How to solve quickly the Sudoku hard puzzle level 4 game 17
Important breakthroughs by advanced Sudoku techniques and valid cell details in solving the Sudoku hard puzzle level 4 game 17 are explained in four stages.
Solution of this difficult Sudoku puzzle will give you a good idea on the techniques and strategy for solving a difficult Sudoku puzzle using some of the advanced Sudoku techniques.
The Sudoku hard puzzle level 4 Game 17
Before going through the solution solve the puzzle first.
Solving Sudoku hard puzzles: Level 4 Game 17 Stage 1: Breakthrough by DSA reduction and Cycle of twins
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 few valid cells are by direct row column scan.
R5C4 5 by scan for 5 in C5, C6.
R9C5 8 by scan for 8 in R7, C4, C6.
With only two incidences of 9 we get a valid digit for 9 by row column scan: R6C5 9 by scan in R4, C6 followed by -- R2C4 9 by scan in C5, C6.
This reduces the possible digit subset DS in C4 to [2,4,7] -- with existing digits [4,7] in R8, valid cell breakthrough R8C4 2 by reduction of [4,7] from DS [2,4,7]. This creates Cycle of twins (4,7) in R7C4, R9C4 in C8 as well as in bottom middle major square.
This Cycle of twins (4,7) reduces the possible digit subset in the major square to [3,6] and with  in R7, R7C5 3 by reduction -- R8C6 6 by exception in bottom middle major square.
Next R3C5 6 in favorable zone of C5 and top middle major square by scan for 6 in R1, C6 -- R1C5 2 by scan for 2 in C6 on the empty cells of top middle major square -- R4C5 4 by exception in C5 -- R5C6 3 by exception in central middle major square.
We have not colored the Cycle (1,4,7) in top middle square yellow as this won't have any effect on any other breakthrough directly by reduction.
Results of the actions taken shown.
Solving Sudoku Hard Puzzles: Level 4 Game 17 Stage 2: Important breakthrough by Triple digit scan, Double digit scan and Cycle of triplets
Highlight color of valid cells at this second stage is light blue.
First positive result by good old row column scan in R2C3 6 by scan in R1, R3, R7.
First breakthrough at this stage is by force-forming the Cycle (3,9) of twins in R8C1, R9C1 by double digit scan for [3,9] in R7 and C2.
This Cycle of twins (3,9) reduces 3 (as it is locked in the Cycle and cannot appear in any cell outside the Cycle in C1) from possible digit subset [3,7] in R6C1 to produce the valid cell R6C1 7.
A surprising breakthrough Cycle of triplets (1,6,9) formed in right middle cells R5C7, R5C8, R5C9 by triple digit scan for [1,6,9] in row R4 eliminating the cells R4C7, R4C8, R4C9 for the three digits and [1,9] in R6 and  in C7 eliminating the cell R6C7 for the three digits.
Thus only the three cells R5C7, R5C8 and R5C9 are left for the three digits [1,6,9]. This Cycle of triplets gives us the next valid cell breakthrough in R6C2 6 by scan for 6 in R6 by the Cycle (1,6,9) debarring of 6, and 6 in C3.
Cycle of triplets (2,4,8) formed by the leftover three digits in R5C1, R5C2, R5C3 in R5.
This helps to form another Cycle of (3,5) in R4C3, R6C3 by leftover two digits [3,5] in C3 and left middle major square.
Observe the creation of the last breakthrough by parallel digit scan for 2 on empty cells of C1 at R6C1 2 by elimination of R1C1 and R1C2 for 2 by 2 in R1 and R2. This leaves the single cell R6C1 for 2. We won't show this valid cell at this stage as it would disturb the Cycle (2,4,8) now.
We'll stop here at this stage for ease of understanding of the earlier breakthroughs because the next breakthrough will result in release of a series of valid cells.
Game status shown below.
Solving Sudoku Hard Puzzles: Level 4 Game 17 Stage 3: Critical breakthrough by hard parallel digit scan, Cycle of twins
Highlight color of valid cells at this third stage, which is not the last stage, is light grayish yellow.
