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Solving Sudoku Hard Puzzles: Level 4 Game 15

Solving Sudoku Hard Puzzles: Level 4 Game 15

How to solve Sudoku hard puzzle level 4 game 15 quickly

Solving Sudoku Hard Puzzles: Level 4 Game 15 quick solution by early breakthroughs using advanced Sudoku techniques. All steps explained.

Solution of this fairly difficult Sudoku puzzle will give you a good idea on the techniques and strategy of solving a difficult Sudoku puzzle using your basic reasoning skills and pattern identification abilities created by experience of solving Sudoku hard puzzles.

The Sudoku hard puzzle level 4 Game 15

Sudoku hard level 4 puzzle game 15

Before going through the solution solve the puzzle first.

Solving Sudoku hard puzzles: Level 4 Game 15 Stage 1: Breakthrough by Double digit scan, Parallel digit scan, Cycle of triplets and possible digit subset analysis or DSA

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 a single digit lock by cross scan.

If we get one, we should get usually a breakthrough in one step or two.

In addition, at this early stage while carrying on row column scan, we remain alert for the possibility of a double digit scan also.

Highlight color of valid cells in this first stage is banana leaf green.

First valid cell by row column scan starting from digit 1 is for digit 4: R8C5 4 by scan for 4 in R7, R9 -- R1C4 4 by scan in R3, C5, C6.

Next valid cell by row column scan is for digit 5: R8C2 5 by scan for 5 in C1, C3. And that's all by row column scan at this point of the game.

Next valid cell R7C4 9 by possible digit DSA analysis by reducing existing digit subset [1,2,3,6] from possible digit subset [1,2,3,6,9] in empty cells of R7 and hence also in R7C4.

Turning focus on empty cells of bottom right major square, possible digit subset in R8C9, R9C9 reduced to [6,9] by double digit scan of [6.9] in C8.

This forms a Cycle of triplets (1,2,3) in R7C8, R8C8, R9C8 and a valid cell R8C9 9 by reduction of 6 from [6,9] by 6 in R8 -- R9C9 6 by exception in bottom right major square.

This gives valid cell R8C3 7 by existing digit subset reduction of [1,3] in C3 from possible digit subset of [1,3,7] in R8 -- R8C1 3 by reduction -- we won't reduce now R8C8 to 1 to preserve the Cycle of triplets.

Reduced possible digit subset in C8 [4,5,7] further reduced by [4.7] to get the valid cell R4C8 5.

By this new incidence of digit 5, three of the four empty cells in C4, R3C4, R4C4, R5C4 are debarred for 5 leaving the single cell R9C4 5. This breakthrough is by parallel scan for 5 on empty cells of C4.

Results of the actions taken shown.

Sudoku hard level 4 puzzle game 15 stage 1

Solving Sudoku Hard Puzzles: Level 4 Game 15 Stage 2: Breakthrough by Cycles of twins, Double digit scan and possible digit subset analysis or DSA

Highlight color of valid cells at this second stage is light blue.

First job is to finish the pending task: with 3 in R8C1, R8C8 1.

Possible digit subset in R7 [1,2,3,6] reduced by [2,6] in bottom middle major square to form the breakthrough Cycle of twins (1,3) in two cells R7C5, R7C6 -- R7C8 2 by reduction of 3 by the Cycle of twins -- R9C8 3 by reduction of 2 and R7C3 6 again by reduction of 2.

Cycle (7,8) formed in R9C5, R9C6 by leftover two digits in bottom middle major square.

Same way Cycle (2,9) formed in bottom left major square.

Both these Cycles of twins would have no effects of reduction on other cells any more.

Nevertheless the twin [2,9] in R9C3 joins with the possible digit subset DS of [2,9] in R6C3 for a new Cycle of twins (2,9) in these two cells and breakthrough valid cell R4C3 8 in C3 by exclusion in C3.

This is followed by R4C4 2 by DS reduction of [7,8] from [2,7,8] in C4 -- R4C9 1 by reduction of [3,6,9] from DS [1,3,6,9] in R4.

Still working on C9 form the breakthrough Cycle of twins (2,4) in R5C9, R6C9 by existing digit reduction of [5,8] from possible digit subset [2,4,5,8] in C9 -- R5C8 4 by reduction of 4 -- R2C8 4 by exception in C8 and reduction of 7.

A new Cycle of twins (3,9) formed in leftover two cells R4C7, R6C7 of right middle major square.

Valid cells R5C4 7 by reduction of 7 in R5 -- R3C4 8 by reduction of 7.

A conventional valuable valid cell in R5C1 1 by row column scan for 1 in R4, R6, C2, C3 -- A new Cycle of twins (3,4) formed in the neighborhood in R5C2, R6C2 by double digit scan for [3,4] in C1 as well as in C3 on the empty cells of left middle major square.

By this twins, the rest three digits [2,6,9] for the rest three cells in C2 get reduced to [2,6] in both R2C2, R3C2 by 9 in R2, R3 resulting in R1C2 9 by exception.

Game status shown below.

Sudoku hard level 4 puzzle game 15 stage 2

Solving Sudoku Hard Puzzles: Level 4 Game 15 Stage 3: Breakthrough by DSA reduction

Highlight color of valid cells at this third stage, which is not the last stage, is light grayish yellow.

A late breakthrough by DS reduction of [2,4,9] from [2,3,4,9] for R5C6 3 -- R5C2 4 by reduction of 3 -- R6C2 3 by reduction of 4 -- R6C7 9 by reduction of 3 -- R4C7 3 by reduction of 9 -- R6C3 2 by reduction of 9 -- R9C3 4 by reduction of 2, R6C9 4 by reduction of 2 -- R5C9 2 by reduction of 4 -- R5C5 9 by exception in R5.

With 9 in R9C3, R9C1 2 by reduction of 9.

Results shown.

Solving Sudoku Hard Puzzles: Level 4 Game 15 stage 3

Solving Sudoku Hard Puzzles: Level 4 Game 15 Final Stage 4: Breakthrough by parallel digit scan breaks the game wide open

Highlight color of valid cells at this fourth and last stage is light cyan.

With 9 in R6C7, R6C1 6 by reduction -- R4C1 9 by reduction -- R4C5 6 by exception.

With 3 in R5C6, R7C6 1 by reduction -- R7C5 3 by reduction.

At this late stage we'll apply a parallel scan for 6 on empty cells of C6 to get a breakthrough R3C6 6 by eliminating R1C6 for 6 by 6 in R1, R6C6 by 6 in R6 and R9C6 by 6 in R9.

This breaks open the game completely.

With 6 in R3C6, R3C2 2 by reduction of 6 -- R2C2 6, R3C7 1 by reduction of 2 -- R3C5 8 by exception in R3 -- R9C5 7 by reduction -- R9C6 8 by reduction -- R6C6 7 by reduction.

R1C5 1 by scan for 1 in R2 -- R2C5 2 by exception in C5 -- R2C7 7 by reduction -- R2C1 8 by reduction -- R1C1 7 by exception in C1, R2C9 5 by reduction -- R1C9 8 by exception in C9 -- R1C7 2 by exception in whole game.

Final solution below.

Solving Sudoku Hard Puzzles: Level 4 Game 15 final stage 4

Early breakthroughs by advanced Sudoku techniques of Parallel digit scan, Double digit scan, Cycles of twins and triplets ensured quick solution to this 15th very hard Sudoku hard puzzle at level 4.

There has been no need to unnecessarily enumerate the possible digit subsets in many empty cells. This speeds up the solution.


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