Learn how to solve very hard expert Sudoku level 5 game 22 quickly simple way
Very hard expert Sudoku level 5 game 22 solved simple way using Sudoku solving techniques of DSA, Cycles, Single digit lock and double digit scan.
- Very Hard Expert Sudoku level 5 game 22
- How to solve very hard expert Sudoku level 5 game 22 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 double 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.
- Single digit lock and how to use it in solving an Expert Sudoku puzzle.
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The following Expert Sudoku game 22 we have found to be very hard to solve. Major Sudoku Solving Techniques have been used for solving the puzzle. 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 Very Hard Expert Sudoku level 5 game 22 Simple Way by Sudoku Solving Techniques Stage 1: Double digit scan, DSA, Cycles
R4C3 5 scan R5, R6. R4C7 8 scan 8 in R5, R6, C7. R6C4 1 scan for 1 in C5, C6, R4.
Cycle (6,9) by double digit scan for [6,9] in C1, C3 on empty cells of bottom left major square.
Cycle (5,7) in R8C1, R9C1 by double digit scan for [5,7] in R7, C3 on empty cells of bottom left major square.
Unexpected breakthrough Cycle (4,6) in R4C5, R5C5 by reduction of [5,7,8] in both cells from DS [4,5,6,7,8] in C5. Major breakthrough R7C5 8 by reduction of [5,7] in R7 from reduced DS [5,7,8] in C5 -- Cycle (5,7) in R2C5, R3C5 in C5 and top middle major square. R2C4 8 scan 8 in R3, C5, C6 -- R9C9 8 scan 8 in R7, C7, C8.
A second major breakthrough R8C3 8 by scan 8 in R7, R9. This won't have been possible without the Cycles [5,7] and [6,9] obtained earlier by double digit scans. R1C1 8 scan 8 in C2, C3, R2. Cycle (4,5,7) in R1C2, R2C2, R3C2 by reduction of 2 in top left major square from DS [2,4,5,7] in C2 -- R5C2 2 residual in C2.
R2C1 3 reduction of 1 in R2 from DS [1,3] -- R1C3 1 residual in top left major square -- R5C1 4 by reduction of 3 in C1 from residual DS [3,4] in left middle major square -- R6C3 3 residual -- R6C9 4 reduction of [2,7] in C9 from residual DS [2,4,7] in R6.
R7C1 1 scan 1 in C3 again because of the two-digit Cycles formed earlier.
R1C6 6 scan 6 in R3 and breakthrough Cycle (4,6) in C5 -- Cycle (3,4) in R3C4, R3C6. With 4 in R5C1, R5C5 6 reduction, R4C5 4. Cycle (4,6) lived its life fully.
How to Solve Very Hard Expert Sudoku level 5 game 22 Stage 2: DSA, Cycles
R6C6 7 scan for 7 in R4 on cells of central middle major square -- Cycle (2,3) in R4C4, R4C6 in R4 and central middle major square -- R4C7 6 residual -- R6C7 2 residual in R6.
R2C8 2 scan on 2 in R1, R3, C7, C9.
R2C9 6 scan on 6 in R1, R3, C7 -- R7C9 9 reduction -- R7C7 4 reduction -- R7C3 2 reduction -- R7C4 6 residual -- R9C3 4 reduction. Cycle (5,7) in R2C5, R2C7 in R2 -- R2C2 4 reduction of [5,7].
R1C8 4 scan on 4 in C7, C9 and Cycle (3,4) in R3.
How to Solve Very Hard Expert Sudoku level 5 game 22 Stage 3: Single digit lock, DSA
Breakthrough Single digit lock on 5 in R8C7, R9C7 in C7 -- R2C7 7 reduction.
This is the last major breakthrough at this late stage of solution. R2C5 5 reduction -- R3C5 7 -- R3C2 5 -- R1C2 7 -- R1C7 3 by reduction of 5 due to single digit lock on 5 in C7 -- R1C9 5.
