Learn how to solve expert Sudoku level 5 game 5 quickly
The very hard expert Sudoku game 5 is solved using advanced Sudoku techniques and expert strategy so that it is easy to follow how to solve expert Sudoku.
- Very hard expert Sudoku level 5 game 5
- Solving very hard expert Sudoku level 5 game 5 by advanced Sudoku techniques and strategy
- Expert Sudoku Strategy and techniques for quick solution
- What is a Cycle of twins or triplets and how to use it in solving an expert Sudoku puzzle.
- How to find a naked single in a valid cell by Possible Digit Subset Analysis (DSA).
- Single digit lock and how to use it in solving an expert Sudoku puzzle.
- Expert Sudoku technique of double digit scan.
- Expert Sudoku breakthrough technique by digit pattern of X wing.
- Expert Sudoku technique of parallel scan for a single digit on a row or a column.
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The following is the Sudoku puzzle that should engage your mind for some time. The Rs are the row labels, Cs are the column labels and this we define as the stage 1 marked on top left corner.
We'll first solve the Sudoku hard using strategies and techniques for solving Sudoku hard puzzles.
The strategy and techniques for quickly solving Sudoku hard are explained with examples in the sections after the solution.
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Following is the solution of the puzzle explained step by step in details.
Please spend your time fruitfully on the game trying to solve it before going through the solutions.
Let us solve our Expert level 5 Sudoku hard puzzle now.
We'll show the puzzle again for ease of understanding.
To follow the details accurately, you should better have the game actually with you written on paper, or better still—created in a spreadsheet.
Single possible valid cell R4C9 5 by row scan R5, R6.
But two single digit locks.
First on 1 in top middle major square, C6 cells R2C6, R3C6 by cross-scan for 1 in R1, C4.
And second on 3 in right middle major square C7 cells R5C7, R6C7 by single scan for 3 in C8.
Time for DSA on promising cells.
R3C4 4 by DSA reduction of [3,8,9] from possible digits in the cell DS [3,4,8,].
This breakthrough is possible by the single digit lock of 1 in the major square.
Parallel scan for 8 on empty cells of C1 -- 8 in top left major square eliminates the three cells in C1 and top left major square for 8 and 8 in R5 leaves R4C1 for 8.
R8C3 8 scan R9, C1, C2.
Double digit scan for [4,8] in R4, R5 -- Cycle (4,8) in R6C5, R6C6.
R4C5 1 by scan for 1 -- 1 in C4, single digit lock on 1 in C6.
Cycle (1,9) in C2, bottom left major square cells R7C2, R8C2 by double digit scan for [1,9] in R9 -- another Cycle (3,5) formed in R9C2, R9C3. This last Cycle reduces 3 in R9C9 DS to [2,4] and helps to form a new Cycle (2,4,7,9) in R1C9, R2C9, R3C9 and R9C9 -- Forms a new Cycle (1,3) in R7C9, R8C9.
Cycle (2,4,7,9) also in top right major square -- R2C8 8 by reduction of [5,6] in R2 by the Cycle in the major square.
By DSA on three empty cells of C5, single digit lock on 3 is created R7C5, R8C5.
Single digit lock on 4 created in R1C9, R2C9 in C9 and top right major square.
At the first stage itself much ground covered without disturbing the digit formations needed for explanation.
Result of the steps taken shown below.
Solving Very hard expert Sudoku level 5 game 5 Stage 3
Going is hard at every step and the first valid cell find needed the use os a single digit lock.
Single digit lock on 4 in R1C9, R2C9 supports a second single digit lock on 4 in R8C8, R9C8 by scan 4 in R7, C7 -- R9C9 2 by reduction of 4.
R2C7 2 only cell left for 2 -- R6C8 2 scan R5, C7, C9 -- R5C8 1.
Cycle (3,7) in R6C2, R6C7 -- R6C3 1 by scan 1 in R4, R5 and in Cycle (1,9) in C2.
DS in R6C2 [3,7], But, DS in R2C2 is also [3,7] by DSA reduction of [2,5,6] from DS [2,3,5,6,7] in 5 empty cells of C2 -- Cycle (3,7) in R2C2, R6C2 -- R9C2 5 by reduction -- R9C3 3 reduction.
R7C6 2 scan R8, R9, C4, C5 -- R8C6 5 scan R9, C4, C5.
For ease of understanding let's close at this point and show you the status below.
Solving Very hard expert Sudoku level 5 game 5 Stage 4
All empty cells are filled with possible digit subsets each involved in one or more than one Cycle with no opportunity for a breakthrough by easier means of DSA, row column scan and so on.
