How to solve quickly the Sudoku hard puzzle level 4 game 16
Difficult breakthroughs as well as finding each valid cell when solving the Sudoku hard puzzle level 4 game 16 both are explained clearly in four stages.
Solution of this 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 16
Before going through the solution solve the puzzle first.
Solving Sudoku hard puzzles: Level 4 Game 16 Stage 1: Important breakthrough by Parallel digit scan or Cycle of quadruplets
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.
R4C4 1 by two column scan for 1 in C4, C5 on empty cells of central middle major square.
R4C7 2 by scan for 2 in R5, R6 on empty cells of right middle major square -- R1C8 5 by scan for 5 in R2, R3, C9 on empty cells of top right major square -- R9C4 5 by scan for 5 in R7, C4, on empty cells of bottom middle major square. And R2C2 9 by scan in R1, C1, C3 for 9 in empty cells of top left major square.
R4C5 9 by DSA reduction of existing digit subset [3,6,8] in C5 from possible digit subset [3,6,8,9] in R4. You may get the valid cell also by simple scan for 9 in R6 on empty cells of central middle square -- R2C4 9 by scan for 9 in R1, R3, C5.
R6C6 3 by scan for 3 in C5 -- R6C5 4 by exception in central middle major square.
An opportunistic valid cell in R1C6 4 by scan for 4 in R3, C2, C3 on empty cells of top middle major square.
A difficult to detect breakthrough by parallel scan for 2 in empty cells of R9 resulting in R9C8 2 by elimination of cells R9C1, R9C2, R9C6 and R9C7 by the presence of 2 in C1, C2, C6 and C7.
Conventionally, you might have had the same breakthrough by laboriously working out the inevitable accompanied Cycle of quadruplets (4,6,7,8) in these four eliminated cells formed by the four leftover digits in the row.
To form the Cycle you would have to form the individual possible digit subsets in the cells first and that would have taken time. That's why, if you can identify a parallel scan opportunity, it would be always be the faster route to a breakthrough.
Question is: how do you detect such an opportunity for a parallel digit scan?
- First, you must be aware of the potential and method forming a valid cell by parallel scan,
- Second, you should gather experience of actually achieving a breakthrough by a parallel scan,
- Third, while enumerating short DSs of a few cells in a row or column, you may foresee a parallel scan,
- And last, for difficult to detect parallel scan opportunity, look for digits that appear in more cells as unique digit.
Results of the actions taken shown.
Solving Sudoku Hard Puzzles: Level 4 Game 16 Stage 2: Critical breakthrough by Parallel digit scan for multiple digits, Cycles of triplets
Highlight color of valid cells at this second stage is light blue.
A surprise triple digit subset of [1,3,5] identified in bottom right major square, and also in C5 and C6 to leave exactly the three cells R8C1, R8C2 and R8C3 in R8 for these three digits. Four cells disallowed for [1,3,5] in R8 are shown red.
This is a breakthrough Cycle of triplets (1,3,5) formed in the three cells and primarily the process is parallel scan for multiple digits on empty cells of R8.
As a result we get the breakthrough valid digit by simple scan: R7C3 2 by scan for 2 in C1, C2 -- R7C4 6 by reduction -- R1C4 3 by reduction -- R3C4 2 by reduction -- R2C5 7 by reduction -- R3C6 6 by reduction. We'll not put the R8C5 2 by exception in C5 for ease of explanation at this stage. That'll be the first thing next stage.
Appreciate how powerful the breakthrough of parallel scan for three digits on empty cells of a row proved to be.
Again a Cycle of triplets (1,3,7) formed in R3C7, R3C8, R3C9 in top right major square, but this time be default three leftover digits in the row.
Would it be useful? Yes, of course. It is already useful in reducing the possible digit subsets of the other four empty cells in top right major square.
Game status shown below.
Solving Sudoku Hard Puzzles: Level 4 Game 16 Stage 3: Breakthrough by DSA reduction and Cycles of twins and triplets
Highlight color of valid cells at this third stage, which is not the last stage, is light grayish yellow.
First thing to do is to put R8C5 2 by exception in C5 that was deferred in the previous stage.
The Cycle of triplets (1,3,7) in top right major square reduces [1,7] from DS [1,6,7]: R1C9 6 -- R2C9 2 by scan for 2 in C7, C8.
This Sudoku hard is all about formation and use of breakthrough Cycles of twins and triplets.
At this third stage the first breakthrough twins of (3,6) is formed in R2C2 and R4C2.
This Cycle of twins reduces [3,6] in the other empty cells of C1 forming a second Cycle of twins (4,7) in R7C1, R9C1 by further reduction of [5] in R7 and R9 from possible digit subset of [4,5,7] in C1.
Finally the last reduction results in the breakthrough valid cell in R7C2 8 by reduction of 6 from reduced DS [6,8] followed by R9C2 6 by exception in bottom left major square.
Newly created Cycle of twins (6,8) in C2 results in further valid cells in R4C2 3 by reduction of [6,8] -- R8C2 1 by reduction -- R8C1 5 by reduction -- R8C3 3 by reduction.
R2C3 6 by reduction -- R4C3 8 by reduction.
Reduction of R2C1 and R4C1 are preserved at this stage for ease of understanding how the previous breakthroughs were achieved.
Results shown.
Solving Sudoku Hard Puzzles: Level 4 Game 16 Final Stage 4: Rest are easy scans, reductions or exceptions
Highlight color of valid cells at this fourth and last stage is light cyan.
With 3 in R4C2, R4C1 6 -- R2C1 3 -- R5C1 1 by exception in C1 -- R6C3 5 by reduction -- R5C3 7 by reduction -- R5C2 4 by reduction -- R1C2 7 by exception in C2 -- R1C3 1 by exception in C3.
R6C7 6 by scan for 6 in R5, C8, C9.
R5C7 5 by scan for 5 in C7, C8.
R5C8 3 by scan for 3 in R6, C9 -- R5C9 9 by exception.
With 3 in R5C8, R3C8 1 by reduction -- R3C9 7 by reduction -- R3C7 3 by exception in R3.
R6C9 1 by scan for 1 in C8 -- R6C8 8 by exception in R6 -- R2C8 4 by reduction -- R2C7 8 by exception in R2.
R7C8 9 by exception in C8 -- R7C6 7 by reduction -- R7C1 4 by reduction -- R9C1 7 by reduction.
With 9 in R9C3, R9C6 8 by reduction -- R8C6 9 by reduction.
R8C7 7 by scan for 7 in R9, C9 -- R8C9 8 by exception in R8 -- R9C7 4 by exception in whole game.
Final solution below.
Early breakthroughs by advanced Sudoku techniques of Parallel digit scan for multiple digits and single digit, Cycles of twins, triplets and quadruplets ensured quick solution to this 16th Sudoku hard puzzle at level 4.
This is mainly a game of forming and using Cycles of twins, triplets and quadruplets that was not easy.
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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