Learn how to solve Expert Sudoku hard level 5 game 25 quickly simple way
Expert Sudoku hard level 5 game 25 solved simple way using Sudoku solving techniques of DSA, parallel scan, Cycles and triple digit scan.
- Expert Sudoku Hard level 5 game 25
- How to solve Expert Sudoku Hard level 5 game 25 simple way by Sudoku solving techniques
- Expert Sudoku Solving Techniques and How to Use Special Digit Patterns.
- Expert Sudoku solving techniques of single digit scan and triple digit scan.
- Expert Sudoku Solving Technique of Possible Digit Subset Analysis (DSA) and how to find a naked single.
- Special digit pattern of Cycle of twins or triplets and how to use it in solving an Expert Sudoku puzzle.
- Expert Sudoku solving technique of parallel scan for a single digit on a row or a column.
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The following Expert Sudoku hard game 25 is solved using Sudoku Solving Techniques. The techniques used are separately explained after the solution, but you must first try your best to solve the puzzle that will enrich your mind.
The Rs are the row labels, Cs are the column labels.
Following is the solution of the puzzle explained in simple way.
Please spend your time fruitfully on the game trying to solve it before going through the solutions.
How to Solve Expert Sudoku Hard level 5 game 25 Simple Way by Sudoku Solving Techniques Stage 1: Triple digit scan, Parallel scan, DSA, Cycles
R1C5 4 scan 4 in R2, R3, C6. Genuine naked single R8C5 5 by reduction of other 8 digits in parent row, column and major square. DS in promising column C5 [3,6,7,8] reduced by [6,7] in bottom middle major square to form breakthrough Cycle (3,8) in the major square in R7C5, R9C5 in C5 and Cycle (6,7) in R2C5, R3C5.
DSA: DS [1,2,4,7,9] in R8 and R8C1 reduced by [1,2,7,9] due to [1,2] in parent major square and [7,9] in affecting column C1 -- naked single R8C1 4 by DSA. R7C4 4 scan 4 in C5, C6, R8.
Triple digit scan for [4,7,9] in R6 on vacant cells of right middle major square leaves just three vacant cells in the major square forming breakthrough Cycle (4,7,9) in R5C7, R5C9 and R4C9. Final major breakthrough comes through [4,9] in C7 resulting in R5C7 7 and Cycles (4,9) and (1,3,6) in the right middle major square.
R9C8 4 scan 4 in R7, R8, C7 and Cycle (4,9) in C9.
Parallel scan for 7 on vacant calls of R8: R8C5, R8C7 eliminated by 7 in bottom middle major square, R8C7 eliminated for 7 by 7 in C7 -- breakthrough R8C2 7.
R4C3 7 scan 7 in C1, C2, R5, R6. R9C9 7 scan 7 in R7, R8, C7.
Cycle (1,2,5) formed in R7C7, R8C7, R9C7 in bottom right major square reducing DS [1,2,5,8,9] by [8,9] in C7. DS [8,9] in R7C9 reduced by Cycle (4,9) in C9 -- breakthrough R7C9 8 -- R7C8 9 residual. Cycle (3,6) formed in C7 in R6C7, R3C7.
How to Solve Expert Sudoku Hard level 5 game 25 Stage 2: DSA, Cycles
R7C5 3 by DS reduction 8 in R7C9 -- R9C5 8. R9C3 9 scan 9 in R7, C1 -- R9C6 1 -- R9C7 5 -- R9C1 3 -- R7C7 2 reduction of 5 -- R8C7 1 residual.
Cycle (5,6) in R7. Cycle (2,9) in R8 and Cycle (2,5,8) in R6.
R5C3 4 scan 4 in C1, C2, R6 -- R5C9 9 reduction -- R4C9 4 residual. R4C2 9 scan 9 in C1, R5.
How to Solve Expert Sudoku hard level 5 game 25 final Stage 3: Cycles, DSA, parallel scan
R4C1 6 by DSA reduction due to 3 in R9C1 -- Cycle (1,3) in R5 -- R5C4 8 by DSA reduction of 3 by Cycle (1,3) -- R5C6 6 reduction.
Breakthrough R3C8 1 by parallel scan for 1 on vacant cells of R3: R3C4, R3C5, R3C6 eliminated for 1 by 1 in top middle major square, R3C3 by 1 in C3 and R3C7 by 1 in C7 -- R6C8 6 by DSA reduction -- R6C7 3 -- R6C9 1 -- R3C7 6.
Residual Cycle (3,5) in R1C9, R2C9 in C9 and top right major square reduces DS in the remaining two cells to [7,8] -- R1C8 8 by reduction of 7 in R1 -- R2C8 7 residual. R3C5 7 scan for 7 in R1, R2, C4, C6.