We must look for identifying more complex digit patterns for the breakthrough.
Such an advanced digit pattern with many possibilities is the X wing formed by 7 in R7C4, R8C4, R7C8 and R8C8.
The basic property of such an X wing formation is: both pairs of rows R7, R8 and columns C4, C8 are blocked for 7. Usually, elimination of 7 from possible digits of these shared cells and columns gives the breakthrough.
Such an opportunity is not present in this case.
The second property of an X wing is: either of the diagonally placed pair of 7s will occur in the final solution, that is, one of the diagonally placed digits will be valid and the other invalid.
The diagonals are shown by connecting the pairs of digits. We'll check which of the diagonals will be valid. This is a new way to use the powerful X wing digit formation.
Note: The reason why we expect this check to be conclusive quickly is the special configuration of the digits 5 and 6 linked with single locks of 9 and 7 in right bottom major square.
The puzzle game at this stage is shown below.
Checking validity of two diagonals of the X wing: Solving very hard Expert Sudoku level 5 game 5
Downgoing left to right diagonal has [7,9] and [4,6,7] at the two ends and left to right upward moving diagonal has [6,7] and [5,7] at its two ends.
One of the diagonals will finally be valid.
Checking first R7C4 7, its partner R8C8 will have 7. The other cells will have R8C4 6 -- R8C7 9. With 7 in R8C8, R7C8 5 -- R7C7 9.
THIS IS THE CONFLICT WE LOOKED FOR.
This happens because simultaneously R7C8 5 and R8C4 6 occurred to create the conflict by their common linking with digit 9 in R7C7 and R8C7 respectively.
And this invalid digit combinations have occurred by the downgoing left to right diagonal of X wing on 7.
This confirms that the valid diagonal of the X wing will be the second one: R8C4 7 and R7C8 7.
Result of the check shown below.
Solving very hard expert Sudoku level 5 game 5 Stage 5
All valid cells in this stage are by reduction. With the critical breakthrough that was hard to get, the rest of the cells crashed like a house of cards.
R7C8 7 -- R7C4 8 -- R7C5 3 -- R8C5 4 -- R8C8 6 -- R8C7 9 -- R7C7 5 -- R9C8 4 -- R9C6 6 -- R8C4 7 -- R8C2 1 -- R7C2 9 -- R8C9 3 -- R7C9 1.
With 5 in R7C7, R3C7 6 -- R1C8 5.
With 4 in R8C5, R6C5 8 -- R6C6 4.
Result of steps taken is shown below.
Solving Very hard expert Sudoku level 5 game 5: Final Stage 6
Same at this stage. Rest of valid cells are all by reduction.
With 8 in R7C4, R1C4 3 -- R4C4 6 -- R4C2 2 -- R4C3 7 -- R5C3 9 -- R6C2 3 -- R5C1 6 -- R6C7 7 -- R5C7 3 -- R5C6 7 -- R4C6 3.
With 3 in R6C2, R2C2 7 -- R1C2 6.
With 7 in R2C2, R3C9 7 reduction of single digit lock -- With 9 in C3, R2C3 4 -- R3C3 5 -- R1C3 2 With 4 in R3C3, R2C9 9 -- R1C9 4 -- R2C6 1 -- R3C6 9 -- R1C6 8.
R3C1 1 -- R2C1 3 -- R1C1 9.
Final solution below.
Check for the validity of the solution if you need.
As a strategy we always try first—the row-column scan to find the valid cell at any stage, because that is the most basic and easiest of all techniques.
When easy breaks by row-column scan becomes hard to come by, the next technique is used.
Next easy to use technique used is—identification of single valid digit for a cell by Digit Subset Analysis or DSA in short. This technique is explained in a following concept section.
And wherever possible, Cycles are formed that in any situation are a treasure to have and Cycles play a key role in quick solution. Concept and use of Cycles are explained in a following section.
You may wait for Cycles to form automatically in a column or row.
But a proactive approach of forming a Cycle by DS analysis speeds up the solution process considerably. This is what we call forced creation of Cycles.
The last resort of filling EACH EMPTY CELL with valid digit subsets is to be taken when it is absolutely necessary. Only with all empty cells filled with valid digit subsets, the possible breakthrough points in a hard puzzle can be discovered.
Strategically for faster solution, it is better to delay this time consuming task as much as possible.
In hybrid strategy, a few of the cells of interest are filled with DS of shorter length and analyzed for a breakthrough such as forming a Cycle or a single digit lockdown.
One of the most powerful patterns that we have used for highly positive result each time is the lock of a single digit in a row or column inside a 9 cell square so that the digit is eliminated from all other DSs in the locked row or column outside the 9 cell square.
