How to solve very hard Sudoku level 4 puzzle game 12 quickly by advanced Sudoku techniques
How to solve very hard Sudoku puzzle: level 4 game 12 explains how aggressive early breakthroughs by advanced Sudoku techniques results in quick solution.
The very hard Sudoku level 4 puzzle Game 12
Before going through the solution solve the puzzle first.
How to solve very hard Sudoku level 4 puzzle Game 12 Stage 1: Breakthrough by Single digit locks, Deformed X wing, Double digit scan and Cycles of twins and triplets
Strategy adopted: start with row column scan coupled with aggressive breakthroughs by advanced Sudoku techniques.
As a strategy we always start row column scan from digit 1 and continue till digit 9. In the process we may get valid cells, but otherwise our lookout is for identifying for a single digit lock by cross scan.
If we get one, we should get usually a breakthrough in one step or two.
Otherwise we know that the single digit lock might become useful later and we ignore it for the present.
Highlight color of valid cells in this first stage is banana leaf green.
First few valid digits are all by row column scan,
R9C5 2 by scan for 2 R7, R9 on empty cells of bottom middle major square -- R2C9 2 by scan for 2 in R1, C7, C8 on empty cells of top right major square.
R3C2 2 by scan for 2 in R1, R2, C3 on empty cells of top left major square.
Now we'll show a very special breakthrough by a series of single digit locks.
First single digit lock on 3 formed in R5C8, R6C8 in C8 and right middle major square by cross-scan for 3 in R4, C9 on empty cells of right middle major square.
Second single digit lock on 3 formed in R1C7, R2C7 by column scan for 3 in C8 by the lock on 3 behaving as if actually we have a 3 in C8 and 3 in C9.
Both these single digit locks are twin possibilities of the digit occurring in the respective columns thus debarring the digit 3 in any other empty cell in C7 and C8.
A third single digit twin on 3 is formed in R1C2, R2C1 by scan for 3 in C3 on empty cells of top left major square. By itself, this twin possibilities of digit 3 do not debar any cell to have digit 3.
But when combined with the single digit lock in R1C7, R2C7 it forms a special four single digit pattern like a deformed X. We call this as a deformed X wing in which the two pairs of digit 3 share common row R1 and R2 but not the columns that are not common.
Simply speaking, as one of the two 3s in R1C2 and R1C7 must be there in the final solution, all other empty cells are debarred to have digit 3. In the same way, with 3 in R2C1 and R2C7 forming a lock across different major squares but in the same row, all empty cells of R2 also are debarred to have digit 3.
These effectively expose the naked singles R3C4 3 by scan for 3 on empty cells of top middle major square. This is an important breakthrough.
A byproduct is R3C2 8 by exception as the single digit left in row R3.
Turning our attention to down below, R8C5 9 by scan for 9 in C6 on empty cells of bottom middle major square -- possible digit subset in the major square reduces to [3,5,7] -- breakthrough Cycle of twins (3,7) in R8C6, R9C6 by double digit scan for [3,7] in R7 on empty cells of bottom middle major square -- breakthrough valid cell R7C6 5 by reduction of [3,7].
R4C7 4 by scan for 4 in C8, C9 on empty cells of right middle major square -- Reduced DS [4,6,8] in C6 further reduced by [4,8] in R4 -- R4C6 6 -- R6C6 8 by reduction of 4.
in R6 from further reduced DS [4,8] in C6 -- R5C6 4 by exception as the only digit left in C6.
Again carry out an extraordinarily effective double digit scan for [2,9] in C5 and R5 on empty cells of central middle major square to form the Cycle of twins (2,9) in R4C4, R6C4 out of thin air -- R5C4 7 by a simple scan for 7 in C5 on empty cells of central middle major square.
R2C4 5 as exception in C4 -- Cycle (4,8) of twins created in R2C5, R3C5 and Cycle (1,5,7) of triplets created in R4C5, R5C5, R6C5.
Observe that all the breakthroughs have been achieved without doing practically any possible digit subset enumeration of the empty cells.
Results of the actions taken shown.
How to solve very hard Sudoku level 4 puzzle Game 12 Stage 2: Breakthrough by Single digit lock, Double digit scan and Cycles of twins
Highlight color of valid cells at this second stage is light blue.