R3C6 8 by parallel scan for 8 on vacant cells of R3: R3C3 eliminated by 8 in C3 and R3C4 eliminated by 8 in C4. R2C5 6 scan 6 in C4, C6.
With 6 in R4C1, R7C1 5 -- R7C2 6 -- R1C3 6 scan 6 in R2, R3, C1, C2.
With 3 in R9C1, R5C1 1 -- R5C2 3. DS [2,3,5] in R1C2 reduces by [2,3] in C2: R1C2 5 -- R1C9 3 -- R2C9 5 -- R3C3 3 -- R3C4 5 -- R4C4 3 -- R4C6 5.
R1C6 2 residual -- R2C4 9 -- R8C4 2 -- R8C6 9 -- R2C6 3 residual.
DS in C3 reduces to [2,5] and with 5 in R2, R2C3 2, R6C3 5, R6C2 8, R6C1 2, R2C1 8, R2C2 1.
Final solution shown.
Check for the validity of the solution if you need.
As a strategy always try first—the row-column single digit scan to find the valid cell at any stage, because that is the most basic and easiest of all techniques.
While doing the single digit scan, look out for possible breakthroughs by double digit scan and even triple digit scan. Wherever possible, Cycles are formed that in any situation are valuable digit patterns to have and Cycles play a key role in quick solution.
Possible Digit Subset Analysis or DSA is a general technique that is the basis of finding a unique valid digit for a cell by Reduction, a Cycle or even the valuable digit pattern of a single digit lock. Whenever possible, short length possible digit subsets of 2 or 3 digits are to be formed in vacant cells by DSA.
A Single digit lock and an X wing are comparatively more powerful digit patterns that usually create important breakthroughs.
The last resort of filling EACH EMPTY CELL with valid possible digit subsets by DSA is to be taken when it is absolutely necessary. But,
Strategically for faster solution, it is better to delay this time consuming task as much as possible.
A basic part of overall strategy is,
Whether we search for a breakthrough of a bottleneck or a valid cell identification, our focus usually is on the promising zones, the zones (row, column or a major square) that contain larger number of filled digits including Cycles.
The main strategy should always be to adopt the easier and faster technique and path to the solution by looking for key patterns all the time. Digit lock, Cycles, Valid cell by DSA are some of the key patterns.
The four Sudoku solving techniques and special digit patterns used in solving the puzzle are explained now.
The following is the starting stage for explaining how a rare major breakthrough by a Triple digit scan is achieved.
The result of breakthroughs by single digit scan and triple digit scan are shown.
Digit scans are carried out for a single, double or triple set of digits ON THE VACANT CELLS OF A SUITABLE MAJOR SQUARE.
First condition of a breakthrough by a scan is: the digits scanned for must not exist already in the target major square.
Second condition of a breakthrough by scan is: Final number of eligible vacant cells must equal the number of digits scanned for after taking care of eligible cell eliminations by the presence of the digits scanned for in more than one row or column.
As an example of single digit scan for 4 on vacant cells of top middle major square, the major square does not have 4 in it and presence of 4 in R2, R3 and C6 eliminates seven of the eight vacant cells leaving ONLY cell R1C5 for the digit 4. This is an example of a minor breakthrough R1C5 4 by a single digit scan for 4.
Now observe that the digits [7,9,4] appear in row R6 in central middle major square as if focusing their attention on the vacant cells of the adjacent right middle major square which by chance do not have any of these three digits. So the three cells R6C7, R6C8 and R6C9 become ineligible to have any of these digits leaving fortunately just three vacant cells in the right middle major square to accommodate the three digits [4,7,9].
As a result, Triple digit scan for [4,7,9] in R6 on vacant cells of right middle major square leaves just three vacant cells in the major square forming first the breakthrough Cycle (4,7,9) in R5C7, R5C9 and R4C9. Final major breakthrough comes through presence of digits [4,9] in C7 resulting in R5C7 7 by DS reduction from DS [4,7,9] in the cell.
As byproducts, Cycles (4,9) and (1,3,6) (not shown for clarity of observation) are also formed in the right middle major square. Though rare, never let go an opportunity of a breakthrough by a triple digit scan.
Expert Sudoku Solving Technique of Possible Digit Subset Analysis (DSA) and how to find a naked single
The following is an initial stage of the Sudoku puzzle solution.
The result of finding a unique valid cell or naked cell by digit subset analalysis technique applied on the above game stage shown now.
Digit Subset Analysis or DSA is a concept as well as a technique. By DSA, digits that can occupy a particular cell are identified.
