Let's play the sixth Sudoku game at third level of hardness
This is the 6th game play session at Sudoku third level of hardness. We found this game not particularly difficult, perhaps because we have become used to this level of hardness in the game. But we did find a special turning moment when a new situation rule had to be used to break the final bottleneck. We will go through the specific explanation during the game play.
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 had 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.
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 only digits 7, and 8 and no other digit to occupy the two cells. This is in fact 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 occupancies), 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 this case, 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.
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.
Let us play the game now.
The Sudoku 6th game play at third level of hardness
First valid cells, R6C5 5, scan R4,R5,C6 -- R8C9 5, scan R7,R9,C7 -- R1C2 5, scan C1,C3,R3 -- R2C8 5, scan R1,R3,C7 -- R5C8 6, scan R4,C7 -- R9C1 6, scan C2,C3,R7 -- R6C6 6, scan R4,R5 -- R1C4 6, scan R2,R3,C6 -- R8C5 6, scan C4,C6 -- R5C5 8, DSA [4,7,8] in C5 -- Cycle (1,2) in R5 -- R5C2 3, cycle (1,2) in R5 -- R6C1 4, DSA [1,4.8] in R6 -- R9C8 2, scan R7,C7 -- R8C2 2, scan R7,R9,C3.
Valid cell coloring
The first valid cell is in green color and the rest of the valid cells at this stage are in yellow color. These colors will be carried through unchanged till the end of the game. The valid cells in the next stages will be colored differently. Next stage valid cells will be in sea-green, next to next stage valid cells in light pink-red and the last stage in light blue to differentiate. This use of different colors makes it easier to recover from mistakes and also eases explanation and following the game flow.
The results achieved till now are shown below.
We will use now the cycle (4,7,8) in C2. The first cells: Cycle (4,7,8) formed in C2, R6C2 1 -- R4C3 8 -- R6C7 8 -- R3C8 8, scan R1,C7,C9 -- R7C2 8, scan C1,C3,R9, Cycle (4,7) formed in C2 -- R3C7 9, scan R1,R2,C9 -- R7C8 9, scan C7,C9 -- R8C3 9, scan R7,C2 -- R9C4 9, scan R7,R8,C6.
We will get the main breakthrough in the next stage. For ease of explanation we will go over to the next stage.
The results are shown in the board below.
The first valid cell in this stage is, R5C4 1, as 1 can't appear in rest of the four empty cells in C4. Observe that by normal scan we won't have been able to identify this valid cell especially because of the cell R4C4 where, by the occurrences of digit 1, the DS would have been (1,4,7) but for the earlier formation of cycle (1,3) on the right in R4. This has eliminated the digit 1 from the DS of [1,4,7] in R4C4 so that digit 1 is left only in cell R5C4. In case of the other three empty cells in C4, R8C4, R3C4 and R2C4 digit 1 is already present in the interacting zones.
This is the main breakthrough.
Next valid cells, R5C6 2, DS cancel -- R1C6 7, DS cancel -- R2C5 4, DS cancel -- R3C4 2, DS cancel -- R4C5 7, DS cancel -- R4C4 4. We will close here to create a new last stage to finish the game. The results till now are shown in the game board below.
Next few valid cells in the final stage, R1C9 2, DS cancel -- R2C1 2, scan R3,C3 -- R3C2 4, cycle (3,7) in R3 -- R9C2 7 -- R9C7 1, DS cancel -- R7C1 3, DS cancel -- R7C3 4 -- R7C9 7, DS cancel -- R8C7 3 -- R3C9 3, DS cancel -- R4C9 1 -- R4C8 3 -- R1C8 1 -- R1C3 3 -- R2C7 7 -- R3C1 7 -- R2C3 1 -- R9C6 4.
And the last few cells, R8C6 8, DS cancel -- R8C4 7 -- R2C4 8 -- R2C6 3 -- R7C6 1. Game solved. End.
The final result is shown below.
As usual we leave you here with a new game to solve.
Seventh game at Third level of hardness