Step by step easy to understand solution to Sudoku level 3 game 6
Solution to the Sudoku level 3 game 6 is explained step by step in an easy to understand manner. Each breakthrough by Sudoku techniques explained.
The following Sudoku level 3 puzzle we'll solve this time. First try your hand in solving the puzzle before going through the solution.
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 sixth 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 6 Stage 1: Breakthroughs by intensive use of DSA and Cycles
First valid cell by row column scan is for digit 5: R1C2 5 by scan for 5 in R3, C1, C3 -- R2C8 5 by scan in R1, R3, C7 -- R8C9 5 by scan in R7, C7, C8 -- and lastly, R6C5 5 by scan in R4, R5, C6. Digit 5 fully filled.
Scanning next for digit 6: R9C1 6 by scan for 6 in R7, C2, C3 -- R5C8 6 by scan for 6 in R4, C7 -- R6C6 6 by scan in R4, R5 -- R1C4 6 by scan for 6 in R2, R3, C6 -- and lastly R8C5 6 by scan for 6 in C4, C6. Digit 6 is fully filled.
Opportunistic valid cell hit R6C1 4 by DSA reduction of [1,8] in C1 from possible digit subset DS [1,4,8] in R6 and hence in R6C1.
We'll now showcase how a series of Cycles finally gives a valid cell hit by DSA.
As no easy valid cell hit is available by row column scan or possible digit subset analysis DSA, short length possible digit subsets of length 2 and 3 are filled up in convenient more favorable zones with more cells already with filled digits.
In this process, a Cycle of (1,3,8) is first created in cells R4C3, R4C8 and R4C9 of R4.
Note on the Cycle (1,3,8)
All the digits 1,3 and 8 appear in DSs of only these three cells and in DSs of no other cell of row R4.
Effectively, the three digits are possible in a Cycle restricted between these three cells only.
Effect of a Cycle is: Reduction of all three digits in the Cycle from all cells of parent zones. Usually a Cycle appears in a row or a column, but it can also be a part of a zone.
To form this Cycle of three digits covering three cells following conditions must be true,
- There must not be any fourth digit in any of the three cells in the Cycle, and,
- Each of the three digits must appear at least twice among the three cell possible digit subsets.
We find that this is true for the DSs: [1,8] in R4C3, [1,3,8] in R4C8 and [1,3] in R4C9. That's how the Cycle (1,3,8) is formed spanning the three cells in R4.
Because of this Cycle, remaining two digits [4,7] in the row formed a second Cycle of (4,7) in R4C4, R4C5.
As this second Cycle also belonged to the central middle major square, it reduced the possible digits for the three cells R5C4, R5C5 and R5C6 to [1,2,8].
Now observe that C5 has already the digits [1,2]. This reduces the DS from [1,2,8] to single digit 8 for R5C5: R5C5 8 -- A third Cycle (1,2) formed in R5C4 and R5C6 and this produces another new valid cell hit: R5C2 3 by reduction of 1 because of Cycle (1,2).
Opportunistic scan for 2: R9C8 2 by scan for 2 in R7, C7 -- R8C2 2 by scan for 2 in R7, R9, C3.
Game status at this first stage shown below.
Solution to the Sudoku level 3 game 6 Stage 2: Breakthrough by single digit lock
By CROSS-SCAN for 1 in R3, C1, a single digit lock on 1 is formed in R1C3, R2C3.
Digit 1 is locked in these two cells in column C3 and that reduces 1 from DS [1,8] in R4C3 to produce the breakthrough: R4C3 8 -- R6C2 1 by reduction -- R6C7 8 by reduction.
R3C8 8 by scan for 8 in R1, C7, C9.
An easy scan hit for 9: R3C7 9 by scan for in R1, R2, C9 -- R7C8 9 by scan for 9 in C7, C9 -- R8C3 9 by scan for 9 in R7, C1, C2 -- R9C4 9 by scan in R7, R8, C6. Digit 9 is fully filled.
R7C2 8 by scan for 8 in R9, C1, C3.
DS in bottom middle major square [1,4,7,8] but [1,4] is in R8. This reduces the DS for cells R8C4, R8C6 to [7,8] forming a Cycle and R8C7 3 by reduction.
opportunistic scan for 2: R1C9 2 by scan for 2 in C7, C8.
Cycle (7,8) creates a second Cycle (1,4) in C6 and it creates a valid cell hit -- R5C6 2 by reduction -- R5C4 1 by reduction.
Breakthrough Cycle (1,3) formed in R1.
Result of actions taken at this stage shown below.
A note on breakthrough by Single digit lock
By CROSS-SCAN for 1 in R3, C1, a single digit lock on 1 is formed in R1C3, R2C3. Digit 1 appears only in the possible digit subsets of these two cells in the major square and column C3.
In the final solution, one of these two must be present ensuring that digit 1 is prohibited in all other possible digit subsets of empty cells in C3.
Effectively this is as if digit 1 already exists in C3.
This is an effective single digit lock.
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.
By this single digit lock possible digit subset DS [1,8] is reduced to just a valid cell hit of R4C3 8 by reduction of digit 1.
An effective signle digit lock always will produce a breakthrough.
Solution to Sudoku level 3 game 6 Final Stage 3: No more challenges left
R1C6 7 by reduction -- R8C6 8 by reduction -- R8C4 7 by reduction -- R4C4 4 by reduction -- R4C5 7 by reduction -- R2C5 4 by reduction.
R3C2 4 by scan for 4 in R1, R2, C1 -- R9C2 7 by reduction -- R9C8 1 by reduction -- R9C6 4 by reduction -- R7C6 1 by reduction -- R7C1 3 by reduction -- R7C3 4 by reduction -- R7C9 7 by reduction -- R3C9 3 by reduction -- R4C9 1 by reduction -- R4C8 3 by reduction -- R1C8 1 by reduction -- R1C3 3 by reduction -- R2C3 1 by reduction -- R2C7 7 by reduction -- R2C6 3 by exception in C6.
R3C1 7 by scan for 7 in R2 -- R2C1 2 by exception in C1 -- R2C4 8 by exception in R2 -- R3C4 2 by exception in whole game.
Final solution 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.
Enjoy.
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