The necessity of use of this digit lockdown technique indicates in a way the hardness of the puzzle. This technique is also explained in a following section.
In solving this Sudoku hard, we have mentioned the use of an additional technique of Parallel scan for a single digit on the cells of a row or column. But it has a minor and unimportant role in discovering the concerned valid cell. We'll not elaborate it here.
A rarely encountered powerful pattern is 4 cell single digit lock in a rectangular formation that may be encountered in very hard or expert level Sudoku hard puzzles. Naturally, it is a superset of the more common single digit lock in 2 cells and so is much more effective.
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 and 9 cell square combined) 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 lockdown, Cycles, Valid cell by DSA are some of the key patterns.
Focus when solving a hard Sudoku puzzle should be on using the technique that would produce best results fastest. Easy to say, not so easy to do—comes with practice.
Form of a Cycle:
In a Cycle, the digits involved are locked within the few cells forming the cycle. The locked digits can't appear in any other cell in the corresponding zone outside the few cells forming the cycle.
For example, if a 3 digit cycle (4,7,8) in column C2 is formed with a breakup of, (4,7) in R1C2, (4,7,8) in R5C2 and (7,8) in R6C2, the digits 4, 7 and 8 can't appear in any of the vacant cells in column C2 further.
If we assume 4 in R1C2, you will find R5C2 and R6C2 both to have DSs (7,8) implying either digit 7, or 8 and no other digit to occupy the two cells. This in fact is a two digit cycle in the two cells. Together with 4 in R1C2, the situation conforms to only digits 4, 7 and 8 occupying the set of three cells involved in the cycle.
Alternately if we assume 7 in R1C2 (this cell has only these two possible digit occupancy), by Digit Subset cancellation we get, digit 8 in R6C2 and digit 7 in R5C2 in that order repeating the same situation of only the digits 4,7 and 8 to occupy the set of three cells.
Effectively, the three digits involved cycle within the three cells and can't appear outside this set of three cells.
This property of a cycle limits the occupancy the cycled digits in other cells of the zone involved (which may be a row, a column or a 9 cell square). This generally simplifies the situation and occasionally provides a breakthrough by reducing the number of possible digits in the affected cells.
A number of Cycles are shown below from a Sudoku hard solution stage:
Cycle (1,2,6) in column C1 is over all three 9 cell squares on the left. It affects only the column C1.
Cycle (3,8,9) in top right 9 cell square is also in row R2, so it should affect both the 9 cell square and R2.
But Cycle (3,6,7) in top right 9 cell square is formed only in the 9 cell squares, it affects only the cells in the 9 cell square.
Can you see another Cycle in row R1 apart from Cycle (1,6)? The second Cycle (3,6,7) is formed by the cells R1C2, R1C3 and the far away cell R1C9. This Cycle affects only the row R1.
Can you say which are the affected areas for Cycle (1,6) in R1?
Two cells of this Cycle belong to row R1 as well as to the top middle 9 cell square. So the Cycle affects two areas, the row and the 9 cell square. This will be true for any two digit Cycle.
Use of a cycle:
In the example of cycle described above, if a vacant cell R8C2 in column C2 has a possible DS of (1,4), as digit 4 has already been consumed in the cycle (4,7,8) in the column, only digit 1 can now be placed in R8C2. You get a single valid digit 1 for R8C2.
This is how a new valid cell is obtained using a Cycle that was not visible otherwise.
In any hard Sudoku game solution, creating, analyzing and using the pattern of Cycles play a very important role.
Sometimes when we analyze the DSs in a cell, especially in highly occupied zones with small number of vacant cells, we find only one digit possible for placement in the cell. We call valid cell identification in this way as Digit Subset Analysis.
For example, if in row R4 we have four empty cells, R4C1, R4C3, R4C6 and R4C9 with digits left to be filled up [1,3,5,9] we say, the row R4 has a DS of [1,3,5,9] that can be analyzed for validity in each of the four empty cells.
By the occurrence of digits in other cells if we find in only cell R4C1 all the other three digits 3,5 and 9 eliminated as these are already present in the interacting zones of middle left 9 cell square and the column C1, we can say with confidence that only the left out digit 1 of the DS [1,3,5,9] can occupy the cell R4C1.
While evaluating the valid digit subset or DS of an empty cell, you would analyze not only the digits that are already filled in corresponding row, column and 9 cell square, you must include the Cycles present in the three interest zones also.
This is how we identify a valid cell by Digit Subset Analysis.