How to Solve Very Hard Expert Sudoku level 5 game 22 final stage 4: All easy pickings
R9C7 5 reduction -- R8C7 1 -- R8C8 6 -- R9C8 3 -- R9C1 7 -- R8C1 5 -- R8C2 9 -- R9C2 6 -- R8C6 4 -- R8C4 7 -- R9C4 2 -- R9C6 9 -- R4C4 3 -- R3C4 4 -- R3C6 3 -- R4C6 2 -- R5C7 9 residual in C7.
R5C8 1 -- R5C9 3 -- R3C8 9 -- R3C9 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 are explained now.
Let us use the following Sudoku game as the starting point to explain the two techniques of single digit and double digit scan.
This is an initial stage of the Sudoku puzzle solution.
The result of breakthroughs by single digit scan and double digit scan are shown.
Digit scans are done on the cells of a major square. Single digit scan for 8 in R9C9 in C9 and in R5C3 in R5 blocks all other cells in the target right middle major square except R4C8. This is then the only cell in the major square where digit 8 can be placed. Single digit scan blocks then all other cells in the target major square except ONE cell to be occupied by the scanned digit.
This is the most used and most basic technique to find a unique cell for a digit.
In the same way, digits [1,9] appear in R6 and the pair of digits blocks R6C1, R6C2 and R6C3 for both the digits [1,9] in the target left middle major square. This is the double digit scan.
This leaves ONLY TWO VACANT CELLS for the two scanned digits [1,9] in the major square as well as in column C3.
This digit structure of two possible digits for only two cells in a column, row or a major square is termed as a two-cell Cycle.
Every opportunity of forming Cycles are utilized because, Cycles play a very important role in solving very hard to extremely hard Sudoku Puzzles.
Just as single digit scan on the cells of a major square is done searching for suitably affecting digit in multiple rows and columns, double digit scan can also be done with same two digits appearing in more than one row or column affecting the cells of a specific major square.
Double digit scan invariably provides an important breakthrough. Always look out for an opportunity for a double digit scan.
The concept of double digit scan can be extended to triple digit scan as well though it is rare. In this case a three digit long Cycle is formed.
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 starting position is an advanced stage of the solution. Because of Cycle (1,9) in C2 the DS in C2 is reduced to [2,3,4,5,8] which again is reduced by [2,4,8] in R1 to form DS [3,5] in R1C2 and forms the Cycles (3,5) and (2,4,8) in C2. Important result of these formations is DS [4,8] in R7C2.
The result of forming a single digit lock and breakthrough achieved by the lock is shown.
Shift your attention now to row R8 with DS [1,3,4,6,8] and especially the DSs of cells R8C1 and R8C3. By [1,6] in the parent major square, 8 in C3 and 4 in C1 the DSs in these two cells are reduced to [3,4] and [3,8] respectively. Recall that DS in R7C2 was [4,8]. These three promptly join together to form a breakthrough Cycle (3,4,8) which eliminates any other possibility of digit 3 in the two remaining cells R7C3 and R9C1 in the major square.
The digit 3 is now locked in only two cells R8C1 and R8C3 in R8 and bottom left major square. The two digits are highlighted by larger size and different color. This powerful digit pattern is what we call a single digit lock.
As this lock on 3 eliminates possibility of 3 in all other vacant cells of the row R8, especially 3 in DS [1,3] of R8C9, it provides the breakthrough of R8C9 1. Now double digit scan for [4,6] in R9, C9 produces Cycle (4,6) in bottom right major square and DS [3,7,9] in R9C7 reduces by [3,9] in C7 to R9C7 7. This is the second major breakthrough as it causes easily obtained valid cell chain starting from R5C7 4.
A Single digit lock is a very important digit pattern that may not only provide a major breakthrough but also may join suitably with a second single digit lock to form the more powerful and advanced digit pattern of X wing.
As a rule, while solving a hard Sudoku puzzle, the Cycles and single digit locks are identified and marked for immediate or later use.
To go through the solution of this Expert Sudoku puzzle once more, click here.
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