11th Sudoku game at 3rd level of hardness and its solution
This is the 11th game play session at Sudoku third level of hardness.The specialty of this game has been use of large number of cycles that helped at the start.
Interesting has been the four cell single digit rectangular lockdown cycle, a special structure that provided the much needed breakthrough in this specially difficult Sudoku game.
The game below is followed by three sections on Strategies and concepts that you may skip if you so desire.
The step by step solution follows the concept sections and at the end are given the list of links of all our Sudoku game plays.
The 11th Sudoku game
The following is the game that should intensively engage your mind for some time.
You may go through the next two sections for learning strategies and techniques of Sudoku solving in brief. Or, you may skip.
The step by step solution follows these two concept sections. Please spend your time fruitfully on the game before going through the solutions explained fully.
Overall strategy adopted and techniques used
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.
Hardness level being higher now, easy breaks by row-column scan are few and far between. We have to use in general the method of enumerating small length (2 or 3 digit long) Digit Subsets possible to be placed in favorable cells and writing down the DSs in the cells. This act of writing down the DSs in empty cells helps later in identifying a valid digit by Digit Subset cancellation.
Occasionally, these DSs give rise to Cycles in the corresponding row, column or 9-cell square immediately simplifying the situation. Or sometimes we are able to pinpoint a valid cell by analyzing the DSs in empty cells of a zone with respect to other interacting zones when we find only one digit is left for placement in the cell, rest eliminated. We call this technique as Digit Subset Analysis or DSA in short.
The third routine technique that we are using now is to look for a digit locked in one column or row of a 9 cell square where at least one of the rest two adjacent 9 cell squares does not have the digit in it yet in a single cell. It means from the DSs of the empty cells in the adjacent 9-cell squares in the specific column, the locked out digit will be cancelled and breakthrough can come this way thus reducing the size of the DSs. We call this as digit lockdown technique.
Structure and use of a cycle
Form of a cycle:
In a Cycle the digits involved are locked within the few cells forming the cycles, they 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) generally simplifying the situation and occasionally providing a breakthrough.
Use of a cycle:
In the example of cycle 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.
This is how a new valid cell is broken through which otherwise we were not able to find out in any other way.
In any hard Sudoku game solution, creating, analyzing and using the structure of Cycles play a very important role.
How a valid cell is identified by Digit Subset Analysis or DSA in short
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.
This is how we identify a valid cell by Digit Subset Analysis.
You may also refer to our first and second game play sessions at level 2 where we first explained use of a cycle and DSA.
Let us solve the game now.
The Sudoku 11th game play at third level of hardness
We'll show the game board again for convenience of understanding.
To follow the details accurately, you should better have the game actually with you written on paper, or still better—created in a spreadsheet.
The first valid cell filled is 3 in R6C9 by cross-scanning rows R4, R5 and column C7 each of which has a 3 in it. This combination of rows and columns containing 3 has eliminated the 5 cells R4C7, R4C8, R4C9, R5C8 and R6C7 for filling up of 3. The only possible cell left for 3 is then R6C9. This first fill is colored turquoise blue.
Next 3 direct fills by cross-scan are—
R6C2 1 by cross-scan C1,C3,R5 -- R7C4 1 by cross-scan C5,C6 -- R4C1 4 by cross-scan R5,R6,C3.
The 5th fill is by DS analysis—
R2C8 4 DSA analysis of R2, C8 and top right square.
The 6th fill is by cross-scan of two rows and two columns—
R7C9 4 by cross-scan rows R8,R9 and columns C7,C8.
First cycle (2,8) is formed now on the two cells R1C9 and R8C9 of column C9 by Digit subset analysis on row R1, top right square, column C9 and row R8, bottom right square, column C9 respectively.
Digit subset DS in two remaining cells in C9, R4C9 and R9C9 is (1,7) as the digits (2,8) are blocked by the cycle. This gives us next two fills—
R9C9 1 by scan in bottom right square and DS (1,7) in R9C9 -- R4C9 7 by exclusion (only digit left).
Second Cycle (5,8) formed on R1C4 and R5C4 as 7 and 9 of DS [5,7,8,9] in C4 gets cancelled by 7 and 9 in both R1 and R5. This immediately gives us the next two fills—
R2C4 9 by cancellation of 7 in DS [7,9] by 7 in top middle square -- R6C4 7 by exclusion in C4.
Third Cycle (6,8) formed on R2C1 and R2C5 by scan R2, C1 and top middle square. This gives us next fill—
R2C7 7 by exclusion in R2.
No other fill could be obtained easily and so all the empty cells are populated with possible digit subsets (DSs) by suitable cross-scanning. This exercise is time-taking but could not be delayed further.
Analysis of the digit subsets of all cells
The cell-fill results along with cycles and DS population of empty cells are shown in stage 2 of Sudoku 3rd level 11 board below,
This is the result of all the analysis and actions we had taken starting with turquoise blue colored cell R6C9 filled with 3 in Stage 1.
Along with cell filling by valid digits, 3 cycles have been created. Those are shown in yellow color.
If you want, you may continue with the DS analysis directly by skipping the next section.
On filling up of every empty cell DS or full DS population
We have not discussed the filling up of every empty cell with their valid digit subsets or DSs.