First valid cell at this stage is an opportunistic R9C8 5 by scan for 5 in R7, C7, C9 on empty cells of bottom right major square Single digit lock on 8 created in R7C7, R8C7 in C7 and bottom right major square by cross-scan for 8 in R9 and C8.
This takes part in breakthrough valid cell R1C9 8 by scan for 8 in C7 by the single digit lock on 8 and 8 in C8.
R8C9 6 by cross-scan for 6 in R7, C7 on empty cells of bottom middle major square -- breakthrough Cycle (3,6) of twins created in R5C8, R6C8 by double digit scan for [3,6] in R4 and C9 -- a second Cycle (7,9) of twins created as a result in R4C9, R6C9 by a second double digit scan for [7,9] in R5 -- breakthrough valid digit R5C9 1 by exception of only digit left in the major square.
With 8 in R1C9, R1C5 4 by reduction -- R2C5 8 by reduction.
Game status shown below.
How to solve very hard Sudoku level 4 puzzle Game 12 Stage 3: Breakthrough by Double digit scan, Single digit lock and Cycle of twins
Highlight color of valid cells at this third stage, which is not the last stage, is light grayish yellow.
A special breakthrough at the start by forming a Cycle (1,7) of twins in R4C2, R6C3 by double digit scan for [1,7] in R5 and C1 on empty cells of left middle major square.
This effectively eliminates a digit 6 in DS of R6C3 and thus creates a single digit lock on 6 in R5C1, R6C1 by scan for 6 in C2 -- breakthrough R9C3 6 by scan for 6 in R8, lock on 6 in C1 and 6 in C2 on empty cells of bottom left major square -- R9C1 3 by DSA reduction of [1,3,7] from possible digit subset [1,3,7,9] in R9 -- R9C6 7 byeduction of 3 -- R8C6 3 by reduction of 7.
R1C2 3 by scan for 3 in C1, C3 on empty cells of top left major square.
On the other side down below, DS [1,7,9] in C8 reduced by 7 in R7 tin R7C8 forms DS [1,9]. It joins DS [1,9] of leftover two digits in R9 to form a Cycle of twins (1,9) in R9C7, R7C8.
This reduces the DS in rest two cells in bottom right major square to [7,8] and with 7 in R7, breakthrough R7C7 8 by reduction -- R8C7 7 by exception in C7.
DS in C7 reduced to [1,3] and with 1 in R2, R2C7 3 by reduction of 1 -- R1C3 5 by scan for 5 in R2 -- DS in C3 reduced to [1,4,7] -- R8C3 4 by reduction of [1,7] from DS [1,4,7] in R8C3 -- R2C1 4 by scan for 4 in C3 -- R2C3 7 by exception as the only digit left in top left major square -- R6C3 1 by exception in C3 -- R6C5 3 by reduction of 1 -- R4C5 1 by reduction -- R4C2 7 by reduction -- R4C9 9 by reduction -- R6C9 7 by reduction.
With 9 in R4C9, R4C4 2 by reduction -- R6C4 9 by reduction.
With 5 in R5C5, R5C2 8 by reduction.
All breakthrough have been taken care of.
How to solve hard Sudoku level 4 puzzle Game 12 Final Stage 4: No more challenges left
Highlight color of valid cells at this fourth and last stage is light cyan.
R1C8 7 by scan for 7 in R2, C7 on empty cells of top right major square.
DS in C8 reduced to [1,9] -- R2C8 9 by reduction of 1 from [1,9] in R2.
R1C7 1 by scan for 1 in R2 on empty cells of top right major square -- R9C9 9 by reduction of 1 -- R2C7 3 by exception in C7.
With 9 in R9C7, R9C2 1, R7C8 1 by reduction of 9 -- R7C2 9 by reduction.
R8C2 5 by exception in C2 -- R8C1 8 by exception in R8.
With 3 in R6C5, R6C8 6 by reduction -- R5C8 3 by reduction -- R5C1 6 by exception in R5 -- R6C1 2 by exception in R6 -- R4C1 5 by exception in whole game.
Final solution below.
Highlight background colors of valid cells are used for keeping track of stages in which the valid cells are obtained.
Early breakthroughs by advanced Sudoku techniques of Deformed X wing, Parallel digit scan, Single digit lock, Double digit scan, Cycle of twins and triplets ensured quick solution to this 12th very hard Sudoku level 4 puzzle game.
It is satisfying that there has been no need to unnecessarily enumerate the possible digit subsets in many empty cells and this saved a lot of time.
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