This is an essential and very important function for identifying all other digit patterns possible in a target cell. When no easy unique possible digit in any cell can be identified, the only way to move ahead in solving the puzzle is to carry out DSA for PROMISING CELLS. The simplest type of promising cell is the cell with smallest number of possible digits DS of 2 or 3 digits (not 4 digits at first).
To identify a promising cell, identify first a row, column or major square with maximum number of already occupying digits. In the example above, row R4 is such a row with possible digit subset in the four empty cells [1,2,7,9].
Next, the cell R4C4 is easily identified as a promising cell as, [2,9] in Column C4 affecting the cell REDUCES the possible digit subset or DS for the cell to just [1,7]. Moving ahead, the third cell R4C6 gets DS [1,7,9].
And finally, for the fourth empty cell in the row R4C9, [1,9] in its parent major square and 7 in parent column C9 combine to form [1,7,9] to be reduced from the DS [1,2,7,9]. Result is, R4C9 2, a valid unique digit for a cell.
REDUCTION is a fundamental process in solving Sudoku puzzles.
Naked Single: By definition, a naked single is a digit that only can occupy a specific cell. If you analyze possible digits in R4C9 ignoring the earlier process of reduction from DS [1,2,7,9] in row R4, you will find only digit 2 can occupy the cell.
You may adopt this process of identifying a naked single WITHOUT taking help of smaller possible digit subsets in vacant cells, but this process is easier only occasionally.
Special digit pattern of Cycle of twins or triplets and how to use it in solving an Expert Sudoku puzzle
The following is an initial stage of the Sudoku puzzle solution.
The result of a breakthrough unique valid digit by forming 2 digit (twin) and 3 digit (triplet) Cycles is shown.
The digits [1,9] in R6 affect the possible digit subsets or DSs of vacant cells of left middle major square leaving only two cells of the square R4C2, R5C2 for the two digits [1,9]. It is not certain which of these two cells will finally be occupied by 1 or 9 but we can confidently say none other than these two digits are the only eligible candidates for occupying these two cells.
Thus a Cycle (1,9) is formed in these two cells restricting any other cell in the parent column and major square from having any of these two digits.
If we place 1 in R4C2, automatically R5C2 must have 9 and if we place 9 in R4C2 the cell R5C2 must have 1. Potentially these two digits Cycle between these two cells till their final positions are determined. That is why we can place the DS [1,2] in both the cells blocking any other cell of parent major square and column C2 from having these two digits.
This is a two digit Cycle and is the most frequently occurring one.
The direct positive result is formation of a second Cycle (3,5,7) in the three remaining vacant cells by exactly three remaining digits in the major square. This is a three-digit Cycle debarring all other vacant cells of parent row R6 to have these three digits. Result is formation of a third Cycle (4,8) in R6C5, R6C6 and a unique valid digit 6 in R6C9 as the REDUCED DS [4,6,8] in R6C9 is further reduced by [4,8] in C9 and right middle major square combined.
This breakthrough won't have been possible without the Cycle (3,5,7) in R6.
The main function of a Cycle is to REDUCE the length of possible digit subsets or DSs in affected parent zones and with each DS length reduction, certainty of getting a unique valid cell in the whole set of 81 squares increases.
The following is an initial stage of the Sudoku puzzle solution.
Result of carrying out parallel scan for digit 2 on the vacant cells of row R8 is shown.
In the relatively empty 81 square puzzle, digit 2 in left bottom major square and right bottom major square debar every vacant cell in the squares from having digit 2. With keen interest we observe, out of six vacant cells in row R8, four cells cannot have digit 2. If we can debar any of the two other remaining cells R8C5, R8C6 from having digit 2, we will get a unique valid digit breakthrough.
This actually happens, as 2 in R1C6 in C6 eliminates the fifth vacant cell R8C6 from having digit 2. We have achieved the breakthrough of unique valid digit 2 in R8C5 as if out of thin air.
In essence, a PARALLEL DIGIT SCAN for digit 2 is done on the empty cells of row R8. Even if digit 2 in R9C3 were in R2C3 or R3C3, the result would have been the same.
A parallel digit scan is done on vacant cells of a row or column (and NOT on vacant cells of a major square). Fortuitous presence of the digit scanned in other cells, all except one cell of the target row or column scanned are debarred from having the digit scanned for. This is an advanced and powerful Sudoku Solving Technique often providing a major unexpected breakthrough.
Observe, you could also have achieved the breakthrough by forming the five-cell long Cycle (1,3,4,6,8) in R8. But that would have been laborious. If you are aware of the possibility, a parallel scan will give you a quick and clean breakthrough.
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