Step by step easy to understand solution to Sudoku level 3 game 7
Solution to the Sudoku level 3 game 7 is explained step by step in an easy to understand manner. Each breakthrough by Sudoku techniques explained.
We'll solve the following Sudoku level 3 puzzle this time. But, first try to solve the puzzle before going through the solution.
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.
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 seventh 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 7 Stage 1: Breakthrough by single digit lock
For digit 1 there is no valid cell by row column scan, but for digit 2 we get success: R2C2 by scan for 2 in R1, R3. That's the only one for 2 at this point of time.
Digit 3 has three filled cells but no success.
Success with 4: R3C5 4 by scan for 4 in R1, R2, C4.
No more direct success by row column scan for 4, but we do have a more important success of the single digit lock on 4 in R5C8, R5C9 by cross-scan for 4 in R6 and C7.
This single digit lock on 4 acting as if 4 is already there in R5 then contributes for the next valid cell by row column scan for 4: R4C6 4 by scan in R5, R6, C4, C5 -- R9C1 4 by scan for 4 in R7, C2, C3 -- R8C8 4 by scan for 4 in R7, R9.
Next there is an interesting valid cell hit for 4 essentially by scan but we'll explain it in next stage.
For now continuing with increasing digit row column scan, we'll scan for 5: R1C3 5 by scan for 5 in R3, C2 -- R4C1 5 by scan in R5, C3 -- R9C8 5 by scan for 5 in R7, C7.
Game status at this first stage shown below.
A note on breakthrough by Single digit lock
By CROSS-SCAN for 4 in R6 and C7, a single digit lock on 4 is formed in R5C8, R5C9. Digit 4 appears only in the possible digit subsets of these two cells in the right middle major square and row R5.
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 R5.
Effectively this is as if digit 4 already exists in R4 and so it participates in row column scan for 4 carried out on the empty cells of central middle major square.
Result is the breakthrough R4C6 4 by scan for 4 in R5 by single digit lock, R6, C4 and C5.
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.
An effective single digit lock always will produce a breakthrough.
Solution to the Sudoku level 3 game 7 Stage 2: Breakthroughs by multiple single digit lock
The function of the single digit lock on 4 in R5C8, R5C9 is over. Now it be broken up by scan for 4 in C8 resulting in R5C9 4. Digit 4 fully filled.
Now we'll see how effective single digit lock scan can be!
Single digit lock on 8 by CROSS-SCAN FOR 8 IN R9 AND C8 -- R8C3 8 by scanning back for 8 in R7 by the lock on 8 and 8 in R9 as well as 8 in C2.
Always look for and use a single digit lock by a cross-scan. It's a never-failing valuable resource.
We'll again get a valid cell by another single lock on 2 this time by single column scan for 2 in C2 -- 2 possible only in R7C1, R7C2 -- R9C7 2 by scan by lock on 2 in R7.
Let's now get an easy valid cell by possible digit analysis DSA in R7C8 6.
The possible digit subset DS in four empty cells in C8 is [3,6,7,9] and out of these four, three digits [3,7,9] are eliminated by their presence in bottom right major square -- R4C7 6 by scan for 6 in R6, C8.
Again by DSA reduction of [1,6,8] from DS [1,6,8,9] in C9: R6C9 9 -- R5C8 3 by reduction -- R6C7 7 by exception in right middle major square.
Result of actions taken at this stage shown below.
Solution to the Sudoku level 3 game 7 Stage 3: Breakthroughs by DSA reduction
R2C7 8 by DSA reduction of [6,7,9] in C7 from possible digit subset [6,7,8,9] in four empty cells of R2 -- R7C9 8 by reduction of 8 of the single digit lock partner in R7C7 -- R7C7 1 by exception in bottom right major square -- R3C7 3 by exception in C7.
Cycle (3,9) formed in R3C3, R4C3 by DSA reduction -- Cycle (1,2) in R6C3, R7C3.
With 3 in R3C7, R3C3 9 by reduction -- R4C3 3 by reduction.
R3C1 7 by DSA reduction of [3,6] in C1 from possible digit subset DS [3,6,7] in top left major square -- R3C2 6 by DS elimination of 3 in R3 from [3,6] in top left major square -- R1C2 3 by exception in top left major square -- R3C9 1 by reduction -- R3C4 8 by exception in R3 -- R1C9 6 by reduction.
Game status shown.
Finding a valid cell by Sudoku technique of possible digit subset analysis or DSA
We'll understand how we get R2C7 8 by DSA reduction. of [6,7,9] in C7 from possible digit subset [6,7,8,9] in four empty cells of R2
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 case, our target cell is R2C7 in the target row R2. In row R2 the possible digit subset for the four empty cells is the DS [6,7,8,9]. This must also be the DS in R2C7 to start with.
This is the starting point of DSA analysis.
In the second stage the unique set of existing filled digit subset in the two other parent zones of the cell R2C7 is formed.
In the intersecting column C7 itself the DS [6,7,9] of existing digits identified that reduces the DS in R2C7 to the unique single digit 8.
This is how a valid cell by DSA is achieved.
Observe that this process starts with a target zone and a target cell that where we expect to get a positive result. At the next stage only we analyze the effect of the existing digits in the other two parent zones of the target cells on its possible digit subset DS.
Solution to Sudoku level 3 game 7 Final Stage 4: Easy to find valid cells
With 1 in R7C7, R7C3 2 by reduction -- R7C1 9 by reduction -- R5C1 2 by reduction -- R6C3 1 by reduction -- R5C2 9 by reduction and exception in left middle major square.
R9C2 7 by scan for 7 in R8 -- R8C2 1 by exception in C2.
R5C5 8 by scan for 8 in R4, R6, C4, C6.
R6C4 5 by DSA reduction of [1,6,7,9] from possible digit subset DS [1,5,6,7,9] in C4 -- in the same way R8C4 6 by DSA reduction of [1,7,9] from possible digit subset [1,6,7,9] in C4 -- R5C4 1 by DSA reduction of [7,9] from [1,7,9] in C4 -- R9C4 9 by DSA reduction of 7 from [7,9] in C4 -- R4C4 7 by exception in C4 -- R4C5 9 by exception in R4 -- R5C6 6 by exception in R5 -- R9C5 1 by exception in R9.
R8C5 5 by exception in R8.
R6C5 2 by scan in C6 -- R6C6 3 by exception in R6 -- R7C6 7 by reduction -- R75 3 by reduction.
R1C6 1 by scan for 1 in R2, C4, C5 -- R2C6 9 by exception in C6 -- R2C8 7 by reduction -- R1C8 9 by exception and reduction -- R1C5 7 by exception in R1 -- R2C5 6 by exception in the 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.
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