Step by step easy and quick solution to Medium Hard Sudoku level 3 game 14
Medium hard Sudoku level 3 game 14 solved using simple but powerful Sudoku techniques step by step so that you understand clearly how to solve it yourself.
We'll now solve the medium Sudoku level 3 puzzle. First try to solve it before going through the solution.
You may skip the next section if you are already aware of the strategy. To skip click here.
Strategy of solving a medium hard Sudoku 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 get the chance.
Evaluation of possible digit subsets for all empty cells is often prescribed to do first. We strongly recommend to avoid this time-consuming activity as far as possible and instead go on filling up the empty cells by unique valid digits using appropriate Sudoku techniques.
The Sudoku techniques for solving this 14th medium hard Sudoku level 3 puzzle game are briefly but clearly explained along with the step by step solution. This medium hard Sudoku needed use of advanced technique of single digit lock.
Let's go through the solution of the game.
Solution to Medium Hard Sudoku level 3 game 14 Stage 1: Breakthrough by Cycles and Single digit lock
Valid digits by single digit scans: R9C9 1 scan 1 in R7, R8, C7, C8 -- R6C1 1 scan 1 in R4, R5 -- R1C4 1 scan 1 in R2, R3, C5 -- R9C1 3 scan 3 in R7, R8C3 -- R2C4 4 scan 4 in R3, C5 -- R9C5 8 scan 8 in C6.
R1C5 9 scan 9 in C4 -- R3C4 7 leftover.
R1C1 5 scan 5 in C2, C3, R2.
Breakthrough by Single digit lock on 2 in R1C9, R3C9 by scan 2 in R2, C7 -- R8C8 2 scan 2 in R9, C7, and lock on 2 in C9.
(You may skip) Note on Single digit lock: A single digit lock can form in exactly two cells, in this case, it occurs in R1C9, R3C9. Such a lock behaves as if digit 2 is already present in the zone it appears. If in a row or column, it can take part in scans (as it does here). In any case it removes the digit locked from possible digits in any other cell in the zones it affects. For example, because of this lock on 2, digit 2 cannot appear in any other cell in the parent major square as well as the parent column C9 (or the parent row, as the case may be). Such an effective single digit lock is created by a single row (or column) scan or generally in a cross-scan by a row and and an intersecting column. This is a valuable digit pattern to have, as it may pair up with a second lock on same digit to form advanced breakthrough digit pattern of X wing.
R4C6 3 scan 3 in R5, R6, C4, C5. Cycle (5,8) in R4C4, R6C4 in C4 as two leftover digits.
Cycle (6,7) in R4C5, R6C5 in C5 as two leftover digits -- R6C6 4 as leftover.
(You may skip) Note on Cycle (6,7): The two digits 6 and 7 can occupy and so block the two cells R4C5, R6C5. The pair of digits cycle between these two cells, because, as of now, we are not sure in which cell either will apear in the solution. Only thing we know that in C5 the two digits cannor appear in any other cell except these two. That is the advantage of Cycles: a Cycle reduces the length of possible digits in the zone it affects. That way, a Cycle reduces the overall uncertainty in the game.
Single digit lock on 2 in R1C3, R3C3 scan 2 in C2, R2 -- R7C1 2 scan for 2 in R8, R9, and lock on 2 in C3.
That's all for this stage. Results shown.
Solution to the Sudoku level 3 game 14 Stage 2: Breakthrough by DSA technique
R1C7 6 by reduction of [2,4,7] from possible digits [2,4,6,7] -- R1C2 4 by reduction -- R1C9 7 scan 7 in R2, R3 -- R1C3 2 leftover in R1 -- R3C9 2 scan 2 in R1, R2, C7 -- R3C7 5 scan 5 in R1, R2 -- R3C3 9 leftover in R3.
R2C2 3 scan 3 in C1 -- R2C1 6 leftover in top left major square.
R2C9 9 scan 9 in C8 -- R2C8 8 leftover in R2.
R9C7 4 by reduction of [8,9] from possible digits [4,8,9] in C7 -- R7C7 8 reduction -- R7C3 4 reduction -- R6C7 9 leftover in C7.
DSA technique: Reduction of one or more than one digit from the cell(s) of interest is the chief mechanism for solving a Sudoku game. It can be done by a row column scan, by an effective single digit lock or an X wing. But when none of these are available, you need to get down to analyze which of the 9 digits light up (or look on, or affect) the cell. The rest form the possible digits for the cell by DSAnalysis. Rest assured, possible digit subset analysis or DSA is the bread and butter activity that you always go on doing when playing Sudoku.
Result of actions taken at this stage shown below.
Solution to Hard Sudoku level 3 game 14 Final Stage 3: It's easy hereon
R9C3 6 by reduction of [5,9] from possible digits [5,6,9] in R9 -- R9C8 5, R8C9 6, R9C6 9, R8C6 5 by reduction.
R4C3 8 leftover -- R4C4 5, R6C4 8 reduction. R6C9 5 scan 5 in R4, R5, C8.
R5C9 8 scan 8 in R4, C8 -- R5C2 6 reduction -- R6C8 6 scan 6 in R5, C9 -- R6C5 7, R4C5 6 by reduction.
R5C8 7 leftover in C8 -- R5C1 4 reduction -- R4C9 4 leftover in C9.
R4C1 7 scan 7 in C2 -- R4C2 9 leftover in R4 -- R8C2 leftover in C2 -- R8C1 9 leftover among the 81 cell game.
Final solution shown below.
Check for the validity of the solution if you need.
End note
Degree of difficulty at this third level of Sudoku medium hard puzzles is less than the proper Sudoku hard puzzles at level 4 mainly because of fairly large number of 31 filled cells in this game.
A mainstream hard Sudoku puzzle at level 4 will always have number of filled cells not more than 25 or at most 26. Even a difference of 1 filled cell may make a lot of difference in hardness of Sudoku puzzles when number of filled cells is at this low range.
This particular game you may have found easy to solve. Even then, it gives good practice.
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Enjoy playing Sudoku.