As mentioned earlier, first breakthrough at this stage is in R5C1 2 by eliminating the two other cells in C1, R1C1, R2C1 for 2 by parallel digit scan for 2 on empty cells of C1 by 2 in R1, R2.
This results in R5C3 8 by reduction of 2 and R5C2 4 by reduction of [2,8].
Next valid cell is in R8C2 8 by row column scan for 8 in R7, R9 and Cycle of twins (4,8) in C1.
It helps next breakthrough R7C2 5 by scan for 5 in R9, C1 and Cycle of twins (3,5) in C3.
Now it is time to create a hard-earned critical breakthrough in R4C7 8 by parallel scan for 8 on empty cells of C7: R7C7, R8C7, R9C7 eliminated by 8 in bottom right major square and R3C7, R5C7 and R6C7 eliminated by 8 in R3, R5 and R6 respectively. All in all, SIX CELL IN C7 ARE ELIMINATED FOR DIGIT 8 TO GET THE BREAKTHROUGH.
This apparently is a very hard to discover breakthrough by parallel digit scan.
Alternately, you might have had the same breakthrough by laboriously working out the inevitable accompanied Cycle of sextuplets (1,3,4,5,7,9) in these six eliminated cells formed by the six leftover digits in the column.
Unfortunately, to form the Cycle you would have had to form the INDIVIDUAL POSSIBLE DIGIT SUBSETS in the six cells first and that would have taken lots of time. That's why, if you can identify the parallel scan opportunity, it would be a much faster breakthrough.
Question is: how do you detect such an opportunity for a parallel digit scan in general?
- First, you must be aware of the potential and method of 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 relatively more cells as unique digit.
In this special case, digit 8 in bottom right major square and a few other incidences of 8 alerted us to explore the possibility of a real breakthrough.
With this breakthrough it follows, R4C9 7 by scan for 7 in R5, R6 and C8 -- R3C7 7 by scan for 7 in C8, C9 -- R3C6 1 by reduction of 7 -- R1C2 1 by scan for 1 in R3, C1.
R3C2 2 by reduction of 1 -- R1C3 7 by reduction -- R1C2 1 by reduction.
With 7 in R1C3, R1C6 4 by reduction -- R2C6 7 by reduction.
With 2 in R2C2, R9C2 7 by reduction -- R9C4 4 by reduction -- R7C4 7 by reduction -- R7C3 2 by reduction -- R3C8 3 by exception in R3.
There are other valid cell possibilities by reduction, but we'll stop here to keep the configuration undisturbed for ease of understanding the actions at this stage.
Solving Sudoku Hard Puzzles: Level 4 Game 17 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 4 in R1C6, R1C1 8 by reduction -- R2C1 4 by reduction.
R7C7 4 by scan for 4 in R9, C8, C9 -- R7C8 1 by exception in R7.
R2C9 1 by scan for 1 in R1, C2 -- R2C8 8 by exception in R2.
With 3 in R3C8, R4C8 5 by reduction -- R6C7 3 by reduction -- R6C3 5 by reduction -- R4C3 3 by reduction.
With 1 in R7C8, R5C8 6 by reduction -- R5C9 9 by reduction -- R5C7 1 by reduction -- R1C8 9 by scan for 9 in C9 -- R1C9 5 by exception in R1 -- R9C8 2 by exception in C8.
With 5 in R1C9 and 9 in R5C9, R8C9 3 by reduction of [5,9] -- R9C9 6 by exception in C9 -- R8C1 9 by reduction -- R9C1 3 by reduction -- R9C7 9 by exception in R9 -- R8C7 5 by exception in the whole game.
Final solution below.
Breakthroughs by advanced Sudoku techniques of Difficult Parallel digit scan, Triple digit scan, Double digit scan, Cycles of twins and triplets ensured quick solution to this 17th Sudoku hard puzzle at level 4.
This Sudoku hard by no means is an easy one to solve even though it may seem otherwise because of the extraordinary breakthroughs.
Unnecessary enumeration of possible digit subsets for empty cells are avoided as far as possible.
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