Identifying a valid digit in a cell by DSA is like a bread and butter technique. It is possibly the most heavily used technique after the simplest row-column scan.
Though DSA may not be considered as an advanced technique it often provides a much required breakthrough. So always look for a valid cell by DSA.
An example of a breakthrough at the late stage of Sudoku hard puzzle solution by DSA is shown below.
We'll do DSA on cell R7C5. The possible digit subset or DS in column C5 and hence in cell R7C5 is [5,7,9], but the two digits [5,9] both are present in row R7.
So eliminating these two from the three digit DS for R7C5, we get the single valid digit 7 for R7C5 --- R7C5 7.
This is a breakthrough even at this late stage.
Occasionally, after evaluating valid DSs for a number of empty cells, you may find that,
A single digit appears only in the DSs of two or three cells in a 9 cell square, in a column or a row, and in no other DSs in the 9 cell square.
This is what we call as single digit lock.
If it happens in a row (or a column) inside a 9 cell square, the digit cannot appear in any other cell in the row (or the column) outside the square.
This eliminates all occurrences of the locked digit from the DSs in the row (or the column) outside the 9 cell square. Usually it creates a much needed breakthrough. It is a very powerful pattern.
Single digit lock - Conditions for single digit lock - how to identify it
Two conditions for single digit lock pattern,
- the digit can be placed in only two or three cells of a column or a row, AND,
- the locking cells must also be in SAME 9 cell square.
The third desired condition is,
- The lock to be effective, the locked digit should not be present as a single cell candidate in both the adjacent two 9 cell squares through which the locked column or row passes.
The following shows an example of single digit lock of 5 in cells R7C1 and R9C1.
How a single digit lockdown is formed
Look at columns C1, C2 and C3 in the bottom left 9 cell square R7R8R9-C1C2C3. Out of 3 empty cells, the cell R7C3 is debarred for placing digit 5 as column C3 has a 5 and it lights up the cell for digit 5.
5 can appear only in two cells in column C1, R7C1 and R9C1 and in no other cell in the 9 cell square or the column C1.
It is locked inside these two cells in C1 and 9 cell parent square.
How a Sudoku single digit lock is used - What it does
The locked digit 5 eliminates itself from the DSs of the other two empty cells R5C1 and R6C1 and a new Cycle (2,3) is created in C1.
Focus again on the bottom left 9 cell square. With Cycle (2,3) in C1, another Cycle (5,9) is formed in the two cells of the 9 cell square. As a result, digit 1 becomes the only digit left and cell R7C3 only cell left for it in the 9 cell square.
Still more happens. With 1 in C3 now, digit 9 now must occupy the cell R6C3.
These two single digit candidates obtained by the single digit lock of 5 affects other cells and breaks the bottleneck.
As a strategy, always form a single digit lock as soon as it is discovered.
You may think, what is the point of it, what would it achieve after all!
Well, in a similar situation in the process of solving a hard Sudoku puzzle game, the reduced DS in R9C1 formed a cycle (3,7) in column C1 and helped to pinpoint a valid digit 4 in cell R2C1 and that started a deluge of valid cell finds. This proved to be the key turning point in the whole game.
This technique sounds simple, but being aware of its existence and identifying it would always result in an important breakthrough. This digit pattern usually occurs in very hard Sudoku.
We will explain this advanced Sudoku hard technique on the following situation in a Sudoku hard game,
Notice the two highlighted digits [1,6] appearing in both row R4 and C5. Together these two result in DIRECT FORMATION OF CYCLE (1,6) in central middle 9 cell square.
This is a double digit scan simultaneously on a row and a column.
Now observe a second set of highlighted double digits [3,9] in C5 which DIRECTLY FORMS TWO CYCLES (4,7,8) AND (3,9) IN CENTRAL MIDDLE 9 CELL SQUARE.
This is a double digit scan on a single zone of C5.
Finally, with 3 in C4, R4C4 9 and R4C6 3.
Together these two double digit scans have produced two valid cells and two Cycles. It is a major breakthrough early in the Sudoku hard game.
This special and not often occurring digit pattern involves A SINGLE DIGIT SHARING SAME TWO ROWS AND TWO COLUMNS.
This can be imagined as an advanced form of single digit lock over one pair of rows and one pair of columns in the formation of a rectangle. That's why we call it a single digit rectangle.
This digit formation is commonly known as X wing.
The following figure shows an X wing formation on digit 4 at an advanced stage of solution.
At this advanced stage, possible digit subsets for all empty cells are evaluated after the valid cells and easily formed Cycles are created. By this approach, time waste is minimal.
Interestingly, if you examine closely you will find that all the possible digit subsets are involved in one or more than one Cycle.