You must be quite aware of how to do this by now.
For example, to evaluate the DS for cell R6C3, look at the row R6 (as the column C3 has number of unique digits 4 less than that of 5 in R6) and form DS for the row as, [2,5,6,9]. These are the digits possible for the four empty cells of row R6. For each of the 4 empty cells, as a start you would use DS as [2,5,6,9].
Next, check in the middle bottom square, the home of cell R6. The digits are exactly same as in R6. No new cancellation achieved.
End with scanning crossing column C3 to cancel the only additional digit 2 from [2,5,6,9] and finalize the DS for cell R6C3 as [5,6,9]. That's how you evaluate the DSs for each empty cell.
It may seem to be very time-consuming, but you have to do it if you want to solve the puzzle. With practice, speed of evaluating DSs improves a lot. But you have to be careful and double check every cell DS because a single mistake at this stage may cost you the solution and unending wasted attempts.
On to our all empty cell DS analysis program again.
First a minor point—the digit 6 colored red is locked in two cells R1C5 and R2C5 that belong to the top middle square and the column C5. As a result, digit 6 cannot appear in any of the remaining cells of C5. So we have crossed out 6 in cells R4C5 and R6C5 reducing thus the number of digits in each of these two DSs. This is the identification and use of digit lockdown technique opportunity.
You would realize shortly that this structure and its property is only a special case of the more comprehensive larger structure of 4 cell single digit rectangular lockdown.
In this case, the most important structure formed is the single digit cycle of digit 2 in the four orange colored cells—R4C5, R4C7, R6C5 and R6C7.
You have to identify this rare structure specially looking for it because each of the four cells would usually be populated more digits other than 2.
Formation of single digit 4 cell rectangular lockdown
The specialty of this powerful structure is—both the columns C5, C7 and rows R4, R6 are locked down for digit occupancy by 2.
No cells other than the four can have 2 in two of these four cells in two rows and two columns.
This happens because—2 is locked in the two cells of C4 and central-middle square. In this square, in no other cell 2 can be placed. The same is true for the two cells with 2 in column C7.
This structure blocks both rows and columns involved because—if you put 2 in R4C5, it can be placed only in R6C7, the diagonally placed cell in the four cell structure. And so it blocks both the rows and columns for 2.
Again if you put 2 in R4C7, you can put it only in R6C5, blocking both rows and columns—
Either way the pairs of columns and rows are blocked for 2 excluding the allowed zone of the four cells.
As a result, you have to cancel 2 in R6C1, only in which 2 appears other than the four cells in the pairs of rows and columns.
This creates the breakthrough in middle-left square—
R5C2 2 by single occupancy in the home square.
This completely breaks open the puzzle and filling of all the empty cells is quick and easy.
The formality of completing the game puzzle—final stage
We'll mention the process of filling the next three cells—
R9C1 2 by cancellation of 2 in two home square cells of C2 and then by exclusion in home square -- Creation of Cycle (5,9) in R7C8 and R9C8 -- By cancellation R7C7 8 and R8C9 2.
We'll just mention the digits in all the rest of the empty cells. The process in all cases are very simple and straightforward—
R1C9 8, R3C7 1, R3C8 3, R1C8 2, R5C8 6, R4C8 1,
R1C4 5, R1C1 6, R2C5 6, R2C1 8, R1C5 3, R3C6 8, R3C5 4,
R8C6 3, R4C6 6, R5C6 5, R9C6 9, R9C8 5, R7C8 9, R9C2 7,
R9C5 8, R8C5 5, R7C5 7, R5C4 8, R8C2 9, R8C3 8, R3C3 7,
R1C2 4, R3C2 5, R3C1 9, R6C1 5, R6C7 2, R4C7 5, R6C5 9, and,
R4C5 2, R6C3 6, R4C3 9, R7C3 5, R7C2 6.
The solved final game board is shown below.
End note on Problem solving in Sudoku
Any puzzle solving involves essentially problem solving. The general steps are,
- First stage analysis and breaking it down into smaller chunks if possible as well as adapt the strategy of solving this type of problem,
- Solving the easier component problems so that the main problem size and complexity is reduced,
- Detailed information collection, that is, defining the problem in more details as far as possible,
- Second stage analysis of structure of problem (in this case of Sudoku) and information content,
- Key pattern identification,
- Use of the key pattern to create the breakthrough,
- Repeating the last four steps (steps 4, 5, 6 and 7) for finally solving the problem.
As this Sudoku problem is large and complex, you had to go through at least the first six steps—the seventh step had been trivial for this game board. One key breakthrough has broken open the problem completely—no second key pattern identification and use was required.
But for solving still more complex puzzles or real life problems, you may have to repeat last three steps.
Checkout and tally these seven steps with the process of solving this puzzle.
This is what we call the life cycle of Problem Solving, an independent and enormously powerful overlay subject standing on its own.
We leave you here this time with one new game of 3rd level hardness to solve.
Twelfth game at Third level of hardness
Other Sudoku game plays at third level hardness
Sudoku third level game play 11
Assorted Interesting Sudoku game plays
These Sudoku game solutions are collected from various sources and are found to be interesting. You can get these Sudoku solutions at Interesting Sudoku not classified at any hardness difficulty level.