by reduction of [5,9] from DS [4,5,9] in three empty cells of R2 -- R2C1 5 by reduction -- R2C4 9 by exception in R2.

Cycle (1,4,7) in cells R3C2, R7C2 and R8C2 by reduction of [5,8] from DS [1,4,5,7,8] in the column C2 -- R6C2 8 by reduction of 1 because of this Cycle -- R4C2 5 by reduction -- R4C6 8 by reduction -- R6C4 1 by reduction -- R9C4 4 by reduction -- R8C4 7 by reduction -- R7C4 5 by reduction of 1 from DS [1,5] in R7C4, R7C6 -- R7C6 1 by exception in bottom middle major square.

Result of actions taken at this stage shown below.

How to solve Sudoku level 3 hard Sudoku Game 8 Stage 2

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

We'll understand two things:

  • How we have got the Cycle (1,4,7) in cells of C2 by DSA reduction, and,
  • What are the characteristics of a Cycle such as Cycle (1,4,7).

DSA reduction technique is the short form of possible Digit Analysis Technique on empty cells of a zone that may be a row, a column or even a major square. Our objective would always be to get a unique digit in a valid cell, but if we don't get it we'll at least get short length possible digit subsets in the cells where we expect a valid cell.

In this game we have got such possible digit subsets [1,4,7] in all three cells R3C2, R7C2 and R8C2 of column C2 by reduction of [5,8] from DS [1,4,5,7,8] in the column C2. Observe how it happened.

First we have formed the possible digit subset of [1,4,5,7,8] in five empty cells of column C2. These are the digits that are still missing in C2.

At the second phase, we have checked the other two parent zones (other than C2) or each of these three cells to see how many of these five digits [1,4,5,7,8] are present in the two zones combined for each cell.

For R3C2, the two parent zones other than C2 are R3 and top left major square. Observe that combined effect of all digits present in these two zones is to reduce the DS [1,4,5,7,8] by [5,8]. Row R3 doesn't have any of these two, but the top left major square has both. As a result, the possible digit subset for R3C2 reduces to [1,4,7].

This is the mechanism of finding a valid cell or a short length DS by possible digit subset analysis or DSA technique.

Similarly for R7C2 with DS [1,4,5,7,8], [5,8] exists in both the parent zones R7 and bottom left major square. Thus the possible digit subset for R7C2 reduces to [1,4,7].

For R8C2 with possible digit subset [1,4,5,7,8], [5,8] again exists in both R8 and parent bottom left major square, reducing the possible digit subset DS in R8C2 also to [1,4,7].

Characteristics of the Cycle (1,4,7)

Instead a direct valid cell hits of unique digits we have got same DS [1,4,7] in three cells of the same column C2. What does it signify?

We call this digit pattern as the Cycle of (1,4,7) in column C2. Effectively, these three digits can be placed only in these three cells and in no other cell. If one of these cells, say R3C2 gets finally 1 and another cell R7C2 gets finally 7, the third cell in the Cycle must have the the third digit 4 of the Cycle. Similarly for other possibilities in the final solution for these cells.

This as if these three digits Cycle through the three cells. That's why the name Cycle.

Okay, but what is the effect of a Cycle on the other digits in the empty cells of C2?

There we get finally the breakthrough: R6C2 8 by reduction of 1 from DS [1,8] because of this Cycle of [1,4,7].

This is the basic result-bearing effect of a Cycle - the Cycled digits are eliminated from the possible digit subsets of all other empty cells in the zones parent to the whole Cycle, in this case, column C2.

What are the conditions a Cycle must satisfy?

A proper Cycle of possible digit subsets or DSs in empty cells of a zone have the following characteristics:

  • Number of digits in the Cycle must be same as number of cells involved in the Cycle. There may be two, three, four or higher digit long Cycles, two digit Cycles being more common and easier to identify.
  • Each digit in a Cycle must appear in at least two cells involved in a Cycle.
  • A Cycle may be formed in a column, a row or even in the cells of a major square without being in a row or column. In this last case, the Cycle will reduce the digits only in the DSs of the major square of the Cycle.
  • The digits involved in a Cycle cannot appear in any cell outside the Cycle, thus reducing the DSs of empty cells in the Cycled zone. For example, in this case of Cycle (1,4,7), digit 1 is reduced from the DS [1,8] in R6C2 giving a valid cell hit R6C2 8.
  • A proper Cycle cannot contain a second smaller length Cycle. For example, if DSs of R7C2 and R8C2 were just [4,7] these two cells would have formed a smaller length Cycle reducing the DS of R3C2 to a direct valid cell hit of R3C2 1.
  • A valid cell hit by a Cycle formation always would have an alternate way to get the valid cell hit by a Parallel scan. For example in this case we have got the valid cell hit R6C2 8 because of the formation of the Cycle (1,4,7) in C2. Alternately we could have got the same hit of R6C2 8 by parallel scan for 8 on empty cells of C2. 8 in top left major square and in bottom left major square eliminate three of the five empty cells in C2 for 8. Also 8 in R4 eliminates the fourth empty cell R4C2 for 8 leaving the single cell R6C2 for 8. Often we don't form the Cycle and get a valid cell hit just by a quick bit of parallel scan.

Cycles are valuable assets to have for reducing the overall uncertainty in a Sudoku game.

Solution to Sudoku level 3 game 8 Final Stage 3: Easy to find valid cells

With 4 in R9C4, R9C3 1 by reduction. With 5 in R7C4, R5C4 3 by reduction -- R5C6 5 by reduction. With 1 in R9C3, R5C3 4 by reduction -- R5C1 1 by reduction -- R3C2 1 by scan for 1 in C1.

With 7 in R8C4, R8C2 4 -- R7C2 7 -- R8C7 1 by exception -- R7C8 4 by exception -- R6C8 5 by reduction.

Possible digit subset DS [1,3] in R1C8, R3C8 in C8 and with 1 in C3, R3C8 3 by reduction -- R1C8 1 by exception -- R3C6 7 by reduction -- R1C6 3 by reduction.

R3C4 8 by exception in top middle major square.

R1C7 8 by scan R3, C9. R1C9 4 by DSA reduction of [5,9] in R1 from DS [4,5,9] in C9 and hence in R1C9 -- R3C1 4 by scan for 4 in R1 -- R1C1 7 by exception in R1, C1 and top left major square.

With 4 in R1C9, R6C9 9 by reduction -- R6C7 4 by reduction -- R3C9 5 by exception in C9 -- R3C7 9 by exception in whole game.

Final solution shown below.

How to solve Sudoku level 3 hard Sudoku Game 8 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.


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