The second more important aspect of this puzzle status is,
You cannot find any more valid cell breakthrough by the more frequently used techniques of row column scan (hidden singles), DSA (naked singles), Cycles (naked and hidden groups), single digit lock (direct interaction), double digit scan or parallel scan.
In this situation especially, you need to identify the more complex breakthrough digit patterns of X wing, XY wing or Swordfish. Specialty of all these next advanced level digit patterns is,
Usually all of X wing, XY wing or Swordfish digit patterns usually involve 3, 4 or more cells sharing one common digit.
Observe that we have indeed such a digit pattern formation of 9 in four cells R1C1, R7C1, R1C4 and R7C4.
This single digit of 9 is shared between two columns C1, C4 and two rows R1, R7.
What's so special about this formation?
The special effect of the X wing formation is simply,
The commonly shared digit cannot appear in any cell of these rows and columns except these four cells.
As a result all instances of commonly shared digit in these other cells are eliminated from the possible digit subsets, and this REDUCTION INVARIABLY PRODUCES A CRITICAL BREAKTHROUGH.
In our example game, the X wing cause reduction of 9 in possible digit subset [3,9] in cell R1C2 and causes the most critical breakthrough of the puzzle in the form of,
In fact, with this valid cell breakthrough rest of the cells are all reduced to valid cells just by reduction.
Notice that we have joined the diagonally opposite pairs of digit 9 by two lines forming a large X. One of the two diagonals will actually occur blocking all appearances of 9 in the two columns and rows.
Let's see a second example in the following figure.
Here also a stage is reached when no easy breakthrough could be seen. This is a sure sign of breakthrough by more advanced techniques of X wing, XY wing, Sword fish or likes that involve a single digit shared between more than 2 cells.
Though spotting such a digit pattern spanning at least 4 cells is not easy, if you know what to look for, you can find the pattern quickly. Also you need to practice.
In the above figure digit 4 is shared between two columns C3, C7 and two rows R2, R3. The property that makes this digit configuration a valuable and proper X wing is:
In each column, digit 4 can occupy only these two cells and no other cell. That is, digit 4 is locked in the two cells in both C3 and C7.
No doubt that this is a case of two numbers of SINGLE DIGIT LOCKS OVER TWO COLUMNS SHARING THE ROWS ALSO.
Effect is two-fold. First, digit 4 will have to be removed from all possible digit subsets in the cells of R2, R3, C3 and C7 as well as in cells of their home major squares, top right and top left.
Result is the CRITICAL BREAKTHROUGH OF FIRST FORMING CYCLE (2,6) IN R2C4, R2C8 AND THEN THE VALID CELL R2C6 8 BY REDUCTION OF (2,6) BY THE CYCLE IN R2.
That single valid cell resulted in ALL THE REST OF THE VALID CELLS TILL THE FINAL SOLUTION.
A parallel scan is carried out for a specific digit on the empty cells of a promising row (or column). Because of presence of the specific digit in the interconnecting columns (or rows) for all empty cells of the scanned row (or column) except one, the valid cell for scanned digit can be identified as this cell.
The digit pattern and the technique to identify a breakthrough valid cell by parallel scan is shown in the figure below,
The parallel scan for digit 6 is done in this case on the empty cells of R1. Out of 4empty cells R1C4, R1C6, R1C7 and R1C9, digit 6 is disallowed in the first two by 6 in top middle major square and disallowed in R1C7 by 6 in C7.
This leaves only the single cell R1C9 where digit 6 can be placed. That becomes the valid cell for digit 6.
Observe that as a result a Cycle (4,7,9) is formed in the rest of the three empty cells in R1.
If you could have identified the Cycle before parallel scan, you could automatically have got the valid cell without parallel scan. That's the interesting property of parallel scan, if you can spot one, you would be sure to find an equivalent Cycle as a result.
To us, valid call by parallel scan is easier and faster.
To go through the solution of this Sudoku hard once more, click here.
Finding valid cells has been hard all through the intial stages. After the series of these hard to get valid cells, the puzzle game finally became locked with possible digits in all the empty cells involved in one or more than one Cycle.
It was not easy to discover how to breakthrough this deadlock. An X wing identified but it didn't result in any reduction in possible digits in any other cell that would have given us the breakthrough.
The situation forced deeper analysis of the digit linking of the X wing digits and adopt the new method of testing validity of two diagonals. It has been easy to invalidate one diagonal and that has been the most important breakthrough.
Overall, this is a very hard puzzle to solve.
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Enjoy solving and learning to solve Expert Sudoku puzzles.