Let's play the 9th Sudoku game at 3rd level of hardness
This is the 9th game play session at Sudoku third level of hardness. In this game every step in the beginning was hard-going and finding the breakthrough was not easy. We needed to use a new technique. But now when we know and have used this pattern, we won't forget it in a hurry. In fact we are incorporating this new technique as a routine step in our overall strategy of Sudoku game solving.
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 will employ from 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. At least the major breakthrough in this game came this way. We name this new technique 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 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 9th game play at third level of hardness
First valid cells, R7C3 3, scan C1,C2,R8,R9 -- R2C8 8, DSA [1,2,3,4,8] in R2 -- R4C8 9, cycle (1,6) in right middle 9 cell square, DSA [2,8,9] in R4C8 -- Digit 6 in either R8C5 or R9C5, scan R7C6 for 6. This disallows digit 6 in any other cell in C5 thus cancelling 6 in R5C5. This is locking a digit occurrence within a 9 cell square in one column or row so that it is equivalent to the digit appearing in the 9 cell square in one of the three cells in the marked row or column and it cannot further appear in any other cell in the marked row or column.
This locking down of 6 in C5 bottom most two cells and cancellation of 6 in R5C5 created the cycle (1,7) in R5 that provided the critical breakthrough. It was all plain sailing after this point.
To show the use of cycles properly we will close at this stage and go over to the next stage. We have shown this new structure of DS resulting in locking of a single digit by 6 by "..6..' in the cell meaning there may be other possible digits but 6 is a must occupant in the cell.
Valid cell coloring
The first stage valid cells are colored light blue, second stage sea-green, and the solved stage cells colored yellow. 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.
Observe how cross scanning for digit 6 in R7, C6, the digit 6 got locked down into two cells (instead of one cell as a valid cell, but still it produced a great desired result). The special usefulness happens because in one of the 9 cell square just above this bottom middle 9 cell square 6 is not present and so appeared in the DSs such as in R5C5. Because of this locking down of 6 now, in the DS of R5C5 has to be canceled giving rise to the all important cycle (1,7) in R5.
The first valid cells at this stage are, Digit 6 cancelled in R5C5 forming a cycle (1,7) in R5 -- R5C7 6, cycle (1,7) in R5 -- R6C8 1, DS cancel -- R9C9 6, scan R7,C7 -- R8C5 6, scan R7,R9 -- R1C8 6, scan C7,C9,R3 -- R7C8 5, DSA [5,7] in C8 -- R3C8 7 -- R3C7 9, scan R1,R2 -- R7C9 9, scan C7 -- R2C7 3, scan R1,C9 -- R3C5 3, scan R1,R2,C4 3 over -- R4C9 8, cycle (2,4) in C9 -- R6C7 2 -- R1C7 5 -- R9C7 1, DSA [1,8] in C7 -- R8C7 8 -- R7C2 8, scan R8,R9 -- R8C2 9, cycle (1,5) in R8 -- R9C5 9, scan R7,C6 9 over -- Cycle (2,7) in R9, scan C2, R9C2 4 -- R9C6 5 -- R3C2 5, DSA [1,5,6] in C2 -- R6C2 6, DSA [1,6] in C2 -- R4C2 1.
Again we will close this stage here to show use of cycles properly. The results are shown in the game board below.
This final stage starts with, R5C3 7, DS cancel -- R5C5 1, DS cancel -- R9C3 2, DS cancel -- R9C1 7, DS cancel -- R3C8 8, cycle (2,4) in R3 -- R1C3 8, scan R2,C1 -- R6C5 8, scan C4,C6,R4 8 over -- R1C4 1, scan R3,C5 -- R2C3 1, scan R1,R3 -- R8C3 5, DS cancel -- R8C1 1 -- R4C3 4 -- R6C1 5 -- R1C5 7, scan R2,R3 -- R6C4 4, DSA [4,7] in R6 -- R6C6 7 -- R7C4 7, scan C5,C6 7 over -- R7C5 2, DSA [1,2] in R7.
And the last few cells, R7C6 1 -- R4C6 2 -- R4C4 6, DSA [5,6] in R4 -- R4C5 5 -- R2C5 4 -- R3C4 2 -- R3C1 4 -- R1C1 2 -- R1C9 4 -- R2C9 2. Game solved. End.
The final result is shown below.
We leave you here with a new game to solve.
Tenth game at Third level of hardness